\displaystyle \textbf{Question 1: }~\text{A couple has two children. Find the probability that both are boys,} \\ \text{if it is known that (i) one of the children is a boy, (ii) the older child is a boy.} \\ \text{[CBSE 2010, 2014]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }B_i\text{ and }G_i\text{ stand for the }i^{\text{th}}\text{ child being a boy and girl respectively.}
\displaystyle \text{Then the sample space is }S=\{B_1B_2,B_1G_2,G_1B_2,G_1G_2\}.
\displaystyle \text{Let }A=\text{both the children are boys},
\displaystyle B=\text{one of the children is a boy},
\displaystyle C=\text{the older child is a boy}.
\displaystyle A=\{B_1B_2\}.
\displaystyle B=\{B_1B_2,B_1G_2,G_1B_2\}.
\displaystyle C=\{B_1B_2,B_1G_2\}.
\displaystyle A\cap B=\{B_1B_2\}.
\displaystyle A\cap C=\{B_1B_2\}.
\displaystyle \text{(i) Required probability }=P(A\mid B)=\frac{P(A\cap B)}{P(B)}=\frac{\frac{1}{4}}{\frac{3}{4}}=\frac{1}{3}.
\displaystyle \text{(ii) Required probability }=P(A\mid C)=\frac{P(A\cap C)}{P(C)}=\frac{\frac{1}{4}}{\frac{2}{4}}=\frac{1}{2}.

\displaystyle \textbf{Question 2: }~\text{Consider a random experiment in which a coin is tossed and if} \\ \text{the coin shows head it is tossed again, but if it shows a tail then a die is tossed.} \\ \text{[CBSE 2010, 2014]}
\displaystyle \text{If eight possible outcomes are equally likely, find the probability that the die shows a} \\ \text{number greater than four, if it is known that the first throw of the coin results in a tail.}
\displaystyle \text{Answer:}
\displaystyle \text{The sample space associated with the given random experiment is}
\displaystyle S=\{(H,H),(H,T),(T,1),(T,2),(T,3),(T,4),(T,5),(T,6)\}.
\displaystyle \text{Let }A=\text{the event that the die shows a number greater than four}
\displaystyle \text{and }B=\text{the event that the first throw of the coin results in a tail}.
\displaystyle A=\{(T,5),(T,6)\}.
\displaystyle B=\{(T,1),(T,2),(T,3),(T,4),(T,5),(T,6)\}.
\displaystyle \text{Required probability }=P(A\mid B)=\frac{P(A\cap B)}{P(B)}=\frac{2}{6}=\frac{1}{3}.

\displaystyle \textbf{Question 3: }~\text{If }A\text{ and }B\text{ are independent events associated with a random experiment,} \\ \text{then prove that}
\displaystyle \text{(i) }\overline{A}\text{ and }B\text{ are independent events},
\displaystyle \text{(ii) }A\text{ and }\overline{B}\text{ are independent events},
\displaystyle \text{(iii) }\overline{A}\text{ and }\overline{B}\text{ are also independent events}.  \hspace{3.0cm} \text{[CBSE 2010, 2014]}
\displaystyle \text{Answer:}
\displaystyle \text{Since }A\text{ and }B\text{ are independent events, therefore,}
\displaystyle P(A\cap B)=P(A)P(B). \text{... ... i}
\displaystyle \text{(i) It is evident from the Venn diagram that }A\cap B\text{ and }\overline{A}\cap B\text{ are mutually} \\ \text{exclusive events such that }(A\cap B)\cup(\overline{A}\cap B)=B.

\displaystyle \text{Therefore, by addition theorem on probability,}
\displaystyle P(A\cap B)+P(\overline{A}\cap B)=P(B).
\displaystyle \Rightarrow P(\overline{A}\cap B)=P(B)-P(A\cap B).
\displaystyle =P(B)-P(A)P(B).
\displaystyle =P(B)\{1-P(A)\}.
\displaystyle =P(B)P(\overline{A}).
\displaystyle \text{Thus, }P(\overline{A}\cap B)=P(\overline{A})P(B).
\displaystyle \text{Hence, }\overline{A}\text{ and }B\text{ are independent events}.
\displaystyle \text{(ii) It is clear from the Venn diagram that }A\cap\overline{B}\text{ and }A\cap B\text{ are mutually} \\ \text{exclusive events such that }(A\cap\overline{B})\cup(A\cap B)=A.
\displaystyle \text{So, by addition theorem on probability,}
\displaystyle P(A\cap\overline{B})+P(A\cap B)=P(A).
\displaystyle \Rightarrow P(A\cap\overline{B})=P(A)-P(A\cap B).
\displaystyle =P(A)-P(A)P(B).
\displaystyle =P(A)\{1-P(B)\}.
\displaystyle =P(A)P(\overline{B}).
\displaystyle \text{Thus, }P(A\cap\overline{B})=P(A)P(\overline{B}).
\displaystyle \text{Hence, }A\text{ and }\overline{B}\text{ are independent events}.
\displaystyle \text{(iii) We have to show that }\overline{A}\text{ and }\overline{B}\text{ are independent events}.
\displaystyle \text{For this it is sufficient to prove that }P(\overline{A}\cap\overline{B})=P(\overline{A})P(\overline{B}).
\displaystyle \text{Now, }\overline{A}\cap\overline{B}=\overline{A\cup B}.
\displaystyle P(\overline{A}\cap\overline{B})=1-P(A\cup B).
\displaystyle =1-\{P(A)+P(B)-P(A\cap B)\}.
\displaystyle =1-\{P(A)+P(B)-P(A)P(B)\}.
\displaystyle =\{1-P(A)\}\{1-P(B)\}.
\displaystyle =P(\overline{A})P(\overline{B}).
\displaystyle \text{Hence, }\overline{A}\text{ and }\overline{B}\text{ are independent events}. \quad \text{Q.E.D.}

\displaystyle \textbf{Quetion 4: }~\text{If }A\text{ and }B\text{ are two independent events such that }P(\overline{A}\cap B)=\frac{2}{15}\text{ and }P(A\cap\overline{B})=\frac{1}{6},\text{ then find }P(A)\text{ and }P(B).
\displaystyle \text{Answer:}
\displaystyle \text{Let }P(A)=x\text{ and }P(B)=y.
\displaystyle \text{It is given that }A\text{ and }B\text{ are independent events such that}
\displaystyle P(\overline{A}\cap B)=\frac{2}{15}\text{ and }P(A\cap\overline{B})=\frac{1}{6}.
\displaystyle \Rightarrow P(\overline{A})P(B)=\frac{2}{15}\text{ and }P(A)P(\overline{B})=\frac{1}{6}.
\displaystyle \Rightarrow \{1-P(A)\}P(B)=\frac{2}{15}\text{ and }P(A)\{1-P(B)\}=\frac{1}{6}.
\displaystyle \Rightarrow (1-x)y=\frac{2}{15}\text{ and }x(1-y)=\frac{1}{6}.
\displaystyle \Rightarrow y-xy=\frac{2}{15}\text{... ... i}
\displaystyle \text{and }x-xy=\frac{1}{6}\text{... ... ii}
\displaystyle \text{Subtracting (i) from (ii), we obtain}
\displaystyle x-y=\frac{1}{6}-\frac{2}{15}.
\displaystyle \Rightarrow x-y=\frac{1}{30}.
\displaystyle \Rightarrow x=y+\frac{1}{30}. \text{... ... iii}
\displaystyle \text{Putting }x=y+\frac{1}{30}\text{ in (i), we obtain}
\displaystyle y-\left(y+\frac{1}{30}\right)y=\frac{2}{15}.
\displaystyle \Rightarrow y-y^{2}-\frac{1}{30}y=\frac{2}{15}.
\displaystyle \Rightarrow y^{2}-\frac{29}{30}y+\frac{2}{15}=0.
\displaystyle \Rightarrow 30y^{2}-29y+4=0.
\displaystyle \Rightarrow 30y^{2}-24y-5y+4=0.
\displaystyle \Rightarrow 6y(5y-4)-1(5y-4)=0.
\displaystyle \Rightarrow (6y-1)(5y-4)=0.
\displaystyle \Rightarrow y=\frac{1}{6}\text{ or }y=\frac{4}{5}.
\displaystyle \textbf{Case I: }y=\frac{1}{6}.
\displaystyle \text{Putting }y=\frac{1}{6}\text{ in (iii), we obtain }x=\frac{1}{5}.
\displaystyle \textbf{Case II: }y=\frac{4}{5}.
\displaystyle \text{Putting }y=\frac{4}{5}\text{ in (iii), we obtain }x=\frac{5}{6}.
\displaystyle \text{Thus, }P(A)=\frac{1}{5}\text{ and }P(B)=\frac{1}{6}\text{ or }P(A)=\frac{5}{6}\text{ and }P(B)=\frac{4}{5}.

\displaystyle \textbf{Question 5: }~\text{Probabilities of solving a specific problem independently by }\\ A\text{ and }B\text{ are }\frac{1}{2}\text{ and }\frac{1}{3}\text{ respectively.}
\displaystyle \text{If both try to solve the problem independently, find the probability that}
\displaystyle \text{(i) the problem is solved}\qquad \text{(ii) exactly one of them solves the problem}.
\displaystyle \text{Answer:}
\displaystyle \text{Let }E\text{ be the event that the problem is solved by }A\text{ and }F\text{ be the event that the} \\ \text{problem is solved by }B.
\displaystyle \text{It is given that }P(E)=\frac{1}{2}\text{ and }P(F)=\frac{1}{3}.
\displaystyle \text{(i) The problem is solved if at least one of }A\text{ and }B\text{ solves the problem. Therefore,}
\displaystyle \text{Required probability }=P(E\cup F).
\displaystyle =1-P(\overline{E})P(\overline{F}).
\displaystyle =1-\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right).
\displaystyle =\frac{2}{3}.
\displaystyle \text{(ii) Required probability }=P(E)+P(F)-2P(E\cap F).
\displaystyle =P(E)+P(F)-2P(E)P(F).
\displaystyle =\frac{1}{2}+\frac{1}{3}-2\times\frac{1}{2}\times\frac{1}{3}.
\displaystyle =\frac{1}{2}.

\displaystyle \textbf{Question 6: }~\text{A can hit a target }4\text{ times in }5\text{ shots, }B\text{ }3\text{ times in }4\text{ shots,} \\ \text{and }C\text{ }2\text{ times in }3\text{ shots. Calculate the probability that}
\displaystyle \text{(i) }A,B,C\text{ all may hit}\qquad \\ \text{(ii) }B,C\text{ may hit and }A\text{ may not},
\displaystyle \text{(iii) any two of }A,B\text{ and }C\text{ will hit the target}\qquad \\ \text{(iv) none of them will hit the target}. \qquad \text{[CBSE 2005]}
\displaystyle \text{Answer:}
\displaystyle \text{Solution: Consider the following events:}
\displaystyle E=\text{A hits the target},\;F=\text{B hits the target},\;\text{and }G=\text{C hits the target}.
\displaystyle \text{We have }P(E)=\frac{4}{5},\;P(F)=\frac{3}{4}\text{ and }P(G)=\frac{2}{3}.
\displaystyle \text{(i) Required probability }=P(A,B,C\text{ all may hit}) \\ =P(E\cap F\cap G).
\displaystyle =P(E)P(F)P(G).
\displaystyle =\frac{4}{5}\times\frac{3}{4}\times\frac{2}{3}=\frac{2}{5}.
\displaystyle \text{(ii) Required probability }=P(B,C\text{ may hit and }A\text{ may not}) \\ =P(\overline{E}\cap F\cap G).
\displaystyle =P(\overline{E})P(F)P(G).
\displaystyle =\left(1-\frac{4}{5}\right)\times\frac{3}{4}\times\frac{2}{3}=\frac{1}{10}.
\displaystyle \text{(iii) Required probability }=P(\text{any two of }A,B,C\text{ will hit the target}).
\displaystyle =P(E\cap F\cap\overline{G})\cup P(\overline{E}\cap F\cap G)\cup P(E\cap\overline{F}\cap G).
\displaystyle =P(E\cap F\cap\overline{G})+P(\overline{E}\cap F\cap G)+P(E\cap\overline{F}\cap G).
\displaystyle =P(E)P(F)P(\overline{G})+P(\overline{E})P(F)P(G)+P(E)P(\overline{F})P(G).
\displaystyle =\frac{4}{5}\times\frac{3}{4}\times\frac{1}{3}+\frac{1}{5}\times\frac{3}{4}\times\frac{2}{3}+\frac{4}{5}\times\frac{1}{4}\times\frac{2}{3}=\frac{13}{30}.
\displaystyle \text{(iv) Required probability }=P(\text{none of }A,B,C\text{ will hit the target}) \\ =P(\overline{E}\cap\overline{F}\cap\overline{G}).
\displaystyle =P(\overline{E})P(\overline{F})P(\overline{G}).
\displaystyle =\frac{1}{5}\times\frac{1}{4}\times\frac{1}{3}=\frac{1}{60}.

\displaystyle \textbf{Question 6: }~A\text{ speaks truth in }60\%\text{ of the cases and }B\text{ in }90\%\text{ of the cases.} \text{In what percentage} \\ \text{of cases are they likely to (i) contradict each other in stating the same fact? (ii) agree in stating} \\ \text{the same fact?} \qquad \text{[CBSE 2003, 2013]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }E\text{ be the event that }A\text{ speaks truth and }F\text{ be the event that } \\ B\text{ speaks truth. Then }E\text{ and }F\text{ are independent events such that}
\displaystyle P(E)=\frac{60}{100}=\frac{3}{5}\text{ and }P(F)=\frac{90}{100}=\frac{9}{10}.
\displaystyle \text{(i) }A\text{ and }B\text{ will contradict each other in narrating the same fact in the following} \\ \text{mutually exclusive ways:}
\displaystyle \text{(I) }A\text{ speaks truth and }B\text{ tells a lie i.e. }E\cap\overline{F},\qquad \text{(II) }A\text{ tells} \\ \text{a lie and }B\text{ speaks truth i.e. }\overline{E}\cap F.
\displaystyle \therefore P(A\text{ and }B\text{ contradict each other})=P(\text{I or II}).
\displaystyle =P\bigl((E\cap\overline{F})\cup(\overline{E}\cap F)\bigr).
\displaystyle =P(E\cap\overline{F})+P(\overline{E}\cap F).
\displaystyle =P(E)P(\overline{F})+P(\overline{E})P(F).
\displaystyle =\frac{3}{5}\times\left(1-\frac{9}{10}\right)+\left(1-\frac{3}{5}\right)\times\frac{9}{10}.
\displaystyle =\frac{3}{5}\times\frac{1}{10}+\frac{2}{5}\times\frac{9}{10}=\frac{42}{100}.
\displaystyle \text{Hence, in }42\%\text{ cases }A\text{ and }B\text{ are likely to contradict each other}.
\displaystyle \text{(ii) }A\text{ and }B\text{ will agree in stating the same fact in the following mutually exclusive ways:}
\displaystyle \text{(I) }A\text{ and }B\text{ both speak truth},\qquad \text{(II) }A\text{ and }B\text{ both tell a lie}.
\displaystyle \therefore P(A\text{ and }B\text{ agree})=P\bigl((E\cap F)\cup(\overline{E}\cap\overline{F})\bigr).
\displaystyle =P(E\cap F)+P(\overline{E}\cap\overline{F}).
\displaystyle =P(E)P(F)+P(\overline{E})P(\overline{F}).
\displaystyle =\frac{3}{5}\times\frac{9}{10}+\frac{2}{5}\times\frac{1}{10}=\frac{29}{50}=\frac{58}{100}.
\displaystyle \text{Hence, }A\text{ and }B\text{ will agree in }58\%\text{ cases}.

\displaystyle \textbf{Question 7: }~\text{A bag }A\text{ contains }4\text{ black and }6\text{ red balls and bag } B\text{ contains }7\text{ black} \\ \text{and }3\text{ red balls. A die is thrown.}
\displaystyle \text{If }1\text{ or }2\text{ appears on it, then bag }A\text{ is chosen, otherwise bag }B.
\displaystyle \text{If two balls are drawn at random (without replacement) from the selected bag, find} \\ \text{the probability of one of them being red and another black.} \qquad \text{[CBSE 2015]}
\displaystyle \text{Answer:}
\displaystyle \text{Consider the following events:}
\displaystyle E_{1}=\text{getting }1\text{ or }2\text{ on the die},\qquad E_{2}=\text{getting any number other} \\ \text{than }1\text{ and }2\text{ on the die}.
\displaystyle A=\text{getting }1\text{ red and }1\text{ black ball from the selected bag}.
\displaystyle P(E_{1})=\frac{2}{6}=\frac{1}{3},\qquad P(E_{2})=\frac{4}{6}=\frac{2}{3}.
\displaystyle P(A\mid E_{1})=\frac{{}^{4}C_{1}\times{}^{6}C_{1}}{{}^{10}C_{2}}=\frac{8}{15}.
\displaystyle P(A\mid E_{2})=\frac{{}^{7}C_{1}\times{}^{3}C_{1}}{{}^{10}C_{2}}=\frac{7}{15}.
\displaystyle \text{Required probability }=P(A)=P(E_{1})P(A\mid E_{1})+P(E_{2})P(A\mid E_{2}).
\displaystyle =\frac{1}{3}\times\frac{8}{15}+\frac{2}{3}\times\frac{7}{15}.
\displaystyle =\frac{22}{45}.

