\displaystyle \textbf{Question 1: } \text{Evaluate } P(A \cup B), \text{ if } 2P(A)=P(B)=\frac{5}{13} \text{ and }  \\ P\left(\frac{A}{B}\right)=\frac{2}{3}. \text{ (ISC 2024)}
\displaystyle \text{Answer:}
\displaystyle \text{Given, } 2P(A)=P(B)=\frac{5}{13}
\displaystyle \therefore P(A)=\frac{5}{26},\quad P(B)=\frac{5}{13}
\displaystyle \text{and } P\left(\frac{A}{B}\right)=\frac{2}{5}
\displaystyle \text{We know that } P\left(\frac{A}{B}\right)=\frac{P(A\cap B)}{P(B)}
\displaystyle \Rightarrow \frac{2}{5}=\frac{P(A\cap B)}{\frac{5}{13}}
\displaystyle \Rightarrow P(A\cap B)=\frac{2}{13}
\displaystyle \therefore P(A\cup B)=P(A)+P(B)-P(A\cap B)
\displaystyle =\frac{5}{26}+\frac{5}{13}-\frac{2}{13}
\displaystyle =\frac{5}{26}+\frac{10}{26}-\frac{4}{26}
\displaystyle =\frac{11}{26}
\\

\displaystyle \textbf{Question 2: } \text{A jewellery seller has precious gems in white and red colour which he has }
\displaystyle \text{put in three boxes. The distribution } \text{of these gems is shown in the table given below } \quad \text{ISC 2024}
\displaystyle \begin{array}{|c|c|c|} \hline \text{Box} & \text{White} & \text{Red}\\ \hline \text{I} & 1 & 2\\ \text{II} & 2 & 3\\ \text{III} & 3 & 1\\ \hline \end{array}
\displaystyle \text{He wants to gift two gems to his mother. So, he asks her to select one box at random }
\displaystyle \text{and pick out any two gems one after the other without replacement from the}
\displaystyle \text{selected box. The mother selects one white and one red gem Calculate the probability }
\displaystyle \text{that the gems drawn are from } \text{Box II.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }A=\text{one gem is white and the other is red}
\displaystyle E_{1}=\text{both gems are from Box I}
\displaystyle E_{2}=\text{both gems are from Box II}
\displaystyle E_{3}=\text{both gems are from Box III}
\displaystyle P(E_{1})=P(E_{2})=P(E_{3})=\frac{1}{3}
\displaystyle \text{Required probability}=P\left(\frac{E_{2}}{A}\right)
\displaystyle =\frac{P\left(\frac{A}{E_{2}}\right)P(E_{2})}{P\left(\frac{A}{E_{1}}\right)P(E_{1})+P\left(\frac{A}{E_{2}}\right)P(E_{2})+P\left(\frac{A}{E_{3}}\right)P(E_{3})}
\displaystyle \text{[using Bayes' theorem]}
\displaystyle =\frac{\frac{{}^{2}C_{1}\times{}^{3}C_{1}}{{}^{5}C_{2}}\times\frac{1}{3}}{\frac{{}^{1}C_{1}\times{}^{2}C_{1}}{{}^{3}C_{2}}\times\frac{1}{3}+\frac{{}^{2}C_{1}\times{}^{3}C_{1}}{{}^{5}C_{2}}\times\frac{1}{3}+\frac{{}^{3}C_{1}\times{}^{1}C_{1}}{{}^{4}C_{2}}\times\frac{1}{3}}
\displaystyle =\frac{\frac{6}{10}\times\frac{1}{3}}{\frac{2}{3}\times\frac{1}{3}+\frac{6}{10}\times\frac{1}{3}+\frac{3}{6}\times\frac{1}{3}}
\displaystyle =\frac{\frac{6}{10}}{\frac{2}{3}+\frac{6}{10}+\frac{3}{6}}
\displaystyle =\frac{\frac{6}{10}}{\frac{20+18+15}{30}}
\displaystyle =\frac{6}{10}\times\frac{30}{53}
\displaystyle =\frac{18}{53}
\displaystyle \text{Hence, the probability that the gems drawn are from Box II is } \frac{18}{53}
\\

\displaystyle \textbf{Question 3: } \text{A primary school teacher wants to teach the concept of  'larger number'}
\displaystyle \text{to the students of Class II in a simple and engaging manner. To teach this concept, he}
\displaystyle \text{conducts an activity in his class where he asks the children to select two numbers}
\displaystyle \text{from a set given as } 2, 3, 4, 5 \text{ one after the other without replacement. All the}
\displaystyle \text{outcomes of this activity are systematically tabulated in the form of ordered pairs}
\displaystyle \text{as shown in the table given below for better understanding. \hspace{1cm} ISC 2024}
\displaystyle \begin{array}{c|cccc} & 2 & 3 & 4 & 5 \\ \hline 2 & (2,2) & (2,3) & (2,4) & (2,5) \\ 3 & (3,2) & (3,3) & (3,4) & (3,5) \\ 4 & (4,2) & (4,3) & (4,4) & (4,5) \\ 5 & (5,2) & (5,3) & (5,4) & (5,5) \end{array}
\displaystyle \text{(i) Complete the table given above carefully. \hspace{2.5cm} (ii) Find the total number of}
\displaystyle \text{ordered pairs having exactly one larger number in each case.}
\displaystyle \text{(iii) Let the random variable } X \text{ denote the larger of the two numbers in an ordered}
\displaystyle \text{pair. Now, complete the probability distribution table for } X \text{ given below.}
\displaystyle \begin{array}{c|ccc} X & 3 & 4 & 5 \\ \hline P(X=x) & & & \end{array}
\displaystyle \text{(iv) Find the value of } P(X<5). \hspace{2cm} \text{(v) Calculate the expected value of the}
\displaystyle \text{given probability distribution.}
\displaystyle \text{Answer:}
\displaystyle \text{Given set of numbers is }A=\{2,3,4,5\}
\displaystyle \text{Total number of ordered pairs }=2^{|A|}=2^{4}=16
\displaystyle \text{(i) All outcomes of this cutting are tabulated in the form of ordered pairs}
\displaystyle \text{(ii) The total number of pairs having one larger number}
\displaystyle =\text{Total number of ordered pairs}-\text{Number of ordered pairs having same number}
\displaystyle =16-4=12
\displaystyle \text{(iii) Let }X\text{ be a random variable denoting the larger of two numbers in ordered pair.}
\displaystyle \text{Total number of ordered pairs }=12
\displaystyle \text{Number of ordered pairs having }3\text{ as larger number }=2
\displaystyle P(X=3)=\frac{2}{12}=\frac{1}{6}
\displaystyle \text{Number of ordered pairs having }4\text{ as larger number }=4
\displaystyle P(X=4)=\frac{4}{12}=\frac{1}{3}
\displaystyle \text{Number of ordered pairs having }5\text{ as larger number }=6
\displaystyle P(X=5)=\frac{6}{12}=\frac{1}{2}
\displaystyle \text{(iv) }P(X\leq5)=P(X=2)+P(X=3)+P(X=4)
\displaystyle =0+\frac{1}{6}+\frac{1}{3}=\frac{1}{2}
\displaystyle \text{(v) Expected value of probability distribution}
\displaystyle E(X)=\sum x_i p_i
\displaystyle =2\cdot P(X=2)+3\cdot P(X=3)+4\cdot P(X=4)+5\cdot P(X=5)
\displaystyle =0+3\cdot\frac{1}{6}+4\cdot\frac{1}{3}+5\cdot\frac{1}{2}
\displaystyle =\frac{1}{2}+\frac{4}{3}+\frac{5}{2}
\displaystyle =\frac{3+8+15}{6}=\frac{26}{6}=\frac{13}{3}
\displaystyle \text{Hence, the expected value of probability distribution is }\frac{13}{3}
\displaystyle \text{Answer: }\frac{13}{3}
\\

\displaystyle \textbf{Question 4: } \text{Teena is practising for an upcoming Rifle Shooting tournament.}
\displaystyle \text{The probability of her shooting the target in the 1st, 2nd, 3rd and 4th shots are } \\  0.4,0.3,0.2 \text{ and } 0.1   \text{respectively. Find the probability of at least one shot of Teena} \\ \text{hitting the target.}\text{ISC 2024}
\displaystyle \text{Answer:}
\displaystyle \text{Given probability of her shooting the target in}
\displaystyle P(\text{1st shot})=0.4
\displaystyle P(\text{2nd shot})=0.3
\displaystyle P(\text{3rd shot})=0.2
\displaystyle \text{and } P(\text{4th shot})=0.1
\displaystyle \therefore \text{Probability of at least one shot of Teena hitting the target}
\displaystyle =1-\text{Probability of none of the shots hitting the target}
\displaystyle =1-[(1-0.4)(1-0.3)(1-0.2)(1-0.1)]
\displaystyle =1-(0.6\times 0.7\times 0.8\times 0.9)
\displaystyle =1-0.3024
\displaystyle =0.6976
\\

\displaystyle \textbf{Question 5: } \text{A bag contains } 19 \text{ tickets, numbered from } 1 \text{ to } 19. \text{ Two tickets} \\ \text{are drawn randomly in succession with replacement.} \text{Find the probability that} \\ \text{both tickets drawn are even numbers.} \quad \text{ISC 2023}
\displaystyle \text{Answer:}
\displaystyle \text{There are } 9 \text{ even numbers from } 1 \text{ to } 19.
\displaystyle \text{Probability that both tickets drawn are even numbers}
\displaystyle =\frac{9}{19}\times\frac{8}{18}=\frac{4}{19}
\\

\displaystyle \textbf{Question 6: } \text{The probability of the event } A \text{ occurring is } \frac{1}{3} \text{ and of the event } B \\ \text{ occurring is } \frac{1}{2}.  \text{If } A \text{ and } B \text{ are independent events, then find the probability of neither } \\ A \text{ nor } B \text{ occurring. (ISC 2023)}
\displaystyle \text{Answer:}
\displaystyle \text{Given, } P(A)=\frac{1}{3}\text{ and }P(B)=\frac{1}{2}
\displaystyle \text{where } A \text{ and } B \text{ are independent events}
\displaystyle \therefore P(A\cap B)=P(A)\cdot P(B)=\frac{1}{3}\times\frac{1}{2}=\frac{1}{6}
\displaystyle \therefore P\left((A\cup B)'\right)=1-P(A\cup B)
\displaystyle =1-[P(A)+P(B)-P(A\cap B)]
\displaystyle =1-\left[\frac{1}{3}+\frac{1}{2}-\frac{1}{6}\right]
\displaystyle =1-\frac{4}{6}=\frac{2}{6}=\frac{1}{3}
\\

\displaystyle \textbf{Question 7: } \text{A problem in Mathematics is given to three students } A, B \text{ and } C. \text{ Their} \\ \text{chances of solving the problem are } \frac{1}{2}, \frac{1}{3} \text{ and } \frac{1}{4}, \text{ respectively.}
\displaystyle \text{(a) Find the probability that exactly two students will solve the problem.}
\displaystyle \text{(b) Find the probability that at least two of them will solve the problem.} \quad \text{ISC 2023}
\displaystyle \text{Answer:}
\displaystyle \text{Consider the events}
\displaystyle A=\text{Problem solved by } A
\displaystyle B=\text{Problem solved by } B
\displaystyle C=\text{Problem solved by } C
\displaystyle \text{Given, } P(A)=\frac{1}{2},\ P(B)=\frac{1}{3},\ P(C)=\frac{1}{4}
\displaystyle \therefore P(A')=\frac{1}{2},\ P(B')=\frac{2}{3},\ P(C')=\frac{3}{4}
\displaystyle \text{(a) } P(\text{exactly two students will solve the problem})
\displaystyle =P(ABC')+P(AB'C)+P(A'BC)
\displaystyle =P(A)P(B)P(C')+P(A)P(B')P(C)+P(A')P(B)P(C)
\displaystyle =\frac{1}{2}\times\frac{1}{3}\times\frac{3}{4}+\frac{1}{2}\times\frac{2}{3}\times\frac{1}{4}+\frac{1}{2}\times\frac{1}{3}\times\frac{1}{4}
\displaystyle =\frac{3+2+1}{24}=\frac{6}{24}=\frac{1}{4}
\displaystyle \text{(b) } P(\text{at least two of them will solve the problem})
\displaystyle =P(ABC')+P(AB'C)+P(A'BC)+P(ABC)
\displaystyle =P(A)P(B)P(C')+P(A)P(B')P(C)+P(A')P(B)P(C)+P(A)P(B)P(C)
\displaystyle =\frac{1}{2}\times\frac{1}{3}\times\frac{3}{4}+\frac{1}{2}\times\frac{2}{3}\times\frac{1}{4}+\frac{1}{2}\times\frac{1}{3}\times\frac{1}{4}+\frac{1}{2}\times\frac{1}{3}\times\frac{1}{4}
\displaystyle =\frac{3+2+1+1}{24}=\frac{7}{24}
\\

\displaystyle \textbf{Question 8: } \text{In a company, } 15\% \text{ of the employees are graduates and } 85\% \text{ are non-graduates.}
\displaystyle \text{As per the annual report of the company, } 80\% \text{ of the graduate employees and } 10\% \text{ of the}
\displaystyle \text{non-graduate employees are in administrative positions. Find the probability that an employee}
\displaystyle \text{selected at random from those working in administrative positions will be a graduate.}
\displaystyle \quad \text{ISC 2023}
\displaystyle \text{Answer:}
\displaystyle \text{Consider the event}
\displaystyle E_{1}:\text{ Employees are graduates},\quad E_{2}:\text{ Employees are non-graduates}
\displaystyle A:\text{ Employees are in administrative position}
\displaystyle P(E_{1})=15\%=\frac{15}{100},\quad P(E_{2})=85\%=\frac{85}{100}
\displaystyle P\left(\frac{A}{E_{1}}\right)=80\%=\frac{80}{100},\quad P\left(\frac{A}{E_{2}}\right)=10\%=\frac{10}{100}
\displaystyle \text{Required probability }=P\left(\frac{E_{1}}{A}\right)
\displaystyle =\frac{P(E_{1})P\left(\frac{A}{E_{1}}\right)}{P(E_{1})P\left(\frac{A}{E_{1}}\right)+P(E_{2})P\left(\frac{A}{E_{2}}\right)}
\displaystyle =\frac{\frac{15}{100}\times\frac{80}{100}}{\frac{15}{100}\times\frac{80}{100}+\frac{85}{100}\times\frac{10}{100}}
\displaystyle =\frac{1200}{1200+850}=\frac{1200}{2050}=\frac{24}{41}
\\

