\displaystyle \textbf{Question 1: } \text{Bhuvan is preparing chapati and curry for a dinner party at his home.}
\displaystyle \text{Let event } A=\text{Bhuvan prepares chapati well. Let event } B=\text{Bhuvan cooks curry well.}
\displaystyle \text{Given, } P(A)=0.40, P(B)=0.30 \text{ and } P(A \text{ and } B)=0.20.
\displaystyle \text{Which one of the following is true about the events } A \text{ and } B 
\displaystyle \text{(a) They are mutually exclusive but not independent.}
\displaystyle \text{(b) They are independent but not mutually exclusive.}
\displaystyle \text{(c) They are both mutually exclusive and independent.}
\displaystyle \text{(d) They are neither mutually exclusive nor independent.}
\displaystyle \text{Answer:}
\displaystyle \text{(d) Here, } P(A)\cdot P(B)=0.4\times 0.3=0.12\neq P(A\cap B)
\displaystyle \therefore A \text{ and } B \text{ are not independent and } P(A\cap B)\neq 0
\displaystyle \therefore A \text{ and } B \text{ are not mutually exclusive}
\\

\displaystyle \textbf{Question 2: } \text{A company has estimated that the probabilities of success for three} \\ \text{products introduced in the market are } \frac{1}{3}, \frac{2}{5} \text{ and } \frac{2}{3} \text{ respectively.} \text{Assuming independence,} \\ \text{find the probability that at least one product is successful.}   
\displaystyle \text{Answer:}
\displaystyle \text{Let } P(A)=\frac{1}{3},\ P(B)=\frac{2}{5},\ \text{and } P(C)=\frac{2}{3}
\displaystyle \therefore P(\text{at least one product is successful})
\displaystyle =1-P(\text{none of the products is successful})
\displaystyle =1-\left(\frac{2}{3}\times \frac{3}{5}\times \frac{1}{3}\right)
\displaystyle =1-\frac{2}{15}=\frac{13}{15}
\\

\displaystyle \textbf{Question 3: } \text{A speaks truth in } 60\% \text{ of cases and } B \text{ speaks truth in } 90\% \text{ of the cases.}
\displaystyle \text{In what percentage of cases are they likely to contradict each other in stating the} \\ \text{same fact?}   
\displaystyle \text{Answer:}
\displaystyle \text{ Let } E=\text{ event of } A \text{ speaking truth}
\displaystyle \text{and } F=\text{ event of } B \text{ speaking truth}
\displaystyle \therefore P(E)=\frac{60}{100}=\frac{6}{10},\quad P(F)=\frac{90}{100}=\frac{9}{10}
\displaystyle \text{Probability of } A \text{ and } B \text{ contradicting each other}
\displaystyle =P(EF'+E'F)
\displaystyle =P(E)P(F')+P(E')P(F)
\displaystyle =P(E)[1-P(F)]+[1-P(E)]P(F)
\displaystyle =\frac{6}{10}\left(1-\frac{9}{10}\right)+\left(1-\frac{6}{10}\right)\frac{9}{10}
\displaystyle =\frac{6}{10}\times\frac{1}{10}+\frac{4}{10}\times\frac{9}{10}
\displaystyle =\frac{6+36}{100}=\frac{42}{100}
\displaystyle \therefore A \text{ and } B \text{ are likely to contradict each other in } 42\% \text{ cases}
\\

\displaystyle \textbf{Question 4: } \text{A bag contains } 9 \text{ red, } 7 \text{ white and } 4 \text{ black balls. If two balls are drawn }
\displaystyle \text{at random without replacement, the probability that both balls are red will be}
\displaystyle \text{(a) } \frac{11}{95} \quad \text{(b) } \frac{15}{95} \quad \text{(c) } \frac{18}{95} \quad \text{(d) } \frac{18}{23}
\displaystyle \text{Answer:}
\displaystyle \text{7. (c) There are } 20 \text{ balls in the bag}
\displaystyle \text{Total possible outcomes of getting two balls}={}^{20}C_{2}=190
\displaystyle \text{Favourable outcomes of getting both red balls}={}^{9}C_{2}=36
\displaystyle \therefore \text{Required probability}=\frac{36}{190}=\frac{18}{95}
\\

\displaystyle \textbf{Question 5: } \text{Two cards are drawn out randomly from a pack of } 52 \text{ cards one after} \\ \text{the other, without replacement.} \text{The probability of first card being a king and second} \\ \text{card not being a king is}
\displaystyle \text{(a) } \frac{48}{663} \quad \text{(b) } \frac{24}{663} \quad \text{(c) } \frac{12}{663} \quad \text{(d) } \frac{4}{663} 
\displaystyle \text{Answer:}
\displaystyle \text{(a) Required probability}
\displaystyle =P(\text{first card is a king and second card is not a king})
\displaystyle =\frac{{}^{4}C_{1}}{{}^{52}C_{1}}\times\frac{{}^{48}C_{1}}{{}^{51}C_{1}}
\displaystyle =\frac{4}{52}\times\frac{48}{51}=\frac{48}{663}
\\

\displaystyle \textbf{Question 6: } \text{If two balls are drawn from a bag containing } 3 \text{ white, } 4 \text{ black and } 5
\displaystyle \text{red balls. Then, the probability that the drawn balls are of different colours is}
\displaystyle \text{(a) } \frac{1}{66} \quad \text{(b) } \frac{3}{66} \quad \text{(c) } \frac{19}{66} \quad \text{(d) } \frac{47}{66} 
\displaystyle \text{Answer:}
\displaystyle \text{(d) There are } 3 \text{ white, } 4 \text{ black and } 5 \text{ red balls in a bag}
\displaystyle \therefore \text{Total number of balls}=3+4+5=12
\displaystyle \text{Required probability}=P(\text{two drawn balls are of different colours})
\displaystyle =\frac{{}^{3}C_{1}\times{}^{4}C_{1}+{}^{3}C_{1}\times{}^{5}C_{1}+{}^{4}C_{1}\times{}^{5}C_{1}}{{}^{12}C_{2}}
\displaystyle =\frac{3\times4+3\times5+4\times5}{66}
\displaystyle =\frac{12+15+20}{66}=\frac{47}{66}
\\