\displaystyle \textbf{Question 8: }~\text{In a bolt factory, machines }A,B\text{ and }C\text{ manufacture respectively }\\ 25\%,35\%\text{ and }40\%\text{ of the total bolts.} \text{Of their output }5\%,4\%\text{ and }2\%\text{ percent are} \\ \text{respectively defective bolts. A bolt is drawn at random from the product.} \text{If the bolt} \\ \text{drawn is found to be defective, what is the probability that it is manufactured by} \\ \text{machine }B\text{?} \qquad \text{[CBSE 2008, 2015]}
\displaystyle \text{Answer:}
\displaystyle \text{Solution: Let }E_{1},E_{2},E_{3}\text{ and }A\text{ be the events defined as follows:}
\displaystyle E_{1}=\text{bolt is manufactured by machine }A,\;E_{2}=\text{bolt is manufactured by machine }B,\;E_{3}=\text{bolt is manufactured by machine }C,\;A=\text{bolt is defective}.
\displaystyle P(E_{1})=\frac{25}{100},\;P(E_{2})=\frac{35}{100},\;P(E_{3})=\frac{40}{100}.
\displaystyle P(A\mid E_{1})=\frac{5}{100},\;P(A\mid E_{2})=\frac{4}{100},\;P(A\mid E_{3})=\frac{2}{100}.
\displaystyle \text{Required probability }=P(E_{2}\mid A).
\displaystyle =\frac{P(E_{2})P(A\mid E_{2})}{P(E_{1})P(A\mid E_{1})+P(E_{2})P(A\mid E_{2})+P(E_{3})P(A\mid E_{3})}.
\displaystyle =\frac{\frac{35}{100}\times\frac{4}{100}}{\frac{25}{100}\times\frac{5}{100}+\frac{35}{100}\times\frac{4}{100}+\frac{40}{100}\times\frac{2}{100}}.
\displaystyle =\frac{140}{125+140+80}=\frac{140}{345}=\frac{28}{69}.

\displaystyle \textbf{Question 9: }~\text{A company has two plants to manufacture scooters. Plant I manufactures }\\ 70\%\text{ of the scooters and Plant II manufactures }30\%. \text{At Plant I, }80\%\text{ of the scooters are rated} \\ \text{as of standard quality and at Plant II, }90\%\text{ are rated as of standard quality. A scooter is} \\ \text{chosen at random and is found to be of standard quality. What is the probability that it} \\ \text{has come from Plant II?} \qquad \text{[CBSE 2000, 04, 05]}
\displaystyle \text{Answer:}
\displaystyle \text{Solution: Let }E_{1},E_{2}\text{ and }A\text{ be the following events:}
\displaystyle E_{1}=\text{Plant I is chosen},\;E_{2}=\text{Plant II is chosen},\;A=\text{scooter is of standard quality}.
\displaystyle P(E_{1})=\frac{70}{100},\;P(E_{2})=\frac{30}{100},\;P(A\mid E_{1})=\frac{80}{100},\;P(A\mid E_{2})=\frac{90}{100}.
\displaystyle \text{By Bayes' theorem,}
\displaystyle P(E_{2}\mid A)=\frac{P(E_{2})P(A\mid E_{2})}{P(E_{1})P(A\mid E_{1})+P(E_{2})P(A\mid E_{2})}.
\displaystyle =\frac{\frac{30}{100}\times\frac{90}{100}}{\frac{70}{100}\times\frac{80}{100}+\frac{30}{100}\times\frac{90}{100}}.
\displaystyle =\frac{27}{56+27}=\frac{27}{83}.

\displaystyle \textbf{Question 10: }~\text{An insurance company insured }2000\text{ scooter drivers, }4000\text{ car drivers} \\ \text{and }6000\text{ truck drivers.} \text{The probabilities of an accident involving a scooter driver, car} \\ \text{driver and a truck driver are }0.01,0.03\text{ and }0.15\text{ respectively. One of the insured} \\ \text{persons meets with an accident. What is the probability that he is a scooter driver?} \qquad \\ \text{[CBSE 2000, 02, 08, 12, 14]}
\displaystyle \text{Answer:}
\displaystyle \text{Solution: Let }E_{1},E_{2},E_{3}\text{ and }A\text{ be the events defined as follows:}
\displaystyle E_{1}=\text{person chosen is a scooter driver},\;E_{2}=\text{person chosen is a car driver},\;E_{3}=\text{person chosen is a truck driver},\;A=\text{person meets with an accident}.
\displaystyle P(E_{1})=\frac{2000}{12000}=\frac{1}{6},\;P(E_{2})=\frac{4000}{12000}=\frac{1}{3},\;P(E_{3})=\frac{6000}{12000}=\frac{1}{2}.
\displaystyle P(A\mid E_{1})=0.01,\;P(A\mid E_{2})=0.03,\;P(A\mid E_{3})=0.15.
\displaystyle \text{By Bayes' rule,}
\displaystyle P(E_{1}\mid A)=\frac{P(E_{1})P(A\mid E_{1})}{P(E_{1})P(A\mid E_{1})+P(E_{2})P(A\mid E_{2})+P(E_{3})P(A\mid E_{3})}.
\displaystyle =\frac{\frac{1}{6}\times0.01}{\frac{1}{6}\times0.01+\frac{1}{3}\times0.03+\frac{1}{2}\times0.15}.
\displaystyle =\frac{1}{52}.

\displaystyle \textbf{Question 11: }~\text{A card from a pack of }52\text{ cards is lost. From the remaining cards of the} \\ \text{pack, two cards are drawn and are found to be hearts.} \text{Find the probability of the missing} \\ \text{card to be a heart.} \qquad \text{[CBSE 2000, 2010]}
\displaystyle \text{Answer:}
\displaystyle \text{Solution: Let }E_{1},E_{2},E_{3},E_{4}\text{ and }A\text{ be the events as defined below:}
\displaystyle E_{1}=\text{missing card is a heart},\;E_{2}=\text{missing card is a spade},\;E_{3}=\text{missing card is a club},\\ E_{4}=\text{missing card is a diamond},\;A=\text{drawing two heart cards from the remaining cards}.
\displaystyle P(E_{1})=P(E_{2})=P(E_{3})=P(E_{4})=\frac{13}{52}=\frac{1}{4}.
\displaystyle P(A\mid E_{1})=\frac{{}^{12}C_{2}}{{}^{51}C_{2}},\;P(A\mid E_{2})=\frac{{}^{13}C_{2}}{{}^{51}C_{2}},\;P(A\mid E_{3})=\frac{{}^{13}C_{2}}{{}^{51}C_{2}},\;P(A\mid E_{4})=\frac{{}^{13}C_{2}}{{}^{51}C_{2}}.
\displaystyle \text{Required probability }=P(E_{1}\mid A).
\displaystyle =\frac{P(E_{1})P(A\mid E_{1})}{P(E_{1})P(A\mid E_{1})+P(E_{2})P(A\mid E_{2})+P(E_{3})P(A\mid E_{3})+P(E_{4})P(A\mid E_{4})}.
\displaystyle =\frac{\frac{1}{4}\times\frac{{}^{12}C_{2}}{{}^{51}C_{2}}}{\frac{1}{4}\times\frac{{}^{12}C_{2}}{{}^{51}C_{2}}+\frac{1}{4}\times\frac{{}^{13}C_{2}}{{}^{51}C_{2}}+\frac{1}{4}\times\frac{{}^{13}C_{2}}{{}^{51}C_{2}}+\frac{1}{4}\times\frac{{}^{13}C_{2}}{{}^{51}C_{2}}}.
\displaystyle =\frac{{}^{12}C_{2}}{{}^{12}C_{2}+3{}^{13}C_{2}}=\frac{66}{66+78+78+78}=\frac{11}{50}.

\displaystyle \textbf{Question 12: }~\text{A card from a pack of }52\text{ cards is lost. From the remaining cards of the} \\ \text{pack, two cards are drawn and are found to be hearts.} \text{Find the probability of the missing} \\ \text{card to be a heart.} \qquad \text{[CBSE 2000, 2010]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }E_{1},E_{2},E_{3},E_{4}\text{ and }A\text{ be the events as defined below:}
\displaystyle E_{1}=\text{missing card is a heart},\\ E_{2}=\text{missing card is a spade},\\ E_{3}=\text{missing card is a club},\\ E_{4}=\text{missing card is a diamond},\\ A=\text{drawing two heart cards from the remaining cards}.
\displaystyle P(E_{1})=P(E_{2})=P(E_{3})=P(E_{4})=\frac{13}{52}=\frac{1}{4}.
\displaystyle P(A\mid E_{1})=\frac{{}^{12}C_{2}}{{}^{51}C_{2}},\;P(A\mid E_{2})=\frac{{}^{13}C_{2}}{{}^{51}C_{2}},\;P(A\mid E_{3})=\frac{{}^{13}C_{2}}{{}^{51}C_{2}},\;P(A\mid E_{4})=\frac{{}^{13}C_{2}}{{}^{51}C_{2}}.
\displaystyle \text{Required probability }=P(E_{1}\mid A).
\displaystyle =\frac{P(E_{1})P(A\mid E_{1})}{P(E_{1})P(A\mid E_{1})+P(E_{2})P(A\mid E_{2})+P(E_{3})P(A\mid E_{3})+P(E_{4})P(A\mid E_{4})}.
\displaystyle =\frac{{}^{12}C_{2}}{{}^{12}C_{2}+3{}^{13}C_{2}}=\frac{66}{66+78+78+78}=\frac{11}{50}.

\displaystyle \textbf{Question 13: }~\text{Suppose a girl throws a die. If she gets }5\text{ or }6,\text{ she tosses a coin} \\ \text{three times and notes the number of heads.}\text{If she gets }1,2,3\text{ or }4,\text{ she tosses a} \\ \text{coin once and notes whether head or tail is obtained.}\text{If she obtained exactly} \\ \text{one head, what is the probability that she threw a }1,2,3\text{ or }4\text{ with the die?} \qquad \\ \text{[CBSE 2012]}
\displaystyle \text{Answer:}
\displaystyle \text{Consider the following events:}
\displaystyle E_{1}=\text{getting }5\text{ or }6\text{ in a single throw of a die},\\ E_{2}=\text{getting }1,2,3\text{ or }4\text{ in a single throw of a die},\\ A=\text{getting exactly one head}.
\displaystyle P(E_{1})=\frac{2}{6}=\frac{1}{3},\;P(E_{2})=\frac{4}{6}=\frac{2}{3}.
\displaystyle P(A\mid E_{1})={}^{3}C_{1}\left(\frac{1}{2}\right)^{1}\left(\frac{1}{2}\right)^{2}=\frac{3}{8}.
\displaystyle P(A\mid E_{2})=\frac{1}{2}.
\displaystyle \text{Required probability }=P(E_{2}\mid A).
\displaystyle =\frac{P(E_{2})P(A\mid E_{2})}{P(E_{1})P(A\mid E_{1})+P(E_{2})P(A\mid E_{2})}.
\displaystyle =\frac{\frac{2}{3}\times\frac{1}{2}}{\frac{1}{3}\times\frac{3}{8}+\frac{2}{3}\times\frac{1}{2}}=\frac{8}{11}.

\displaystyle \textbf{Question 14: }~\text{Given three identical boxes I, II and III, each containing two coins.} \text{In box I} \\ \text{both coins are gold, in box II both are silver coins and in box III there is one gold and one} \\ \text{silver coin.}\text{A person chooses a box at random and takes out a coin. If the coin is gold, what} \\ \text{is the probability that the other coin in the box is also gold?} \qquad \text{[CBSE 2011]}
\displaystyle \text{Answer:}
\displaystyle \text{Consider the following events:}
\displaystyle E_{1}=\text{Box I is chosen},\;E_{2}=\text{Box II is chosen},\;E_{3}=\text{Box III is chosen},\\ A=\text{coin drawn is gold}.
\displaystyle P(E_{1})=P(E_{2})=P(E_{3})=\frac{1}{3}.
\displaystyle P(A\mid E_{1})=1,\;P(A\mid E_{2})=0,\;P(A\mid E_{3})=\frac{1}{2}.
\displaystyle \text{Required probability }=P(E_{1}\mid A).
\displaystyle =\frac{P(E_{1})P(A\mid E_{1})}{P(E_{1})P(A\mid E_{1})+P(E_{2})P(A\mid E_{2})+P(E_{3})P(A\mid E_{3})}.
\displaystyle =\frac{\frac{1}{3}\times1}{\frac{1}{3}\times1+\frac{1}{3}\times0+\frac{1}{3}\times\frac{1}{2}}=\frac{2}{3}.

\displaystyle \textbf{Question 15: }~\text{Bag I contains }3\text{ red and }4\text{ black balls and Bag II contains }4\text{ red and }\\ 5\text{ black balls.} \text{One ball is transferred from Bag I to Bag II and then a ball is drawn from Bag II.} \\ \text{The ball so drawn is found to be red in colour.} \text{Find the probability that the transferred ball} \\ \qquad \text{is black. [CBSE 2012]}
\displaystyle \text{Answer:}
\displaystyle \text{Consider the following events:}
\displaystyle E_{1}=\text{ball transferred from Bag I to Bag II is red},\\ E_{2}=\text{ball transferred from Bag I to Bag II is black},\\ A=\text{ball drawn from Bag II is red}.
\displaystyle P(E_{1})=\frac{3}{7},\;P(E_{2})=\frac{4}{7},\;P(A\mid E_{1})=\frac{5}{10}=\frac{1}{2},\;P(A\mid E_{2})=\frac{4}{10}=\frac{2}{5}.
\displaystyle \text{Required probability }=P(E_{2}\mid A).
\displaystyle =\frac{P(E_{2})P(A\mid E_{2})}{P(E_{1})P(A\mid E_{1})+P(E_{2})P(A\mid E_{2})}.
\displaystyle =\frac{\frac{4}{7}\times\frac{2}{5}}{\frac{3}{7}\times\frac{1}{2}+\frac{4}{7}\times\frac{2}{5}}=\frac{16}{31}.

\displaystyle \textbf{Question 16: }~\text{Suppose that }5\%\text{ of men and }0.25\%\text{ of women have grey hair. A grey-haired} \\ \text{person is selected at random.}\text{What is the probability of this person being male? Assume that} \\ \text{there are equal numbers of males and females.} \qquad \text{[CBSE 2011]}
\displaystyle \text{Answer:}
\displaystyle \text{Consider the following events:}
\displaystyle E_{1}=\text{person selected is male},\;E_{2}=\text{person selected is female},\;A=\text{person selected is grey haired}.
\displaystyle P(E_{1})=P(E_{2})=\frac{1}{2},\;P(A\mid E_{1})=\frac{5}{100},\;P(A\mid E_{2})=\frac{1}{400}.
\displaystyle \text{Required probability }=P(E_{1}\mid A).
\displaystyle =\frac{P(E_{1})P(A\mid E_{1})}{P(E_{1})P(A\mid E_{1})+P(E_{2})P(A\mid E_{2})}.
\displaystyle =\frac{\frac{1}{2}\times\frac{5}{100}}{\frac{1}{2}\times\frac{5}{100}+\frac{1}{2}\times\frac{1}{400}}=\frac{20}{21}.

\displaystyle \textbf{Question 17: }~\text{A bag contains }4\text{ balls. Two balls are drawn at random without} \\ \text{replacement and are found to be white.}\text{What is the probability that all balls are white?} \qquad \\ \text{[CBSE 2010, 2016]}
\displaystyle \text{Answer:}
\displaystyle \text{Since two balls drawn are white, the possible cases are:}
\displaystyle \text{(i) the bag contains two white balls and two balls of other colour},
\displaystyle \text{(ii) the bag contains three white balls and one ball of other colour},
\displaystyle \text{(iii) the bag contains all white balls}.
\displaystyle \text{Consider the following events:}
\displaystyle E_{1}=\text{there are two white and two other colour balls in the bag},
\displaystyle E_{2}=\text{there are three white and one other colour ball in the bag},
\displaystyle E_{3}=\text{there are all white balls in the bag},\;A=\text{drawing two white balls from the bag}.
\displaystyle P(E_{1})=P(E_{2})=P(E_{3})=\frac{1}{3}.
\displaystyle P(A\mid E_{1})=\frac{{}^{2}C_{2}}{{}^{4}C_{2}}=\frac{1}{6},\;P(A\mid E_{2})=\frac{{}^{3}C_{2}}{{}^{4}C_{2}}=\frac{1}{2},\;P(A\mid E_{3})=\frac{{}^{4}C_{2}}{{}^{4}C_{2}}=1.
\displaystyle \text{Required probability }=P(E_{3}\mid A).
\displaystyle =\frac{P(E_{3})P(A\mid E_{3})}{\sum P(E_{i})P(A\mid E_{i})}.
\displaystyle =\frac{\frac{1}{3}\times1}{\frac{1}{3}\times\frac{1}{6}+\frac{1}{3}\times\frac{1}{2}+\frac{1}{3}\times1}=\frac{3}{5}.

\displaystyle \textbf{Question 18: }~\text{A bag contains }3\text{ red and }7\text{ black balls. Two balls are selected at random} \\ \text{one-by-one without replacement.}\text{If the second selected ball happens to be red, what is the} \\ \text{probability that the first selected ball is also red?} \qquad \text{[CBSE 2014]}
\displaystyle \text{Answer:}
\displaystyle \text{Consider the following events:}
\displaystyle E_{1}=\text{first ball drawn is red and second is of any colour},
\displaystyle E_{2}=\text{first ball drawn is black and second is of any colour},\;A=\text{second ball drawn is red}.
\displaystyle P(E_{1})=\frac{3}{10},\;P(E_{2})=\frac{7}{10},\;P(A\mid E_{1})=\frac{2}{9},\;P(A\mid E_{2})=\frac{3}{9}.
\displaystyle \text{Required probability }=P(E_{1}\mid A).
\displaystyle =\frac{P(E_{1})P(A\mid E_{1})}{P(E_{1})P(A\mid E_{1})+P(E_{2})P(A\mid E_{2})}.
\displaystyle =\frac{\frac{3}{10}\times\frac{2}{9}}{\frac{3}{10}\times\frac{2}{9}+\frac{7}{10}\times\frac{3}{9}}=\frac{2}{9}.