\displaystyle \textbf{Question 9: } \text{A box contains } 30 \text{ fruits, out of which } 10 \text{ are rotten. Two}
\displaystyle \text{fruits are selected at random one by one without replacement from}
\displaystyle \text{the box. Find the probability distribution of the number of unspoiled}
\displaystyle \text{fruits. Also, find the mean of the probability distribution. \hspace{1cm} ISC 2023}
\displaystyle \text{Answer:}
\displaystyle \text{It is given that out of }30\text{ fruits, }10\text{ are rotten}
\displaystyle \therefore\ \text{Number of unspoiled fruits }=30-10=20
\displaystyle \text{2 fruits are drawn without replacement}
\displaystyle \text{Let }X\text{ be the random variable denoting number of unspoiled fruits}
\displaystyle P(X=0)=P(\text{2 spoiled and 0 unspoiled})
\displaystyle =\frac{10}{30}\times\frac{9}{29}=\frac{3}{29}
\displaystyle P(X=1)=P(\text{1 spoiled and 1 unspoiled})
\displaystyle =\frac{10}{30}\times\frac{20}{29}+\frac{20}{30}\times\frac{10}{29}
\displaystyle =2\times\frac{20\times10}{30\times29}=\frac{40}{87}
\displaystyle P(X=2)=P(\text{0 spoiled and 2 unspoiled})
\displaystyle =\frac{20}{30}\times\frac{19}{29}=\frac{38}{87}
\displaystyle \text{Answer: }P(X=0)=\frac{3}{29},\ P(X=1)=\frac{40}{87},\ P(X=2)=\frac{38}{87}
\displaystyle \text{Therefore, the required probability distribution is}
\displaystyle \begin{array}{c|ccc}X&0&1&2\\ \hline P(X)&\frac{3}{29}&\frac{40}{87}&\frac{38}{87}\end{array}
\displaystyle \therefore\ \text{Mean}=\sum_{i=1}^{n}P_iX_i
\displaystyle =0\times\frac{3}{29}+1\times\frac{40}{87}+2\times\frac{38}{87}
\displaystyle =0+\frac{40}{87}+\frac{76}{87}
\displaystyle =\frac{116}{87}
\\

\displaystyle \textbf{Question 10: } \text{A bag contains } 5 \text{ white, } 7 \text{ red and } 4 \text{ black balls.}
\displaystyle \text{If four balls are drawn one by one with replacement, what is the probability that} \\ \text{none is white? (ISC 2020)}
\displaystyle \text{Answer:}
\displaystyle \text{There are } 5 \text{ white, } 7 \text{ red and } 4 \text{ black balls in a bag}
\displaystyle \therefore \text{Total number of balls}=5+7+4=16
\displaystyle \text{Probability that none is white when four balls are drawn with replacement}
\displaystyle =\frac{11}{16}\times\frac{11}{16}\times\frac{11}{16}\times\frac{11}{16}
\displaystyle =\left(\frac{11}{16}\right)^{4}
\\

\displaystyle \textbf{Question 11: } \text{Let } A \text{ and } B \text{ be two events such that}
\displaystyle P(A)=\frac{1}{2}, P(B)=p \text{ and } P(A \cup B)=\frac{3}{5}
\displaystyle \text{Find } p, \text{ if } A \text{ and } B \text{ are independent events. (ISC 2020)}
\displaystyle \text{Answer:}
\displaystyle \text{Given, } P(A)=\frac{1}{2},\ P(B)=p \text{ and }P(A\cup B)=\frac{3}{5}
\displaystyle \text{where } A \text{ and } B \text{ are independent events}
\displaystyle \therefore P(A\cap B)=P(A)\times P(B)=\frac{1}{2}\times p=\frac{p}{2}
\displaystyle \text{We know that } P(A\cup B)=P(A)+P(B)-P(A\cap B)
\displaystyle \Rightarrow \frac{3}{5}=\frac{1}{2}+p-\frac{p}{2}
\displaystyle \Rightarrow \frac{p}{2}=\frac{3}{5}-\frac{1}{2}=\frac{1}{10}
\displaystyle \Rightarrow p=\frac{1}{5}
\\

\displaystyle \textbf{Question 12: } \text{Three persons } A, B \text{ and } C \text{ shoot to hit a target. Their probabilities}
\displaystyle \text{of hitting the target } \frac{5}{6}, \frac{4}{5} \text{ and } \frac{3}{4},   \text{respectively. Find the probability that (ISC 2020)}
\displaystyle \text{(i) exactly two persons hit the target.}
\displaystyle \text{(ii) atleast one person hits the target.}
\displaystyle \text{Answer:}
\displaystyle \text{Given, } P(A)=\frac{5}{6},\ P(B)=\frac{4}{5},\ P(C)=\frac{3}{4}
\displaystyle \therefore P(A')=\frac{1}{6},\ P(B')=\frac{1}{5},\ P(C')=\frac{1}{4}
\displaystyle \text{(i) Probability of exactly two persons hitting the target}
\displaystyle =P(ABC')+P(AB'C)+P(A'BC)
\displaystyle =P(A)P(B)P(C')+P(A)P(B')P(C)+P(A')P(B)P(C)
\displaystyle =\frac{5}{6}\times\frac{4}{5}\times\frac{1}{4}+\frac{5}{6}\times\frac{1}{5}\times\frac{3}{4}+\frac{1}{6}\times\frac{4}{5}\times\frac{3}{4}
\displaystyle =\frac{20}{120}+\frac{15}{120}+\frac{12}{120}=\frac{47}{120}
\displaystyle \text{24. (ii) Probability of at least one person hitting the target}
\displaystyle =1-\text{Probability that none hits the target}
\displaystyle =1-P(A'B'C')
\displaystyle =1-P(A')P(B')P(C')
\displaystyle =1-\frac{1}{6}\times\frac{1}{5}\times\frac{1}{4}
\displaystyle =1-\frac{1}{120}=\frac{119}{120}
\\

\displaystyle \textbf{Question 13: } \text{If events } A \text{ and } B \text{ are independent, such}   \text{that } P(A)=\frac{3}{5} \text{ and } \\ P(B)=\frac{2}{3}, \text{ then find } P(A \cup B). \text{ (ISC 2019)}
\displaystyle \text{Answer:}
\displaystyle \text{Given, } P(A)=\frac{3}{5}\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)P(B)
\displaystyle \text{Now, } P(A\cup B)=P(A)+P(B)-P(A\cap B)
\displaystyle =\frac{3}{5}+\frac{2}{3}-\frac{3}{5}\times\frac{2}{3}
\displaystyle =\frac{9+10-6}{15}=\frac{13}{15}
\\

\displaystyle \textbf{Question 14: } \text{Two balls are drawn from an urn containing } 3 \text{ white, } 5 \text{ red and } 2 \text{ black balls,}
\displaystyle \text{one by one without replacement. What is the probability that at least one ball is red?}
\displaystyle \quad \text{ISC 2019}
\displaystyle \text{Answer:}
\displaystyle \text{Let } A=\text{event of not getting a red ball in first draw}
\displaystyle \text{and }B=\text{event of not getting a red ball in second draw}
\displaystyle \text{Required probability}=P(\text{at least one ball is red})
\displaystyle =1-P(\text{none is red})
\displaystyle =1-P(A\cap B)
\displaystyle =1-P(A)P\left(\frac{B}{A}\right) \quad (i)
\displaystyle P(A)=\text{probability of getting another colour ball in first draw}
\displaystyle =\frac{5}{10}=\frac{1}{2}
\displaystyle \text{When another colour ball is drawn first, there are }4\text{ other colour balls and }5\text{ red balls left}
\displaystyle \therefore P\left(\frac{B}{A}\right)=\frac{4}{9}
\displaystyle \text{Substituting in Eq. (i), we get}
\displaystyle \text{Required probability}=1-\frac{1}{2}\times\frac{4}{9}
\displaystyle =1-\frac{2}{9}=\frac{7}{9}
\\

\displaystyle \textbf{Question 15: } \text{Given three identical boxes } A,B \text{ and } C, \text{ box } A \text{ contains } 2 \text{ gold and } 1 \text{ silver coin,}
\displaystyle \text{box } B \text{ contains } 1 \text{ gold and } 2 \text{ silver coins and box } C \text{ contains } 3 \text{ silver coins.}
\displaystyle \text{A person chooses a box at random and takes out a coin. If the coin drawn is of silver,}
\displaystyle \text{find the probability that it has been drawn from the box which has the remaining two coins also of silver.}
\displaystyle \quad \text{ISC 2019}
\displaystyle \text{Answer:}
\displaystyle \text{Given boxes }A,B,C\text{ are identical}
\displaystyle P(A)=P(B)=P(C)=\frac{1}{3}
\displaystyle \text{Let }E=\text{event of getting one silver coin}
\displaystyle P\!\left(\frac{E}{A}\right)=\frac{1}{3},\quad P\!\left(\frac{E}{B}\right)=\frac{2}{3},\quad P\!\left(\frac{E}{C}\right)=1
\displaystyle \text{Using Bayes' theorem, } P\!\left(\frac{C}{E}\right)= \frac{P(C)P\!\left(\frac{E}{C}\right)} {P(A)P\!\left(\frac{E}{A}\right)+P(B)P\!\left(\frac{E}{B}\right)+P(C)P\!\left(\frac{E}{C}\right)}
\displaystyle = \frac{\frac{1}{3}\times1} {\frac{1}{3}\times\frac{1}{3}+\frac{1}{3}\times\frac{2}{3}+\frac{1}{3}\times1}
\displaystyle = \frac{\frac{1}{3}} {\frac{1+2+3}{9}} =\frac{\frac{1}{3}}{\frac{6}{9}} =\frac{1}{2}
\\

\displaystyle \textbf{Question 16: } \text{In a race, the probabilities of } A \text{ and } B \text{ winning the race are } \frac{1}{3} \text{ and } \frac{1}{6}, \\ \text{ respectively.}   \text{Find the probability of neither of them winning the race.} \quad \text{ISC 2018}
\displaystyle \text{Answer:}
\displaystyle \text{Given, } P(A)=\frac{1}{3}\text{ and }P(B)=\frac{1}{6}
\displaystyle \therefore P(A')=1-\frac{1}{3}=\frac{2}{3}\text{ and }P(B')=1-\frac{1}{6}=\frac{5}{6}
\displaystyle \therefore \text{Required probability}=P(A')\times P(B')=\frac{2}{3}\times\frac{5}{6}=\frac{5}{9}
\\

\displaystyle \textbf{Question 17: } \text{A speaks truth in } 60\% \text{ of the cases, while } B \text{ in } 40\% \text{ of the cases. In what }
\displaystyle \text{percent of cases are they likely to }  \text{contradict each other in stating the same fact? (ISC 2018)}
\displaystyle \text{Answer:}
\displaystyle \text{Given, } P(A)=60\%=\frac{60}{100}
\displaystyle \therefore P(A')=\frac{40}{100}
\displaystyle \text{and } P(B)=40\%=\frac{40}{100},\ P(B')=\frac{60}{100}
\displaystyle \text{Required probability}=P(A\cap B')+P(A'\cap B)
\displaystyle =P(A)P(B')+P(A')P(B)
\displaystyle =\frac{60}{100}\times\frac{60}{100}+\frac{40}{100}\times\frac{40}{100}
\displaystyle =\frac{3600+1600}{10000}=\frac{5200}{10000}=\frac{52}{100}=52\%
\\

\displaystyle \textbf{Question 18: } \text{If } A \text{ and } B \text{ are events such that}
\displaystyle P(A)=\frac{1}{2}, P(B)=\frac{1}{3} \text{ and } P(A\cap B)=\frac{1}{4}, \text{ then find}
\displaystyle \text{(a) } P\left(\frac{A}{B}\right) \quad \text{(b) } P\left(\frac{B}{A}\right) \quad \text{ISC 2018}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }P(A)=\frac{1}{2},\ P(B)=\frac{1}{3},\ P(A\cap B)=\frac{1}{4}
\displaystyle \text{(a) }P\left(\frac{A}{B}\right)=\frac{P(A\cap B)}{P(B)}
\displaystyle =\frac{\frac{1}{4}}{\frac{1}{3}}=\frac{3}{4}
\displaystyle \text{(b) }P\left(\frac{B}{A}\right)=\frac{P(A\cap B)}{P(A)}
\displaystyle =\frac{\frac{1}{4}}{\frac{1}{2}}=\frac{1}{2}
\\

\displaystyle \textbf{Question 19: } \text{A problem is given to three students whose chances of solving it are }
\displaystyle \frac{1}{4}, \frac{1}{5} \text{ and } \frac{1}{3}, \text{ respectively. Find the probability that the problem is solved. }
\displaystyle \text{(ISC 2017, 03)}
\displaystyle \text{Answer:}
\displaystyle \text{Required probability}=P(\text{at least one solves the problem})
\displaystyle =1-P(\text{none solves})
\displaystyle =1-\left(1-\frac{1}{4}\right)\left(1-\frac{1}{5}\right)\left(1-\frac{1}{3}\right)
\displaystyle =1-\frac{3}{4}\times\frac{4}{5}\times\frac{2}{3}
\displaystyle =1-\frac{2}{5}=\frac{3}{5}
\\

\displaystyle \textbf{Question 20: } \text{In a class of } 60 \text{ students, } 30 \text{ opted for Mathematics, 32 opted for Biology}
\displaystyle \text{and } 24 \text{ opted for both } \text{Mathematics and Biology. If one of these students is selected}
\displaystyle \text{at random, find the probability that (ISC 2017)}
\displaystyle \text{(i) the student opted for Mathematics or Biology.}
\displaystyle \text{(ii) the student has opted neither Mathematics nor } \text{Biology.}
\displaystyle \text{(iii) the student has opted Mathematics but not Biology.}
\displaystyle \text{Answer:}
\displaystyle \text{Let } A=\text{student opted Mathematics},\ B=\text{student opted Biology}
\displaystyle \text{Total students}=60
\displaystyle \text{Students opting Mathematics}=30\Rightarrow P(A)=\frac{30}{60}=\frac{1}{2}
\displaystyle \text{Students opting Biology}=32\Rightarrow P(B)=\frac{32}{60}=\frac{8}{15}
\displaystyle \text{Number of students who opted for both Mathematics and Biology}=24
\displaystyle \therefore P(A\cap B)=\frac{\text{Number of students who opted for both Mathematics and Biology}}{\text{Total number of students}}
\displaystyle =\frac{24}{60}=\frac{2}{5}
\displaystyle \text{Now,}
\displaystyle \text{(i) Probability that the selected student opted for Mathematics or Biology}
\displaystyle =P(A\cup B)=P(A)+P(B)-P(A\cap B)
\displaystyle =\frac{1}{2}+\frac{8}{15}-\frac{2}{5}
\displaystyle =\frac{15+16-12}{30}=\frac{19}{30}
\displaystyle \text{(ii) Probability that the selected student opted neither for Mathematics nor Biology}
\displaystyle =P(A'\cap B')
\displaystyle =P((A\cup B)')
\displaystyle =1-P(A\cup B)
\displaystyle =1-\frac{19}{30}=\frac{11}{30}
\displaystyle \text{(iii) Probability that the selected student opted for Mathematics but not Biology}
\displaystyle =P(A\cap B')=P(A)-P(A\cap B)
\displaystyle =\frac{1}{2}-\frac{2}{5}
\displaystyle =\frac{5-4}{10}=\frac{1}{10}
\\