\displaystyle \textbf{Question 7: } \text{A and B appeared for an interview for two vacancies. The probability}
\displaystyle \text{of } A\text{'s selection is } \frac{4}{5} \text{ and } B\text{'s selection is } \frac{1}{3}. \text{ Find the probability that none of them}
\displaystyle \text{Answer:}
\displaystyle \text{Given that, } P(A)=\frac{4}{5}\text{ and }P(B)=\frac{1}{3}
\displaystyle \therefore P(A')=\frac{1}{5}\text{ and }P(B')=\frac{2}{3}
\displaystyle \text{Probability that none is selected}
\displaystyle =P(A'\cap B')=P(A')P(B')
\displaystyle =\frac{1}{5}\times\frac{2}{3}=\frac{2}{15}
\\

\displaystyle \textbf{Question 8: } \text{If } A \text{ and } B \text{ are independent events such that } \\  P(A)=\frac{1}{4} \text{ and } P(B)=\frac{2}{3}.  \text{Find } P(A \cap B).
\displaystyle \text{Answer:}
\displaystyle \text{Given, } P(A)=\frac{1}{4}\text{ and }P(B)=\frac{2}{3}
\displaystyle \text{Since } A \text{ and } B \text{ are independent events}
\displaystyle \therefore P(A\cap B)=P(A)\cdot P(B)=\frac{1}{4}\times\frac{2}{3}=\frac{1}{6}
\\

\displaystyle \textbf{Question 9: } \text{Given that the events } A \text{ and } B \text{ are such that } P(A)=\frac{1}{2},
\displaystyle P(A \cup B)=\frac{3}{5} \text{ and } P(B)=k. \text{ Find } k, \text{ if}
\displaystyle \text{(a) } A \text{ and } B \text{ are mutually exclusive.}
\displaystyle \text{(b) } A \text{ and } B 
\displaystyle \text{Answer:}
\displaystyle \text{Given, } P(A)=\frac{1}{2},\ P(A\cup B)=\frac{3}{5}\text{ and }P(B)=k
\displaystyle \text{(a) Since } A \text{ and } B \text{ are mutually exclusive events}
\displaystyle \therefore P(A\cap B)=0
\displaystyle \text{Now, } P(A\cup B)=P(A)+P(B)-P(A\cap B)
\displaystyle \Rightarrow \frac{3}{5}=\frac{1}{2}+k
\displaystyle \Rightarrow k=\frac{3}{5}-\frac{1}{2}=\frac{1}{10}
\displaystyle \text{(b) Since } A \text{ and } B \text{ are independent events}
\displaystyle \therefore P(A\cap B)=P(A)\times P(B)=\frac{1}{2}k
\displaystyle \text{Now, } P(A\cup B)=P(A)+P(B)-P(A\cap B)
\displaystyle \Rightarrow \frac{3}{5}=\frac{1}{2}+k-\frac{1}{2}k
\displaystyle \Rightarrow \frac{1}{2}k=\frac{3}{5}-\frac{1}{2}=\frac{1}{10}
\displaystyle \Rightarrow k=\frac{1}{5}
\\

\displaystyle \textbf{Question 10: } \text{A candidate takes three tests in succession and the probability of} \\ \text{passing the first test is } \frac{1}{2}.   \text{The probability of passing each succeeding test is } \frac{1}{2} \text{ or } \frac{1}{4} \\ \text{depending on whether he passes or fails in the preceding one.}   \text{The candidate is selected} \\ \text{if he passes at least two tests. Find the probability that the candidate is selected.}
\displaystyle \text{Answer:}
\displaystyle \text{Candidate will be selected in the following cases}
\displaystyle \text{I. PPP \quad [P means passed]}
\displaystyle \text{II. PPF \quad [F means failed]}
\displaystyle \text{III. PFP}
\displaystyle \text{IV. FPP}
\displaystyle \therefore P(\text{I})=\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}=\frac{1}{8}
\displaystyle P(\text{II})=\frac{1}{2}\times\frac{1}{2}\times\left(1-\frac{1}{2}\right)=\frac{1}{8}
\displaystyle P(\text{III})=\frac{1}{2}\times\left(1-\frac{1}{2}\right)\times\frac{1}{2}=\frac{1}{8}
\displaystyle P(\text{IV})=\left(1-\frac{1}{2}\right)\times\frac{1}{2}\times\frac{1}{2}=\frac{1}{8}
\displaystyle \therefore \text{Probability of success}=\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}=\frac{4}{8}=\frac{1}{2}
\\

\displaystyle \textbf{Question 11: } \text{Two horses are considered for a race. The probability of selection of} \\ \text{the first horse is } \frac{1}{5} \text{ and that of the second is } \frac{2}{3}.
\displaystyle \text{Find the probability that:}
\displaystyle \text{(i) both will be selected}
\displaystyle \text{(ii) only one of them will be selected}
\displaystyle \text{(iii) none of them will be selected}
\displaystyle \text{(iv) at least one of them will be selected} 
\displaystyle \text{Answer:}
\displaystyle \text{Let } A \text{ and } B \text{ be the events of selecting first and second horse}
\displaystyle \therefore P(A)=\frac{1}{5},\quad P(B)=\frac{2}{3}
\displaystyle \text{(i) } P(\text{both selected})=P(AB)=P(A)P(B)=\frac{1}{5}\times\frac{2}{3}=\frac{2}{15}
\displaystyle \text{(ii) } P(\text{only one selected})=P(A\cap B')+P(A'\cap B)
\displaystyle =P(A)P(B')+P(A')P(B)
\displaystyle =\frac{1}{5}\left(1-\frac{2}{3}\right)+\left(1-\frac{1}{5}\right)\frac{2}{3}
\displaystyle =\frac{1}{5}\times\frac{1}{3}+\frac{4}{5}\times\frac{2}{3}
\displaystyle =\frac{1}{15}+\frac{8}{15}=\frac{9}{15}=\frac{3}{5}
\displaystyle \text{(iii) } P(\text{none selected})=P(A'\cap B')=P(A')P(B')
\displaystyle =\left(1-\frac{1}{5}\right)\left(1-\frac{2}{3}\right)=\frac{4}{5}\times\frac{1}{3}=\frac{4}{15}
\displaystyle \text{(iv) } P(\text{at least one selected})=1-P(\text{none selected})
\displaystyle =1-\frac{4}{15}=\frac{11}{15}
\\