\displaystyle \textbf{Question 19: }~\text{A man is known to speak truth }3\text{ out of }4\text{ times. He throws a die and reports} \\ \text{that it is a six. }\text{Find the probability that it is actually a six.} \text{[CBSE 2005, 2011, 2014, 2017]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }E_{1},E_{2}\text{ and }A\text{ be the events defined as follows:}
\displaystyle E_{1}=\text{six occurs},\;E_{2}=\text{six does not occur},\;A=\text{the man reports that it is a six}.
\displaystyle P(E_{1})=\frac{1}{6},\;P(E_{2})=\frac{5}{6}.
\displaystyle P(A\mid E_{1})=\frac{3}{4},\;P(A\mid E_{2})=\frac{1}{4}.
\displaystyle \text{Required probability }=P(E_{1}\mid A).
\displaystyle \text{By Bayes' theorem,}
\displaystyle P(E_{1}\mid A)=\frac{P(E_{1})P(A\mid E_{1})}{P(E_{1})P(A\mid E_{1})+P(E_{2})P(A\mid E_{2})}.
\displaystyle =\frac{\frac{1}{6}\times\frac{3}{4}}{\frac{1}{6}\times\frac{3}{4}+\frac{5}{6}\times\frac{1}{4}}=\frac{3}{8}.

\displaystyle \textbf{Question 20. }\text{If }P(A\cap B)=\frac{1}{8}\text{ and }P(\overline{A})=\frac{3}{4},\text{ then }P\left(\frac{B}{A}\right)\text{ is equal to} \hspace{2.2cm}\text{[CBSE 2023]}
\displaystyle \text{(a) }\frac{1}{2} \qquad \text{(b) }\frac{1}{3}
\displaystyle \text{(c) }\frac{1}{6} \qquad \text{(d) }\frac{2}{3}
\displaystyle \text{Answer:}
\displaystyle \text{(a) Given, }P(A\cap B)=\frac{1}{8}\text{ and }P(\overline{A})=\frac{3}{4}
\displaystyle \text{Now, }P\left(\frac{B}{A}\right)=\frac{P(A\cap B)}{P(A)}=\frac{P(A\cap B)}{1-P(\overline{A})}
\displaystyle =\frac{1/8}{1-3/4}=\frac{1/8}{1/4}=\frac{4}{8}=\frac{1}{2}
\\

\displaystyle \textbf{Question 21. }\text{The events }E\text{ and }F\text{ are independent. If }P(E)=0.3\text{ and}
\displaystyle P(E\cup F)=0.5,\text{ then }P(E/F)-P(F/E)\text{ equals to} \hspace{2.2cm}\text{[CBSE 2023]}
\displaystyle \text{(a) }\frac{1}{7} \qquad \text{(b) }\frac{2}{7}
\displaystyle \text{(c) }\frac{3}{35} \qquad \text{(d) }\frac{1}{70}
\displaystyle \text{Answer:}
\displaystyle \text{(d) Given, }P(E)=0.3\text{ and }P(E\cup F)=0.5
\displaystyle \text{Let }P(F)=x
\displaystyle \text{We know that}
\displaystyle P(E\cup F)=P(E)+P(F)-P(E\cap F)
\displaystyle =P(E)+P(F)-P(E)\cdot P(F)
\displaystyle \Rightarrow 0.5=0.3+x-0.3x
\displaystyle \Rightarrow 0.7x=0.5-0.3
\displaystyle \Rightarrow x=\frac{0.2}{0.7}=\frac{2}{7}
\displaystyle \therefore P(F)=\frac{2}{7}
\displaystyle \text{Now, }P(E/F)-P(F/E)
\displaystyle =\frac{P(E\cap F)}{P(F)}-\frac{P(F\cap E)}{P(E)}
\displaystyle =\frac{P(E\cap F)\cdot P(E)-P(F\cap E)\cdot P(F)}{P(E)\cdot P(F)}
\displaystyle =\frac{P(E\cap F)[P(E)-P(F)]}{P(E\cap F)}
\displaystyle =P(E)-P(F)
\displaystyle =\frac{3}{10}-\frac{2}{7}=\frac{21-20}{70}=\frac{1}{70}
\\

\displaystyle \textbf{Question 22. }\text{For two events }A\text{ and }B,\text{ if }P(A)=0.4,\ P(B)=0.8\text{ and}
\displaystyle P(B/A)=0.6,\text{ then }P(A\cup B)\text{ is equal to} \hspace{2.2cm}\text{[CBSE 2023]}
\displaystyle \text{(a) }0.24 \qquad \text{(b) }0.3
\displaystyle \text{(c) }0.48 \qquad \text{(d) }0.96
\displaystyle \text{Answer:}
\displaystyle \text{(d) Given, }P(A)=0.4,\ P(B)=0.8\text{ and }P(B/A)=0.6
\displaystyle \text{We know that}
\displaystyle P\left(\frac{B}{A}\right)=\frac{P(B\cap A)}{P(A)}=\frac{P(A\cap B)}{P(A)}
\displaystyle \Rightarrow 0.6=\frac{P(A\cap B)}{0.4}\Rightarrow P(A\cap B)=0.24
\displaystyle \text{Now, }P(A\cup B)=P(A)+P(B)-P(A\cap B)
\displaystyle =0.4+0.8-0.24=0.96
\\

\displaystyle \textbf{Question 23. }\text{If the sum of numbers obtained on throwing a pair of}
\displaystyle \text{dice is }9,\text{ then the probability that number obtained on}   \text{one of the} \\ \text{dice is }4,\text{ is} \hspace{0.2cm}\text{[CBSE 2023]}
\displaystyle \text{(a) }\frac{1}{9} \qquad \text{(b) }\frac{4}{9}
\displaystyle \text{(c) }\frac{1}{18} \qquad \text{(d) }\frac{1}{2}
\displaystyle \text{Answer:}
\displaystyle \text{(d) In two throw of a dice the sample space }S\text{ is given by, }n(S)=6\times 6=36
\displaystyle \text{Let }E=\text{ Event of getting a sum of }9
\displaystyle E=\{(3,6),(4,5),(5,4),(6,3)\}
\displaystyle n(E)=4
\displaystyle \text{Let }F=\text{ Event of getting a }4\text{ on one of the dice}
\displaystyle F=\{(4,1),(4,2),(4,3),(4,4),(4,5),(4,6),(1,4),(2,4),(3,4),(5,4),(6,4)\}
\displaystyle n(F)=11
\displaystyle \therefore n(E\cap F)=2
\displaystyle \text{Required probability}=P\left(\frac{F}{E}\right)=\frac{P(F\cap E)}{P(E)}
\displaystyle =\frac{2/36}{4/36}=\frac{1}{2}
\\

\displaystyle \textbf{Question 24. }\text{The probability that }A\text{ speaks the truth is }\frac{4}{5}\text{ and that of}
\displaystyle B\text{ speaking the truth is }\frac{3}{4}.\text{ The probability that they contradict each other in stating}
\displaystyle \text{the same fact is} \hspace{2.2cm}\text{[CBSE 2023]}
\displaystyle \text{(a) }\frac{7}{20} \qquad \text{(b) }\frac{1}{5}
\displaystyle \text{(c) }\frac{3}{20} \qquad \text{(d) }\frac{4}{5}
\displaystyle \text{Answer:}
\displaystyle \text{(a) The probability of speaking truth of A is, }P(A)=\frac{4}{5}
\displaystyle \text{The probability of not speaking truth of A is}
\displaystyle P(\overline{A})=1-\frac{4}{5}=\frac{1}{5}
\displaystyle \text{The probability of speaking truth of B is, }P(B)=\frac{3}{4}
\displaystyle \text{The probability of not speaking truth of B is}
\displaystyle P(\overline{B})=1-\frac{3}{4}=\frac{1}{4}
\displaystyle \text{Now, the probability of they contradict each other}
\displaystyle =P(A)\cdot P(\overline{B})+P(\overline{A})\cdot P(B)
\displaystyle =\frac{4}{5}\times\frac{1}{4}+\frac{1}{5}\times\frac{3}{4}
\displaystyle =\frac{1}{5}+\frac{3}{20}=\frac{7}{20}
\\

\displaystyle \textbf{Question 25. }\text{Five fair coins are tossed simultaneously. The}
\displaystyle \text{probability of the events that atleast one head comes } \text{up is equal to} \hspace{2.2cm}\text{[CBSE 2023]}
\displaystyle \text{(a) }\frac{27}{32} \qquad \text{(b) }\frac{5}{32}
\displaystyle \text{(c) }\frac{31}{32} \qquad \text{(d) }\frac{1}{32}
\displaystyle \text{Answer:}
\displaystyle \text{(c) Total number of outcomes}=2^5=32
\displaystyle \text{Probability of getting all tails}=\frac{1}{32}
\displaystyle \therefore \text{Probability of getting atleast one head}
\displaystyle =1-\text{Probability of getting all tails}
\displaystyle =1-\frac{1}{32}=\frac{31}{32}
\\

\displaystyle \textbf{Question 26. }\text{For any two events }A\text{ and }B,\text{ if }P(\overline{A})=\frac{1}{2},\ P(B)=\frac{2}{3}\text{ and}
\displaystyle P(A\cap B)=\frac{1}{4},\text{ then }P\left(\frac{\overline{A}}{\overline{B}}\right)\text{ equals to} \hspace{2.2cm}\text{[CBSE 2023]}
\displaystyle \text{(a) }\frac{3}{8} \qquad \text{(b) }\frac{5}{8}
\displaystyle \text{(c) }\frac{1}{8} \qquad \text{(d) }\frac{1}{4}
\displaystyle \text{Answer:}
\displaystyle \text{(b) Given, }P(\overline{A})=\frac{1}{2},\ P(\overline{B})=\frac{2}{3}\text{ and }P(A\cap B)=\frac{1}{4}
\displaystyle \therefore P(A)=1-\frac{1}{2}=\frac{1}{2}\text{ and }P(B)=1-\frac{2}{3}=\frac{1}{3}
\displaystyle \text{Now, }P(A\cup B)=P(A)+P(B)-P(A\cap B)
\displaystyle =\frac{1}{2}+\frac{1}{3}-\frac{1}{4}
\displaystyle =\frac{6+4-3}{12}=\frac{7}{12}
\displaystyle \text{We have to find,}
\displaystyle P\left(\frac{\overline{A}}{\overline{B}}\right)=\frac{P(\overline{A}\cap \overline{B})}{P(\overline{B})}=\frac{P(\overline{A\cup B})}{P(\overline{B})}
\displaystyle =\frac{1-P(A\cup B)}{P(\overline{B})}
\displaystyle =\frac{1-\frac{7}{12}}{\frac{2}{3}}=\frac{5/12}{2/3}=\frac{5}{12}\times\frac{3}{2}=\frac{5}{8}
\\

\displaystyle \textbf{Question 27. }\text{Assertion (A) Two coins are tossed simultaneously. The probability of }
\displaystyle \text{getting two heads, if it is known }   \text{that at least one head comes up, is }\frac{1}{3}. \hspace{0.2cm}\text{[CBSE 2023]}
\displaystyle \text{Reason (R) Let }E\text{ and }F\text{ be two events with a random } \text{experiment, then }
\displaystyle P(F/E)=\frac{P(E\cap F)}{P(E)}.
\displaystyle \text{(a) Both Assertion and Reason are correct and Reason is the correct explanation of Assertion.}
\displaystyle \text{(b) Both Assertion and Reason are correct but Reason is not the correct explanation of Assertion.}
\displaystyle \text{(c) Assertion is correct but Reason is incorrect.}
\displaystyle \text{(d) Both Assertion and Reason are incorrect.}
\displaystyle \text{Answer:}
\displaystyle \text{(a) If two coins are tossed simultaneously, then sample space is given by}
\displaystyle S=\{(H,H),(H,T),(T,H),(T,T)\}
\displaystyle \text{Now, let A be the event of getting two heads .}
\displaystyle \therefore P(A)=\frac{1}{4}
\displaystyle \text{and B be the event of getting atleast one head.}
\displaystyle \therefore P(B)=\frac{3}{4}
\displaystyle \text{Thus, }P\left(\frac{A}{B}\right)=\frac{P(A\cap B)}{P(B)}=\frac{1/4}{3/4}=\frac{1}{4}\times\frac{4}{3}=\frac{1}{3}
\displaystyle \text{Hence, Assertion is true.}
\displaystyle \text{Reason is also true.}
\\

\displaystyle \textbf{Question 28. }A\text{ and }B\text{ are independent events such that } P(A\cap\overline{B})=\frac{1}{4}\text{ and }
\displaystyle P(\overline{A}\cap B)=\frac{1}{6}.\text{ Find }P(A)\text{ and }   P(B). \hspace{0.2cm}\text{[CBSE 2023]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, A and B are two independent events such that}
\displaystyle P(A\cap\overline{B})=\frac{1}{4}\text{ and }P(\overline{A}\cap B)=\frac{1}{6}
\displaystyle \text{Now,}
\displaystyle P(A\cap\overline{B})=\frac{1}{4}
\displaystyle \Rightarrow P(A)\cdot P(\overline{B})=\frac{1}{4}
\displaystyle \Rightarrow P(A)[1-P(B)]=\frac{1}{4}\qquad [\because P(B)+P(\overline{B})=1]
\displaystyle \Rightarrow P(A)-P(A)P(B)=\frac{1}{4}\qquad \cdots(i)
\displaystyle \text{and }P(\overline{A}\cap B)=\frac{1}{6}\Rightarrow P(\overline{A})\cdot P(B)=\frac{1}{6}
\displaystyle \Rightarrow [1-P(A)]P(B)=\frac{1}{6}
\displaystyle \Rightarrow P(B)-P(A)P(B)=\frac{1}{6}\qquad \cdots(ii)
\displaystyle \text{On subtracting Eq. (ii) from Eq. (i), we get}
\displaystyle P(A)-P(B)=\frac{1}{4}-\frac{1}{6}\Rightarrow P(A)=\frac{1}{12}+P(B)
\displaystyle \text{Now, on substituting this value in Eq. (ii), we get}
\displaystyle P(B)-\left(\frac{1}{12}+P(B)\right)P(B)=\frac{1}{6}
\displaystyle \text{Let }P(B)=x,\text{ then }x-\left(\frac{1+12x}{12}\right)x=\frac{1}{6}
\displaystyle \Rightarrow 12x-x-12x^2=2\Rightarrow 12x^2-11x+2=0
\displaystyle \Rightarrow 12x^2-8x-3x+2=0
\displaystyle \Rightarrow (4x-1)(3x-2)=0
\displaystyle \Rightarrow x=\frac{1}{4}\text{ or }x=\frac{2}{3}\Rightarrow P(B)=\frac{1}{4}\text{ or }P(B)=\frac{2}{3}
\displaystyle [\text{put }x=P(B)]
\displaystyle \text{Now, if }P(B)=\frac{1}{4},\text{ then}
\displaystyle P(A)=\frac{1}{12}+\frac{1}{4}=\frac{1+3}{12}=\frac{4}{12}=\frac{1}{3}
\displaystyle \text{and if }P(B)=\frac{2}{3},\text{ then }P(A)=\frac{1}{12}+\frac{2}{3}=\frac{1+8}{12}=\frac{9}{12}=\frac{3}{4}
\\

\displaystyle \textbf{Question 29. }A\text{ and }B\text{ throw a die alternately till one of them gets a}
\displaystyle '6'\text{ and wins the game. Find their respective } \text{probabilities of wining, if }A
\displaystyle \text{starts the game first.} \hspace{2.2cm}\text{[CBSE 2023]}
\displaystyle \text{Answer:}
\displaystyle \text{Let A be the event that player A gets 6 and B be the event that players B gets 6 when a} \\ \text{die is thrown, the probability of getting 6 is }\frac{1}{6}
\displaystyle \therefore P(A)=\frac{1}{6},\ P(A')=1-P(A)=\frac{5}{6}
\displaystyle \text{Similarly, }P(B)=\frac{1}{6},\ P(B')=\frac{5}{6}
\displaystyle \text{It is given that A starts the game and he will win in the following mutually exclusive ways.}
\displaystyle \text{(i) Player A wins at the first throw.}
\displaystyle \text{(ii) Player A wins in the third throw when A and B fail in the first and second throw and so on.}
\displaystyle \therefore P(A\text{ wins})=P(A)+P(A')P(B')P(A)
\displaystyle +P(A')P(B')P(A')P(B')P(A)+\cdots
\displaystyle =\frac{1}{6}+\frac{5}{6}\times\frac{5}{6}\times\frac{1}{6}+\frac{5}{6}\times\frac{5}{6}\times\frac{5}{6}\times\frac{5}{6}\times\frac{1}{6}+\cdots
\displaystyle =\frac{1}{6}\left[1+\left(\frac{5}{6}\right)^2+\left(\frac{5}{6}\right)^4+\cdots\right]
\displaystyle =\frac{1}{6}\left[\frac{1}{1-\left(\frac{5}{6}\right)^2}\right]
\displaystyle \left[\because \text{ it is an infinite GP with }a=1,\ r=\frac{25}{36}\right]
\displaystyle =\frac{1}{6}\left[\frac{1}{\frac{11}{36}}\right]=\frac{6}{11}
\displaystyle P(B\text{ wins})=1-P(A\text{ wins})=1-\frac{6}{11}=\frac{5}{11}
\\

\displaystyle \textbf{Question 30. }\text{There are two coins. One of them is a biased coin such that}
\displaystyle P(\text{head}):P(\text{tail})=1:3\text{ and the other coin is a fair coin. A coin is selected at}
\displaystyle \text{random and tossed once. If the coin showed head, then find the probability that it }
\displaystyle \text{is biased coin.} \hspace{2.2cm}\text{[CBSE 2023]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }E_1=\text{Event of selecting biased coin}
\displaystyle E_2=\text{Event of selecting fair coin,}
\displaystyle \text{and }A=\text{Event of getting head}
\displaystyle \text{Then, }P(E_1)=P(E_2)=\frac{1}{2}
\displaystyle P\left(\frac{A}{E_1}\right)=\frac{1}{4},\quad P\left(\frac{A}{E_2}\right)=\frac{1}{2}
\displaystyle \text{By Baye's theorem,}
\displaystyle P\left(\frac{E_1}{A}\right)=\frac{P(E_1)\cdot P\left(\frac{A}{E_1}\right)}{P(E_1)\cdot P\left(\frac{A}{E_1}\right)+P(E_2)\cdot P\left(\frac{A}{E_2}\right)}
\displaystyle =\frac{\frac{1}{2}\cdot\frac{1}{4}}{\frac{1}{2}\cdot\frac{1}{4}+\frac{1}{2}\cdot\frac{1}{2}}=\frac{\frac{1}{8}}{\frac{1}{8}+\frac{1}{4}}=\frac{1}{8}\times\frac{8}{3}=\frac{1}{3}
\\