\displaystyle \textbf{Question 21: } \text{Bag } A \text{ contains } 1 \text{ white, } 2 \text{ blue and } 3 \text{ red balls, Bag } B \text{ contains } 3 \text{ white, } 3 \text{ blue}
\displaystyle \text{and } 2 \text{ red balls and Bag } C \text{ contains } 2 \text{ white, } 3 \text{ blue and } 4 \text{ red balls. One bag is selected at random}
\displaystyle \text{and then two balls are drawn from the selected bag. Find the probability that the balls drawn}
\displaystyle \text{are white and red.} \quad \text{ISC 2017}
\displaystyle \text{Answer:}
\displaystyle \text{Let }E_{1},E_{2},E_{3}\text{ be events of selecting bags }A,B,C
\displaystyle P(E_{1})=P(E_{2})=P(E_{3})=\frac{1}{3}
\displaystyle \text{20. Let }E_{1}:\text{ Bag A is chosen},\ E_{2}:\text{ Bag B is chosen},\ E_{3}:\text{ Bag C is chosen}
\displaystyle P(E_{1})=P(E_{2})=P(E_{3})=\frac{1}{3}
\displaystyle \text{Let }E:\text{ drawing two balls (one white and one red)}
\displaystyle P\!\left(\frac{E}{E_{1}}\right) =\frac{{}^{1}C_{1}\times{}^{3}C_{1}}{{}^{6}C_{2}} =\frac{1\times3}{15} =\frac{1}{5}
\displaystyle \text{[Bag A: }1\text{ white, }2\text{ blue, }3\text{ red]}
\displaystyle P\!\left(\frac{E}{E_{2}}\right) =\frac{{}^{3}C_{1}\times{}^{2}C_{1}}{{}^{8}C_{2}} =\frac{3\times2}{28} =\frac{3}{14}
\displaystyle \text{[Bag B: }3\text{ white, }3\text{ blue, }2\text{ red]}
\displaystyle P\!\left(\frac{E}{E_{3}}\right) =\frac{{}^{2}C_{1}\times{}^{4}C_{1}}{{}^{9}C_{2}} =\frac{2\times4}{36} =\frac{2}{9}
\displaystyle \text{[Bag C: }2\text{ white, }3\text{ blue, }4\text{ red]}
\displaystyle \text{By total probability,}
\displaystyle P(E)=P(E_{1})P\!\left(\frac{E}{E_{1}}\right) +P(E_{2})P\!\left(\frac{E}{E_{2}}\right) +P(E_{3})P\!\left(\frac{E}{E_{3}}\right)
\displaystyle =\frac{1}{3}\cdot\frac{1}{5} +\frac{1}{3}\cdot\frac{3}{14} +\frac{1}{3}\cdot\frac{2}{9}
\displaystyle =\frac{1}{15}+\frac{1}{14}+\frac{2}{27}
\displaystyle =\frac{378+405+420}{5670} =\frac{1203}{5670} =\frac{401}{1890}
\\

\displaystyle \textbf{Question 22: } \text{A fair die is rolled. If face } 1 \text{ turns up, a ball is drawn from Bag } A.
\displaystyle \text{If face } 2 \text{ or } 3 \text{ turns up, a ball is drawn from Bag } B \text{ and if face } 4,5 \text{ or } 6 \text{ turns up,}
\displaystyle \text{a ball is drawn from Bag } C. \text{ Bag } A \text{ contains } 3 \text{ red and } 2 \text{ white balls, Bag } B
\displaystyle \text{contains } 3 \text{ red and } 4 \text{ white balls and Bag } C \text{ contains } 4 \text{ red and } 5 \text{ white balls.}
\displaystyle \text{The die is rolled, a bag is picked and a ball is drawn. If the drawn ball is red,}
\displaystyle \text{what is the probability that it is drawn from Bag } B? \quad \text{ISC 2017}
\displaystyle \text{Answer:}
\displaystyle \text{Now, let }E:\text{ the drawn ball is red}
\displaystyle P(E_{1})=\frac{1}{6},\quad P(E_{2})=\frac{2}{6}=\frac{1}{3},\quad P(E_{3})=\frac{3}{6}=\frac{1}{2}
\displaystyle P\!\left(\frac{E}{E_{1}}\right)=\frac{3}{5},\quad P\!\left(\frac{E}{E_{2}}\right)=\frac{3}{7},\quad P\!\left(\frac{E}{E_{3}}\right)=\frac{4}{9}
\displaystyle \text{Required probability }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) +P(E_{3})P\!\left(\frac{E}{E_{3}}\right)}
\displaystyle = \frac{\frac{1}{3}\cdot\frac{3}{7}} {\frac{1}{6}\cdot\frac{3}{5} +\frac{1}{3}\cdot\frac{3}{7} +\frac{1}{2}\cdot\frac{4}{9}}
\displaystyle = \frac{\frac{1}{7}} {\frac{1}{10}+\frac{1}{7}+\frac{2}{9}}
\displaystyle = \frac{\frac{1}{7}} {\frac{63+90+140}{630}} =\frac{1}{7}\cdot\frac{630}{293} =\frac{90}{293}
\\

\displaystyle \textbf{Question 23: } \text{An urn contains } 25 \text{ balls of which } 10 \text{ balls are red and the}
\displaystyle \text{remaining are green. A ball is drawn at random from the urn, the}
\displaystyle \text{colour is noted and the ball is replaced. If } 6 \text{ balls are drawn in this}
\displaystyle \text{way, find the probability that: \hspace{1cm} (ISC 2017)}
\displaystyle \text{(i) all the balls are red. \hspace{2cm} (ii) not more than } 2 \text{ balls are green.}
\displaystyle \text{(iii) number of red balls and green balls are equal.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }p\text{ denote the probability of drawing red ball and }q
\displaystyle \text{denote the probability of drawing non-red, i.e. green ball, then}
\displaystyle p=\frac{10}{25}=\frac{2}{5}\text{ and }q=\frac{3}{5}
\displaystyle \text{Let }X\text{ be a random variable that denotes the number of red balls in }6\text{ trials.}
\displaystyle \text{Clearly, }X\sim B\left(6,\frac{2}{5}\right)
\displaystyle \therefore\ P(X=x)={}^{6}C_x\left(\frac{2}{5}\right)^x\left(\frac{3}{5}\right)^{6-x}
\displaystyle \text{where, }x=0,1,2,\ldots,6
\displaystyle \text{Now,}
\displaystyle \text{(i) }P(\text{getting all red balls})
\displaystyle =P(X=6)={}^{6}C_6\left(\frac{2}{5}\right)^6\left(\frac{3}{5}\right)^{6-6}=\left(\frac{2}{5}\right)^6
\displaystyle \text{(ii) }P(\text{getting not more than }2\text{ green balls})
\displaystyle =P(X\geq4)=P(X=4)+P(X=5)+P(X=6)
\displaystyle ={}^{6}C_4\left(\frac{2}{5}\right)^4\left(\frac{3}{5}\right)^2+{}^{6}C_5\left(\frac{2}{5}\right)^5\left(\frac{3}{5}\right)^1
\displaystyle \quad +{}^{6}C_6\left(\frac{2}{5}\right)^6\left(\frac{3}{5}\right)^0
\displaystyle =15\left(\frac{2}{5}\right)^4\left(\frac{3}{5}\right)^2+6\left(\frac{2}{5}\right)^5\left(\frac{3}{5}\right)+\left(\frac{2}{5}\right)^6
\displaystyle \text{(iii) }P(\text{getting equal number of red and green balls})
\displaystyle =P(X=3)
\displaystyle ={}^{6}C_3\left(\frac{2}{5}\right)^3\left(\frac{3}{5}\right)^{6-3}=20\left(\frac{2}{5}\right)^3\left(\frac{3}{5}\right)^3
\displaystyle \text{Answer: }P(X=6)=\left(\frac{2}{5}\right)^6
\displaystyle P(X\geq4)=15\left(\frac{2}{5}\right)^4\left(\frac{3}{5}\right)^2+6\left(\frac{2}{5}\right)^5\left(\frac{3}{5}\right)+\left(\frac{2}{5}\right)^6
\displaystyle P(X=3)=20\left(\frac{2}{5}\right)^3\left(\frac{3}{5}\right)^3
\\

\displaystyle \textbf{Question 24: } \text{A pair of dice is thrown. What is the probability of getting an even }
\displaystyle \text{number on the first die or a total of } 8\text{? (ISC 2016)}
\displaystyle \text{Answer:}
\displaystyle \text{Total outcomes}=6\times 6=36
\displaystyle \text{Favourable outcomes (even on first die or total }=8)
\displaystyle =\{(2,1),(2,2),(2,3),(2,4),(2,5),(2,6),(4,1),(4,2),(4,3),(4,4),(4,5),(4,6),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6),(3,5),(5,3)\}
\displaystyle =20
\displaystyle \therefore P=\frac{20}{36}=\frac{5}{9}
\\

\displaystyle \textbf{Question 25: } \text{A pair of dice is thrown. If the two numbers appearing on them are}
\displaystyle \text{different, find the probability that the sum of the numbers appearing }  \text{is } 6. \text{ (ISC 2016)}
\displaystyle \text{Answer:}
\displaystyle \text{Total outcomes (distinct pairs)}=30
\displaystyle \text{Favourable pairs (sum }=6)=\{(1,5),(5,1),(2,4),(4,2)\}
\displaystyle \therefore P=\frac{4}{30}=\frac{2}{15}
\\

\displaystyle \textbf{Question 26: } \text{An urn contains } 10 \text{ white and } 3 \text{ black balls while another urn contains } 
\displaystyle 3 \text{ white} \text{and } 5 \text{ black balls. Two balls are drawn from the first urn and put into the }
\displaystyle \text{second urn and then a ball is drawn from the second urn. Find the probability that }
\displaystyle \text{the ball drawn from the second urn is a white ball. ISC 2016}
\displaystyle \text{Answer:}
\displaystyle \text{Here, there are three cases:}
\displaystyle \text{Case I: Two white balls are transferred from Urn I to Urn II}
\displaystyle P(\text{transferring two white balls from Urn I})=\frac{{}^{10}C_{2}}{{}^{13}C_{2}}
\displaystyle =\frac{45}{78}=\frac{15}{26}
\displaystyle \text{Now, in Urn II, white balls}=3+2=5,\quad \text{black balls}=5
\displaystyle P(\text{getting a white ball from Urn II})=\frac{{}^{5}C_{1}}{{}^{10}C_{1}}=\frac{5}{10}=\frac{1}{2}
\displaystyle \therefore \text{Probability of these events occurring together}
\displaystyle =\frac{15}{26}\times\frac{1}{2}=\frac{15}{52}
\displaystyle \text{Case II: One white and one black ball are transferred from Urn I to Urn II}
\displaystyle P(\text{transferring one white and one black ball from Urn I})=\frac{{}^{10}C_{1}\times{}^{3}C_{1}}{{}^{13}C_{2}}
\displaystyle =\frac{30}{78}=\frac{5}{13}
\displaystyle \text{Now, in Urn II, white balls}=3+1=4,\quad \text{black balls}=5+1=6
\displaystyle P(\text{getting a white ball from Urn II})=\frac{{}^{4}C_{1}}{{}^{10}C_{1}}=\frac{4}{10}=\frac{2}{5}
\displaystyle \therefore \text{Probability of these events occurring together}
\displaystyle =\frac{5}{13}\times\frac{2}{5}=\frac{2}{13}
\displaystyle \text{Case III: Two black balls are transferred from Urn I to Urn II}
\displaystyle P(\text{transferring two black balls from Urn I to Urn II})=\frac{{}^{3}C_{2}}{{}^{13}C_{2}}
\displaystyle =\frac{3}{78}=\frac{1}{26}
\displaystyle \text{Now, in Urn II, white balls}=3,\quad \text{black balls}=5+2=7
\displaystyle P(\text{getting a white ball from Urn II})=\frac{{}^{3}C_{1}}{{}^{10}C_{1}}=\frac{3}{10}
\displaystyle \therefore \text{Probability of these events occurring together}
\displaystyle =\frac{1}{26}\times\frac{3}{10}=\frac{3}{260}
\displaystyle \text{Since all three cases are mutually exclusive,}
\displaystyle \text{Required probability of drawing a white ball}
\displaystyle =\frac{15}{52}+\frac{2}{13}+\frac{3}{260}
\displaystyle =\frac{75+40+3}{260}=\frac{118}{260}=\frac{59}{130}
\\

\displaystyle \textbf{Question 27: } \text{A committee of } 4 \text{ persons has to be chosen from } 8 \text{ boys } \text{and } 6 \text{ girls, }
\displaystyle \text{consisting of at least one girl. Find the probability that the committee consists of more}
\displaystyle \text{ girls than } \text{boys. (ISC 2016)}
\displaystyle \text{Answer:}
\displaystyle \text{Given, number of boys}=8\text{ and number of girls}=6
\displaystyle \text{A committee of }4\text{ persons is to be formed}
\displaystyle \text{If the committee has more girls than boys, then possible cases are:}
\displaystyle \text{(i) }0\text{ boys and }4\text{ girls, or (ii) }1\text{ boy and }3\text{ girls}
\displaystyle \text{Number of favourable ways}={}^{8}C_{0}\times{}^{6}C_{4}+{}^{8}C_{1}\times{}^{6}C_{3}
\displaystyle =1\times15+8\times20
\displaystyle =15+160=175
\displaystyle \text{Total number of ways}={}^{14}C_{4}=1001
\displaystyle \therefore \text{Required probability}=\frac{175}{1001}
\\

\displaystyle \textbf{Question 28: } \text{In an automobile factory, certain parts are to be fixed into the chassis in a section}
\displaystyle \text{before it moves into another section. On a given day, one of the three persons } A,B \text{ and } C
\displaystyle \text{carries out this task. } A \text{ has } 45\% \text{ chance, } B \text{ has } 35\% \text{ chance and } C \text{ has } 20\%
\displaystyle \text{chance of doing the task. The probability that } A,B \text{ and } C \text{ will take more than the}
\displaystyle \text{allotted time is } \frac{1}{6}, \frac{1}{10} \text{ and } \frac{1}{20} \text{ respectively. If it is found that the time taken}
\displaystyle \text{is more than the allotted time, then what is the probability that it was done by } B
\displaystyle A \text{ has done the task?} \quad \text{ISC 2016}
\displaystyle \text{Answer:}
\displaystyle \text{Let }E_{1},E_{2},E_{3}\text{ and }A\text{ be defined as:}
\displaystyle E_{1}:\text{ A has done the task},\ E_{2}:\text{ B has done the task},\ E_{3}:\text{ C has done the task}
\displaystyle A:\text{ person takes more than allotted time}
\displaystyle P(E_{1})=\frac{45}{100}=\frac{9}{20},\quad P(E_{2})=\frac{35}{100}=\frac{7}{20},\quad P(E_{3})=\frac{20}{100}=\frac{4}{20}
\displaystyle P\!\left(\frac{A}{E_{1}}\right)=\frac{1}{6},\quad P\!\left(\frac{A}{E_{2}}\right)=\frac{1}{10},\quad P\!\left(\frac{A}{E_{3}}\right)=\frac{1}{20}
\displaystyle \text{By Bayes' theorem,}
\displaystyle P\!\left(\frac{E_{1}}{A}\right) =\frac{P(E_{1})P\!\left(\frac{A}{E_{1}}\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{9}{20}\cdot\frac{1}{6}} {\frac{9}{20}\cdot\frac{1}{6} +\frac{7}{20}\cdot\frac{1}{10} +\frac{4}{20}\cdot\frac{1}{20}}
\displaystyle = \frac{0.075}{0.075+0.035+0.01} =\frac{0.075}{0.12} =0.625
\\