\displaystyle \textbf{Question 12: } \text{A bag contains } 3 \text{ red and } 4 \text{ white balls and another bag contains } \\  2 \text{ red and } 3 \text{ white balls.}   \text{If one ball is drawn from the first bag and } 2 \text{ balls are} \\ \text{drawn from the second bag, then find the probability}   \text{that all three balls are of} \\ \text{the same colour.} 
\displaystyle \text{Answer:}
\displaystyle \text{Bag I: } 3 \text{ red, } 4 \text{ white; \quad Bag II: } 2 \text{ red, } 3 \text{ white}
\displaystyle \text{Required probability }=P(\text{all three balls are of same colour})
\displaystyle =P(\text{1 red from I and 2 red from II})+P(\text{1 white from I and 2 white from II})
\displaystyle =\left(\frac{{}^{3}C_{1}}{{}^{7}C_{1}}\times\frac{{}^{2}C_{2}}{{}^{5}C_{2}}\right)+\left(\frac{{}^{4}C_{1}}{{}^{7}C_{1}}\times\frac{{}^{3}C_{2}}{{}^{5}C_{2}}\right)
\displaystyle =\left(\frac{3}{7}\times\frac{1}{10}\right)+\left(\frac{4}{7}\times\frac{3}{10}\right)
\displaystyle =\frac{3}{70}+\frac{12}{70}=\frac{15}{70}=\frac{3}{14}
\\

\displaystyle \textbf{Question 13: } \text{Bag } A \text{ contains } 4 \text{ white balls and } 3 \text{ black balls, while Bag } B
\displaystyle \text{ contains } 3 \text{ white balls}  \text{ and } 5 \text{ black balls. Two balls are drawn from Bag } A \text{ and placed in Bag  B.}
\displaystyle \text{Then, what is the probability of drawing a white ball from Bag } B? 
\displaystyle \text{Answer:}
\displaystyle \text{There are three cases:}
\displaystyle \text{Case I: Two white balls from Bag A}
\displaystyle \text{Case II: Two black balls from Bag A}
\displaystyle \text{Case III: One white and one black from Bag A}
\displaystyle \text{Required probability}
\displaystyle =\frac{{}^{4}C_{2}}{{}^{7}C_{2}}\times\frac{5}{10}+\frac{{}^{3}C_{2}}{{}^{7}C_{2}}\times\frac{3}{10}+\frac{{}^{4}C_{1}\times{}^{3}C_{1}}{{}^{7}C_{2}}\times\frac{4}{10}
\displaystyle =\frac{6}{21}\times\frac{1}{2}+\frac{3}{21}\times\frac{3}{10}+\frac{12}{21}\times\frac{4}{10}
\displaystyle =\frac{1}{7}+\frac{3}{70}+\frac{8}{35}
\displaystyle =\frac{10+3+16}{70}=\frac{29}{70}
\\

\displaystyle \textbf{Question 14: } A, B \text{ and } C \text{ throw a die one after the other in the same order till one of }
\displaystyle \text{them gets a } 6 \text{ and wins the game.} \text{Find their respective probability of winning, if } A \text{ starts}
\displaystyle \text{the game. }
\displaystyle \text{Answer:}
\displaystyle \text{Given, } P(A)=\frac{1}{6},\ P(\overline{A})=\frac{5}{6},\ P(B)=\frac{1}{6},\ P(\overline{B})=\frac{5}{6}
\displaystyle P(C)=\frac{1}{6},\ P(\overline{C})=\frac{5}{6}
\displaystyle \text{Game problem: } A\text{ starts the game}
\displaystyle P(A\text{ wins})=P(\text{first throw})+P(\text{fourth throw})+P(\text{seventh throw})+\cdots
\displaystyle =\frac{1}{6}+\left(\frac{5}{6}\right)^{3}\frac{1}{6}+\left(\frac{5}{6}\right)^{6}\frac{1}{6}+\cdots
\displaystyle =\frac{1}{6}\left[1+\left(\frac{5}{6}\right)^{3}+\left(\frac{5}{6}\right)^{6}+\cdots\right]
\displaystyle =\frac{1}{6}\left[\frac{1}{1-\left(\frac{5}{6}\right)^{3}}\right]
\displaystyle =\frac{1}{6}\times\frac{216}{216-125}=\frac{36}{91}
\displaystyle P(B\text{ wins})=P(\text{second throw})+P(\text{fifth throw})+P(\text{eighth throw})+\cdots
\displaystyle =\frac{5}{6}\cdot\frac{1}{6}+\left(\frac{5}{6}\right)^{4}\frac{1}{6}+\left(\frac{5}{6}\right)^{7}\frac{1}{6}+\cdots
\displaystyle =\frac{5}{36}\left[1+\left(\frac{5}{6}\right)^{3}+\left(\frac{5}{6}\right)^{6}+\cdots\right]
\displaystyle =\frac{5}{36}\times\frac{1}{1-\left(\frac{5}{6}\right)^{3}}
\displaystyle =\frac{5}{36}\times\frac{216}{91}=\frac{30}{91}
\displaystyle P(C\text{ wins})=1-[P(A)+P(B)]
\displaystyle =1-\left(\frac{36}{91}+\frac{30}{91}\right)=\frac{25}{91}
\\