\displaystyle \textbf{Question 31. }\text{In answering a question on a multiple choice test, a student either knows the}
\displaystyle \text{answer or guesses. Let }\frac{3}{5}\text{ be the probability that he knows the answer and }\frac{2}{5}
\displaystyle \text{be the probability that he guesses. Assuming that a student who guesses at the answer}
\displaystyle \text{will be correct with probability }\frac{1}{3},\text{ what is the probability that the student knows}
\displaystyle \text{the answer, given that he answered it correctly?} \hspace{2.2cm}\text{[CBSE 2023; All India 2015 C]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }E_1:\text{ Event that the student knows the answer}
\displaystyle E_2:\text{ Event that the student guesses the answer}
\displaystyle E:\text{ Event that the answer is correct}
\displaystyle \text{Here, }E_1\text{ and }E_2\text{ are mutually exclusive and exhaustive events.}
\displaystyle \therefore P(E_1)=\frac{3}{5}\text{ and }P(E_2)=\frac{2}{5}
\displaystyle \text{Now, }P\left(\frac{E}{E_1}\right)=P(\text{the student answered correctly, given he knows the answer})
\displaystyle =1
\displaystyle P\left(\frac{E}{E_2}\right)=P(\text{the student answered correctly, given he guesses})
\displaystyle =\frac{1}{3}
\displaystyle \text{The probability that the student knows the answer given that he} \\ \text{answered it correctly is given by }P\left(\frac{E_1}{E}\right).
\displaystyle \text{By using Baye's theorem, we get}
\displaystyle P\left(\frac{E_1}{E}\right)=\frac{P\left(\frac{E}{E_1}\right)\cdot P(E_1)}{P\left(\frac{E}{E_1}\right)P(E_1)+P\left(\frac{E}{E_2}\right)P(E_2)}
\displaystyle =\frac{1\times\frac{3}{5}}{1\times\frac{3}{5}+\frac{1}{3}\times\frac{2}{5}}
\displaystyle =\frac{\frac{3}{5}}{\frac{3}{5}+\frac{2}{15}}=\frac{3\times15}{5\times11}=\frac{9}{11}
\\

\displaystyle \textbf{Question 32. }\text{A pair of dice is thrown and the sum of the numbers appearing on the }
\displaystyle \text{dice is observed to be }7.\text{ Find the }   \text{probability that the number }5\text{ has appeared on atleast}
\displaystyle \text{one die.} \hspace{2.2cm}\text{[CBSE 2022 (Term II)]}
\displaystyle \text{Answer:}
\displaystyle \text{We know that when a pair of dice is thrown, total possible outcomes} \\ =6\times 6=36
\displaystyle \text{Let A be the event of getting sum as }7.
\displaystyle \text{Possible outcomes }=(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)
\displaystyle \therefore P(A)=\frac{6}{36}=\frac{1}{6}
\displaystyle \text{Let B be the event of getting number }5\text{ on atleast one dice.}
\displaystyle \text{Possible outcomes }=(1,5),(2,5),(3,5),(4,5),(5,1),(5,2),(5,3),(5,4),(5,5),(5,6),(6,5)
\displaystyle \text{Hence, }P(B)=\frac{11}{36}
\displaystyle (A\cap B)=\text{ Getting sum as }7\text{ with atleast one die showing }5
\displaystyle =(2,5),(5,2)
\displaystyle \text{Hence, }P(A\cap B)=\frac{2}{36}
\displaystyle \therefore P\left(\frac{B}{A}\right)=\frac{P(A\cap B)}{P(A)}=\frac{2/36}{1/6}=\frac{1}{3}
\\

\displaystyle \textbf{Question 33. }\text{The probability that }A\text{ hits the target is }\frac{1}{3}\text{ and the } \text{probability that }
\displaystyle B\text{ hits it is }\frac{2}{5}.\text{ If both try to hit the }   \text{target independently, find the probability that the}
\displaystyle \text{target is hit.} \hspace{2.2cm}\text{[CBSE 2022 (Term II)]}
\displaystyle \text{Answer:}
\displaystyle \text{Let E and F be the events defined as follows}
\displaystyle E:A\text{ hits the target}
\displaystyle \text{and }F:B\text{ hits the target}
\displaystyle P(E)=\frac{1}{3},\ P(\overline{E})=1-P(E)=\frac{2}{3}
\displaystyle P(F)=\frac{2}{5},\ P(\overline{F})=1-P(F)=\frac{3}{5}
\displaystyle P(\text{the target is hit})=1-P(\text{the target is not hit})
\displaystyle =1-\left(\frac{2}{3}\times\frac{3}{5}\right)=1-\frac{6}{15}
\displaystyle =\frac{9}{15}=\frac{3}{5}
\\

\displaystyle \textbf{Question 34. }\text{A die is thrown once. Let }A\text{ be the event that the number obtained is }
\displaystyle \text{greater than }3.\text{ Let }B\text{ be the event }   \text{that the number obtained is less than }5.\text{ Then, } \\   P(A\cup B)\text{ is} \hspace{0.2cm}\text{[CBSE 2020]}
\displaystyle \text{(a) }\frac{2}{5} \qquad \text{(b) }\frac{3}{5}
\displaystyle \text{(c) }0 \qquad \text{(d) }1
\displaystyle \text{Answer:}
\displaystyle \text{(d) We have, }A=\{4,5,6\}\text{ and }B=\{1,2,3,4\}
\displaystyle \text{Now, }A\cap B=\{4\}
\displaystyle \text{Now, }P(A\cup B)=P(A)+P(B)-P(A\cap B)
\displaystyle =\frac{3}{6}+\frac{4}{6}-\frac{1}{6}=\frac{6}{6}=1
\\

\displaystyle \textbf{Question 35. }\text{A card is picked at random from a pack of }52\text{ playing cards. Given that}
\displaystyle \text{the picked card is a queen, the }   \text{probability of this card to be a card of spade is} \hspace{0.2cm}\text{[CBSE 2020]}
\displaystyle \text{(a) }\frac{1}{3} \qquad \text{(b) }\frac{4}{13}
\displaystyle \text{(c) }\frac{1}{4} \qquad \text{(d) }\frac{1}{2}
\displaystyle \text{Answer:}
\displaystyle \text{(c) Let A be the event that card drawn is a spade and B}
\displaystyle \text{be the event that card drawn is a queen. We have a}
\displaystyle \text{total of 13 spades and 4 queen and one queen is from spade.}
\displaystyle \therefore P(A)=\frac{13}{52}=\frac{1}{4},\ P(B)=\frac{4}{52}=\frac{1}{13}\text{ and }P(A\cap B)=\frac{1}{52}
\displaystyle \therefore P\left(\frac{A}{B}\right)=\frac{P(A\cap B)}{P(B)}=\frac{1/52}{1/13}=\frac{1}{4}
\\

\displaystyle \textbf{Question 36. }\text{Three dice are thrown simultaneously. The probability of obtaining }
\displaystyle \text{a total score of }5\text{ is} \hspace{2.2cm}\text{[CBSE 2020]}
\displaystyle \text{(a) }\frac{5}{216} \qquad \text{(b) }\frac{1}{6}
\displaystyle \text{(c) }\frac{1}{36} \qquad \text{(d) }\frac{1}{49}
\displaystyle \text{Answer:}
\displaystyle \text{(c) Three dice are thrown simultaneously favourable}
\displaystyle \text{outcomes are }\{(1,1,3),(1,3,1),(3,1,1),(1,2,2),(2,1,2),(2,2,1)\}
\displaystyle =6
\displaystyle \text{Total number of outcomes}=6^3=216
\displaystyle \therefore \text{Required probability}=\frac{6}{216}=\frac{1}{36}
\\

\displaystyle \textbf{Question 37. }\text{A bag contains }3\text{ white, }4\text{ black and }2\text{ red balls. If 2 balls are drawn at }
\displaystyle \text{random (without replacement), }   \text{then the probability that both the balls are white, is} \\ \hspace{0.2cm}\text{[CBSE 2020]}
\displaystyle \text{(a) }\frac{1}{18} \qquad \text{(b) }\frac{1}{36}
\displaystyle \text{(c) }\frac{1}{12} \qquad \text{(d) }\frac{1}{24}
\displaystyle \text{Answer:}
\displaystyle \text{(c) We have, 3 white, 4 black and 2 red balls.}
\displaystyle \text{Total number of balls }=(3+4+2)=9
\displaystyle \text{Two balls are drawn at random (without replacement).}
\displaystyle \text{Then, the probability that both the balls are white}
\displaystyle =\frac{3}{9}\times\frac{2}{8}=\frac{6}{72}=\frac{1}{12}
\\

\displaystyle \textbf{Question 38. }\text{From the set }\{1,2,3,4,5\},\text{ two numbers }a\text{ and }b\ (a\neq b)
\displaystyle \text{are chosen at random. The probability that }\frac{a}{b}\text{ is an}   \text{ integer, is} \hspace{2.2cm}\text{[CBSE 2020]}
\displaystyle \text{(a) }\frac{1}{3} \qquad \text{(b) }\frac{1}{4}
\displaystyle \text{(c) }\frac{1}{2} \qquad \text{(d) }\frac{3}{5}
\displaystyle \text{Answer:}
\displaystyle \text{(c) We have set of numbers }\{1,2,3,4,5\}
\displaystyle \text{Sample space of choosing two numbers}
\displaystyle ={}^{5}C_{2}=\frac{5\times4}{1\times2}=10
\displaystyle \text{Favourable outcomes are }\left(\frac{2}{1},\frac{3}{1},\frac{4}{1},\frac{5}{1},\frac{4}{2}\right)=5
\displaystyle \therefore \text{Required probability}=\frac{5}{10}=\frac{1}{2}
\\

\displaystyle \textbf{Question 39. }\text{Given two independent events }A\text{ and }B\text{ such that}
\displaystyle P(A)=0.3\text{ and }P(B)=0.6,\text{ find }P(A'/B). \hspace{2.2cm}\text{[CBSE 2020]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }P(A)=0.3\text{ and }P(B)=0.6
\displaystyle \text{Now, }P(A'\cap B')=P(A\cup B)'
\displaystyle =1-P(A\cup B)
\displaystyle =1-[P(A)+P(B)-P(A\cap B)]
\displaystyle =1-[0.3+0.6-(0.3\times0.6)]
\displaystyle [\because A\text{ and }B\text{ are independent events }\therefore P(A\cap B)=P(A)P(B)]
\displaystyle =1-[0.9-0.18]
\displaystyle =1-\{0.72\}=0.28
\\

\displaystyle \textbf{Question 40. }\text{Three distinct numbers are chosen randomly from the first 50 natural}
\displaystyle \text{numbers. Find the probability that all } \text{the three numbers are divisible by both 2 and 3.}
\displaystyle \hspace{2.2cm}\text{[CBSE 2020]}
\displaystyle \text{Answer:}
\displaystyle \text{Three distinct numbers are chosen from first }50\text{ natural numbers in }{}^{50}C_{3}\text{ ways.}
\displaystyle \text{Total numbers which is divisible by }2\text{ and }3\text{ from first }50\text{ natural numbers}
\displaystyle \text{is }\{6,12,18,24,30,36,42,48\}=8
\displaystyle \therefore \text{Required probability}=\frac{{}^{8}C_{3}}{{}^{50}C_{3}}
\displaystyle =\frac{8\times7\times6}{50\times49\times48}
\displaystyle =\frac{1}{350}
\\

\displaystyle \textbf{Question 41. }\text{Suppose that }5\text{ men out of }100\text{ and }25\text{ women out of }1000\text{ are good}
\displaystyle \text{orators. Assuming that there are equal number of men and women, find the probability of}
\displaystyle \text{choosing a good orator.} \hspace{2.2cm}\text{[Delhi 2020]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }E_1\text{ be the event that selected person is men's and}
\displaystyle E_2\text{ be the event that selected person is women, }E_1\text{ and }E_2\text{ are mutually exclusive and}
\displaystyle \text{exhaustive eventmoreover }P(E_1)=P(E_2)=\frac{1}{2}
\displaystyle \text{Let }E\text{ be the event that selected person is good orator.}
\displaystyle \therefore P\left(\frac{E}{E_1}\right)=\frac{5}{100}=\frac{1}{20}
\displaystyle \text{and }P\left(\frac{E}{E_2}\right)=\frac{25}{1000}=\frac{1}{40}
\displaystyle \text{The probability of choosing a good orator,}
\displaystyle P(E)=P(E_1)\times P\left(\frac{E}{E_1}\right)+P(E_2)\times P\left(\frac{E}{E_2}\right)
\displaystyle =\frac{1}{2}\times\frac{1}{20}+\frac{1}{2}\times\frac{1}{40}=\frac{2+1}{2\times40}=\frac{3}{80}
\\

\displaystyle \textbf{Question 42. }\text{In a shop }X,\ 30\text{ tins of ghee of type }A\text{ and }40\text{ tins of ghee of type }B
\displaystyle \text{which look alike, are kept for sale. While in shop }Y,\text{ similar }50\text{ tins of ghee of}
\displaystyle \text{type }A\text{ and }60\text{ tins of ghee of type }B\text{ are there. One tin of ghee is purchased}
\displaystyle \text{from one of the randomly selected shop and is found to be of type }B.\text{ Find the}
\displaystyle \text{probability that it is purchased from shop }Y. \hspace{2.2cm}\text{[All India 2020]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }E_1=\text{Getting ghee from shop }X,
\displaystyle E_2=\text{Getting ghee from shop }Y
\displaystyle \text{and }A=\text{Getting type }B\text{ ghee}
\displaystyle \therefore P(E_1)=P(E_2)=\frac{1}{2}
\displaystyle [\because \text{ both shop have equal chances}]
\displaystyle P\left(\frac{A}{E_1}\right)=\text{Probability that type }B\text{ ghee is purchased from shop }X
\displaystyle =\frac{40}{70}=\frac{4}{7}
\displaystyle P\left(\frac{A}{E_2}\right)=\text{Probability that type }B\text{ ghee is purchased from shop }Y
\displaystyle =\frac{60}{110}=\frac{6}{11}
\displaystyle \text{Now, by Baye's theorem, we get}
\displaystyle P\left(\frac{E_2}{A}\right)=\frac{P(E_2)P\left(\frac{A}{E_2}\right)}{P(E_1)P\left(\frac{A}{E_1}\right)+P(E_2)P\left(\frac{A}{E_2}\right)}
\displaystyle =\frac{\frac{1}{2}\times\frac{6}{11}}{\frac{1}{2}\times\frac{4}{7}+\frac{1}{2}\times\frac{6}{11}}=\frac{\frac{6}{11}}{\frac{4}{7}+\frac{6}{11}}
\displaystyle =\frac{\frac{6}{11}}{\frac{44+42}{77}}=\frac{42}{86}=\frac{21}{43}
\\

\displaystyle \textbf{Question 43. }\text{If }P(\text{not }A)=0.7,\ P(B)=0.7\text{ and }P(B/A)=0.5,\text{ then } \text{find }P(A/B).
\displaystyle \hspace{2.2cm}\text{[CBSE 2019]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }P(A')=0.7,\ P(B)=0.7\text{ and }P\left(\frac{B}{A}\right)=0.5
\displaystyle \text{Clearly, }P(A)=1-P(A')=1-0.7=0.3
\displaystyle \text{Now, }P\left(\frac{B}{A}\right)=\frac{P(A\cap B)}{P(A)}
\displaystyle \Rightarrow 0.5=\frac{P(A\cap B)}{0.3}\Rightarrow P(A\cap B)=0.15
\displaystyle \therefore P\left(\frac{A}{B}\right)=\frac{P(A\cap B)}{P(B)}=\frac{0.15}{0.7}\Rightarrow P\left(\frac{A}{B}\right)=\frac{3}{14}
\\

\displaystyle \textbf{Question 44. }\text{A die marked }1,2,3\text{ in red and }4,5,6\text{ in green is tossed. Let A be the event}
\displaystyle \text{ 'number is even' and }B\text{ be the }   \text{event 'number is marked red'. Find whether the events}
\displaystyle A\text{ and }B\text{ are independent or not.} \hspace{2.2cm}\text{[CBSE 2019]}
\displaystyle \text{Or}
\displaystyle \text{A die, whose faces are marked }1,2,3\text{ in red and }4,5,6 \text{ in green is tossed. Let A be the}
\displaystyle \text{event 'number }   \text{obtained is even' and }B\text{ be the event 'number obtained } \text{is red'. Find if }
\displaystyle A\text{ and }B\text{ are independent events.} \hspace{2.2cm}\text{[CBSE 2017]}
\displaystyle \text{Answer:}
\displaystyle \text{When a die is thrown, the sample space is}
\displaystyle S=\{1,2,3,4,5,6\}
\displaystyle \Rightarrow n(S)=6
\displaystyle \text{Also, }A:\text{ number is even and }B:\text{ number is red.}
\displaystyle \therefore A=\{2,4,6\}\text{ and }B=\{1,2,3\}\text{ and }A\cap B=\{2\}
\displaystyle \Rightarrow n(A)=3,\ n(B)=3\text{ and }n(A\cap B)=1
\displaystyle \text{Now, }P(A)=\frac{n(A)}{n(S)}=\frac{3}{6}=\frac{1}{2}
\displaystyle P(B)=\frac{n(B)}{n(S)}=\frac{3}{6}=\frac{1}{2}
\displaystyle \text{and }P(A\cap B)=\frac{n(A\cap B)}{n(S)}=\frac{1}{6}
\displaystyle \text{Now, }P(A)\times P(B)=\frac{1}{2}\times\frac{1}{2}=\frac{1}{4}\ne\frac{1}{6}=P(A\cap B)
\displaystyle \therefore P(A\cap B)\ne P(A)\times P(B)
\displaystyle \text{Thus, A and B are not independent events.}
\\