\displaystyle \textbf{Question 29: } \text{A card is drawn from a well shuffled pack of playing cards. What is the }
\displaystyle \text{probability that it is either a spade } \text{or an ace or both? (ISC 2015)}
\displaystyle \text{Answer:}
\displaystyle \text{Let } E=\text{drawing a spade},\ F=\text{drawing an ace}
\displaystyle n(E)=13,\ n(F)=4,\ n(E\cap F)=1
\displaystyle n(S)=52
\displaystyle \therefore P(E)=\frac{13}{52}=\frac{1}{4},\quad P(F)=\frac{4}{52}=\frac{1}{13}
\displaystyle P(E\cap F)=\frac{1}{52}
\displaystyle \text{Required probability}=P(E\cup F)
\displaystyle =P(E)+P(F)-P(E\cap F)
\displaystyle =\frac{1}{4}+\frac{1}{13}-\frac{1}{52}=\frac{16}{52}=\frac{4}{13}
\\

\displaystyle \textbf{Question 30: } \text{Three persons } A, B \text{ and } C \text{ shoot to hit a target. If } A \text{ hits the target}
\displaystyle \text{four times in five trials, } B \text{ hits three times } \text{in four trials and C hits it two times in} 
\displaystyle \text{three trials, find } \text{the probability that}
\displaystyle \text{(i) exactly two persons hit the target.}
\displaystyle \text{(ii) atleast two persons hit the target.}
\displaystyle \text{(iii) none hit the target. (ISC 2015)}
\displaystyle \text{Answer:}
\displaystyle \text{Given, } P(A)=\frac{4}{5},\ P(B)=\frac{3}{4},\ P(C)=\frac{2}{3}
\displaystyle P(\overline{A})=1-P(A)=1-\frac{4}{5}=\frac{1}{5}
\displaystyle P(\overline{B})=1-P(B)=1-\frac{3}{4}=\frac{1}{4}
\displaystyle P(\overline{C})=1-P(C)=1-\frac{2}{3}=\frac{1}{3}
\displaystyle \text{(i) Probability that exactly two persons hit the target}
\displaystyle =P(A\cap B\cap \overline{C})+P(A\cap \overline{B}\cap C)+P(\overline{A}\cap B\cap C)
\displaystyle =P(A)P(B)P(\overline{C})+P(A)P(\overline{B})P(C)+P(\overline{A})P(B)P(C)
\displaystyle =\frac{4}{5}\cdot\frac{3}{4}\cdot\frac{1}{3}+\frac{4}{5}\cdot\frac{1}{4}\cdot\frac{2}{3}+\frac{1}{5}\cdot\frac{3}{4}\cdot\frac{2}{3}
\displaystyle =\frac{12}{60}+\frac{8}{60}+\frac{6}{60}=\frac{26}{60}=\frac{13}{30}
\displaystyle \text{(ii) Probability that at least two persons hit the target}
\displaystyle =P(\text{exactly two})+P(A\cap B\cap C)
\displaystyle =\frac{13}{30}+P(A)P(B)P(C)
\displaystyle =\frac{13}{30}+\frac{4}{5}\cdot\frac{3}{4}\cdot\frac{2}{3}
\displaystyle =\frac{13}{30}+\frac{12}{30}=\frac{25}{30}=\frac{5}{6}
\displaystyle \text{(iii) Probability that none of them hit the target}
\displaystyle =P(\overline{A}\cap \overline{B}\cap \overline{C})
\displaystyle =P(\overline{A})P(\overline{B})P(\overline{C})
\displaystyle =\frac{1}{5}\cdot\frac{1}{4}\cdot\frac{1}{3}=\frac{1}{60}
\\

\displaystyle \textbf{Question 31: } \text{Box I contains two white and three black balls, Box II contains four white and one black}
\displaystyle \text{balls and Box III contains three white and four black balls. A die having three red, two yellow}
\displaystyle \text{and one green face is thrown to select the box. If a red face turns up, we pick Box I, if a}
\displaystyle \text{yellow face turns up we pick Box II, otherwise we pick Box III. Then, a ball is drawn from}
\displaystyle \text{the selected box. If the ball drawn is white, what is the probability that the die had turned}
\displaystyle \text{up with a red face?} \quad \text{ISC 2015}
\displaystyle \text{Answer:}
\displaystyle \text{Let }B\text{ and }W\text{ denote black and white balls.}
\displaystyle \text{Given, Bag I}=\{2W,3B\}
\displaystyle \text{Bag II}=\{4W,1B\}
\displaystyle \text{and Bag III}=\{3W,4B\}
\displaystyle \text{Let }E_{1}=\text{die shows a red face}
\displaystyle E_{2}=\text{die shows a yellow face}
\displaystyle E_{3}=\text{die shows a green face}
\displaystyle \text{Total number of faces on a die}=6
\displaystyle \therefore P(E_{1})=\frac{3}{6}=\frac{1}{2},\quad P(E_{2})=\frac{2}{6}=\frac{1}{3},\quad P(E_{3})=\frac{1}{6}
\displaystyle \text{Let }E=\text{event of drawing a white ball}
\displaystyle P\left(\frac{E}{E_{1}}\right)=\frac{2}{5},\quad P\left(\frac{E}{E_{2}}\right)=\frac{4}{5},\quad P\left(\frac{E}{E_{3}}\right)=\frac{3}{7}
\displaystyle \text{Required probability that the die turned up red, given that a white ball is selected}
\displaystyle =P\left(\frac{E_{1}}{E}\right)
\displaystyle =\frac{P(E_{1})P\left(\frac{E}{E_{1}}\right)}{P(E_{1})P\left(\frac{E}{E_{1}}\right)+P(E_{2})P\left(\frac{E}{E_{2}}\right)+P(E_{3})P\left(\frac{E}{E_{3}}\right)}
\displaystyle =\frac{\frac{1}{2}\times\frac{2}{5}}{\frac{1}{2}\times\frac{2}{5}+\frac{1}{3}\times\frac{4}{5}+\frac{1}{6}\times\frac{3}{7}}
\displaystyle =\frac{\frac{1}{5}}{\frac{1}{5}+\frac{4}{15}+\frac{1}{14}}
\displaystyle =\frac{\frac{1}{5}}{\frac{42+56+15}{210}}
\displaystyle =\frac{\frac{1}{5}}{\frac{113}{210}}=\frac{42}{113}
\\

\displaystyle \textbf{Question 32: } \text{An urn contains } 2 \text{ white and } 2 \text{ black balls. A ball is drawn at random.}
\displaystyle \text{If it is white, it is not replaced into the urn. Otherwise, it is replaced with another } 2 \text{ balls}
\displaystyle \text{of the same colour. The process is repeated. Find the probability that the third ball drawn}
\displaystyle \text{is black.} \quad \text{ISC 2015}
\displaystyle \text{Answer:}
\displaystyle \text{Let }E_{pq}\text{ denote the event that first ball has colour }p
\displaystyle \text{and second ball has colour }q
\displaystyle \text{Let }B\text{ denote the event that third ball drawn is black}
\displaystyle P(E_{ww})=\frac{2}{4}\times\frac{1}{3}=\frac{1}{6}
\displaystyle P(E_{wb})=\frac{2}{4}\times\frac{2}{3}=\frac{1}{3}
\displaystyle P(E_{bw})=\frac{2}{4}\times\frac{2}{5}=\frac{1}{5}
\displaystyle P(E_{bb})=\frac{2}{4}\times\frac{3}{5}=\frac{3}{10}
\displaystyle P\left(\frac{B}{E_{ww}}\right)=\frac{2}{2}=1,\quad P\left(\frac{B}{E_{wb}}\right)=\frac{3}{4}
\displaystyle P\left(\frac{B}{E_{bw}}\right)=\frac{3}{4},\quad P\left(\frac{B}{E_{bb}}\right)=\frac{4}{6}=\frac{2}{3}
\displaystyle \text{Now, by total probability,}
\displaystyle P(B)=P(E_{ww})P\left(\frac{B}{E_{ww}}\right)+P(E_{wb})P\left(\frac{B}{E_{wb}}\right)
\displaystyle \quad +P(E_{bw})P\left(\frac{B}{E_{bw}}\right)+P(E_{bb})P\left(\frac{B}{E_{bb}}\right)
\displaystyle =\frac{1}{6}\times1+\frac{1}{3}\times\frac{3}{4}+\frac{1}{5}\times\frac{3}{4}+\frac{3}{10}\times\frac{2}{3}
\displaystyle =\frac{1}{6}+\frac{1}{4}+\frac{3}{20}+\frac{1}{5}
\displaystyle =\frac{10+15+9+12}{60}=\frac{46}{60}=\frac{23}{30}
\\

\displaystyle \textbf{Question 33: } \text{Five dice are thrown simultaneously. If the occurrence of an odd}
\displaystyle \text{number on a single die is considered a success, find the probability}
\displaystyle \text{of maximum three successes. \hspace{1cm} (ISC 2015)}
\displaystyle \text{Answer:}
\displaystyle \text{Let }p=\text{probability of occurrence of an odd number in a single die.}
\displaystyle \therefore\ p=\frac{3}{6}=\frac{1}{2}\qquad[\because\ \text{odd numbers on a die are }\{1,3,5\}]
\displaystyle \Rightarrow q=1-\frac{1}{2}=\frac{1}{2}
\displaystyle \text{Here, }n=5
\displaystyle \text{Using Bernoulli's theorem,}
\displaystyle P(X=r)={}^{n}C_rp^rq^{n-r}
\displaystyle \therefore\ \text{The probability of maximum three success}
\displaystyle =P(X=0)+P(X=1)+P(X=2)+P(X=3)
\displaystyle ={}^{5}C_0p^0q^5+{}^{5}C_1p^1q^4+{}^{5}C_2p^2q^3+{}^{5}C_3p^3q^2
\displaystyle ={}^{5}C_0\left(\frac{1}{2}\right)^0\left(\frac{1}{2}\right)^5+{}^{5}C_1\left(\frac{1}{2}\right)^1\left(\frac{1}{2}\right)^4
\displaystyle \quad +{}^{5}C_2\left(\frac{1}{2}\right)^2\left(\frac{1}{2}\right)^3+{}^{5}C_3\left(\frac{1}{2}\right)^3\left(\frac{1}{2}\right)^2
\displaystyle =1\left(\frac{1}{2}\right)^5+5\left(\frac{1}{2}\right)^5+10\left(\frac{1}{2}\right)^5+10\left(\frac{1}{2}\right)^5
\displaystyle =\frac{1}{2^5}(1+5+10+10)=\frac{26}{32}=\frac{13}{16}
\displaystyle \text{Answer: }\frac{13}{16}
\\

\displaystyle \textbf{Question 34: } \text{A bag contains } 20 \text{ balls numbered } 1 \text{ to } 20. \text{ One ball is drawn at }
\displaystyle \text{random from the bag. What is the probability that the ball drawn is marked with }
\displaystyle \text{a number which is } \text{multiple of } 3 \text{ or } 4\text{? (ISC 2014)}
\displaystyle \text{Answer:}
\displaystyle \text{Total number of balls}=20
\displaystyle A=\{3,6,9,12,15,18\},\quad B=\{4,8,12,16,20\}
\displaystyle A\cap B=\{12\}
\displaystyle \therefore P(A\cup B)=P(A)+P(B)-P(A\cap B)
\displaystyle =\frac{6}{20}+\frac{5}{20}-\frac{1}{20}=\frac{10}{20}=\frac{1}{2}
\\

\displaystyle \textbf{Question 35: } \text{A bag contains } 5 \text{ white and } 4 \text{ black balls and another } \text{bag contains } 7 
\displaystyle \text{ white and } 9 \text{ black balls. A ball is drawn from the first bag and two balls drawn from}
\displaystyle \text{ the second } \text{bag. What is the probability of drawing one white and } \text{two black balls? \\ (ISC 2014)}
\displaystyle \text{Answer:}
\displaystyle \text{Given, Bag I: 5 white and 4 black balls}
\displaystyle \text{Bag II: 7 white and 9 black balls}
\displaystyle \text{Case I: 1 black ball from Bag I, and 1 black and 1 white ball from Bag II}
\displaystyle P_1=\frac{{}^{4}C_{1}}{{}^{9}C_{1}}\times\frac{{}^{7}C_{1}\cdot{}^{9}C_{1}}{{}^{16}C_{2}}
\displaystyle =\frac{4}{9}\times\frac{7\cdot 9}{120}=\frac{4}{9}\times\frac{63}{120}=\frac{252}{1080}=\frac{7}{30}
\displaystyle \text{Case II: 1 white ball from Bag I and 2 black balls from Bag II}
\displaystyle P_2=\frac{{}^{5}C_{1}}{{}^{9}C_{1}}\times\frac{{}^{9}C_{2}}{{}^{16}C_{2}}
\displaystyle =\frac{5}{9}\times\frac{36}{120}=\frac{5}{9}\times\frac{3}{10}=\frac{1}{6}
\displaystyle \text{Required probability}=P_1+P_2=\frac{7}{30}+\frac{1}{6}
\displaystyle =\frac{7}{30}+\frac{5}{30}=\frac{12}{30}=\frac{2}{5}
\\

\displaystyle \textbf{Question 36: } \text{In a college, } 70\% \text{ students pass in Physics, } 75\% \text{ pass in Mathematics}
\displaystyle \text{and } 10\% \text{ students fail in both. One student is chosen at random. What is the probability that}
\displaystyle \text{(i) he passes in Physics and Mathematics?}
\displaystyle \text{(ii) he passes Mathematics given that he passes in Physics?}
\displaystyle \text{(iii) he passes in Physics given that he passes in Mathematics?}
\displaystyle \quad \text{ISC 2014}
\displaystyle \text{Answer:}
\displaystyle \text{Consider the following events:}
\displaystyle A:\text{ student passed in Physics}
\displaystyle B:\text{ student passed in Mathematics}
\displaystyle P(A)=\frac{70}{100},\quad P(B)=\frac{75}{100}
\displaystyle P(A'\cap B')=\text{probability that student failed in both}=\frac{10}{100}
\displaystyle P(A'\cap B')=P((A\cup B)')=1-P(A\cup B)
\displaystyle \Rightarrow P(A\cup B)=1-\frac{10}{100}=\frac{90}{100}
\displaystyle \text{(i) Probability of passing in Physics and Mathematics}
\displaystyle =P(A\cap B)=P(A)+P(B)-P(A\cup B)
\displaystyle =\frac{70}{100}+\frac{75}{100}-\frac{90}{100}
\displaystyle =\frac{55}{100}=\frac{11}{20}
\displaystyle \text{(ii) Probability that student passes in Mathematics given Physics}
\displaystyle P\left(\frac{B}{A}\right)=\frac{P(A\cap B)}{P(A)}
\displaystyle =\frac{\frac{55}{100}}{\frac{70}{100}}=\frac{55}{70}=\frac{11}{14}
\displaystyle \text{(iii) Probability that student passes in Physics given Mathematics}
\displaystyle P\left(\frac{A}{B}\right)=\frac{P(A\cap B)}{P(B)}
\displaystyle =\frac{\frac{55}{100}}{\frac{75}{100}}=\frac{55}{75}=\frac{11}{15}
\\