\displaystyle \textbf{Question 15: } \text{50 tickets in a box are numbered } 00,01,02,\ldots,49. \text{ One ticket is drawn}
\displaystyle \text{randomly from the box. Find the probability of the ticket having the product of its digits } 7, 
\displaystyle \text{ given that the sum of the digits is } 8? 
\displaystyle \text{Answer:}
\displaystyle \text{Total number of cases}={}^{50}C_{1}=50
\displaystyle \text{Let } A=\text{event of selecting ticket with sum of digits }8
\displaystyle \text{Favourable cases to }A=\{08,17,26,35,44\}
\displaystyle \text{Let } B=\text{event of selecting ticket with product of digits }7
\displaystyle \text{Favourable case to }B=\{17\}
\displaystyle P\left(\frac{B}{A}\right)=\frac{P(A\cap B)}{P(A)}
\displaystyle =\frac{\frac{1}{50}}{\frac{5}{50}}=\frac{1}{5}
\\

\displaystyle \textbf{Question 16: } \text{If } A \text{ and } B \text{ are two events such that } P(A)>0 \text{ and}
\displaystyle P(B)\ne 1, \text{ then } P\left(\frac{\overline{A}}{\overline{B}}\right) \text{ is }   
\displaystyle \text{(a) } 1-P\left(\frac{\overline{A}}{\overline{B}}\right) \quad \text{(b) } 1-P\left(\frac{A}{B}\right)
\displaystyle \text{(c) } \frac{1-P(A\cup B)}{P(\overline{B})} \quad \text{(d) } \frac{P(\overline{A})}{P(B)}
\displaystyle \text{Answer:}
\displaystyle \text{(c) Since }P(A)>0\text{ and }P(B)\neq 1
\displaystyle P\left(\frac{\overline{A}}{\overline{B}}\right)=\frac{P(\overline{A}\cap\overline{B})}{P(\overline{B})}
\displaystyle =\frac{1-P(A\cup B)}{P(\overline{B})}
\\

\displaystyle \textbf{Question 17: } \text{If } A \text{ and } B \text{ are two events such that } P(A)=\frac{4}{5} \text{ and}
\displaystyle P(B/A)=\frac{7}{8}, \text{ then } P(A\cap B) \text{ is equal to}
\displaystyle \text{(a) } \frac{7}{40} \quad \text{(b) } \frac{21}{40} \quad \text{(c) } \frac{32}{35} \quad \text{(d) } \frac{7}{10}
\displaystyle \text{Answer:}
\displaystyle \text{(d) Given, }P(A)=\frac{4}{5}\text{ and }P\left(\frac{B}{A}\right)=\frac{7}{8}
\displaystyle \text{We know that }P\left(\frac{B}{A}\right)=\frac{P(A\cap B)}{P(A)}
\displaystyle \Rightarrow P(A\cap B)=P\left(\frac{B}{A}\right)P(A)
\displaystyle =\frac{7}{8}\times\frac{4}{5}=\frac{7}{10}
\\

\displaystyle \textbf{Question 18: } \text{If } P(A)=\frac{4}{5} \text{ and } P(A\cap B)=\frac{7}{10}, \text{ then the value of}
\displaystyle P\left(\frac{B}{A}\right) \text{ is } 
\displaystyle \text{(a) } \frac{1}{8} \quad \text{(b) } \frac{4}{7} \quad \text{(c) } \frac{5}{7} \quad \text{(d) } \frac{7}{8}
\displaystyle \text{Answer:}
\displaystyle \text{(d) Given, }P(A)=\frac{4}{5}\text{ and }P(A\cap B)=\frac{7}{10}
\displaystyle P\left(\frac{B}{A}\right)=\frac{P(A\cap B)}{P(A)}
\displaystyle =\frac{\frac{7}{10}}{\frac{4}{5}}=\frac{7}{10}\times\frac{5}{4}=\frac{7}{8}
\\

\displaystyle \textbf{Question 19: } \text{There are two boxes, Box 1 and Box 2. Each box has } 8 \text{ red, } 5 \text{ green and }\\  7 \text{ blue balls.}   \text{Sahil takes one ball from Box 1 and places it in Box 2. Then, he draws two} \\ \text{balls from Box 2 simultaneously.}
\displaystyle \text{(a) What is the probability of Sahil selecting a green ball from Box 1?}
\displaystyle \text{(b) After Sahil transfers a ball from Box 1 to Box 2, what is the probability of drawing one}
\displaystyle \text{red ball and one green ball simultaneously from Box 2?}
\displaystyle \text{(c) Given that the balls drawn from Box 2 are red and green, what is the probability that}
\displaystyle \text{the ball transferred from Box 1 to Box 2 was green?}
\displaystyle \text{(d) Prove that the probability mentioned in (c) is less than } 35\%.   
\displaystyle \text{Answer:}
\displaystyle \text{Bag I: }8R,\ 5G,\ 7B \qquad \text{Bag II: }8R,\ 5G,\ 7B
\displaystyle E_{1}:\text{ red ball is transferred from Bag I to Bag II}
\displaystyle E_{2}:\text{ green ball is transferred from Bag I to Bag II}
\displaystyle E_{3}:\text{ blue ball is transferred from Bag I to Bag II}
\displaystyle \text{(a) }P(E_{2})=\frac{5}{20}=\frac{1}{4}
\displaystyle \text{(b) Let }A:\text{ drawing one red ball and one green ball from Bag II}
\displaystyle P(E_{1})=\frac{8}{20}=\frac{2}{5},\quad P(E_{2})=\frac{5}{20}=\frac{1}{4},\quad P(E_{3})=\frac{7}{20}
\displaystyle P\left(\frac{A}{E_{1}}\right)=\frac{{}^{9}C_{1}\times{}^{5}C_{1}}{{}^{21}C_{2}}=\frac{9\times5}{210}=\frac{45}{210}
\displaystyle P\left(\frac{A}{E_{2}}\right)=\frac{{}^{8}C_{1}\times{}^{6}C_{1}}{{}^{21}C_{2}}=\frac{8\times6}{210}=\frac{48}{210}
\displaystyle P\left(\frac{A}{E_{3}}\right)=\frac{{}^{8}C_{1}\times{}^{5}C_{1}}{{}^{21}C_{2}}=\frac{8\times5}{210}=\frac{40}{210}
\displaystyle P(A)=P(E_{1})P\left(\frac{A}{E_{1}}\right)+P(E_{2})P\left(\frac{A}{E_{2}}\right)+P(E_{3})P\left(\frac{A}{E_{3}}\right)
\displaystyle =\left(\frac{8}{20}\times\frac{45}{210}\right)+\left(\frac{5}{20}\times\frac{48}{210}\right)+\left(\frac{7}{20}\times\frac{40}{210}\right)
\displaystyle =\frac{360+240+280}{20\times210}=\frac{880}{4200}=\frac{22}{105}\approx0.21
\displaystyle \text{(c) }P\left(\frac{E_{2}}{A}\right)=\frac{P(E_{2})P\left(\frac{A}{E_{2}}\right)}{P(A)}
\displaystyle =\frac{\frac{5}{20}\times\frac{48}{210}}{\frac{22}{105}}=\frac{240}{4200}\times\frac{105}{22}=\frac{6}{105}\times\frac{105}{22}=\frac{6}{22}=\frac{3}{11}\approx0.27
\displaystyle \text{(d) The required probability }=0.27\ (27\%),\ \text{which is less than }35\%
\\