\displaystyle \textbf{Question 45. }\text{There are three coins. One is a two-headed coin, another is a biased coin that}
\displaystyle \text{comes up heads }75\%\text{ of the time and the third is an unbiased coin. One of the three}
\displaystyle \text{coins is chosen at random and tossed. If it shows head, what is the probability that}
\displaystyle \text{it is the two-headed coin?} \hspace{2.2cm}\text{[All India 2019]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }E_1=\text{Event of selecting two headed coin}
\displaystyle E_2=\text{Event of selecting biased coin}
\displaystyle E_3=\text{Event of selecting unbiased coin}
\displaystyle A=\text{Event of getting head}
\displaystyle \text{Then, }P(E_1)=P(E_2)=P(E_3)=\frac{1}{3}
\displaystyle P\left(\frac{A}{E_1}\right)=1,\ P\left(\frac{A}{E_2}\right)=\frac{75}{100}=\frac{3}{4},\ P\left(\frac{A}{E_3}\right)=\frac{1}{2}
\displaystyle \text{By Baye's theorem,}
\displaystyle P\left(\frac{E_1}{A}\right)=\frac{P(E_1)\cdot P\left(\frac{A}{E_1}\right)}{P(E_1)\cdot P\left(\frac{A}{E_1}\right)+P(E_2)\cdot P\left(\frac{A}{E_2}\right)+P(E_3)\cdot P\left(\frac{A}{E_3}\right)}
\displaystyle =\frac{\frac{1}{3}\times1}{\frac{1}{3}\times1+\frac{1}{3}\times\frac{3}{4}+\frac{1}{3}\times\frac{1}{2}}
\displaystyle =\frac{\frac{1}{3}}{\frac{1}{3}+\frac{1}{4}+\frac{1}{6}}=\frac{1}{3}\times\frac{12}{9}=\frac{4}{9}
\\

\displaystyle \textbf{Question 46. }\text{A bag contains }5\text{ red and }4\text{ black balls, while a second bag contains }3
\displaystyle \text{red and }6\text{ black balls. One of the two bags is selected at random and two balls are}
\displaystyle \text{drawn at random (without replacement), both of which are found to be red. Find the}
\displaystyle \text{probability that the balls were drawn from the second bag.} \hspace{2.2cm}\text{[All India 2019]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }E_1,E_2\text{ and }A\text{ denote the following events}
\displaystyle E_1=\text{First bag is chosen, }E_2=\text{Second bag is chosen}
\displaystyle \text{and }A=\text{Two balls drawn at random are red.}
\displaystyle \text{Since, one of the bag is chosen at random.}
\displaystyle \therefore P(E_1)=P(E_2)=\frac{1}{2}
\displaystyle \text{If }E_1\text{ has already occurred, i.e., first bag is chosen.}
\displaystyle \text{Therefore, the probability of drawing two red balls in this case}
\displaystyle =P\left(\frac{A}{E_1}\right)=\frac{{}^{5}C_{2}}{{}^{9}C_{2}}=\frac{10}{36}
\displaystyle \text{Similarly, }P\left(\frac{A}{E_2}\right)=\frac{{}^{3}C_{2}}{{}^{9}C_{2}}=\frac{3}{36}
\displaystyle \text{We are required to find }P\left(\frac{E_2}{A}\right)
\displaystyle \text{By Baye's theorem,}
\displaystyle P\left(\frac{E_2}{A}\right)=\frac{P(E_2)\cdot P\left(\frac{A}{E_2}\right)}{P(E_1)\cdot P\left(\frac{A}{E_1}\right)+P(E_2)\cdot P\left(\frac{A}{E_2}\right)}
\displaystyle =\frac{\frac{1}{2}\times\frac{3}{36}}{\frac{1}{2}\times\frac{10}{36}+\frac{1}{2}\times\frac{3}{36}}=\frac{\frac{3}{72}}{\frac{10}{72}+\frac{3}{72}}=\frac{3}{13}
\\

\displaystyle \textbf{Question 47. }\text{A manufacturer has three machine operators }A,\ B\text{ and }C.\text{ The first operator }A
\displaystyle \text{produces }1\%\text{ of defective items, whereas the other two operators }B\text{ and }C\text{ produce }5\%\text{ and }7\%
\displaystyle \text{defective items, respectively. }A\text{ is on the job for }50\%\text{ of the time, }B\text{ on the job }30\%
\displaystyle \text{of the time and }C\text{ on the job for }20\%\text{ of the time. All the items are put into one stockpile}
\displaystyle \text{and then one item is chosen at random from this and is found to be defective. What is the}
\displaystyle \text{probability that it was produced by }A? \hspace{2.2cm}\text{[Delhi 2019]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }A:\text{ Event that item produced by operator A}
\displaystyle B:\text{ Event that item produced by operator B}
\displaystyle C:\text{ Event that item produced by operator C}
\displaystyle D:\text{ Event that product produced is defective}
\displaystyle \text{We need to find out the probability that item is produced by operator} \\ \text{A if it is defective i.e., }P(A/D).
\displaystyle \text{So, }P(A/D)=\frac{P(A)\cdot P(D/A)}{P(A)\cdot P(D/A)+P(B)\cdot P(D/B)+P(C)\cdot P(D/C)}\qquad \cdots(i)
\displaystyle \text{[by Baye's theorem]}
\displaystyle P(A)=\text{Probability of item is produced by operator A}
\displaystyle =50\%=\frac{50}{100}=0.5
\displaystyle P(B)=\text{Probability of item is produced by operator B}
\displaystyle =30\%=\frac{30}{100}=0.3
\displaystyle P(C)=\text{Probability of item is produced by operator C}
\displaystyle =20\%=\frac{20}{100}=0.2
\displaystyle P(D/A)=\text{Probability of a defective item produced by operator A}
\displaystyle =1\%=\frac{1}{100}=0.01
\displaystyle P(D/B)=\text{Probability of a defective item produced by operator B}
\displaystyle =5\%=\frac{5}{100}=0.05
\displaystyle P(D/C)=\text{Probability of a defective item produced by operator C}
\displaystyle =7\%=\frac{7}{100}=0.07
\displaystyle \text{Putting these values in the Eq. (i), we get}
\displaystyle P(A/D)=\frac{0.5\times0.01}{0.5\times0.01+0.3\times0.05+0.2\times0.07}
\displaystyle =\frac{0.005}{0.005+0.015+0.014}=\frac{0.005}{0.034}=\frac{5}{34}
\displaystyle \text{Therefore, required probability}=\frac{5}{34}
\\

\displaystyle \textbf{Question 48. }\text{A black and a red die are rolled together. Find the conditional probability of }
\displaystyle \text{obtaining the sum }8,\text{ given }   \text{that the red die resulted in a number less than }4. \hspace{0.2cm}\text{[CBSE 2018]}
\displaystyle \text{Answer:}
\displaystyle \text{Let us denote the numbers on black die by }B_1,B_2,\ldots,B_6\text{ and the numbers} \\ \text{on red die by }R_1,R_2,\ldots,R_6.
\displaystyle \text{Then, we get the following sample space}
\displaystyle S=\{(B_1,R_1),(B_1,R_2),\ldots,(B_1,R_6),(B_2,R_1),(B_2,R_2),\ldots,(B_2,R_6),\ldots, \\ (B_6,R_1),(B_6,R_2),\ldots,(B_6,R_6)\}
\displaystyle \text{Clearly, }n(S)=36
\displaystyle \text{Now, let }A\text{ be the event that sum of number obtained on the die is } \\ 8\text{ and }B\text{ be the event that red die shows a number less than }4.
\displaystyle \text{Then,}
\displaystyle A=\{(B_2,R_6),(B_6,R_2),(B_3,R_5),(B_5,R_3),(B_4,R_4)\}
\displaystyle \text{and }B=\{(B_1,R_1),(B_1,R_2),(B_1,R_3),(B_2,R_1),(B_2,R_2),(B_2,R_3),\ldots,(B_6,R_1), \\ (B_6,R_2),(B_6,R_3)\}
\displaystyle \Rightarrow A\cap B=\{(B_6,R_2),(B_5,R_3)\}
\displaystyle \text{Now, required probability,}
\displaystyle P\left(\frac{A}{B}\right)=\frac{P(A\cap B)}{P(B)}=\frac{2/36}{18/36}=\frac{2}{18}=\frac{1}{9}
\\

\displaystyle \textbf{Question 48. }\text{Evaluate }P(A\cup B),\text{ if }2P(A)=P(B)=\frac{5}{13}\text{ and } \\   P(A/B)=\frac{2}{5}. \hspace{0.2cm}\text{[CBSE 2018C]}
\displaystyle \text{Answer:}
\displaystyle \text{We have, }2P(A)=P(B)=\frac{5}{13}
\displaystyle \Rightarrow P(A)=\frac{5}{26},\ P(B)=\frac{5}{13}\text{ and }P(A/B)=\frac{2}{5}
\displaystyle \therefore P\left(\frac{A}{B}\right)=\frac{P(A\cap B)}{P(B)}
\displaystyle \frac{2}{5}=\frac{P(A\cap B)}{5/13}\Rightarrow P(A\cap B)=\frac{2}{5}\times\frac{5}{13}=\frac{2}{13}
\displaystyle P(A\cup B)=P(A)+P(B)-P(A\cap B)
\displaystyle =\frac{5}{26}+\frac{5}{13}-\frac{2}{13}=\frac{5+10-4}{26}=\frac{11}{26}
\\

\displaystyle \textbf{Question 50. }\text{Prove that if }E\text{ and }F\text{ are independent events, then the events}
\displaystyle  E\text{ and }F'\text{ are also independent.} \hspace{2.2cm}\text{[CBSE 2017]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, E and F are independent events, therefore}
\displaystyle P(E\cap F)=P(E)P(F)\qquad \cdots(1)
\displaystyle \text{Now, we have}
\displaystyle P(E\cap F')+P(E\cap F)=P(E)
\displaystyle \Rightarrow P(E\cap F')=P(E)-P(E\cap F)
\displaystyle \Rightarrow P(E\cap F')=P(E)-P(E)P(F)
\displaystyle \text{[using Eq. (1)]}
\displaystyle \Rightarrow P(E\cap F')=P(E)[1-P(F)]
\displaystyle \Rightarrow P(E\cap F')=P(E)P(F')
\displaystyle \therefore E\text{ and }F'\text{ are also independent events.}
\displaystyle \text{Hence proved}
\\

\displaystyle \textbf{Question 51. }\text{Suppose a girl throws a die. If she gets }1\text{ or }2,\text{ she tosses a coin three}
\displaystyle \text{times and notes the number of tails. If she gets }3,4,5\text{ or }6,\text{ she tosses a coin}
\displaystyle \text{once and notes whether a 'head' or 'tail' is obtained. If she obtained exactly}
\displaystyle \text{one 'tail', what is the probability that she threw }3,4,5\text{ or }6\text{ with the die?}
\displaystyle \hspace{2.2cm}\text{[CBSE 2018]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }E_1\text{ be the event that the girl gets }1\text{ or }2,
\displaystyle E_2\text{ be the event that the girl gets }3,4,5\text{ or }6
\displaystyle \text{and }A\text{ be the event that the girl gets exactly a 'tail'.}
\displaystyle \text{Then, }P(E_1)=\frac{2}{6}=\frac{1}{3}\text{ and }P(E_2)=\frac{4}{6}=\frac{2}{3}
\displaystyle P\left(\frac{A}{E_1}\right)=P(\text{getting exactly one tail when a coin is tossed three times}) \\ =\frac{3}{8}
\displaystyle P\left(\frac{A}{E_2}\right)=P(\text{getting exactly a tail when a coin is tossed once})  =\frac{1}{2}
\displaystyle \text{Now, required probability}
\displaystyle P\left(\frac{E_2}{A}\right)=\frac{P(E_2)\cdot P\left(\frac{A}{E_2}\right)}{P(E_1)\cdot P\left(\frac{A}{E_1}\right)+P(E_2)\cdot P\left(\frac{A}{E_2}\right)}
\displaystyle =\frac{\frac{2}{3}\cdot\frac{1}{2}}{\frac{1}{3}\cdot\frac{3}{8}+\frac{2}{3}\cdot\frac{1}{2}}
\displaystyle =\frac{\frac{1}{3}}{\frac{1}{8}+\frac{1}{3}}=\frac{8}{11}
\\

\displaystyle \textbf{Question 52. }\text{Two groups are competing for the positions of the Board of Directors of a}
\displaystyle \text{corporation. The probabilities that the first and second group will win are }0.6
\displaystyle \text{and }0.4,\text{ respectively. Further, if the first group wins, the probability of}
\displaystyle \text{introducing a new product is }0.7\text{ and the corresponding probability is }0.3
\displaystyle \text{if the second group wins. Find the probability that the new product introduced}
\displaystyle \text{was by the second group.} \hspace{2.2cm}\text{[CBSE 2018C]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }E_1\text{ and }E_2\text{ denote the events that first and second group will win. Then,}
\displaystyle P(E_1)=0.6\text{ and }P(E_2)=0.4
\displaystyle \text{Let }E\text{ be the event of introducing the new product.}
\displaystyle \text{Then, }P\left(\frac{E}{E_1}\right)=0.7\text{ and }P\left(\frac{E}{E_2}\right)=0.3
\displaystyle \text{Now, we have to find the probability that new product is introduced by second event.}
\displaystyle \therefore P\left(\frac{E_2}{E}\right)=\frac{P(E_2)P\left(\frac{E}{E_2}\right)}{P(E_1)P\left(\frac{E}{E_1}\right)+P(E_2)P\left(\frac{E}{E_2}\right)}
\displaystyle =\frac{0.4\times0.3}{0.6\times0.7+0.4\times0.3}=\frac{0.12}{0.42+0.12}=\frac{0.12}{0.54}
\displaystyle =0.22
\\

\displaystyle \textbf{Question 53. }A\text{ and }B\text{ throw a pair of dice alternately. } A\text{ wins the game,}
\displaystyle \text{if he gets a total of }7\text{ and }B\text{ wins the } \text{game, if he gets a total of }10.
\displaystyle \text{ If }A\text{ starts the game, then }  \text{find the probability that }B\text{ wins.} \hspace{2.2cm}\text{[CBSE 2016]}
\displaystyle \text{Answer:}
\displaystyle \text{Here, }n(S)=6\times6=36
\displaystyle \text{Let A = Event of getting a sum of }7\text{ in pair of dice}
\displaystyle =\{(1,6),(2,5),(3,4),(6,1),(5,2),(4,3)\}
\displaystyle \Rightarrow n(A)=6
\displaystyle \text{and B = Event of getting a sum of }10\text{ in pair of dice}
\displaystyle =\{(4,6),(5,5),(6,4)\}
\displaystyle \Rightarrow n(B)=3
\displaystyle \therefore P(A)=\frac{n(A)}{n(S)}=\frac{6}{36}=\frac{1}{6}
\displaystyle \text{and }P(B)=\frac{n(B)}{n(S)}=\frac{3}{36}=\frac{1}{12}
\displaystyle \Rightarrow P(\overline{A})=1-\frac{1}{6}=\frac{5}{6}
\displaystyle \text{and }P(\overline{B})=1-\frac{1}{12}=\frac{11}{12}
\displaystyle \text{Now, the probability that if A starts the game, then B wins}
\displaystyle \therefore P(B\text{ wins})=P(\overline{A}\cap B)+P(\overline{A}\cap\overline{B}\cap\overline{A}\cap B)
\displaystyle +P(\overline{A}\cap\overline{B}\cap\overline{A}\cap\overline{B}\cap\overline{A}\cap B)+\cdots
\displaystyle =P(\overline{A})P(B)+P(\overline{A})P(\overline{B})P(\overline{A})P(B)
\displaystyle +P(\overline{A})P(\overline{B})P(\overline{A})P(\overline{B})P(\overline{A})P(B)+\cdots
\displaystyle [\because \text{events are independent}]
\displaystyle =\frac{5}{6}\times\frac{1}{12}+\frac{5}{6}\times\frac{11}{12}\times\frac{5}{6}\times\frac{1}{12}
\displaystyle +\frac{5}{6}\times\frac{11}{12}\times\frac{5}{6}\times\frac{11}{12}\times\frac{5}{6}\times\frac{1}{12}+\cdots
\displaystyle =\frac{5}{72}+\frac{5}{72}\times\frac{55}{72}+\frac{5}{72}\times\left(\frac{55}{72}\right)^2+\cdots
\displaystyle =\frac{5}{72}\left[1+\frac{55}{72}+\left(\frac{55}{72}\right)^2+\cdots\right]=\frac{5}{72}\left[\frac{1}{1-\frac{55}{72}}\right]
\displaystyle \left[\because \text{sum of infinite GP series is }\frac{a}{1-r}\right]
\displaystyle =\frac{5}{72}\left[\frac{1}{\frac{17}{72}}\right]=\frac{5}{17}
\\