\displaystyle \textbf{Question 37: } \text{In a bolt factory, three machines, } A,B \text{ and } C \text{ manufacture } 25\%,35\%
\displaystyle \text{and } 40\% \text{ of the total production, respectively. Of their respective outputs, } 5\%,4\%
\displaystyle \text{and } 2\% \text{ are defective. A bolt is drawn at random from the total production and}
\displaystyle \text{it is found to be defective. Find the probability that it was manufactured by machine } C.
\displaystyle \quad \text{ISC 2014}
\displaystyle \text{Answer:}
\displaystyle \text{Let }E_{1},E_{2},E_{3}\text{ be events of manufacturing bolt by machines }A,B,C
\displaystyle \text{respectively, and let }D\text{ be the event of manufacturing a defective bolt}
\displaystyle P(E_{1})=25\%=\frac{25}{100},\quad P(E_{2})=35\%=\frac{35}{100}
\displaystyle \text{and }P(E_{3})=40\%=\frac{40}{100}
\displaystyle \text{Probability of defective bolt produced by machine }A,
\displaystyle P\left(\frac{D}{E_{1}}\right)=5\%=\frac{5}{100}
\displaystyle \text{Probability of defective bolt produced by machine }B,
\displaystyle P\left(\frac{D}{E_{2}}\right)=4\%=\frac{4}{100}
\displaystyle \text{Probability of defective bolt produced by machine }C,
\displaystyle P\left(\frac{D}{E_{3}}\right)=2\%=\frac{2}{100}
\displaystyle \left[\because E_{1},E_{2},E_{3}\text{ are mutually exclusive and exhaustive events}\right]
\displaystyle \text{Now, probability that a defective bolt is produced by machine }C,
\displaystyle P\left(\frac{E_{3}}{D}\right)=\frac{P(E_{3})\cdot P\left(\frac{D}{E_{3}}\right)}{P(E_{1})\cdot P\left(\frac{D}{E_{1}}\right)+P(E_{2})\cdot P\left(\frac{D}{E_{2}}\right)+P(E_{3})\cdot P\left(\frac{D}{E_{3}}\right)}
\displaystyle =\frac{\frac{40}{100}\times\frac{2}{100}}{\frac{25}{100}\times\frac{5}{100}+\frac{35}{100}\times\frac{4}{100}+\frac{40}{100}\times\frac{2}{100}}
\displaystyle =\frac{\frac{80}{10000}}{\frac{125}{10000}+\frac{140}{10000}+\frac{80}{10000}}
\displaystyle =\frac{80}{125+140+80}=\frac{80}{345}=\frac{16}{69}
\displaystyle \text{Hence, required probability }=\frac{16}{69}
\\

\displaystyle \textbf{Question 38: } \text{On dialling certain telephone numbers, assume that on an average,}
\displaystyle \text{one telephone number out of five is busy. Ten telephone numbers}
\displaystyle \text{are randomly selected and dialled. Find the probability that at least}
\displaystyle \text{three of them will be busy. \hspace{1cm} (ISC 2014)}
\displaystyle \text{Answer:}
\displaystyle \text{We have, the probability of a telephone number is busy }(p)=\frac{1}{5}
\displaystyle \text{The probability of a telephone number is not busy }(q)=1-\frac{1}{5}=\frac{4}{5}
\displaystyle \text{Let }x=\text{Number of calls dialling}
\displaystyle \text{Given, the total number of dialling, }n=10
\displaystyle \text{The probability of at least three dialling out of }10\text{ is busy.}
\displaystyle P(x\geq3)=P(x=3)+P(x=4)+\cdots+P(x=10)
\displaystyle =1-[P(x=0)+P(x=1)+P(x=2)]
\displaystyle =1-\left[{}^{10}C_0p^0q^{10}+{}^{10}C_1p^1q^9+{}^{10}C_2p^2q^8\right]
\displaystyle =1-\left[{}^{10}C_0\left(\frac{1}{5}\right)^0\left(\frac{4}{5}\right)^{10}\right.
\displaystyle \left.\quad +{}^{10}C_1\left(\frac{1}{5}\right)^1\left(\frac{4}{5}\right)^9+{}^{10}C_2\left(\frac{1}{5}\right)^2\left(\frac{4}{5}\right)^8\right]
\displaystyle =1-\left[\left(\frac{4}{5}\right)^{10}+10\times\frac{4^9}{5^{10}}+45\times\frac{4^8}{5^{10}}\right]
\displaystyle =1-\frac{4^8}{5^{10}}\left[(4)^2+10\times4+45\right]
\displaystyle =1-\frac{4^8}{5^{10}}[16+40+45]
\displaystyle =1-\frac{4^8}{5^{10}}\times101
\displaystyle =1-\frac{65536}{9765625}\times101
\displaystyle =1-\frac{6619136}{9765625}
\displaystyle =\frac{9765625-6619136}{9765625}=\frac{3146489}{9765625}=0.32
\displaystyle \text{Answer: }0.32
\\

\displaystyle \textbf{Question 39: } \text{If two balls are drawn from a bag containing three red balls and four }
\displaystyle \text{blue balls, find the probability that}
\displaystyle \text{(i) they are of the same colour.}
\displaystyle \text{(ii) they are of different colours. (ISC 2013)}
\displaystyle \text{Answer:}
\displaystyle \text{A bag has 3 red and 4 blue balls}
\displaystyle \text{(i) } P(\text{same colour})=P(\text{both red})+P(\text{both blue})
\displaystyle =\frac{{}^{3}C_{2}}{{}^{7}C_{2}}+\frac{{}^{4}C_{2}}{{}^{7}C_{2}}
\displaystyle =\frac{3}{21}+\frac{6}{21}=\frac{9}{21}=\frac{3}{7}
\displaystyle \text{(ii) } P(\text{different colours})=\frac{{}^{3}C_{1}\times{}^{4}C_{1}}{{}^{7}C_{2}}
\displaystyle =\frac{12}{21}=\frac{4}{7}
\\

\displaystyle \textbf{Question 40: } \text{Bag } A \text{ contains three red and four white balls; bag B contains two}
\displaystyle \text{red and three white balls. If one ball is } \text{drawn from bag } A \text{ and two balls from bag } B,
\displaystyle \text{ find the} \text{ probability that}
\displaystyle \text{(i) one ball is red and two balls are white.}
\displaystyle \text{(ii) all the three balls are of the same colour. (ISC 2013)}
\displaystyle \text{Answer:}
\displaystyle \text{(i) Probability of getting 1 red and 2 white balls}
\displaystyle =P(\text{1 white from A and 1 white and 1 red from B})
\displaystyle +P(\text{1 red from A and 2 white from B})
\displaystyle =\frac{{}^{4}C_{1}}{{}^{7}C_{1}}\times\frac{{}^{2}C_{1}\cdot{}^{3}C_{1}}{{}^{5}C_{2}}+\frac{{}^{3}C_{1}}{{}^{7}C_{1}}\times\frac{{}^{3}C_{2}}{{}^{5}C_{2}}
\displaystyle =\frac{4}{7}\times\frac{6}{20}+\frac{3}{7}\times\frac{3}{10}
\displaystyle =\frac{24}{140}+\frac{9}{70}=\frac{24+18}{140}=\frac{42}{140}=\frac{3}{10}
\displaystyle \text{(ii) Probability that all balls are of the same colour}
\displaystyle =P(\text{all white})+P(\text{all red})
\displaystyle =\frac{{}^{4}C_{1}}{{}^{7}C_{1}}\times\frac{{}^{3}C_{2}}{{}^{5}C_{2}}+\frac{{}^{3}C_{1}}{{}^{7}C_{1}}\times\frac{{}^{2}C_{2}}{{}^{5}C_{2}}
\displaystyle =\frac{4}{7}\times\frac{3}{10}+\frac{3}{7}\times\frac{1}{10}
\displaystyle =\frac{12}{70}+\frac{3}{70}=\frac{15}{70}=\frac{3}{14}
\\

\displaystyle \textbf{Question 41: } \text{Three persons, Aman, Bipin and Mohan attempt a Mathematics problem }
\displaystyle \text{independently. The odds in favour } \text{of Aman and Mohan solving the problem are }
\displaystyle  3:2 \text{ and } 4:1,   \text{respectively and the odds against Bipin solving the problem are }
\displaystyle 2:1.  \text{ Find (ISC 2013)}
\displaystyle \text{(i) the probability that all the three persons will solve } \text{the problem.}
\displaystyle \text{(ii) the probability that problem will be solved.}
\displaystyle \text{Answer:}
\displaystyle \text{Odds in favour of Aman}=3:2
\displaystyle P_1=\frac{3}{5},\quad P_1'=1-\frac{3}{5}=\frac{2}{5}
\displaystyle \text{Odds in favour of Mohan}=4:1
\displaystyle P_2=\frac{4}{5},\quad P_2'=1-\frac{4}{5}=\frac{1}{5}
\displaystyle \text{Odds against Bipin}=2:1
\displaystyle P_3=\frac{1}{3},\quad P_3'=1-\frac{1}{3}=\frac{2}{3}
\displaystyle \text{(i) Probability that all three solve the problem}
\displaystyle =P_1\times P_2\times P_3=\frac{3}{5}\times\frac{4}{5}\times\frac{1}{3}=\frac{4}{25}
\displaystyle \text{53. (ii) Probability that problem will be solved}
\displaystyle =1-P(\text{problem will not be solved})
\displaystyle =1-P_1'P_2'P_3'
\displaystyle =1-\frac{2}{5}\times\frac{1}{5}\times\frac{2}{3}
\displaystyle =1-\frac{4}{75}=\frac{71}{75}
\\

\displaystyle \textbf{Question 42: } \text{For } A,B \text{ and } C \text{ the chances of being selected as the manager of a firm}
\displaystyle \text{are } 4:1:2, \text{ respectively. The probabilities for them to introduce a radical change}
\displaystyle \text{in the marketing strategy are } 0.3,0.8 \text{ and } 0.5, \text{ respectively. If a change takes place,}
\displaystyle \text{then find the probability that it is due to the appointment of } B.
\displaystyle \quad \text{ISC 2013}
\displaystyle \text{Answer:}
\displaystyle \text{Let }E_{1},E_{2},E_{3}\text{ be events of A, B and C being selected as managers,}
\displaystyle \text{and }S\text{ be the event that the selected person introduces a radical change.}
\displaystyle \text{Given ratio }A:B:C=4:1:2\Rightarrow \text{total}=7
\displaystyle \therefore P(E_{1})=\frac{4}{7},\ P(E_{2})=\frac{1}{7},\ P(E_{3})=\frac{2}{7}
\displaystyle P\left(\frac{S}{E_{1}}\right)=0.3,\quad P\left(\frac{S}{E_{2}}\right)=0.8,\quad P\left(\frac{S}{E_{3}}\right)=0.5
\displaystyle \text{By Bayes' theorem,}
\displaystyle P\left(\frac{E_{2}}{S}\right)=\frac{P(E_{2})\cdot P\left(\frac{S}{E_{2}}\right)}{P(E_{1})\cdot P\left(\frac{S}{E_{1}}\right)+P(E_{2})\cdot P\left(\frac{S}{E_{2}}\right)+P(E_{3})\cdot P\left(\frac{S}{E_{3}}\right)}
\displaystyle =\frac{\frac{1}{7}\times0.8}{\frac{4}{7}\times0.3+\frac{1}{7}\times0.8+\frac{2}{7}\times0.5}
\displaystyle =\frac{\frac{0.8}{7}}{\frac{1.2}{7}+\frac{0.8}{7}+\frac{1}{7}}
\displaystyle =\frac{0.8}{1.2+0.8+1}=\frac{0.8}{3}=\frac{8}{30}=\frac{4}{15}
\\

\displaystyle \textbf{Question 43: } \text{Three persons } A, B \text{ and } C \text{ shoot to hit a target. If in } \text{trials } A \text{ hits the}
\displaystyle \text{target } 4 \text{ times in } 5 \text{ shots, } B \text{ hits } 3 \text{ times } \text{in } 4 \text{ shots and } C \text{ hits } 2 \text{ times in } 3 \text{ trials. Find the}
\displaystyle \text{probability that}
\displaystyle \text{(i) exactly two persons hit the target.}
\displaystyle \text{(ii) atleast two persons hit the target. (ISC 2012)}
\displaystyle \text{Answer:}
\displaystyle \text{Given, } P(A)=\frac{4}{5},\ P(B)=\frac{3}{4}\text{ and }P(C)=\frac{2}{3}
\displaystyle \therefore P(A')=1-\frac{4}{5}=\frac{1}{5},\quad P(B')=1-\frac{3}{4}=\frac{1}{4}
\displaystyle \text{and } P(C')=1-\frac{2}{3}=\frac{1}{3}
\displaystyle \text{(i) Probability that exactly two persons hit the target}
\displaystyle =P(ABC')+P(AB'C)+P(A'BC)
\displaystyle =P(A)P(B)P(C')+P(A)P(B')P(C)+P(A')P(B)P(C)
\displaystyle =\frac{4}{5}\times\frac{3}{4}\times\frac{1}{3}+\frac{4}{5}\times\frac{1}{4}\times\frac{2}{3}+\frac{1}{5}\times\frac{3}{4}\times\frac{2}{3}
\displaystyle =\frac{1}{5}+\frac{2}{15}+\frac{1}{10}=\frac{6+4+3}{30}=\frac{13}{30}
\displaystyle \text{(ii) Probability that at least two persons hit the target}
\displaystyle =P(ABC')+P(AB'C)+P(A'BC)+P(ABC)
\displaystyle =\frac{13}{30}+P(A)P(B)P(C)
\displaystyle =\frac{13}{30}+\frac{4}{5}\times\frac{3}{4}\times\frac{2}{3}
\displaystyle =\frac{13}{30}+\frac{2}{5}=\frac{13+12}{30}=\frac{25}{30}=\frac{5}{6}
\\

\displaystyle \textbf{Question 44: } \text{A box contains } 30 \text{ bolts and } 40 \text{ nuts. Half of the bolts and half of the }
\displaystyle \text{nuts are rusted. If two items are } \text{drawn at random from the box, then what is the}
\displaystyle \text{probability that either both are rusted or both are }  \text{bolts? (ISC 2012)}
\displaystyle \text{Answer:}
\displaystyle \text{Given, the box contains }30\text{ bolts and }40\text{ nuts}
\displaystyle \text{Rusted bolts}=15,\quad \text{rusted nuts}=20
\displaystyle \text{Total rusted bolts and nuts}=15+20=35
\displaystyle \text{Let } A=\text{rusted bolts and nuts},\quad B=\text{bolts}
\displaystyle P(\text{both items are rusted or both are bolts})=P(A\cup B)
\displaystyle =P(A)+P(B)-P(A\cap B)
\displaystyle =\frac{{}^{35}C_{2}}{{}^{70}C_{2}}+\frac{{}^{30}C_{2}}{{}^{70}C_{2}}-\frac{{}^{15}C_{2}}{{}^{70}C_{2}}
\displaystyle =\frac{35\times34+30\times29-15\times14}{70\times69}
\displaystyle =\frac{1190+870-210}{4830}
\displaystyle =\frac{1850}{4830}=\frac{185}{483}
\\

\displaystyle \textbf{Question 45: } \text{A bag contains } 8 \text{ red and } 5 \text{ white balls. Two }\text{successive draws of } 3
\displaystyle \text{ balls are made at random from } \text{the bag without replacements. Find the probability that}
\displaystyle \text{the first draw yields } 3 \text{ white balls and the second draw }  3 \text{ red balls. (ISC 2012)}
\displaystyle \text{Answer:}
\displaystyle \text{Given, a bag contains }8\text{ red and }5\text{ white balls}
\displaystyle P(\text{first draw yields }3\text{ white balls and second draw yields }3\text{ red balls})
\displaystyle =\frac{{}^{5}C_{3}}{{}^{13}C_{3}}\times\frac{{}^{8}C_{3}}{{}^{10}C_{3}} \quad [\text{without replacement}]
\displaystyle =\frac{10}{286}\times\frac{56}{120}
\displaystyle =\frac{5}{143}\times\frac{7}{15}
\displaystyle =\frac{7}{429}
\\