\displaystyle \textbf{Question 20: } \text{A student answers a multiple choice question with } 5 \text{ alternatives, of which }
\displaystyle \text{exactly one is correct. The probability that he knows the correct answer is } \frac{1}{5}. \text{If he does} 
\displaystyle \text{not know}\text{ the correct answer, he randomly ticks one answer. Given that he has answere}
\displaystyle \text{ the question correctly, find the probability that he did not tick the answer randomly.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }E_{1}=\text{student does not know the answer}
\displaystyle E_{2}=\text{student knows the answer}
\displaystyle E=\text{student answers correctly}
\displaystyle \therefore P(E_{1})=1-\frac{1}{5}=\frac{4}{5},\quad P(E_{2})=\frac{1}{5}
\displaystyle P\left(\frac{E}{E_{2}}\right)=1,\quad P\left(\frac{E}{E_{1}}\right)=\frac{1}{5}
\displaystyle \text{The probability that the student did not tick the answer randomly}
\displaystyle \text{given that the student ticked the answer correctly is}
\displaystyle P\left(\frac{E_{2}}{E}\right)=\frac{P(E_{2})P\left(\frac{E}{E_{2}}\right)}{P(E_{2})P\left(\frac{E}{E_{2}}\right)+P(E_{1})P\left(\frac{E}{E_{1}}\right)}
\displaystyle =\frac{\frac{1}{5}\times 1}{\frac{1}{5}\times 1+\frac{4}{5}\times\frac{1}{5}}
\displaystyle =\frac{\frac{1}{5}}{\frac{1}{5}+\frac{4}{25}}
\displaystyle =\frac{\frac{1}{5}}{\frac{9}{25}}=\frac{1}{5}\times\frac{25}{9}=\frac{5}{9}
\\

\displaystyle \textbf{Question 21: } \text{A bag contains } 6 \text{ red and } 5 \text{ blue balls and another bag contains } 5 \text{ red}
\displaystyle \text{and } 8 \text{ blue balls. A ball is drawn from the first bag and without noticing its colour is placed}
\displaystyle \text{in the second bag. If a ball is drawn from the second bag, then find the probability that the drawn}
\displaystyle \text{ball is red in colour.}
\displaystyle \text{Answer:}
\displaystyle \text{Consider the following events}
\displaystyle E_{1}=\text{transferred ball from Bag I to Bag II is red}
\displaystyle E_{2}=\text{transferred ball from Bag I to Bag II is blue}
\displaystyle A=\text{ball drawn from Bag II is red}
\displaystyle \text{Bag I has }6\text{ red and }5\text{ blue balls}
\displaystyle \text{Bag II has }5\text{ red and }8\text{ blue balls}
\displaystyle \therefore P(E_{1})=\frac{6}{11},\quad P(E_{2})=\frac{5}{11}
\displaystyle P\left(\frac{A}{E_{1}}\right)=\frac{6}{14},\quad P\left(\frac{A}{E_{2}}\right)=\frac{5}{14}
\displaystyle \text{Required probability }P(A)\text{ is given by}
\displaystyle P(A)=P(E_{1})P\left(\frac{A}{E_{1}}\right)+P(E_{2})P\left(\frac{A}{E_{2}}\right)
\displaystyle =\left(\frac{6}{11}\times\frac{6}{14}\right)+\left(\frac{5}{11}\times\frac{5}{14}\right)
\displaystyle =\frac{36}{154}+\frac{25}{154}=\frac{61}{154}
\\

\displaystyle \textbf{Question 22: } \text{A word consists of } 10 \text{ different alphabets, in which there are } \\ 4 \text{ consonants and } 6 \text{ vowels.}  \text{Four alphabets are chosen at random. What is the} \\ \text{ probability that more than one vowel is selected?}
\displaystyle \text{Answer:}
\displaystyle \text{Let } X \text{ denote the number of vowels.}
\displaystyle \text{Required probability}=P(X>1)
\displaystyle =1-[P(X=0)+P(X=1)]
\displaystyle =1-\left[\frac{{}^{4}C_{4}\times{}^{6}C_{0}}{{}^{10}C_{4}}+\frac{{}^{4}C_{3}\times{}^{6}C_{1}}{{}^{10}C_{4}}\right]
\displaystyle =1-\left[\frac{1\times1}{210}+\frac{4\times6}{210}\right]
\displaystyle =1-\left[\frac{1}{210}+\frac{24}{210}\right]
\displaystyle =1-\frac{25}{210}
\displaystyle =\frac{185}{210}=\frac{37}{42}
\\