\displaystyle \textbf{Question 54. }A\text{ and }B\text{ throw a pair of dice alternately, till one of them}
\displaystyle \text{gets a total of }10\text{ and wins the game. Find their respective probabilities of winning, if}
\displaystyle  A\text{ starts first.} \hspace{2.2cm}\text{[CBSE 2016]}
\displaystyle \text{Answer:}
\displaystyle \text{Here, }n(S)=6\times6=36
\displaystyle \text{Let }E=\text{ Event of getting a total }10
\displaystyle =\{(4,6),(5,5),(6,4)\}
\displaystyle \therefore n(E)=3
\displaystyle \therefore P(\text{getting a total of }10)=P(E)=\frac{n(E)}{n(S)}=\frac{3}{36}=\frac{1}{12}
\displaystyle \text{and }P(\text{not getting a total of }10)=P(\overline{E})=1-P(E)=1-\frac{1}{12}=\frac{11}{12}
\displaystyle \text{Thus, }P(A\text{ getting }10)=P(B\text{ getting }10)=\frac{1}{12}
\displaystyle \text{and }P(A\text{ is not getting }10)=P(B\text{ is not getting }10)=\frac{11}{12}
\displaystyle \text{Now, }P(A\text{ winning})=P(A)+P(\overline{A}\cap\overline{B}\cap A)
\displaystyle +P(\overline{A}\cap\overline{B}\cap\overline{A}\cap\overline{B}\cap A)+\cdots
\displaystyle =P(A)+P(\overline{A})P(\overline{B})P(A)
\displaystyle +P(\overline{A})P(\overline{B})P(\overline{A})P(\overline{B})P(A)+\cdots
\displaystyle =\frac{1}{12}+\frac{11}{12}\times\frac{11}{12}\times\frac{1}{12}
\displaystyle +\frac{11}{12}\times\frac{11}{12}\times\frac{11}{12}\times\frac{11}{12}\times\frac{1}{12}+\cdots
\displaystyle =\frac{1}{12}\left[1+\left(\frac{11}{12}\right)^2+\left(\frac{11}{12}\right)^4+\cdots\right]
\displaystyle =\frac{1}{12}\left[\frac{1}{1-\left(\frac{11}{12}\right)^2}\right]
\displaystyle \left[\because \text{the sum of an infinite GP is }S_{\infty}=\frac{a}{1-r}\right]
\displaystyle =\frac{1}{12}\left[\frac{1}{\frac{144-121}{144}}\right]
\displaystyle =\frac{1}{12}\times\frac{144}{23}=\frac{12}{23}
\displaystyle \text{Now, }P(B\text{ winning})=1-P(A\text{ winning})
\displaystyle =1-\frac{12}{23}=\frac{11}{23}
\displaystyle \text{Hence, the probabilities of winning A and B are respectively, }
\displaystyle \frac{12}{23}\text{ and }\frac{11}{23}
\\

\displaystyle \textbf{Question 55. }\text{Three persons }A,\ B\text{ and }C\text{ apply for a job of Manager in a private company.}
\displaystyle \text{Chances of their selection }(A,\ B\text{ and }C)\text{ are in the ratio }1:2:4.\text{ The probabilities that}
\displaystyle A,\ B\text{ and }C\text{ can introduce changes to improve profits of the company are }0.8,\ 0.5
\displaystyle \text{ and }0.3,\text{ respectively. }\text{If the change does not take place, then find the probability }
\displaystyle \text{probability that it is due to the appointment of }C. \hspace{2.2cm}\text{[Delhi 2016]}
\displaystyle \text{Answer:}
\displaystyle \text{Let us define the following events}
\displaystyle A=\text{Selecting person }A
\displaystyle B=\text{Selecting person }B
\displaystyle C=\text{Selecting person }C
\displaystyle P(A)=\frac{1}{1+2+4},\ P(B)=\frac{2}{1+2+4}
\displaystyle \text{and }P(C)=\frac{4}{1+2+4}
\displaystyle \Rightarrow P(A)=\frac{1}{7},\ P(B)=\frac{2}{7}\text{ and }P(C)=\frac{4}{7}
\displaystyle \text{Let }E=\text{ Event to introduce the changes in their profit.}
\displaystyle \text{Also, given }P\left(\frac{E}{A}\right)=0.8,\ P\left(\frac{E}{B}\right)=0.5\text{ and }P\left(\frac{E}{C}\right)=0.3
\displaystyle \Rightarrow P\left(\frac{\overline{E}}{A}\right)=1-0.8=0.2,\ P\left(\frac{\overline{E}}{B}\right)=1-0.5=0.5
\displaystyle \text{and }P\left(\frac{\overline{E}}{C}\right)=1-0.3=0.7
\displaystyle \text{The probability that change does not take place by the appointment of }C,
\displaystyle P\left(\frac{C}{\overline{E}}\right)=\frac{P(C)\cdot P\left(\frac{\overline{E}}{C}\right)}{P(A)\times P\left(\frac{\overline{E}}{A}\right)+P(B)\times P\left(\frac{\overline{E}}{B}\right)+P(C)\times P\left(\frac{\overline{E}}{C}\right)}
\displaystyle =\frac{\frac{4}{7}\times0.7}{\frac{1}{7}\times0.2+\frac{2}{7}\times0.5+\frac{4}{7}\times0.7}
\displaystyle =\frac{2.8}{0.2+1.0+2.8}=\frac{2.8}{4}=0.7
\\

\displaystyle \textbf{Question 56. }\text{A bag }X\text{ contains }4\text{ white balls and }2\text{ black balls, while another bag }Y
\displaystyle \text{contains }3\text{ white balls and }3\text{ black balls. Two balls are drawn (without replacement)}
\displaystyle \text{at random from one of the bags and were found to be one white and one black.}
\displaystyle \text{Find the probability that the balls were drawn from bag }Y. \hspace{1.2cm}\text{[All India 2016]}
\displaystyle \text{Answer:}
\displaystyle \text{Let us define the following events}
\displaystyle E_1:\text{ Bag }X\text{ is selected}
\displaystyle E_2:\text{ Bag }Y\text{ is selected}
\displaystyle \text{and }E:\text{ Getting one white and one black ball in a draw of two balls.}
\displaystyle \text{Here, }P(E_1)=P(E_2)=\frac{1}{2}
\displaystyle [\because \text{probability of selecting each bag is equal}]
\displaystyle \text{Now, }P\left(\frac{E}{E_1}\right)=\text{Probability of drawing one white and one black ball from bag }X
\displaystyle =\frac{{}^{4}C_{1}\times{}^{2}C_{1}}{{}^{6}C_{2}}=\frac{4\times2}{\frac{6\times5}{2\times1}}=\frac{8}{15}
\displaystyle \text{and }P\left(\frac{E}{E_2}\right)=\text{Probability of drawing one white and one black ball from bag }Y
\displaystyle =\frac{{}^{3}C_{1}\times{}^{3}C_{1}}{{}^{6}C_{2}}=\frac{3\times3}{\frac{6\times5}{2\times1}}=\frac{3}{5}
\displaystyle \therefore \text{The probability that the one white and one black balls are drawn from bag }Y,
\displaystyle P\left(\frac{E_2}{E}\right)=\frac{P(E_2)\cdot P\left(\frac{E}{E_2}\right)}{P(E_1)\cdot P\left(\frac{E}{E_1}\right)+P(E_2)\cdot P\left(\frac{E}{E_2}\right)}
\displaystyle \text{[by using Baye's theorem]}
\displaystyle =\frac{\frac{1}{2}\times\frac{3}{5}}{\frac{1}{2}\times\frac{8}{15}+\frac{1}{2}\times\frac{3}{5}}
\displaystyle =\frac{\frac{3}{5}}{\frac{8}{15}+\frac{3}{5}}=\frac{3\times15}{5\times17}=\frac{9}{17}
\\

\displaystyle \textbf{Question 57. }\text{A bag contains }4\text{ balls. Two balls are drawn at random (without }
\displaystyle \text{replacement) and are found to be white. What is the probability that all the balls} \\ \text{in the bag are white?} \hspace{0.2cm}\text{[All India 2016, 2014C]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }A:\text{ Two drawn balls are white}
\displaystyle E_1:\text{ All the balls are white}
\displaystyle E_2:\text{ Three balls are white}
\displaystyle E_3:\text{ Two balls are white}
\displaystyle \text{Since, }E_1,E_2\text{ and }E_3\text{ are mutually exclusive and exhaustive events.}
\displaystyle \therefore P(E_1)=P(E_2)=P(E_3)=\frac{1}{3}
\displaystyle \text{Now, }P\left(\frac{A}{E_1}\right)=\frac{{}^{4}C_{2}}{{}^{4}C_{2}}=1,\ P\left(\frac{A}{E_2}\right)=\frac{{}^{3}C_{2}}{{}^{4}C_{2}}=\frac{3}{6}=\frac{1}{2}
\displaystyle \text{and }P\left(\frac{A}{E_3}\right)=\frac{{}^{2}C_{2}}{{}^{4}C_{2}}=\frac{1}{6}
\displaystyle \therefore \text{Probability that all balls in the bag are white}
\displaystyle P\left(\frac{E_1}{A}\right)=\frac{P(E_1)\cdot P\left(\frac{A}{E_1}\right)}{P(E_1)\cdot P\left(\frac{A}{E_1}\right)+P(E_2)\cdot P\left(\frac{A}{E_2}\right)+P(E_3)\cdot P\left(\frac{A}{E_3}\right)}
\displaystyle =\frac{\frac{1}{3}\times1}{\frac{1}{3}\times1+\frac{1}{3}\times\frac{1}{2}+\frac{1}{3}\times\frac{1}{6}}
\displaystyle =\frac{1}{1+\frac{1}{2}+\frac{1}{6}}=\frac{6}{10}=0.6
\\

\displaystyle \textbf{Question 58. }\text{Bag }A\text{ contains }3\text{ red and }5\text{ black balls, while bag }B\text{ contains }4\text{ red and }4\text{ black balls.}
\displaystyle \text{Two balls are transferred at random from bag }A\text{ to bag }B\text{ and then a ball is drawn}
\displaystyle \text{from bag }B\text{ at random. If the ball drawn from bag }B\text{ is found to be red, find the}
\displaystyle \text{probability that two red balls were transferred from }A\text{ to }B. \hspace{2.2cm}\text{[Foreign 2016]}
\displaystyle \text{Answer:}
\displaystyle \text{Let us define the following events:}
\displaystyle E_1=\text{One red and one black ball is transferred}
\displaystyle E_2=\text{Two red balls are transferred}
\displaystyle E_3=\text{Two black balls are transferred}
\displaystyle E=\text{Drawn ball is red.}
\displaystyle \text{Then, }P(E_1)=\frac{{}^{3}C_{1}\times{}^{5}C_{1}}{{}^{8}C_{2}}=\frac{15}{28}
\displaystyle P(E_2)=\frac{{}^{3}C_{2}}{{}^{8}C_{2}}=\frac{3}{28}
\displaystyle P(E_3)=\frac{{}^{5}C_{2}}{{}^{8}C_{2}}=\frac{10}{28}
\displaystyle P(E/E_1)=\frac{5}{10}
\displaystyle P(E/E_2)=\frac{6}{10}
\displaystyle P(E/E_3)=\frac{4}{10}
\displaystyle \text{Now, required probability, }P(E_2/E)
\displaystyle =\frac{P(E_2)\cdot P(E/E_2)}{P(E_1)\cdot P(E/E_1)+P(E_2)\cdot P(E/E_2)+P(E_3)\cdot P(E/E_3)}
\displaystyle =\frac{\frac{3}{28}\times\frac{6}{10}}{\frac{15}{28}\times\frac{5}{10}+\frac{3}{28}\times\frac{6}{10}+\frac{10}{28}\times\frac{4}{10}}
\displaystyle =\frac{18}{75+18+40}=\frac{18}{133}
\\

\displaystyle \textbf{Question 59. }\text{Probabilities of solving a specific problem } \text{independently by }A\text{ and }B
\displaystyle \text{ are }\frac{1}{2}\text{ and }\frac{1}{3},\text{ respectively. If }   \text{both try to solve problem independently, then find the}
\displaystyle \text{probability that}
\displaystyle \text{(i) problem is solved.}
\displaystyle \text{(ii) exactly one of them solves the problem.} \hspace{2.2cm}\text{[CBSE 2015C, CBSE 2011]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }P(A)=\text{Probability that A solves the problem}
\displaystyle P(B)=\text{Probability that B solves the problem}
\displaystyle P(\overline{A})=\text{Probability that A does not solve the problem}
\displaystyle \text{and }P(\overline{B})=\text{Probability that B does not solve the problem}
\displaystyle \text{According to the question, we have}
\displaystyle P(A)=\frac{1}{2}
\displaystyle \text{Then, }P(\overline{A})=1-P(A)=1-\frac{1}{2}=\frac{1}{2}\qquad [\because P(A)+P(\overline{A})=1]
\displaystyle \text{and }P(B)=\frac{1}{3}
\displaystyle \text{then }P(\overline{B})=1-P(B)=1-\frac{1}{3}=\frac{2}{3}
\displaystyle \text{(i) }P(\text{problem is solved})
\displaystyle =P(A\cap\overline{B})+P(\overline{A}\cap B)+P(A\cap B)
\displaystyle =P(A)\cdot P(\overline{B})+P(\overline{A})\cdot P(B)+P(A)\cdot P(B)
\displaystyle [\because A\text{ and }B\text{ are independent events}]
\displaystyle =\left(\frac{1}{2}\times\frac{2}{3}\right)+\left(\frac{1}{2}\times\frac{1}{3}\right)+\left(\frac{1}{2}\times\frac{1}{3}\right)
\displaystyle =\frac{2}{6}+\frac{1}{6}+\frac{1}{6}=\frac{4}{6}=\frac{2}{3}
\displaystyle \text{(ii) }P(\text{exactly one of them solves the problem})
\displaystyle =P(A)+P(B)-2P(A\cap B)
\displaystyle =P(A)+P(B)-2P(A)\times P(B)
\displaystyle =\frac{1}{2}+\frac{1}{3}-2\times\frac{1}{2}\times\frac{1}{3}
\displaystyle =\frac{1}{2}
\\

\displaystyle \textbf{Question 60. }\text{If }A\text{ and }B\text{ are two independent events such that } \\ P(\overline{A}\cap B)=\frac{2}{15}\text{ and }P(A\cap \overline{B})=\frac{1}{6},\text{ then find }P(A)   \text{and }P(B).\hspace{0.2cm}\text{[Delhi 2015]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }P(A)=p\text{ and }P(B)=q.
\displaystyle \text{Since }A\text{ and }B\text{ are independent,}
\displaystyle P(\overline{A}\cap B)=P(\overline{A})P(B)=(1-p)q=\frac{2}{15} \qquad ...(1)
\displaystyle P(A\cap \overline{B})=P(A)P(\overline{B})=p(1-q)=\frac{1}{6} \qquad ...(2)
\displaystyle \text{From (1), }q-pq=\frac{2}{15}
\displaystyle \text{From (2), }p-pq=\frac{1}{6}
\displaystyle \therefore q-p=\frac{2}{15}-\frac{1}{6}=-\frac{1}{30}
\displaystyle q=p-\frac{1}{30}
\displaystyle \text{Substituting in (2),}
\displaystyle p\left(1-p+\frac{1}{30}\right)=\frac{1}{6}
\displaystyle p\left(\frac{31}{30}-p\right)=\frac{1}{6}
\displaystyle 30p^{2}-31p+5=0
\displaystyle (6p-1)(5p-5)=0
\displaystyle p=\frac{1}{6}\text{ or }p=1
\displaystyle p=1\text{ is not possible since then }P(\overline{A}\cap B)=0\neq\frac{2}{15}
\displaystyle \therefore P(A)=\frac{1}{6}
\displaystyle P(B)=\frac{1}{6}-\frac{1}{30}=\frac{4}{30}=\frac{2}{15}
\displaystyle \therefore P(A)=\frac{1}{6}\text{ and }P(B)=\frac{2}{15}.
\\

\displaystyle \textbf{Question 61. }\text{A bag }A\text{ contains }4\text{ black and }6\text{ red balls and bag }B\text{ contains }7\text{ black and }3
\displaystyle \text{red balls. A die is thrown. If }1\text{ or }2\text{ appears on it, then bag }A\text{ is chosen, otherwise bag }B
\displaystyle \text{is chosen. If two balls are drawn at random (without replacement) from the selected}
\displaystyle \text{bag, then find the probability of one of them being red and another black.} \hspace{0.2cm}\text{[Delhi 2015]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, bag }A=4\text{ black and }6\text{ red balls}
\displaystyle \text{bag }B=7\text{ black and }3\text{ red balls.}
\displaystyle \text{Let }E_1=\text{The event that die show }1\text{ or }2
\displaystyle E_2=\text{The event that die show }3\text{ or }4\text{ or }5\text{ or }6
\displaystyle E=\text{The event that among two drawn balls, one of them is red and other is black}
\displaystyle \text{Here, }P(E_1)=\frac{2}{6},\ P(E_2)=\frac{4}{6}
\displaystyle [\because \text{total number in a die is six}]
\displaystyle \therefore P\left(\frac{E}{E_1}\right)=P(\text{getting one red and one black from bag }A)
\displaystyle =\frac{{}^{4}C_{1}\times{}^{6}C_{1}}{{}^{10}C_{2}}=\frac{4\times6\times2}{10\times9}
\displaystyle \Rightarrow P\left(\frac{E}{E_2}\right)=P(\text{getting one red and one black from bag }B)
\displaystyle =\frac{{}^{7}C_{1}\times{}^{3}C_{1}}{{}^{10}C_{2}}=\frac{7\times3\times2}{10\times9}
\displaystyle \text{Now, by theorem of total probability}
\displaystyle P(E)=P(E_1)\cdot P\left(\frac{E}{E_1}\right)+P(E_2)\cdot P\left(\frac{E}{E_2}\right)
\displaystyle =\frac{2}{6}\left(\frac{4\times6\times2}{10\times9}\right)+\frac{4}{6}\left(\frac{7\times3\times2}{10\times9}\right)
\displaystyle =\frac{4\times6}{6\times10\times9}(4+7)
\displaystyle =\frac{4\times6\times11}{6\times10\times9}=\frac{22}{45}
\\