\displaystyle \textbf{Question 46: } \text{In a class of } 75 \text{ students, } 15 \text{ are above average, } 45 \text{ are average}
\displaystyle \text{and the rest below average achievers. The probability that an above average achieving}
\displaystyle \text{student fails is } 0.005, \text{ that an average achieving student fails is } 0.05 \text{ and}
\displaystyle \text{the probability of a below average achieving student failing is } 0.15. \text{ If a student}
\displaystyle \text{is known to have passed, then what is the probability that he is a below average achiever?}
\displaystyle \quad \text{ISC 2012}
\displaystyle \text{Answer:}
\displaystyle \text{Let }E_{1},E_{2},E_{3}\text{ be events of above average, average and below average students.}
\displaystyle \text{Total students }=75
\displaystyle P(E_{1})=\frac{15}{75}=\frac{1}{5},\quad P(E_{2})=\frac{45}{75}=\frac{3}{5}
\displaystyle P(E_{3})=\frac{75-(15+45)}{75}=\frac{15}{75}=\frac{1}{5}
\displaystyle \text{Let }A=\text{event that a student passes}
\displaystyle P\left(\frac{A}{E_{1}}\right)=1-0.005=\frac{995}{1000}=\frac{199}{200}
\displaystyle P\left(\frac{A}{E_{2}}\right)=1-0.05=\frac{95}{100}=\frac{19}{20}
\displaystyle P\left(\frac{A}{E_{3}}\right)=1-0.15=\frac{85}{100}=\frac{17}{20}
\displaystyle \text{Required probability that a passed student is below average:}
\displaystyle P\left(\frac{E_{3}}{A}\right)=\frac{P(E_{3})\cdot P\left(\frac{A}{E_{3}}\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}{5}\times\frac{17}{20}}{\frac{1}{5}\times\frac{199}{200}+\frac{3}{5}\times\frac{19}{20}+\frac{1}{5}\times\frac{17}{20}}
\displaystyle =\frac{\frac{17}{100}}{\frac{199}{1000}+\frac{57}{100}+\frac{17}{100}}
\displaystyle =\frac{\frac{17}{100}}{\frac{199}{1000}+\frac{570}{1000}+\frac{170}{1000}}
\displaystyle =\frac{\frac{17}{100}}{\frac{939}{1000}}=\frac{17}{100}\times\frac{1000}{939}=\frac{170}{939}
\\

\displaystyle \textbf{Question 47: } \text{The probability that a bulb produced by a factory will fuse in}
\displaystyle 100 \text{ days of use is } 0.05. \text{ Find the probability that out of } 5 \text{ such bulbs,}
\displaystyle \text{after } 100 \text{ days of use: \hspace{1cm} (ISC 2012)}
\displaystyle \text{(i) none fuse. \hspace{2cm} (ii) no more than one fuses.}
\displaystyle \text{(iii) more than one fuses. \hspace{2cm} (iv) at least one fuses.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }p=\text{Probability of fused bulb}=0.05=\frac{5}{100}=\frac{1}{20}
\displaystyle \text{and }q=\text{Probability of not fused bulb}=1-\frac{1}{20}=\frac{19}{20}\quad[\because p+q=1]
\displaystyle \text{Let }X=\text{Number of fused bulbs after }5\text{ days, }n=5
\displaystyle \text{(i) Probability that none bulb is fused,}
\displaystyle P(X=0)={}^{5}C_0\left(\frac{1}{20}\right)^0\left(\frac{19}{20}\right)^5=\left(\frac{19}{20}\right)^5
\displaystyle \text{(ii) Probability that not more than one bulb is fused,}
\displaystyle P(X=0\ \text{or}\ X=1)=P(X=0)+P(X=1)
\displaystyle =\left(\frac{19}{20}\right)^5+{}^{5}C_1\left(\frac{1}{20}\right)^1\left(\frac{19}{20}\right)^4
\displaystyle =\left(\frac{19}{20}\right)^5+5\times\frac{1}{20}\times\left(\frac{19}{20}\right)^4
\displaystyle =\left(\frac{19}{20}\right)^4\left(\frac{19}{20}+\frac{5}{20}\right)
\displaystyle =\left(\frac{19}{20}\right)^4\left(\frac{24}{20}\right)=\left(\frac{19}{20}\right)^4\times\frac{6}{5}
\displaystyle \text{Answer: }\left(\frac{19}{20}\right)^4\times\frac{6}{5}
\displaystyle \text{(iii) Probability that more than one bulb is fused,}
\displaystyle P(1<X\leq5)=1-P(X=0\ \text{or}\ X=1)
\displaystyle =1-\frac{6}{5}\left(\frac{19}{20}\right)^4
\displaystyle \text{Answer: }1-\frac{6}{5}\left(\frac{19}{20}\right)^4
\displaystyle \text{(iv) Probability that at least one bulb is fused,}
\displaystyle P(1\leq X\leq5)=1-P(X=0)
\displaystyle =1-\left(\frac{19}{20}\right)^5
\displaystyle \text{Answer: }1-\left(\frac{19}{20}\right)^5
\\

\displaystyle \textbf{Question 48: } \text{A word consists of } 9 \text{ different alphabets, in which there are 4 consonants}
\displaystyle \text{and } 5 \text{ vowels. Three alphabets } \text{are chosen at random. What is the probability that}
\displaystyle \text{more than one vowel will be selected? (ISC 2011)}
\displaystyle \text{Answer:}
\displaystyle \text{A word consists of } 9 \text{ different alphabets: }4\text{ consonants and }5\text{ vowels}
\displaystyle \text{Three alphabets are chosen at random}
\displaystyle \text{Probability that more than one vowel will be selected}
\displaystyle =P(\text{two vowels and one consonant})+P(\text{three vowels})
\displaystyle =\frac{{}^{5}C_{2}\times{}^{4}C_{1}}{{}^{9}C_{3}}+\frac{{}^{5}C_{3}}{{}^{9}C_{3}}
\displaystyle =\frac{10\times4}{84}+\frac{10}{84}=\frac{50}{84}=\frac{25}{42}
\\

\displaystyle \textbf{Question 49: } \text{Aman and Bhuvan throw a pair of dice alternatively. In order to win, }
\displaystyle \text{they have to get a sum of } 8. \text{ Find their respective probabilities of winning, if Aman }
\displaystyle \text{starts the } \text{game. (ISC 2011)}
\displaystyle \text{Answer:}
\displaystyle \text{Total number of ways of throwing a pair of dice}=n(S)=36
\displaystyle \text{Let } E=\text{getting }8\text{ as sum in a pair of dice}
\displaystyle E=\{(2,6),(3,5),(4,4),(5,3),(6,2)\}
\displaystyle P(E)=\frac{n(E)}{n(S)}=\frac{5}{36}
\displaystyle P(\overline{E})=1-P(E)=1-\frac{5}{36}=\frac{31}{36}
\displaystyle \text{Now, }P(\text{Aman wins})
\displaystyle =P(E)+P(\overline{E}\overline{E}E)+P(\overline{E}\overline{E}\overline{E}\overline{E}E)+\cdots
\displaystyle =\frac{5}{36}+\left(\frac{31}{36}\right)^{2}\frac{5}{36}+\left(\frac{31}{36}\right)^{4}\frac{5}{36}+\cdots
\displaystyle =\frac{5}{36}\left[1+\left(\frac{31}{36}\right)^{2}+\left(\frac{31}{36}\right)^{4}+\cdots\right]
\displaystyle =\frac{5}{36}\times\frac{1}{1-\left(\frac{31}{36}\right)^{2}}
\displaystyle =\frac{5}{36}\times\frac{36^{2}}{36^{2}-31^{2}}
\displaystyle =\frac{5}{36}\times\frac{36^{2}}{335}=\frac{36}{67}
\displaystyle P(\text{Bhuwan wins})=1-P(\text{Aman wins})
\displaystyle =1-\frac{36}{67}=\frac{31}{67}
\\

\displaystyle \textbf{Question 50: } \text{A purse contains } 4 \text{ silver and } 5 \text{ copper coins. } \text{A second purse contains } 3
\displaystyle \text{ silver and } 7 \text{ copper coins. } \text{If a coin is taken out at random from one of the purses, then}
\displaystyle \text{what is the probability that it is a copper } \text{coin? (ISC 2011)}
\displaystyle \text{Answer:}
\displaystyle \text{Probability of picking either first bag or second bag}=\frac{1}{2}
\displaystyle P(\text{copper coin from first bag})=\frac{5}{4+5}=\frac{5}{9}
\displaystyle P(\text{copper coin from second bag})=\frac{7}{3+7}=\frac{7}{10}
\displaystyle P(\text{copper coin from either bag})
\displaystyle =P(\text{first bag})P(\text{copper from first bag})
\displaystyle \quad +P(\text{second bag})P(\text{copper from second bag})
\displaystyle =\frac{1}{2}\times\frac{5}{9}+\frac{1}{2}\times\frac{7}{10}
\displaystyle =\frac{5}{18}+\frac{7}{20}
\displaystyle =\frac{100+126}{360}=\frac{226}{360}=\frac{113}{180}
\\

\displaystyle \textbf{Question 51: } \text{Bag } A \text{ contains } 2 \text{ white, } 1 \text{ black and } 3 \text{ red balls, bag } B
\displaystyle \text{contains } 3 \text{ white, } 2 \text{ black and } 4 \text{ red balls and bag } C \text{ contains } 4 \text{ white, } 3 \text{ black}
\displaystyle \text{and } 2 \text{ red balls. One bag is chosen at random and } 2 \text{ balls are drawn at random}
\displaystyle \text{from that bag. If the randomly drawn balls happen to be red and black, then}
\displaystyle \text{what is the probability that both balls come from bag } B? \quad \text{ISC 2011}
\displaystyle \text{Answer:}
\displaystyle \text{Let }E_1,E_2\text{ and }E_3\text{ be the events for chosen bags }A,B\text{ and }C\text{ respectively.}
\displaystyle \text{So, }P(E_1)=P(E_2)=P(E_3)=\frac{1}{3}.
\displaystyle \text{Bag }A\text{ contains }2\text{ white, }1\text{ black and }3\text{ red balls.}
\displaystyle \text{Bag }B\text{ contains }3\text{ white, }2\text{ black and }4\text{ red balls.}
\displaystyle \text{Bag }C\text{ contains }4\text{ white, }3\text{ black and }2\text{ red balls.}
\displaystyle \text{If }A\text{ be the event of choosing two balls, i.e. one red and one black,}
\displaystyle P\!\left(\frac{A}{E_1}\right)=\frac{{}^{3}C_1\times{}^{1}C_1}{{}^{6}C_2}=\frac{3\times1}{15}=\frac{1}{5}.
\displaystyle P\!\left(\frac{A}{E_2}\right)=\frac{{}^{4}C_1\times{}^{2}C_1}{{}^{9}C_2}=\frac{4\times2}{36}=\frac{2}{9}.
\displaystyle \text{and }P\!\left(\frac{A}{E_3}\right)=\frac{{}^{2}C_1\times{}^{3}C_1}{{}^{9}C_2}=\frac{2\times3}{36}=\frac{1}{6}.
\displaystyle \text{Probability that both balls drawn from bag }B\text{ having red and black balls}
\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)+P(E_3)\cdot P\!\left(\frac{A}{E_3}\right)}
\displaystyle =\frac{\frac{1}{3}\times\frac{2}{9}}{\frac{1}{3}\times\frac{1}{5}+\frac{1}{3}\times\frac{2}{9}+\frac{1}{3}\times\frac{1}{6}}
\displaystyle =\frac{\frac{2}{27}}{\frac{1}{3}\left(\frac{1}{5}+\frac{2}{9}+\frac{1}{6}\right)}
\displaystyle =\frac{\frac{2}{27}}{\frac{1}{3}\left(\frac{18+20+15}{90}\right)}
\displaystyle =\frac{\frac{2}{27}}{\frac{53}{270}}=\frac{2}{27}\times\frac{270}{53}=\frac{20}{53}.
\\

\displaystyle \textbf{Question 52: } \text{A fair die is thrown once. What is the probability that either an even }
\displaystyle \text{number or a number greater than three }  \text{will turn up? (ISC 2010)}
\displaystyle \text{Answer:}
\displaystyle \text{Let } S=\{1,2,3,4,5,6\}
\displaystyle E_{1}=\text{even numbers}=\{2,4,6\}
\displaystyle E_{2}=\text{numbers greater than }3=\{4,5,6\}
\displaystyle E_{1}\cap E_{2}=\{4,6\}
\displaystyle \text{Probability that either an even number or a number greater than }3\text{ turns up}
\displaystyle =P(E_{1}\cup E_{2})=P(E_{1})+P(E_{2})-P(E_{1}\cap E_{2})
\displaystyle =\frac{3}{6}+\frac{3}{6}-\frac{2}{6}
\displaystyle =1-\frac{1}{3}=\frac{2}{3}
\\

\displaystyle \textbf{Question 53: } \text{Akhil and Vijay appear for an interview for two vacancies. The probability }
\displaystyle \text{of Akhil's selection is } \frac{1}{4} \text{ and} \text{Vijay's selection is } \frac{2}{3}. \text{ Find the probability that only one}
\displaystyle \text{of them will be selected. (ISC 2010)}
\displaystyle \text{Answer:}
\displaystyle \text{Consider the following events:}
\displaystyle A=\text{Akhil is selected},\quad B=\text{Vijay is selected}
\displaystyle P(A)=\frac{1}{4},\quad P(B)=\frac{2}{3}
\displaystyle \text{Probability that only one of them is selected for interview}
\displaystyle =P(A\cap B')+P(B\cap A')
\displaystyle =P(A)P(B')+P(B)P(A')
\displaystyle =\frac{1}{4}\left(1-\frac{2}{3}\right)+\frac{2}{3}\left(1-\frac{1}{4}\right)
\displaystyle =\frac{1}{4}\times\frac{1}{3}+\frac{2}{3}\times\frac{3}{4}
\displaystyle =\frac{1}{12}+\frac{6}{12}=\frac{7}{12}
\\