\displaystyle \textbf{Question 23: } \text{Let } X \text{ denote the number of hours you study during a}
\displaystyle \text{randomly selected school day. The probability that } X \text{ can take}
\displaystyle \text{the values } x \text{ has the following form, where } k \text{ is some}
\displaystyle \text{unknown constant.}
\displaystyle P(X=x)=\begin{cases} 0.1, & \text{if } x=0\\ kx, & \text{if } x=1 \text{ or } 2\\ k(5-x), & \text{if } x=3 \text{ or } 4\\ 0, & \text{otherwise} \end{cases}
\displaystyle \text{(i) Find the value of } k.
\displaystyle \text{(ii) What is the probability that you study:}
\displaystyle \text{(a) at least two hours?} \quad \text{(b) exactly two hours?}
\displaystyle \text{(c) at most } 2 \text{ hours?}
\displaystyle \text{Answer:}
\displaystyle \text{The probability distribution table for given function }X\text{ is}
\displaystyle \begin{array}{c|cccccc} X & 0 & 1 & 2 & 3 & 4 & \text{Otherwise} \\ \hline P(X) & 0.1 & k & 2k & 2k & k & 0 \end{array}
\displaystyle \text{(i) We know that } \sum P(X)=1
\displaystyle \Rightarrow 0.1+k+2k+2k+k+0=1
\displaystyle \Rightarrow 6k=1-0.1
\displaystyle \Rightarrow k=\frac{0.9}{6}=0.15
\displaystyle \text{(ii) (a) At least two hours,}
\displaystyle P(X\geq 2)=1-P(X<2)
\displaystyle =1-P(X=0)-P(X=1)
\displaystyle =1-0.1-0.15
\displaystyle =1-0.25=0.75
\displaystyle \text{(b) Exactly two hours,}
\displaystyle P(X=2)=2k
\displaystyle =2(0.15)=0.30
\displaystyle \text{(c) At most two hours,}
\displaystyle P(X\leq 2)=P(X=0)+P(X=1)+P(X=2)
\displaystyle =0.1+0.15+0.30
\displaystyle =0.55
\\

\displaystyle \textbf{Question 24: } \text{An insurance company insured } 1000 \text{ scooter drivers, } 2000 \text{ car drivers}
\displaystyle \text{and } 4000 \text{ truck drivers. The probability of accidents by scooter, car and truck drivers}
\displaystyle \text{are } 0.02,0.05 \text{ and } 0.03, \text{ respectively. If one of the insured persons meets with an accident,}
\displaystyle \text{find the probability that he is a truck driver.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }E_{1}=\text{scooter drivers are insured},\quad E_{2}=\text{car drivers are insured},\quad E_{3}=\text{truck drivers are insured}
\displaystyle P(E_{1})=\frac{1000}{7000}=\frac{1}{7},\quad P(E_{2})=\frac{2000}{7000}=\frac{2}{7},\quad P(E_{3})=\frac{4000}{7000}=\frac{4}{7}
\displaystyle \text{Let }A=\text{event that drivers meet with an accident}
\displaystyle P\!\left(\frac{A}{E_{1}}\right)=0.02,\quad P\!\left(\frac{A}{E_{2}}\right)=0.05,\quad P\!\left(\frac{A}{E_{3}}\right)=0.03
\displaystyle \text{By Bayes' theorem, } P\!\left(\frac{E_{3}}{A}\right)= \frac{P(E_{3})P\!\left(\frac{A}{E_{3}}\right)} {P(E_{1})P\!\left(\frac{A}{E_{1}}\right)+P(E_{2})P\!\left(\frac{A}{E_{2}}\right)+P(E_{3})P\!\left(\frac{A}{E_{3}}\right)}
\displaystyle = \frac{\frac{4}{7}\times0.03} {\frac{1}{7}\times0.02+\frac{2}{7}\times0.05+\frac{4}{7}\times0.03}
\displaystyle = \frac{0.12/7}{(0.02+0.10+0.12)/7} =\frac{0.12}{0.24} =\frac{1}{2}
\\

\displaystyle \textbf{Question 25: } \text{In a bolt factory, machines } X,Y \text{ and } Z \text{ manufacture } 20\%,35\% \\ \text{and } 45\% \text{ respectively of the total output.}   \text{Of their output, } 8\%,6\% \text{ and } 5\% \text{ respectively} \\ \text{are defective bolts. One bolt is drawn at random}   \text{from the product and is found to be} \\ \text{defective. What is the probability that it was manufactured in machine } Y?
\displaystyle \text{Answer:}
\displaystyle \text{Consider the following events}
\displaystyle E_{1}=\text{bolt manufactured in machine }X,\ E_{2}=\text{bolt manufactured in machine }Y,\ E_{3}=\text{bolt manufactured in machine }Z
\displaystyle A=\text{bolt drawn is defective}
\displaystyle P(E_{1})=\frac{20}{100}=\frac{1}{5},\quad P(E_{2})=\frac{35}{100}=\frac{7}{20},\quad P(E_{3})=\frac{45}{100}=\frac{9}{20}
\displaystyle P\!\left(\frac{A}{E_{1}}\right)=\frac{8}{100}=\frac{2}{25},\quad P\!\left(\frac{A}{E_{2}}\right)=\frac{6}{100}=\frac{3}{50},\quad P\!\left(\frac{A}{E_{3}}\right)=\frac{5}{100}=\frac{1}{20}
\displaystyle \text{Required probability } 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)+P(E_{3})P\!\left(\frac{A}{E_{3}}\right)}
\displaystyle = \frac{\frac{7}{20}\times\frac{3}{50}} {\frac{1}{5}\times\frac{2}{25}+\frac{7}{20}\times\frac{3}{50}+\frac{9}{20}\times\frac{1}{20}}
\displaystyle = \frac{\frac{21}{1000}} {\frac{2}{125}+\frac{21}{1000}+\frac{9}{400}}
\displaystyle = \frac{\frac{21}{1000}} {\frac{16+21+22.5}{1000}} =\frac{21}{59.5} =\frac{42}{119} =\frac{6}{17}
\\