\displaystyle \textbf{Question 62. }\text{Three machines }E_1,\ E_2\text{ and }E_3\text{ in a certain factory producing electric bulbs, produce}
\displaystyle 50\%,\ 25\%\text{ and }25\%\text{ respectively, of the total daily output of electric bulbs. It is known}
\displaystyle \text{that }4\%\text{ of the bulbs produced by each of machines }E_1\text{ and }E_2\text{ are defective}
\displaystyle \text{and that }5\%\text{ of those produced by machine }E_3\text{ are defective. If one bulb is picked up}
\displaystyle \text{at random from a day's production, calculate the probability that it is defective.} \hspace{0.2cm}\text{[CBSE 2015]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }A_1:\text{ Event that the bulb is produced by machine }E_1
\displaystyle A_2:\text{ Event that the bulb is produced by machine }E_2
\displaystyle A_3:\text{ Event that the bulb is produced by machine }E_3
\displaystyle A:\text{ Event that the picked up bulb is defective}
\displaystyle \text{Here, }P(A_1)=50\%=\frac{50}{100}=\frac{1}{2}
\displaystyle P(A_2)=25\%=\frac{25}{100}=\frac{1}{4}
\displaystyle P(A_3)=25\%=\frac{25}{100}=\frac{1}{4}
\displaystyle \text{Also, }P\left(\frac{A}{A_1}\right)=4\%=\frac{4}{100}=\frac{1}{25}
\displaystyle P\left(\frac{A}{A_2}\right)=4\%=\frac{4}{100}=\frac{1}{25}
\displaystyle \text{and }P\left(\frac{A}{A_3}\right)=5\%=\frac{5}{100}=\frac{1}{20}
\displaystyle \therefore \text{The probability that the picked bulb is defective,}
\displaystyle P(A)=P(A_1)\times P\left(\frac{A}{A_1}\right)+P(A_2)\times P\left(\frac{A}{A_2}\right)+P(A_3)\times P\left(\frac{A}{A_3}\right)
\displaystyle =\frac{1}{2}\times\frac{1}{25}+\frac{1}{4}\times\frac{1}{25}+\frac{1}{4}\times\frac{1}{20}
\displaystyle =\frac{1}{50}+\frac{1}{100}+\frac{1}{80}
\displaystyle =\frac{8+4+5}{400}=\frac{17}{400}=0.0425
\\

\displaystyle \textbf{Question 63. }\text{In a factory which manufactures bolts, machines }A,\ B\text{ and }C\text{ manufacture}
\displaystyle 30\%,\ 50\%\text{ and }20\%\text{ of the bolts respectively. Of their outputs, }3\%,\ 4\%\text{ and }1\%
\displaystyle \text{are defective bolts respectively. A bolt is drawn at random from the product and is}
\displaystyle \text{found to be defective. Find the probability that this is not manufactured by machine }B.
\displaystyle \hspace{2.2cm}\text{[All India 2015]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }E_1:\text{ Event that the selected bolt is manufactured by machine }A,
\displaystyle E_2:\text{ Event that the selected bolt is manufactured by machine }B,
\displaystyle E_3:\text{ Event that the selected bolt is manufactured by machine }C,
\displaystyle \text{and }E:\text{ Event that the selected bolt is defective.}
\displaystyle \text{Then, we have }P(E_1)=30\%=\frac{30}{100}
\displaystyle P(E_2)=50\%=\frac{50}{100}
\displaystyle \text{and }P(E_3)=20\%=\frac{20}{100}
\displaystyle \text{Also, given that }3\%,4\%\text{ and }1\%\text{ bolts manufactured by machines } \\ A,B\text{ and }C,\text{ respectively are defective.}
\displaystyle \text{So, }P\left(\frac{E}{E_1}\right)=3\%=\frac{3}{100}
\displaystyle P\left(\frac{E}{E_2}\right)=4\%=\frac{4}{100}
\displaystyle P\left(\frac{E}{E_3}\right)=1\%=\frac{1}{100}
\displaystyle \text{Now, the probability that selected bolt which is defective, is manufactured by machine }B
\displaystyle P\left(\frac{E_2}{E}\right)=\frac{P(E_2)\cdot P\left(\frac{E}{E_2}\right)}{P(E_1)\cdot P\left(\frac{E}{E_1}\right)+P(E_2)\cdot P\left(\frac{E}{E_2}\right)+P(E_3)\cdot P\left(\frac{E}{E_3}\right)}
\displaystyle =\frac{\frac{50}{100}\times\frac{4}{100}}{\frac{30}{100}\times\frac{3}{100}+\frac{50}{100}\times\frac{4}{100}+\frac{20}{100}\times\frac{1}{100}}
\displaystyle =\frac{200}{90+200+20}=\frac{200}{310}
\displaystyle \therefore \text{The probability that selected bolt which is defective, is not manufactured by machine }B
\displaystyle =1-P\left(\frac{E_2}{E}\right)=1-\frac{200}{310}=\frac{110}{310}=\frac{11}{31}
\\

\displaystyle \textbf{Question 64. }\text{A bag contains }4\text{ red and }4\text{ black balls, another bag contains }2\text{ red and }6\text{ black}
\displaystyle \text{balls. One of the two bags is selected at random and two balls are drawn at random}
\displaystyle \text{without replacement from the bag and are found to be both red. Find the probability}
\displaystyle \text{that the balls are drawn from the first bag.} \hspace{2.2cm}\text{[Delhi 2015C]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }B_{1}\text{ and }B_{2}\text{ denote the selection of first and second bag respectively.}
\displaystyle P(B_{1})=P(B_{2})=\frac{1}{2}
\displaystyle \text{Let }E\text{ be the event that both balls drawn are red.}
\displaystyle P(E|B_{1})=\frac{4}{8}\cdot\frac{3}{7}=\frac{3}{14}
\displaystyle P(E|B_{2})=\frac{2}{8}\cdot\frac{1}{7}=\frac{1}{28}
\displaystyle P(B_{1}|E)=\frac{P(B_{1})P(E|B_{1})}{P(B_{1})P(E|B_{1})+P(B_{2})P(E|B_{2})}
\displaystyle =\frac{\frac{1}{2}\cdot\frac{3}{14}}{\frac{1}{2}\cdot\frac{3}{14}+\frac{1}{2}\cdot\frac{1}{28}}
\displaystyle =\frac{\frac{3}{14}}{\frac{3}{14}+\frac{1}{28}}=\frac{6}{7}
\displaystyle \therefore \text{Required probability}=\frac{6}{7}.
\\

\displaystyle \textbf{Question 65. }\text{Assume that each born child is equally likely to be a boy or a girl. If a }
\displaystyle \text{family has two children, then what is }  \text{the conditional probability that both are girls? Given}
\displaystyle \text{(i) the youngest is a girl?}
\displaystyle \text{(ii) atleast one is a girl?} \hspace{2.2cm}\text{[CBSE 2014]}
\displaystyle \text{Answer:}
\displaystyle \text{Let B and b represent elder and younger boy child respectively. Also, G and g} \\ \text{represent elder and younger girl child respectively. If a family has two children,} \\ \text{then all possible cases are}
\displaystyle S=\{Bb,Bg,Gg,Gb\}
\displaystyle n(S)=4
\displaystyle \text{Let us define event }A:\text{ Both children are girls, then }A=\{Gg\}\Rightarrow n(A)=1
\displaystyle \text{(i) Let }E_1:\text{ The event that youngest child is a girl.}
\displaystyle \text{Then, }E_1=\{Bg,Gg\}\text{ and }n(E_1)=2
\displaystyle \text{so }P(E_1)=\frac{n(E_1)}{n(S)}=\frac{2}{4}=\frac{1}{2}
\displaystyle \text{and }A\cap E_1=\{Gg\}
\displaystyle \Rightarrow n(A\cap E_1)=1
\displaystyle \text{so }P(A\cap E_1)=\frac{n(A\cap E_1)}{n(S)}=\frac{1}{4}
\displaystyle \text{Now, }P\left(\frac{A}{E_1}\right)=\frac{P(A\cap E_1)}{P(E_1)}=\frac{1/4}{1/2}=\frac{1}{2}
\displaystyle \therefore \text{Required probability}=\frac{1}{2}
\displaystyle \text{(ii) Let }E_2:\text{ The event that atleast one is girl.}
\displaystyle \text{Then, }E_2=\{Bg,Gg,Gb\}\Rightarrow n(E_2)=3
\displaystyle \text{so }P(E_2)=\frac{n(E_2)}{n(S)}=\frac{3}{4}
\displaystyle \text{and }A\cap E_2=\{Gg\}
\displaystyle \Rightarrow n(A\cap E_2)=1
\displaystyle \text{so }P(A\cap E_2)=\frac{n(A\cap E_2)}{n(S)}=\frac{1}{4}
\displaystyle \text{Now, }P\left(\frac{A}{E_2}\right)=\frac{P(A\cap E_2)}{P(E_2)}=\frac{1/4}{3/4}=\frac{1}{3}
\displaystyle \therefore \text{Required probability}=\frac{1}{3}
\\

\displaystyle \textbf{Question 66. }\text{Consider the experiment of tossing a coin. If the coin shows head, toss }
\displaystyle \text{it again, but if it shows tail, then }   \text{throw a die. Find the conditional probability of the}
\displaystyle \text{event that 'the die shows a number greater than }4\text{',}   \text{ given that 'there is at least one tail'.}  \\ \hspace{0.2cm}\text{[CBSE 2014C]}
\displaystyle \text{Answer:}
\displaystyle \text{The sample space }S\text{ of the experiment is given as}
\displaystyle S=\{(H,H),(H,T),(T,1),(T,2),(T,3),(T,4),(T,5),(T,6)\}
\displaystyle \text{The probabilities of these elementary events are}
\displaystyle P\{(H,H)\}=\frac{1}{2}\times\frac{1}{2}=\frac{1}{4},\quad P\{(H,T)\}=\frac{1}{2}\times\frac{1}{2}=\frac{1}{4}
\displaystyle P\{(T,1)\}=\frac{1}{2}\times\frac{1}{6}=\frac{1}{12},\quad P\{(T,2)\}=\frac{1}{2}\times\frac{1}{6}=\frac{1}{12}
\displaystyle P\{(T,3)\}=\frac{1}{2}\times\frac{1}{6}=\frac{1}{12},\quad P\{(T,4)\}=\frac{1}{2}\times\frac{1}{6}=\frac{1}{12}
\displaystyle P\{(T,5)\}=\frac{1}{2}\times\frac{1}{6}=\frac{1}{12}\text{ and }P\{(T,6)\}=\frac{1}{2}\times\frac{1}{6}=\frac{1}{12}
\displaystyle \text{The outcomes of the experiment can be represented in the following tree diagram.}\displaystyle \text{Consider the following events:}
\displaystyle A=\text{The die shows a number greater than }4\text{ and}
\displaystyle B=\text{There is atleast one tail.}
\displaystyle \text{We have, }A=\{(T,5),(T,6)\},
\displaystyle B=\{(H,T),(T,1),(T,2),(T,3),(T,4),(T,5),(T,6)\}
\displaystyle \text{and }A\cap B=\{(T,5),(T,6)\}
\displaystyle \therefore P(B)=P\{(H,T)\}+P\{(T,1)\}+P\{(T,2)\}
\displaystyle +P\{(T,3)\}+P\{(T,4)\}+P\{(T,5)\}+P\{(T,6)\}
\displaystyle \Rightarrow P(B)=\frac{1}{4}+\frac{1}{12}+\frac{1}{12}+\frac{1}{12}+\frac{1}{12}+\frac{1}{12}+\frac{1}{12}=\frac{3}{4}
\displaystyle \text{and }P(A\cap B)=P\{(T,5)\}+P\{(T,6)\}=\frac{1}{12}+\frac{1}{12}=\frac{1}{6}
\displaystyle \therefore \text{Required probability}
\displaystyle =P\left(\frac{A}{B}\right)=\frac{P(A\cap B)}{P(B)}=\frac{1/6}{3/4}=\frac{4}{18}=\frac{2}{9}
\\

\displaystyle \textbf{Question 67. }\text{A bag contains }3\text{ red and }7\text{ black balls. Two balls are selected at random}
\displaystyle \text{one by one without replacement. If the second selected ball happens to be red, what is}
\displaystyle \text{the probability that the first selected ball is also red?} \hspace{2.2cm}\text{[Delhi 2014C]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }E_1:\text{ First ball is red, }E_2:\text{ First ball is black}
\displaystyle \text{and }A:\text{ Second ball is red}
\displaystyle \text{Total number of balls is }10.
\displaystyle \text{Then, }P(E_1)=\frac{3}{10},\ P(E_2)=\frac{7}{10}
\displaystyle \therefore P\left(\frac{A}{E_1}\right)=\frac{2}{9}\text{ and }P\left(\frac{A}{E_2}\right)=\frac{3}{9}
\displaystyle \text{Then, by Baye's theorem, probability of second selected ball is red} \\ \text{when first selected ball is also red is given by}
\displaystyle P\left(\frac{E_1}{A}\right)=\frac{P(E_1)\cdot P(A/E_1)}{P(E_1)\cdot P(A/E_1)+P(E_2)\cdot P(A/E_2)}
\displaystyle =\frac{\frac{3}{10}\times\frac{2}{9}}{\frac{3}{10}\times\frac{2}{9}+\frac{7}{10}\times\frac{3}{9}}
\displaystyle =\frac{6}{6+21}=\frac{2}{9}
\displaystyle \text{Hence, the probability that the first selected ball is red is }\frac{2}{9}.
\\

\displaystyle \textbf{Question 68. }\text{There are three coins. One is a two headed coin (having head on both faces),}
\displaystyle \text{another is a biased coin that comes up heads }75\%\text{ of the times and third is also a biased}
\displaystyle \text{coin that comes up tails }40\%\text{ of the times. One of the three coins is chosen at random}
\displaystyle \text{and tossed and it shows head. What is the probability that it was the two headed coin?}
\displaystyle \hspace{2.2cm}\text{[All India 2014]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }C_{1},C_{2},C_{3}\text{ denote the selection of the three coins respectively.}
\displaystyle P(C_{1})=P(C_{2})=P(C_{3})=\frac{1}{3}
\displaystyle \text{Let }E\text{ be the event of getting a head.}
\displaystyle P(E|C_{1})=1,\quad P(E|C_{2})=\frac{75}{100}=\frac{3}{4},\quad P(E|C_{3})=\frac{60}{100}=\frac{3}{5}
\displaystyle P(C_{1}|E)=\frac{P(C_{1})P(E|C_{1})}{P(C_{1})P(E|C_{1})+P(C_{2})P(E|C_{2})+P(C_{3})P(E|C_{3})}
\displaystyle =\frac{\frac{1}{3}\cdot1}{\frac{1}{3}\cdot1+\frac{1}{3}\cdot\frac{3}{4}+\frac{1}{3}\cdot\frac{3}{5}}
\displaystyle =\frac{1}{1+\frac{3}{4}+\frac{3}{5}}=\frac{20}{47}
\displaystyle \therefore \text{Required probability}=\frac{20}{47}.
\\

\displaystyle \textbf{Question 69. }\text{An insurance company insured }2000\text{ scooter drivers, }4000\text{ car drivers and }6000
\displaystyle \text{truck drivers. The probabilities of an accident for them are }0.01,\ 0.03\text{ and }0.15\text{ respectively.}
\displaystyle \text{One of the insured persons meets with an accident. What is the probability that he is a}
\displaystyle \text{scooter driver or a car driver?} \hspace{2.2cm}\text{[Foreign 2014; All India 2009C]}
\displaystyle \text{Answer:}
\displaystyle \text{Let us define the events as}
\displaystyle E_1:\text{ Insured person is a scooter driver}
\displaystyle E_2:\text{ Insured person is a car driver}
\displaystyle E_3:\text{ Insured person is a truck driver}
\displaystyle A:\text{ Insured person meets with an accident}
\displaystyle \text{Then, }n(E_1)=2000,\ n(E_2)=4000
\displaystyle \text{and }n(E_3)=6000
\displaystyle \text{Here, total insured person, }n(S)=12000
\displaystyle \text{Now, }P(E_1)=\text{Probability that the insured person is a scooter driver}
\displaystyle =\frac{n(E_1)}{n(S)}=\frac{2000}{12000}=\frac{1}{6}
\displaystyle P(E_2)=\text{Probability that the insured person is a car driver}
\displaystyle =\frac{n(E_2)}{n(S)}=\frac{4000}{12000}=\frac{1}{3}
\displaystyle \text{and }P(E_3)=\text{Probability that the insured person is a truck driver}
\displaystyle =\frac{n(E_3)}{n(S)}=\frac{6000}{12000}=\frac{1}{2}
\displaystyle \text{Also, }P(A/E_1)=\text{Probability that scooter driver meets with an accident}
\displaystyle =0.01
\displaystyle P(A/E_2)=\text{Probability that car driver meets with an accident}
\displaystyle =0.03
\displaystyle \text{and }P(A/E_3)=\text{Probability that truck driver meets with an accident}
\displaystyle =0.15
\displaystyle \text{The probability that the person met with an accident was a scooter driver,}
\displaystyle P(E_1/A)=\frac{P(E_1)\cdot P(A/E_1)}{P(E_1)\cdot P(A/E_1)+P(E_2)\cdot P(A/E_2)+P(E_3)\cdot P(A/E_3)}
\displaystyle \text{[by Baye's theorem]}
\displaystyle =\frac{\frac{1}{6}\times0.01}{\left(\frac{1}{6}\times0.01\right)+\left(\frac{1}{3}\times0.03\right)+\left(\frac{1}{2}\times0.15\right)}
\displaystyle =\frac{\frac{1}{6}}{\frac{1}{6}+1+\frac{15}{2}}=\frac{1}{6}\times\frac{6}{52}=\frac{1}{52}
\displaystyle \text{The probability that the person met with an accident was a car driver, }P(E_2/A)
\displaystyle =\frac{P(E_2)\cdot P(A/E_2)}{P(E_1)P(A/E_1)+P(E_2)P(A/E_2)+P(E_3)P(A/E_3)}
\displaystyle =\frac{\frac{1}{3}\times0.03}{\left(\frac{1}{6}\times0.01\right)+\left(\frac{1}{3}\times0.03\right)+\left(\frac{1}{2}\times0.15\right)}
\displaystyle =\frac{\frac{1}{100}}{\frac{1}{600}+\frac{1}{100}+\frac{15}{200}}=\frac{1}{100}\times\frac{600}{52}=\frac{6}{52}
\displaystyle \text{Hence, the required probability is}
\displaystyle P((E_1\cup E_2)/A)=P(E_1/A)+P(E_2/A)
\displaystyle =\frac{1}{52}+\frac{6}{52}=\frac{7}{52}
\\