\displaystyle \textbf{Question 54: } \text{A factory has three machines } A,B \text{ and } C \text{ producing } 1500,2500
\displaystyle \text{and } 3000 \text{ bulbs per day, respectively. Machine } A \text{ produces } 1.5\% \text{ defective bulbs,}
\displaystyle \text{machine } B \text{ produces } 2\% \text{ defective bulbs and machine } C \text{ produces } 2.5\%
\displaystyle \text{defective bulbs. At the end of the day, a bulb is drawn at random and is found}
\displaystyle \text{to be defective. What is the probability that the defective bulb has been}
\displaystyle \text{produced by machine } B? \quad \text{ISC 2010}
\displaystyle \text{Answer:}
\displaystyle  \text{Given,}
\displaystyle \text{The factory }A\text{ production per day }=1500
\displaystyle \text{The factory }B\text{ production per day }=2500
\displaystyle \text{The factory }C\text{ production per day }=3000
\displaystyle \text{The total production of factories }A,B\text{ and }C=1500+2500+3000=7000
\displaystyle \text{Let }E_1=\text{Event of selecting a bulb produced by }A,
\displaystyle E_2=\text{Event of selecting a bulb produced by }B,
\displaystyle E_3=\text{Event of selecting a bulb produced by }C.
\displaystyle \text{Let }S\text{ be the event of selecting a defective bulb.}
\displaystyle P(S|E_1)=1.5\%=\frac{1.5}{100},\quad P(S|E_2)=2\%=\frac{2}{100},
\displaystyle \text{and }P(S|E_3)=2.5\%=\frac{2.5}{100}.
\displaystyle \text{Also, }P(E_1)=\frac{1500}{7000}=\frac{3}{14},\quad P(E_2)=\frac{2500}{7000}=\frac{5}{14},
\displaystyle \text{and }P(E_3)=\frac{3000}{7000}=\frac{3}{7}.
\displaystyle \therefore\ \text{The probability that the selected bulb produced by }B\text{ is defective,}
\displaystyle P(E_2|S)=\frac{P(E_2)\cdot P(S|E_2)}{P(E_1)\cdot P(S|E_1)+P(E_2)\cdot P(S|E_2)+P(E_3)\cdot P(S|E_3)}
\displaystyle =\frac{\frac{5}{14}\times\frac{2}{100}}{\frac{3}{14}\times\frac{1.5}{100}+\frac{5}{14}\times\frac{2}{100}+\frac{3}{7}\times\frac{2.5}{100}}
\displaystyle =\frac{\frac{5\times2}{14\times100}}{\frac{1}{100}\left(\frac{3}{14}\times1.5+\frac{5}{14}\times2+\frac{3}{7}\times2.5\right)}
\displaystyle =\frac{\frac{10}{1400}}{\frac{1}{100}\left(\frac{4.5}{14}+\frac{10}{14}+\frac{7.5}{7}\right)}
\displaystyle =\frac{\frac{10}{1400}}{\frac{1}{100}\left(\frac{4.5+10+15}{14}\right)}
\displaystyle =\frac{\frac{10}{1400}}{\frac{29.5}{1400}}=\frac{10}{29.5}=\frac{20}{59}.
\\

\displaystyle \textbf{Question 55: } \text{Five bad eggs are mixed with } 10 \text{ good ones. If } 3 \text{ eggs are}
\displaystyle \text{drawn one by one with replacement, then find the probability}
\displaystyle \text{distribution of the number of good eggs drawn. \hspace{1cm} ISC 2010}
\displaystyle \text{Answer:}
\displaystyle \text{Let the random variable be }X.
\displaystyle \text{Since, number of drawing eggs }=3
\displaystyle \text{So, the value of }X\text{ can be }0,1,2\text{ and }3.
\displaystyle \text{For }P(X=0)=P(0\text{ good egg})
\displaystyle =\frac{5}{15}\times\frac{5}{15}\times\frac{5}{15}=\frac{1}{3}\times\frac{1}{3}\times\frac{1}{3}=\frac{1}{27}
\displaystyle \text{For }P(X=1)=P(1\text{ good egg})
\displaystyle =\frac{10}{15}\times\frac{5}{15}\times\frac{5}{15}+\frac{5}{15}\times\frac{10}{15}\times\frac{5}{15}
\displaystyle +\frac{5}{15}\times\frac{5}{15}\times\frac{10}{15}
\displaystyle =\frac{2}{3}\times\frac{1}{3}\times\frac{1}{3}+\frac{1}{3}\times\frac{2}{3}\times\frac{1}{3}+\frac{1}{3}\times\frac{1}{3}\times\frac{2}{3}
\displaystyle =\frac{2}{27}+\frac{2}{27}+\frac{2}{27}=\frac{6}{27}
\displaystyle \text{For }P(X=2)=P(2\text{ good eggs})
\displaystyle =\frac{10}{15}\times\frac{10}{15}\times\frac{5}{15}+\frac{10}{15}\times\frac{5}{15}\times\frac{10}{15}
\displaystyle +\frac{5}{15}\times\frac{10}{15}\times\frac{10}{15}
\displaystyle =\frac{2}{3}\times\frac{2}{3}\times\frac{1}{3}+\frac{2}{3}\times\frac{1}{3}\times\frac{2}{3}+\frac{1}{3}\times\frac{2}{3}\times\frac{2}{3}
\displaystyle =\frac{4}{27}+\frac{4}{27}+\frac{4}{27}
\displaystyle =\frac{12}{27}
\displaystyle \text{For }P(X=3)=P(3\text{ good eggs})
\displaystyle =\frac{10}{15}\times\frac{10}{15}\times\frac{10}{15}
\displaystyle =\frac{2}{3}\times\frac{2}{3}\times\frac{2}{3}=\frac{8}{27}
\displaystyle \therefore\ \text{Required probability distribution is}
\displaystyle \begin{array}{c|cccc}X&0&1&2&3\\ \hline P(X)&\frac{1}{27}&\frac{6}{27}&\frac{12}{27}&\frac{8}{27}\end{array}
\\

\displaystyle \textbf{Question 56: } \text{Five bad eggs are mixed with } 10 \text{ good ones. If } 3 \text{ eggs are}
\displaystyle \text{drawn one by one with replacement, then find the probability}
\displaystyle \text{distribution of the number of good eggs drawn. \hspace{1cm} (ISC 2010)}
\displaystyle \text{Answer:}
\displaystyle \text{Let the random variable be }X.
\displaystyle \text{Since, number of drawing eggs }=3
\displaystyle \text{So, the value of }X\text{ can be }0,1,2\text{ and }3.
\displaystyle \text{For }P(X=0)=P(0\text{ good egg})
\displaystyle =\frac{5}{15}\times\frac{5}{15}\times\frac{5}{15}=\frac{1}{27}
\displaystyle \text{For }P(X=1)=P(1\text{ good egg})
\displaystyle =\frac{10}{15}\times\frac{5}{15}\times\frac{5}{15}+\frac{5}{15}\times\frac{10}{15}\times\frac{5}{15}
\displaystyle +\frac{5}{15}\times\frac{5}{15}\times\frac{10}{15}=\frac{6}{27}
\displaystyle \text{For }P(X=2)=P(2\text{ good eggs})
\displaystyle =\frac{10}{15}\times\frac{10}{15}\times\frac{5}{15}+\frac{10}{15}\times\frac{5}{15}\times\frac{10}{15}
\displaystyle +\frac{5}{15}\times\frac{10}{15}\times\frac{10}{15}=\frac{12}{27}
\displaystyle \text{For }P(X=3)=P(3\text{ good eggs})
\displaystyle =\frac{10}{15}\times\frac{10}{15}\times\frac{10}{15}=\frac{8}{27}
\displaystyle \therefore\ \text{Required probability distribution is}
\displaystyle \begin{array}{c|cccc}X&0&1&2&3\\ \hline P(X)&\frac{1}{27}&\frac{6}{27}&\frac{12}{27}&\frac{8}{27}\end{array}
\\

\displaystyle \textbf{Question 57: } \text{Two horses are considered for a race. The probability of selection of the}
\displaystyle \text{first horse is } \frac{1}{4} \text{ and that of the } \text{second is } \frac{1}{3}. \text{ What is the probability that}
\displaystyle \text{(i) both of them will be selected?}
\displaystyle \text{(ii) only one of them will be selected?}
\displaystyle \text{(iii) none of them will be selected? (ISC 2009)}
\displaystyle \text{Answer:}
\displaystyle \text{Let } A=\text{selection of first horse}
\displaystyle B=\text{selection of second horse}
\displaystyle \therefore P(A)=\frac{1}{4},\quad P(B)=\frac{1}{3}
\displaystyle \text{(i) }P(\text{both selected})=P(A\cap B)=P(A)P(B)
\displaystyle =\frac{1}{4}\times\frac{1}{3}=\frac{1}{12}
\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}{4}\left(1-\frac{1}{3}\right)+\left(1-\frac{1}{4}\right)\frac{1}{3}
\displaystyle =\frac{1}{4}\times\frac{2}{3}+\frac{3}{4}\times\frac{1}{3}
\displaystyle =\frac{1}{6}+\frac{1}{4}=\frac{2+3}{12}=\frac{5}{12}
\displaystyle \text{(iii) }P(\text{none selected})=P(A'\cap B')=P(A')P(B')
\displaystyle =\left(1-\frac{1}{4}\right)\left(1-\frac{1}{3}\right)=\frac{3}{4}\times\frac{2}{3}=\frac{1}{2}
\\

\displaystyle \textbf{Question 58: } \text{An insurance company insured } 4000 \text{ doctors, } 8000 \text{ teachers}
\displaystyle \text{and } 12000 \text{ engineers. The probability of a doctor, a teacher and an engineer}
\displaystyle \text{dying before the age of } 58 \text{ yr are } 0.01,0.03 \text{ and } 0.05, \text{ respectively.}
\displaystyle \text{If one of the insured persons dies before the age of } 58 \text{ yr, then find}
\displaystyle \text{the probability that he is a doctor.}
\displaystyle \quad \text{ISC 2009}
\displaystyle \text{Answer:}
\displaystyle \text{Given, the total number of doctors insured by the insurance company }=4000
\displaystyle \text{The total number of teachers insured by the insurance company }=8000
\displaystyle \text{The total number of engineers insured by the insurance company }=12000
\displaystyle \text{The total number of insured persons }=4000+8000+12000=24000
\displaystyle \text{Let }D\text{ be the event that an insured person die.}
\displaystyle \text{Let }E_1=\text{Insured people are doctors, }E_2=\text{Insured people are teachers}
\displaystyle \text{and }E_3=\text{Insured people are engineers.}
\displaystyle P(E_1)=\frac{4000}{24000}=\frac{1}{6},\quad P(E_2)=\frac{8000}{24000}=\frac{1}{3},\quad P(E_3)=\frac{12000}{24000}=\frac{1}{2}
\displaystyle \text{Now, }P(\text{insured person die and having a doctor})=P(D|E_1)=0.01
\displaystyle P(\text{insured person die and having a teacher})=P(D|E_2)=0.03
\displaystyle P(\text{insured person die and having an engineer})=P(D|E_3)=0.05
\displaystyle \therefore\ P(E_1|D)=\frac{P(E_1)\cdot P(D|E_1)}{P(E_1)\cdot P(D|E_1)+P(E_2)\cdot P(D|E_2)+P(E_3)\cdot P(D|E_3)}
\displaystyle =\frac{\frac{1}{6}\times0.01}{\frac{1}{6}\times0.01+\frac{1}{3}\times0.03+\frac{1}{2}\times0.05}
\displaystyle =\frac{\frac{0.01}{6}}{\frac{0.01}{6}+\frac{0.03}{3}+\frac{0.05}{2}}
\displaystyle =\frac{0.01}{0.01+0.06+0.15}=\frac{0.01}{0.22}=\frac{1}{22}
\displaystyle \text{Answer: }\frac{1}{22}
\\

\displaystyle \textbf{Question 59: } \text{In a single throw of two dice, find the probability of getting a total }
\displaystyle \text{of atmost } 9. \text{ (ISC 2008)}
\displaystyle \text{Answer:}
\displaystyle \text{Total outcomes of a single throw of two dice}=36
\displaystyle \text{Favourable outcomes for getting a total of }10,11,12
\displaystyle =\{(4,6),(5,5),(6,4),(5,6),(6,5),(6,6)\}
\displaystyle \therefore P(\text{getting a total of at least }10)=\frac{6}{36}=\frac{1}{6}
\displaystyle P(\text{getting a total of at most }9)
\displaystyle =1-P(\text{getting a total of at least }10)
\displaystyle =1-\frac{1}{6}=\frac{5}{6}
\\

\displaystyle \textbf{Question 60: } \text{A class consists of } 50 \text{ students out of which there are } 10 \text{ girls.}
\displaystyle \text{In the class, } 2 \text{ girls and } 5 \text{ boys are rank holders in an examination.}
\displaystyle \text{If a student is selected at random from the class and is found to be a rank holder,}
\displaystyle \text{then what is the probability that the student selected is a girl?}
\displaystyle \quad \text{ISC 2008}
\displaystyle \text{Answer:}
\displaystyle \text{Given,}
\displaystyle \text{The total number of students }=50
\displaystyle \text{The total number of girls }=10,\quad \text{boys }=40
\displaystyle \text{Girls who achieve rank }=2,\quad \text{boys who achieve rank }=5
\displaystyle \text{Rank holder students }=2+5=7
\displaystyle \text{Let }R=\text{event that the student achieve rank}
\displaystyle \text{Let }E=\text{event of selecting a girl, }B=\text{event of selecting a boy}
\displaystyle P(E)=\frac{10}{50}=\frac{1}{5},\quad P(B)=\frac{40}{50}=\frac{4}{5}
\displaystyle P(R|E)=\frac{2}{7},\quad P(R|B)=\frac{5}{7}
\displaystyle \therefore\ P(E|R)=\frac{P(E)\cdot P(R|E)}{P(E)\cdot P(R|E)+P(B)\cdot P(R|B)}
\displaystyle =\frac{\frac{1}{5}\times\frac{2}{7}}{\frac{1}{5}\times\frac{2}{7}+\frac{4}{5}\times\frac{5}{7}}
\displaystyle =\frac{\frac{2}{35}}{\frac{2}{35}+\frac{20}{35}}=\frac{2}{22}=\frac{1}{11}
\displaystyle \text{Answer: }\frac{1}{11}
\\

\displaystyle \textbf{Question 61: } \text{Eight coins are thrown simultaneously.}
\displaystyle \text{(i) Show that the probability of getting at least } 6 \text{ heads is } \frac{37}{256}. \quad \text{ISC 2008}
\displaystyle \text{(ii) What is the probability of getting at least } 3 \text{ heads?}
\displaystyle \text{Answer:}
\displaystyle \text{Given,}
\displaystyle \text{The total number of coins }=8=n
\displaystyle \text{The probability of getting heads }=\frac{1}{2}=p
\displaystyle \therefore\ q=1-p=1-\frac{1}{2}=\frac{1}{2}\quad[\because p+q=1]
\displaystyle \text{(i) By using Binomial theorem,}
\displaystyle \text{Required probability }P(\text{at least }6\text{ heads})
\displaystyle =P(X=6)+P(X=7)+P(X=8)
\displaystyle ={}^{8}C_6\left(\frac{1}{2}\right)^6\left(\frac{1}{2}\right)^2+{}^{8}C_7\left(\frac{1}{2}\right)^7\left(\frac{1}{2}\right)^1
\displaystyle +{}^{8}C_8\left(\frac{1}{2}\right)^8
\displaystyle ={}^{8}C_6\left(\frac{1}{2}\right)^8+{}^{8}C_7\left(\frac{1}{2}\right)^8+{}^{8}C_8\left(\frac{1}{2}\right)^8
\displaystyle =\frac{1}{2^8}\left[{}^{8}C_6+{}^{8}C_7+{}^{8}C_8\right]
\displaystyle =\frac{1}{256}\left[28+8+1\right]=\frac{37}{256}
\displaystyle \text{Answer: }\frac{37}{256}
\displaystyle =\frac{8!}{2!6!}\left(\frac{1}{2}\right)^6\left(\frac{1}{2}\right)^2+\frac{8!}{7!1!}\left(\frac{1}{2}\right)^7\left(\frac{1}{2}\right)+\left(\frac{1}{2}\right)^8
\displaystyle =28\left(\frac{1}{2}\right)^8+8\left(\frac{1}{2}\right)^8+\left(\frac{1}{2}\right)^8
\displaystyle =\left(\frac{1}{2}\right)^8(28+8+1)
\displaystyle =\left(\frac{1}{2}\right)^8(37)=\frac{1}{256}\times37=\frac{37}{256}
\displaystyle \text{(ii) Required probability }(\text{at least }3\text{ heads})
\displaystyle =1-\text{Probability of }(\text{at most }2\text{ heads})
\displaystyle =1-[P(x=0)+P(x=1)+P(x=2)]
\displaystyle =1-\left[{}^{8}C_0\left(\frac{1}{2}\right)^0\left(\frac{1}{2}\right)^8+{}^{8}C_1\left(\frac{1}{2}\right)^1\left(\frac{1}{2}\right)^7\right.
\displaystyle \left.+{}^{8}C_2\left(\frac{1}{2}\right)^2\left(\frac{1}{2}\right)^6\right]
\displaystyle =1-\left[\left(\frac{1}{2}\right)^8+8\left(\frac{1}{2}\right)^8+\frac{8\times7}{2}\left(\frac{1}{2}\right)^8\right]
\displaystyle =1-\left(\frac{1}{2}\right)^8\left[1+8+\frac{56}{2}\right]
\displaystyle =1-\left(\frac{1}{2}\right)^8(1+8+28)
\displaystyle =1-\left(\frac{1}{2}\right)^8\times37
\displaystyle =1-\frac{37}{256}=\frac{256-37}{256}=\frac{219}{256}
\displaystyle \text{Answer: }\frac{219}{256}
\\