\displaystyle \textbf{Question 26: } \text{A group of class XI students designed a game for class XII students}
\displaystyle \text{to be played during the farewell programme organized in their school. They kept two}
\displaystyle \text{sets of cards on the stage for the activity. Each set contained four cards numbered}
\displaystyle 1, 2, 3 \text{ and } 4. \text{ Each student of class XII had to draw one card from each set}
\displaystyle \text{simultaneously. The score obtained by a student was defined as the difference between}
\displaystyle \text{the numbers shown on the two cards drawn. The possible outcomes of the score are}
\displaystyle \text{given in the table below for reference.  }
\displaystyle  \text{The numbers on two cards are shown in the table below.}
\displaystyle \text{The numbers on card 1 are written along the top and the numbers on}
\displaystyle \text{card 2 are written along the left-hand side of the table.}
\displaystyle \begin{array}{c|cccc} \text{Card 2 \textbackslash\ Card 1} & 1 & 2 & 3 & 4 \\ \hline 1 & 0 & p & 2 & 3 \\ 2 & 1 & 0 & p & 2 \\ 3 & q & 1 & 0 & p \\ 4 & 3 & q & 1 & 0 \end{array}
\displaystyle \text{(i) Calculate the values of } p \text{ and } q \text{ in the above table.}
\displaystyle \text{(ii) Reshma plays the game once. Let the random variable } S \text{ represent Reshma's score.}
\displaystyle \text{Construct the probability distribution for } S.
\displaystyle \begin{array}{c|ccccccc} S & -3 & -2 & -1 & 0 & 1 & 2 & 3 \\ \hline P(S=s) & & & & & & & \end{array}
\displaystyle \text{(iii) Find the probability that Reshma's score is at most } 2.
\displaystyle \text{(iv) Find Reshma's expected score.}
\displaystyle \text{Answer:}
\displaystyle \text{(i) }p=2-1=1,\ q=3-1=2
\displaystyle \text{(ii) }P(S=0)=\frac{4}{16}=\frac{1}{4}
\displaystyle P(S=1)=\frac{6}{16}=\frac{3}{8}
\displaystyle P(S=2)=\frac{4}{16}=\frac{1}{4}
\displaystyle P(S=3)=\frac{2}{16}=\frac{1}{8}
\displaystyle \text{(iii) }P(S\leq2)=\frac{1}{4}+\frac{3}{8}+\frac{1}{4}=\frac{7}{8}
\displaystyle \text{(iv) Expected score}
\displaystyle =0\cdot\frac{1}{4}+1\cdot\frac{3}{8}+2\cdot\frac{1}{4}+3\cdot\frac{1}{8}
\displaystyle =\frac{3}{8}+\frac{2}{4}+\frac{3}{8}=\frac{3+4+3}{8}=\frac{10}{8}=\frac{5}{4}
\displaystyle \text{Answer: }\frac{5}{4}
\\

\displaystyle \textbf{Question 27: } \text{A biased four-sided die with faces labelled } 1, 2, 3 \text{ and } 4 \text{ is}
\displaystyle \text{rolled and recorded. Let } X \text{ be the result obtained when the die is rolled.}
\displaystyle \text{The probability distribution for } X \text{ is given in the following table,}
\displaystyle \text{where } p \text{ and } q \text{ are constants.}
\displaystyle \begin{array}{c|cccc} x & 1 & 2 & 3 & 4 \\ \hline P(X=x) & p & 0.3 & q & 0.1 \end{array}
\displaystyle \text{For the above probability distribution, it is known that } E(X)=2.
\displaystyle \text{Find } p \text{ and } q. \hspace{2cm} \text{Also, find } P(X>2).
\displaystyle \text{Ajay plays a game with this four-sided die. In this game, he is allowed}
\displaystyle \text{a maximum of five rolls. His score is calculated by adding the results}
\displaystyle \text{of each roll. He wins the game if his score is at least } 10. \text{ After } 3 \text{ rolls,}
\displaystyle \text{Ajay has a score of } 4 \text{ points. Assuming that the rolls of the die are}
\displaystyle \text{independent, find the probability that Ajay wins the game.  }
\displaystyle \text{Answer:}
\displaystyle \text{Here, the probability distribution of }x\text{ is given below}
\displaystyle \begin{array}{c|cccc}x&1&2&3&4\\ \hline P(X=x)&p&0.3&q&0.1\end{array}
\displaystyle \therefore\ \text{Mean}=E(X)=2
\displaystyle E(X)=\sum_{i=1}^{n}X_iP_i
\displaystyle \Rightarrow 2=1\times p+2\times0.3+3q+4\times0.1
\displaystyle \Rightarrow 2=p+0.6+3q+0.4
\displaystyle \Rightarrow 2=p+3q+1
\displaystyle \Rightarrow 1=p+3q\qquad ...(i)
\displaystyle \text{Now, }p+0.3+q+0.1=1
\displaystyle \Rightarrow p+q=1-0.4=0.6\qquad ...(ii)
\displaystyle \text{From Eqs. (i) and (ii), we get}
\displaystyle q=0.2\text{ and }p=0.4
\displaystyle \therefore\ P(X>2)=P(X=3)+P(X=4)
\displaystyle =q+0.1=0.2+0.1
\displaystyle =0.3
\displaystyle \text{Now, }4\text{ points after }3\text{ rolls can be achieved in following ways}
\displaystyle \{(2,1,1),(1,1,2),(1,2,1)\}.
\displaystyle \therefore\ \text{Probability of having }4\text{ points after }3\text{ rolls is}
\displaystyle [P(X=2)\times P(X=1)\times P(X=1)]
\displaystyle +[P(X=1)\times P(X=1)\times P(X=2)]
\displaystyle +[P(X=1)\times P(X=2)\times P(X=1)]
\displaystyle =(0.3\times0.4\times0.4)+(0.4\times0.4\times0.3)+(0.4\times0.3\times0.4)
\displaystyle =0.144
\displaystyle \text{Now, at least }10\text{ points can be scored, if in the last }2\text{ rolls}
\displaystyle \text{at least }6\text{ points are scored.}
\displaystyle \therefore\ \text{This can happen in following ways}
\displaystyle \{(4,3),(3,4),(4,4),(3,3),(4,2),(2,4)\}
\displaystyle \text{Probability of scoring at least }6\text{ points in the last }2\text{ throws is}
\displaystyle [P(X=4)\times P(X=3)]+[P(X=3)\times P(X=4)]
\displaystyle +[P(X=4)\times P(X=4)]+[P(X=3)\times P(X=3)]
\displaystyle +[P(X=4)\times P(X=2)]+[P(X=2)\times P(X=4)]
\displaystyle =0.02+0.02+0.01+0.04+0.03+0.03
\displaystyle =0.15
\displaystyle \text{Now, the probability that Ajay wins the game}
\displaystyle =0.144\times0.15=0.0216
\displaystyle \text{Answer: }0.0216
\\

