\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}.

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