\displaystyle \textbf{Question 70. }\text{A man is known to speak truth }3\text{ out of }5\text{ times. He throws a die and reports}
\displaystyle \text{that it is }1.\text{ Find the probability that it is actually }1. \hspace{2.2cm}\text{[Delhi 2014C]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }E_1=\text{Event that 1 occurs in a die}
\displaystyle E_2=\text{Event that 1 does not occur in a die}
\displaystyle A=\text{Event that the man reports that 1 occurs in a die}
\displaystyle \text{Then, }P(E_1)=\frac{1}{6}\text{ and }P(E_2)=\frac{5}{6}
\displaystyle P\left(\frac{A}{E_1}\right)=\text{P(man reports that 1 occurs when 1 occurs)}=\frac{3}{5}
\displaystyle \text{and }P\left(\frac{A}{E_2}\right)=\text{P(man reports that 1 occurs but 1 does not occur)}=\frac{2}{5}
\displaystyle \text{Thus, by Baye's theorem, we get}
\displaystyle P(\text{get actually 1 when he reports that 1 occur})
\displaystyle =P\left(\frac{E_1}{A}\right)
\displaystyle =\frac{P(E_1)\cdot P\left(\frac{A}{E_1}\right)}{P(E_1)\cdot P\left(\frac{A}{E_1}\right)+P(E_2)\cdot P\left(\frac{A}{E_2}\right)}
\displaystyle =\frac{\frac{1}{6}\times\frac{3}{5}}{\frac{1}{6}\times\frac{3}{5}+\frac{5}{6}\times\frac{2}{5}}
\displaystyle =\frac{3}{13}
\\

\displaystyle \textbf{Question 71. }\text{A card from a pack of }52\text{ playing cards is lost. From the remaining cards of the}
\displaystyle \text{pack three cards are drawn at random (without replacement) and are found to be all}
\displaystyle \text{spades. Find the probability of the lost card being a spade.} \hspace{2.2cm}\text{[Delhi 2014]}
\displaystyle \text{Answer:}
\displaystyle \text{Let us define the following events :}
\displaystyle E_1:\text{ Lost card is a spade card}
\displaystyle E_2:\text{ Lost card is not a spade card}
\displaystyle A:\text{ Drawn cards are spade cards}
\displaystyle \text{Then,}
\displaystyle P(E_1)=\frac{13}{52}=\frac{1}{4}
\displaystyle P(E_2)=\frac{39}{52}=\frac{3}{4}
\displaystyle P(A/E_1)=\frac{{}^{12}C_{3}}{{}^{51}C_{3}}=\frac{220}{20825}
\displaystyle \text{and }P(A/E_2)=\frac{{}^{13}C_{3}}{{}^{51}C_{3}}=\frac{286}{20825}
\displaystyle \text{Now, required probability, }P(E_1/A)
\displaystyle =\frac{P(E_1)\cdot P(A/E_1)}{P(E_1)\cdot P(A/E_1)+P(E_2)\cdot P(A/E_2)}
\displaystyle =\frac{\frac{1}{4}\times\frac{220}{20825}}{\frac{1}{4}\times\frac{220}{20825}+\frac{3}{4}\times\frac{286}{20825}}
\displaystyle =\frac{220}{220+858}=\frac{220}{1078}=\frac{20}{98}=\frac{10}{49}
\\

\displaystyle \textbf{Question 72. }A\text{ speaks truth in }75\%\text{ of the cases, while }B\text{ in }90\%\text{ of the cases. In what }
\displaystyle \text{percent of cases are they likely to }   \text{contradict each other in stating the same fact? Do you}
\displaystyle \text{think that statement of }B\text{ is true?} \hspace{2.2cm}\text{[CBSE 2013]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }A_T:\text{ Event that A speaks truth}
\displaystyle \text{and }B_T:\text{ Event that B speaks truth.}
\displaystyle \text{Given, }P(A_T)=\frac{75}{100},\text{ then }P(\overline{A_T})=1-\frac{75}{100}
\displaystyle \left[\because P(\overline{A})=1-P(A)\right]
\displaystyle =\frac{25}{100}
\displaystyle \text{and }P(B_T)=\frac{90}{100},
\displaystyle \text{then, }P(\overline{B_T})=1-\frac{90}{100}=\frac{10}{100}
\displaystyle \text{Now, }P(A\text{ and }B\text{ are contradict to each other})
\displaystyle =P(A_T\cap\overline{B_T})+P(\overline{A_T}\cap B_T)
\displaystyle =P(A_T)\cdot P(\overline{B_T})+P(\overline{A_T})\cdot P(B_T)
\displaystyle [\because \text{events }A_T\text{ and }B_T\text{ are independent events}]
\displaystyle =\frac{75}{100}\times\frac{10}{100}+\frac{25}{100}\times\frac{90}{100}
\displaystyle =\frac{750+2250}{10000}=\frac{3000}{10000}=\frac{3}{10}
\displaystyle \therefore \text{Percentage of }P(A\text{ and }B\text{ are contradict to each other})
\displaystyle =\frac{3}{10}\times100=30\%
\displaystyle \text{Since, B speaks truth in only }90\%\text{ (i.e. not }100\%)\text{ of the cases, therefore} \\ \text{we think, the statement of B may be false.}
\\

\displaystyle \textbf{Question 73. }P\text{ speaks truth in }70\%\text{ of the cases and }Q\text{ in }80\%\text{ of the cases.}
\displaystyle \text{ In what  percent of cases are they likely to agree }   \text{in stating the same fact? Do you }
\displaystyle \text{think, when they agree, means both are speaking truth?} \hspace{2.2cm}\text{[CBSE 2013]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }P_T:\text{ Event that P speaks truth}
\displaystyle \text{and }Q_T:\text{ Event that Q speaks truth.}
\displaystyle \text{Given, }P(P_T)=\frac{70}{100},\text{ then }P(\overline{P_T})=1-\frac{70}{100}=\frac{30}{100}
\displaystyle \text{and }P(Q_T)=\frac{80}{100},\text{ then }P(\overline{Q_T})=1-\frac{80}{100}=\frac{20}{100}
\displaystyle P(A\text{ and }B\text{ are agree to each other})
\displaystyle =P(P_T\cap Q_T)+P(\overline{P_T}\cap\overline{Q_T})
\displaystyle =P(P_T)\cdot P(Q_T)+P(\overline{P_T})\cdot P(\overline{Q_T})
\displaystyle [\because \text{events }P_T\text{ and }Q_T\text{ are independent events}]
\displaystyle =\frac{70}{100}\times\frac{80}{100}+\frac{30}{100}\times\frac{20}{100}
\displaystyle =\frac{5600}{10000}+\frac{600}{10000}=\frac{6200}{10000}=\frac{62}{100}
\displaystyle \therefore \text{Percentage of }P(A\text{ and }B\text{ are agree to each other})
\displaystyle =\frac{62}{100}\times100=62\%
\displaystyle \text{No, agree does not mean that they are speaking truth.}
\\

\displaystyle \textbf{Question 74. }\text{A speaks truth in }60\%\text{ of the cases, while }B\text{ in }90\% \text{ of the cases.} \\ \text{In what percent of cases are they likely to }   \text{contradict each other in stating} \\ \text{the same fact? In the cases of contradiction do you think, the statement of }B\text{ will } \\  \text{carry more weight as he speaks truth in more number of cases than }A?\hspace{0.2cm}\text{[Delhi 2013]}
\displaystyle \text{Answer:}
\displaystyle P(A\text{ speaks truth})=\frac{60}{100}=\frac{3}{5}
\displaystyle P(B\text{ speaks truth})=\frac{90}{100}=\frac{9}{10}
\displaystyle P(A\text{ speaks false})=\frac{2}{5}
\displaystyle P(B\text{ speaks false})=\frac{1}{10}
\displaystyle \text{They contradict each other when one speaks truth and the other speaks false.}
\displaystyle P(\text{contradiction})=P(A\text{ true and }B\text{ false})+P(A\text{ false and }B\text{ true})
\displaystyle =\frac{3}{5}\cdot\frac{1}{10}+\frac{2}{5}\cdot\frac{9}{10}
\displaystyle =\frac{3}{50}+\frac{18}{50}=\frac{21}{50}
\displaystyle =42\%
\displaystyle \text{In contradiction cases, }P(B\text{ true and }A\text{ false})=\frac{18}{50}
\displaystyle P(A\text{ true and }B\text{ false})=\frac{3}{50}
\displaystyle \text{Since }\frac{18}{50}>\frac{3}{50},\text{ the statement of }B\text{ will carry more weight.}
\\

\displaystyle \textbf{Question 75. }\text{Assume that the chances of a patient having a heart attack is }40\%.\text{ Assuming that a}
\displaystyle \text{meditation and yoga course reduces the risk of heart attack by }30\%\text{ and prescription}
\displaystyle \text{of certain drug reduces its chance by }25\%.\text{ At a time, a patient can choose anyone}
\displaystyle \text{of the two options with equal probabilities. It is given that after going through one of}
\displaystyle \text{the two options, the patient selected at random suffers a heart attack. Find the}
\displaystyle \text{probability that the patient followed a course of meditation and yoga. Interpret the}
\displaystyle \text{result and state which of the above methods is more beneficial for the patient.} \hspace{2.2cm}\text{[Delhi 2013]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }E_1:\text{ The patient follows meditation and yoga.}
\displaystyle E_2:\text{ The patient uses drug.}
\displaystyle \text{Then, }E_1\text{ and }E_2\text{ are mutually exclusive}
\displaystyle \text{and }P(E_1)=P(E_2)=\frac{1}{2}
\displaystyle \text{Also, let }E:\text{ The selected patient suffers a heart attack.}
\displaystyle \text{Then, }P(E/E_1)=\frac{40}{100}\left(1-\frac{30}{100}\right)=\frac{40}{100}\times\frac{70}{100}=\frac{28}{100}
\displaystyle \text{and }P(E/E_2)=\frac{40}{100}\left(1-\frac{25}{100}\right)=\frac{40}{100}\times\frac{75}{100}=\frac{30}{100}
\displaystyle \therefore P(\text{patient who suffers heart attack follows meditation and yoga})
\displaystyle =P(E_1/E)
\displaystyle =\frac{P(E/E_1)\cdot P(E_1)}{P(E/E_1)\cdot P(E_1)+P(E/E_2)\cdot P(E_2)}
\displaystyle \text{[using Baye's theorem]}
\displaystyle =\frac{\frac{28}{100}\times\frac{1}{2}}{\frac{28}{100}\times\frac{1}{2}+\frac{30}{100}\times\frac{1}{2}}
\displaystyle =\frac{14}{29}
\displaystyle \text{Yoga course and meditation are more beneficial for the heart patient.}
\\

\displaystyle \textbf{Question 76. }\text{Suppose a girl throws a die. If she gets }5\text{ or }6,\text{ then she tosses a coin }3\text{ times and}
\displaystyle \text{notes the number of heads. If she gets }1,2,3\text{ or }4,\text{ she tosses a coin once and notes}
\displaystyle \text{whether a head or tail is obtained. If she obtained exactly one head, what is the}
\displaystyle \text{probability that she threw }1,2,3\text{ or }4\text{ with the die?} \hspace{2.2cm}\text{[All India 2012]}
\displaystyle \text{Answer:}
\displaystyle \text{Let us define the events as}
\displaystyle E_1:\text{ Girl gets 5 or 6 on a die}
\displaystyle E_2:\text{ Girl gets 1, 2, 3 or 4 on a die}
\displaystyle A:\text{ She gets exactly one head}
\displaystyle \text{Now, }P(E_1)=\text{Probability of getting 5 or 6 on a die}=\frac{2}{6}=\frac{1}{3}
\displaystyle \text{and }P(E_2)=\text{Probability of getting 1, 2, 3 or 4 on a die}=\frac{4}{6}=\frac{2}{3}
\displaystyle \text{Also, }P(A/E_1)=\text{Probability that girl gets exactly one head when she throws} \\ \text{coin thrice}=\frac{3}{8}
\displaystyle [\because \{HHH,TTT,HHT,HTH,THH,TTH,THT,HTT\}]
\displaystyle P(A/E_2)=\text{Probability that girl gets exactly one head when she throws} \\ \text{coin once}=\frac{1}{2}
\displaystyle \text{The probability that she throws 1, 2, 3 or 4 with the die for getting exactly one head,}
\displaystyle P(E_2/A)=\frac{P(E_2)\cdot P(A/E_2)}{P(E_1)\cdot P(A/E_1)+P(E_2)\cdot P(A/E_2)}
\displaystyle \text{[by Baye's theorem]}
\displaystyle =\frac{\frac{2}{3}\times\frac{1}{2}}{\left(\frac{1}{3}\times\frac{3}{8}\right)+\left(\frac{2}{3}\times\frac{1}{2}\right)}
\displaystyle =\frac{\frac{1}{3}}{\frac{1}{8}+\frac{1}{3}}=\frac{\frac{1}{3}}{\frac{11}{24}}=\frac{1}{3}\times\frac{24}{11}=\frac{8}{11}
\\

\displaystyle \textbf{Question 77. }\text{An insurance company insured }2000\text{ scooter drivers, }4000\text{ car drivers and}
\displaystyle 6000\text{ truck drivers. The probabilities of their meeting an accident are }0.01,\ 0.03\text{ and}
\displaystyle 0.15\text{ respectively. One of the insured persons meets with an accident. What is the}
\displaystyle \text{probability that he is a car driver?} \hspace{2.2cm}\text{[All India 2012C]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }S,\ C,\ T\text{ denote the events that the insured person is a scooter driver, car driver} \\ \text{and truck driver respectively.}
\displaystyle P(S)=\frac{2000}{12000}=\frac{1}{6},\quad P(C)=\frac{4000}{12000}=\frac{1}{3},\quad P(T)=\frac{6000}{12000}=\frac{1}{2}
\displaystyle \text{Let }A\text{ be the event that the insured person meets with an accident.}
\displaystyle P(A|S)=0.01,\quad P(A|C)=0.03,\quad P(A|T)=0.15
\displaystyle P(C|A)=\frac{P(C)P(A|C)}{P(S)P(A|S)+P(C)P(A|C)+P(T)P(A|T)}
\displaystyle =\frac{\frac{1}{3}\times0.03}{\frac{1}{6}\times0.01+\frac{1}{3}\times0.03+\frac{1}{2}\times0.15}
\displaystyle =\frac{0.01}{0.0016667+0.01+0.075}
\displaystyle =\frac{0.01}{0.0866667}
\displaystyle =\frac{3}{26}
\displaystyle \therefore \text{The probability that the person is a car driver is }\frac{3}{26}.
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\displaystyle \textbf{Question 78. }\text{Among the students in a college, it is known that }60\%\text{ reside in hostel and }40\%
\displaystyle \text{are day scholars (not residing in hostel). Previous year results report that }30\%\text{ of all students}
\displaystyle \text{who reside in a hostel attain }A\text{ grade and }20\%\text{ of day scholars attain }A\text{ grade in their}
\displaystyle \text{annual exams. At the end of year, one student is chosen at random from the college and he}
\displaystyle \text{has }A\text{ grade, what is the probability that the student is a hosteler?} \hspace{2.2cm}\text{[Delhi 2012, 2011C]}
\displaystyle \text{Answer:}
\displaystyle \text{Let us define the events as}
\displaystyle E_1:\text{ Students reside in a hostel}
\displaystyle E_2:\text{ Students are day scholars}
\displaystyle A:\text{ Students get A grade}
\displaystyle \text{Then,}
\displaystyle P(E_1)=\text{Probability that student reside in a hostel}=\frac{60}{100}
\displaystyle \text{and }P(E_2)=\text{Probability that students are day scholars}=1-\frac{60}{100}=\frac{40}{100}
\displaystyle \text{Also, }P(A/E_1)=\text{Probability that hostellers get A grade}=\frac{30}{100}
\displaystyle \text{and }P(A/E_2)=\text{Probability that students having day scholars get A grade}=\frac{20}{100}
\displaystyle \text{Then, }P(E_1/A)
\displaystyle =\frac{P(E_1)\cdot P(A/E_1)}{P(E_1)\cdot P(A/E_1)+P(E_2)\cdot P(A/E_2)}
\displaystyle =\frac{\frac{60}{100}\times\frac{30}{100}}{\frac{60}{100}\times\frac{30}{100}+\frac{40}{100}\times\frac{20}{100}}
\displaystyle =\frac{1800}{1800+800}=\frac{18}{26}=\frac{9}{13}
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\displaystyle \textbf{Question 79. }\text{There are two bags, Bag I and Bag II. Bag I contains }4\text{ white and }3\text{ red balls,}
\displaystyle \text{while another Bag II contains }3\text{ white and }7\text{ red balls. One ball is drawn at random}
\displaystyle \text{from one of the bags and it is found to be white. Find the probability that it was drawn from Bag I.}
\displaystyle \hspace{2.2cm}\text{[Delhi 2010]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }E_1\text{ be the event that bag I is chosen,}
\displaystyle E_2\text{ be the event that bag II is chosen}
\displaystyle \text{and }A\text{ be the event that the drawn ball is white.}
\displaystyle \text{Then, }P(E_1)=\frac{1}{2},
\displaystyle P(E_2)=\frac{1}{2},
\displaystyle P(A/E_1)=\frac{4}{7},
\displaystyle \text{and }P(A/E_2)=\frac{3}{10},
\displaystyle \text{Now, required probability }P(E_1/A)
\displaystyle =\frac{P(E_1)\cdot P(A/E_1)}{P(E_1)\cdot P(A/E_1)+P(E_2)\cdot P(A/E_2)}
\displaystyle =\frac{\frac{1}{2}\cdot\frac{4}{7}}{\frac{1}{2}\cdot\frac{4}{7}+\frac{1}{2}\cdot\frac{3}{10}}
\displaystyle =\frac{\frac{4}{7}}{\frac{4}{7}+\frac{3}{10}}=\frac{\frac{4}{7}}{\frac{40+21}{70}}
\displaystyle =\frac{4}{7}\times\frac{70}{61}=\frac{40}{61}
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