\displaystyle \textbf{Question 28: } \text{Dobleron, a chocolate is known for its distinctive shape and its taste.}
\displaystyle \text{The quality assurance team of the brand undertakes quality test in two steps}
\displaystyle \text{Step I: } 30 \text{ out of every } 990 \text{ chocolate bars are randomly picked for physical examination,}
\displaystyle \text{where the shape of the bar is examined.}
\displaystyle \text{Step II: For every } 30 \text{ chocolate bars whose shape is examined, } 1 \text{ chocolate bar is randomly picked}
\displaystyle \text{for chemical examination, where the composition and taste of the chocolate is examined.}
\displaystyle \text{If one chocolate bar is randomly picked out of } 990 \text{ bars, what is the probability that it was not}
\displaystyle \text{chemically examined, provided it was physically examined?  }
\displaystyle \text{Answer:}
\displaystyle \text{Let } A:\text{ event that a chocolate bar is physically examined}
\displaystyle \text{Let } B:\text{ event that a chocolate bar is chemically examined}
\displaystyle \text{We need to find } P\left(\frac{B'}{A}\right), \text{ i.e., probability of } B' \text{ given } A
\displaystyle \text{According to the problem, } 30 \text{ out of every } 990 \text{ chocolate bars are physically examined}
\displaystyle \therefore P(A)=\frac{30}{990}=\frac{1}{33}
\displaystyle \text{In Step II, for every } 30 \text{ chocolate bars physically examined, one is chemically examined}
\displaystyle \therefore P\left(\frac{B}{A}\right)=\frac{1}{30}
\displaystyle \text{Now, probability that a chocolate bar was not chemically examined given it was physically examined is}
\displaystyle P\left(\frac{B'}{A}\right)=1-P\left(\frac{B}{A}\right)=1-\frac{1}{30}=\frac{29}{30}
\\

\displaystyle \textbf{Question 29: } \text{A family has two children. What is the probability that both children are boys,}
\displaystyle \text{given that at least one of them is a boy? }   
\displaystyle \text{Answer:}
\displaystyle \text{Given, a family has two children}
\displaystyle \text{Let }A=\text{event that both children are boys}
\displaystyle A=\{BB\}
\displaystyle \text{Let }B=\text{event that at least one child is a boy}
\displaystyle B=\{BB,BG,GB\}
\displaystyle A\cap B=\{BB\}
\displaystyle P(A)=\frac{1}{4},\quad P(B)=\frac{3}{4},\quad P(A\cap B)=\frac{1}{4}
\displaystyle \text{Required probability}=P\left(\frac{A}{B}\right)
\displaystyle =\frac{P(A\cap B)}{P(B)}=\frac{\frac{1}{4}}{\frac{3}{4}}=\frac{1}{3}
\\

\displaystyle \textbf{Question 30: } \text{On a festival, two friends namely Shanti and Diya got identical gift bags}
\displaystyle \text{containing } 7 \text{ dark, } 8 \text{ milk} \text{and } 5 \text{ white chocolates each.}
\displaystyle \text{(a) What is the probability of Diya drawing a dark chocolate from Shanti's bag?}
\displaystyle \text{(b) While Shanti was away, Diya transferred one chocolate from Shanti's bag to hers without looking.}
\displaystyle \text{What is the probability of drawing a dark chocolate from Diya's bag after the transfer?}
\displaystyle \text{(c) Which one of the following probabilities is higher? Why?}
\displaystyle \text{The probability of drawing a dark chocolate from Diya's bag after the transfer or the probability}
\displaystyle \text{of drawing a dark chocolate from Shanti's bag before the transfer.}   
\displaystyle \text{Answer:}
\displaystyle \text{Let }E_{1}:\text{ Diya draws a dark chocolate from Shanti's bag}
\displaystyle E_{2}:\text{ Diya draws a milk chocolate from Shanti's bag}
\displaystyle E_{3}:\text{ Diya draws a white chocolate from Shanti's bag}
\displaystyle A:\text{ dark chocolate drawn from Diya's bag after transfer}
\displaystyle \text{(a) }P(E_{1})=\frac{7}{20}
\displaystyle \text{(b) }P(E_{1})=\frac{7}{20},\quad P(E_{2})=\frac{8}{20},\quad P(E_{3})=\frac{5}{20}
\displaystyle P\left(\frac{A}{E_{1}}\right)=\frac{8}{21},\quad P\left(\frac{A}{E_{2}}\right)=\frac{7}{21},\quad P\left(\frac{A}{E_{3}}\right)=\frac{7}{21}
\displaystyle P(A)=P(E_{1})P\left(\frac{A}{E_{1}}\right)+P(E_{2})P\left(\frac{A}{E_{2}}\right)+P(E_{3})P\left(\frac{A}{E_{3}}\right)
\displaystyle =\left(\frac{7}{20}\times\frac{8}{21}\right)+\left(\frac{8}{20}\times\frac{7}{21}\right)+\left(\frac{5}{20}\times\frac{7}{21}\right)
\displaystyle =\frac{56+56+35}{420}=\frac{147}{420}=0.35
\displaystyle \text{(c) Probability of drawing a dark chocolate after transfer }=0.35
\displaystyle \text{Probability before transfer }=\frac{7}{20}=0.35
\displaystyle \therefore \text{Both probabilities are equal.}
\\

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