\displaystyle \textbf{Question 62: } \text{The probability that a teacher will give an unannounced test}
\displaystyle \text{during any class meeting is } \frac{1}{5}. \text{ If a student is absent twice,}
\displaystyle \text{find the probability that the student will miss at least one test. \hspace{1cm} (ISC 2008)}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }p=\text{Probability that a teacher will give an unannounced test during class}=\frac{1}{5}
\displaystyle \text{Then, }q=1-p=1-\frac{1}{5}=\frac{4}{5}\qquad[\because p+q=1]
\displaystyle \text{Also, given student is absent twice, i.e. }n=2
\displaystyle P(\text{student misses only one test})=P(X=1)
\displaystyle ={}^{2}C_1p^1q^1\qquad[\because P(X=r)={}^{n}C_rp^rq^{n-r}]
\displaystyle =2\times\left(\frac{1}{5}\right)\times\left(\frac{4}{5}\right)=\frac{8}{25}
\displaystyle P(\text{student misses both the tests})=P(X=2)
\displaystyle ={}^{2}C_2p^2q^0=1\left(\frac{1}{5}\right)^2\left(\frac{4}{5}\right)^0=\frac{1}{25}
\displaystyle P(\text{student misses at least one test})=P(X\geq1)
\displaystyle =P(X=1)+P(X=2)
\displaystyle =\frac{8}{25}+\frac{1}{25}=\frac{9}{25}
\displaystyle \text{Answer: }\frac{9}{25}

\displaystyle \textbf{Question 63: } \text{Kamal and Monica appear for an interview for two vacancies. The }
\displaystyle \text{probability of Kamal's selection is } \frac{1}{3}   \text{ and that of Monica's selection is } \frac{1}{5}. \text{ Find the}
\displaystyle \text{probability that only one of them will be selected. (ISC 2007)}
\displaystyle \text{Answer:}
\displaystyle \text{Let } E_{1}=\text{Kamal's selection after appearing for an interview}
\displaystyle \text{and } E_{2}=\text{Monica's selection after appearing for an interview}
\displaystyle \text{Then, } P(E_{1})=\frac{1}{3}\text{ and }P(E_{2})=\frac{1}{5}
\displaystyle P(\text{either Kamal or Monica is selected})
\displaystyle =P(E_{1}\cap E_{2}')+P(E_{1}'\cap E_{2})
\displaystyle =P(E_{1})P(E_{2}')+P(E_{1}')P(E_{2})
\displaystyle =\frac{1}{3}\left(1-\frac{1}{5}\right)+\left(1-\frac{1}{3}\right)\frac{1}{5}
\displaystyle =\frac{1}{3}\times\frac{4}{5}+\frac{2}{3}\times\frac{1}{5}
\displaystyle =\frac{4}{15}+\frac{2}{15}=\frac{6}{15}=\frac{2}{5}
\\

\displaystyle \textbf{Question 64: } \text{An insurance company insured } 1500 \text{ scooter drivers, } 2500 \text{ car drivers}
\displaystyle \text{and } 4500 \text{ truck drivers. The probability of a scooter, a car and a truck meeting}
\displaystyle \text{with an accident is } 0.01,0.02 \text{ and } 0.04, \text{ respectively. If one of the insured}
\displaystyle \text{persons meets with an accident, then find the probability that he is a scooter driver.}
\displaystyle \quad \text{(ISC 2007)}
\displaystyle \text{Answer:}
\displaystyle \text{Given,}
\displaystyle \text{The total number of insured scooter drivers }=1500
\displaystyle \text{The total number of insured car drivers }=2500
\displaystyle \text{The total number of insured truck drivers }=4500
\displaystyle \text{Total insured drivers }=1500+2500+4500=8500
\displaystyle \text{Let }A=\text{event of drivers who meet with an accident}
\displaystyle \text{Let }E_1,E_2,E_3\text{ be scooter, car and truck drivers respectively}
\displaystyle P(E_1)=\frac{1500}{8500}=\frac{3}{17},\quad P(E_2)=\frac{2500}{8500}=\frac{5}{17},\quad P(E_3)=\frac{4500}{8500}=\frac{9}{17}
\displaystyle P(A|E_1)=0.01,\quad P(A|E_2)=0.02,\quad P(A|E_3)=0.04
\displaystyle \therefore\ 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 =\frac{\frac{3}{17}\times0.01}{\frac{3}{17}\times0.01+\frac{5}{17}\times0.02+\frac{9}{17}\times0.04}
\displaystyle =\frac{3\times0.01}{3\times0.01+5\times0.02+9\times0.04}
\displaystyle =\frac{0.03}{0.03+0.10+0.36}=\frac{0.03}{0.49}=\frac{3}{49}
\displaystyle \text{Answer: }\frac{3}{49}
\\

\displaystyle \textbf{Question 65: } \text{The bag } A \text{ contains } 3 \text{ white and } 2 \text{ black balls while the bag } B
\displaystyle \text{contains } 2 \text{ white and } 5 \text{ black balls. One of the bags is chosen at random}
\displaystyle \text{and a ball is drawn from it. What is the probability that the ball is white?}
\displaystyle \quad \text{ISC 2007}
\displaystyle \text{Answer:}
\displaystyle \text{Bag }A:3\text{ white and }2\text{ black balls.}
\displaystyle \text{Bag }B:2\text{ white and }5\text{ black balls.}
\displaystyle \text{Probability of selecting bag }A=\text{Probability of selecting bag }B=\frac{1}{2}
\displaystyle \text{i.e. }P(A)=P(B)=\frac{1}{2}
\displaystyle \text{Let }W:\text{ Drawing a white ball from any bag.}
\displaystyle \therefore\ \text{Required probability}
\displaystyle =\text{Probability of selecting bag }A
\displaystyle \times\text{Probability of selecting a white ball from bag }A
\displaystyle +\text{Probability of selecting bag }B
\displaystyle \times\text{Probability of selecting a white ball from bag }B
\displaystyle =P(A)\times P(W|A)+P(B)\times P(W|B)
\displaystyle =\frac{1}{2}\times\frac{3}{5}+\frac{1}{2}\times\frac{2}{7}
\displaystyle =\frac{3}{10}+\frac{1}{7}=\frac{21+10}{70}=\frac{31}{70}
\displaystyle \text{Answer: }\frac{31}{70}
\\

\displaystyle \textbf{Question 66: } \text{The probability of } A, B \text{ and } C \text{ solving a problem are } \frac{1}{3}, \frac{2}{7} \text{ and } \frac{3}{8},
\displaystyle \text{ respectively. If all try and solve the } \text{problem simultaneously, find the probability that only}
\displaystyle \text{one of them will solve it. (ISC 2006)}
\displaystyle \text{Answer:}
\displaystyle \text{Let } A=\text{problem solved by } A
\displaystyle B=\text{problem solved by } B
\displaystyle C=\text{problem solved by } C
\displaystyle \text{Given, } P(A)=\frac{1}{3},\ P(B)=\frac{2}{7}\text{ and }P(C)=\frac{3}{8}
\displaystyle P(\text{only one of }A,B,C\text{ solves the problem})
\displaystyle =P(AB'C')+P(A'BC')+P(A'B'C)
\displaystyle =P(A)P(B')P(C')+P(A')P(B)P(C')+P(A')P(B')P(C)
\displaystyle =\frac{1}{3}\times\frac{5}{7}\times\frac{5}{8}+\frac{2}{3}\times\frac{2}{7}\times\frac{5}{8}+\frac{2}{3}\times\frac{5}{7}\times\frac{3}{8}
\displaystyle =\frac{25}{168}+\frac{20}{168}+\frac{30}{168}
\displaystyle =\frac{75}{168}=\frac{25}{56}
\\

\displaystyle \textbf{Question 67: } \text{A company has two plants which manufacture scooters. Plant I}
\displaystyle \text{manufactures } 80\% \text{ of the scooters while plant II manufactures } 20\% \text{ of the scooters.}
\displaystyle \text{At plant I, } 85 \text{ out of } 100 \text{ scooters are rated as being of standard quality,}
\displaystyle \text{while at plant II only } 65 \text{ out of } 100 \text{ scooters are rated as being of standard quality.}
\displaystyle \text{If a scooter is of standard quality, what is the probability that it came from plant I?}
\displaystyle \quad \text{ISC 2006}
\displaystyle \text{Answer:}
\displaystyle \text{Let }A=\text{Scooters manufactured by plant I}
\displaystyle B=\text{Scooters manufactured by plant II}
\displaystyle \text{Let }S\text{ represents the event that scooter manufactured by plants is of standard quality.}
\displaystyle \text{Given, plant I manufactured }=80\%\text{ scooters}
\displaystyle \text{Plant II manufactured }=20\%\text{ scooters}
\displaystyle \text{So, the probabilities of manufactured scooters by plants I and II are}
\displaystyle P(A)=\frac{80}{100}=0.8\text{ and }P(B)=\frac{20}{100}=0.2
\displaystyle \text{Now, the probabilities that the scooter of standard quality manufactured by plants I and II are}
\displaystyle P(S|A)=\frac{85}{100}=0.85\text{ and }P(S|B)=\frac{65}{100}=0.65
\displaystyle \text{The probability of selecting scooters with standard quality manufactures from plant I,}
\displaystyle P(A|S)=\frac{P(A)\cdot P(S|A)}{P(A)\cdot P(S|A)+P(B)\cdot P(S|B)}
\displaystyle \text{[by using Baye's theorem]}
\displaystyle =\frac{0.8\times0.85}{0.8\times0.85+0.2\times0.65}
\displaystyle =\frac{0.68}{0.68+0.13}=\frac{0.68}{0.81}=\frac{68}{81}=0.84
\displaystyle \text{Answer: }0.84
\\

\displaystyle \textbf{Question 68: } \text{Tickets numbered from } 1 \text{ to } 20 \text{ are mixed up together}
\displaystyle \text{ and then a ticket is drawn at random. What is the } \text{probability that the ticket has }
\displaystyle \text{a number which is a multiple of } 3 \text{ or } 7\text{? (ISC 2005)}
\displaystyle \text{Answer:}
\displaystyle \text{Given, tickets are numbered from }1\text{ to }20
\displaystyle A=\text{numbers on tickets which are multiples of }3=\{3,6,9,12,15,18\}
\displaystyle B=\text{numbers on tickets which are multiples of }7=\{7,14\}
\displaystyle A\cap B=\phi
\displaystyle P(\text{ticket has a number which is a multiple of }3\text{ or }7)
\displaystyle =P(A\cup B)=P(A)+P(B)-P(A\cap B)
\displaystyle =\frac{6}{20}+\frac{2}{20}-0
\displaystyle =\frac{8}{20}=\frac{2}{5}
\\

\displaystyle \textbf{Question 69: } \text{A manufacturing firm produces steel pipes in three plants } A,B \text{ and } C
\displaystyle \text{with daily production of } 500,1000 \text{ and } 2000 \text{ units, respectively. The fractions of defective}
\displaystyle \text{steel pipes output produced by the plants } A,B \text{ and } C \text{ are respectively } 0.005,0.008
\displaystyle  \text{ and } 0.010. \text{ If a pipe is selected from a day's total production and found to be defective}
\displaystyle \text{defective, then find } \text{out the probability that it came from the first plant.} \quad \text{ISC 2005}
\displaystyle
\displaystyle \text{Answer:}
\displaystyle \text{Given, per day production from plant }A=500\text{ units,}
\displaystyle \text{per day production from plant }B=1000\text{ units}
\displaystyle \text{and per day production from plant }C=2000\text{ units}
\displaystyle \text{Let the events, }E_1=\text{Selecting the pipe from plant }A
\displaystyle E_2=\text{Selecting the pipe from plant }B
\displaystyle E_3=\text{Selecting the pipe from plant }C
\displaystyle \text{Let }D\text{ be the event of selecting defective pipe.}
\displaystyle \text{Now, }P(E_1)=\frac{500}{500+1000+2000}=\frac{500}{3500}=\frac{1}{7}
\displaystyle P(E_2)=\frac{1000}{3500}=\frac{2}{7}\text{ and }P(E_3)=\frac{2000}{3500}=\frac{4}{7}
\displaystyle \text{Now, the probability that defective pipe is produced from plant }A,\ P(D|E_1)=0.005
\displaystyle \text{The probability that defective pipe is produced from plant }B,\ P(D|E_2)=0.008
\displaystyle \text{The probability that defective pipe is produced from plant }C,\ P(D|E_3)=0.010
\displaystyle \therefore\ \text{The probability that the selected pipe produced from plant }A\text{ is defective is,}
\displaystyle P(E_1|D)=\frac{P(E_1)\cdot P(D|E_1)}{P(E_1)\cdot P(D|E_1)+P(E_2)\cdot P(D|E_2)+P(E_3)\cdot P(D|E_3)}
\displaystyle \text{[by using Baye's theorem]}
\displaystyle =\frac{\frac{1}{7}\times0.005}{\frac{1}{7}\times0.005+\frac{2}{7}\times0.008+\frac{4}{7}\times0.010}
\displaystyle =\frac{\frac{1}{7}\times0.005}{\frac{1}{7}(0.005+2\times0.008+4\times0.010)}
\displaystyle =\frac{0.005}{0.005+0.016+0.040}=\frac{0.005}{0.061}=\frac{5}{61}
\displaystyle \text{Answer: }\frac{5}{61}
\\

\displaystyle \textbf{Question 70: } \text{What is the probability that a leap year has }  53 \text{ Sunday? (ISC 2004)}
\displaystyle \text{Answer:}
\displaystyle \text{In a leap year, there are }366\text{ days}
\displaystyle \text{i.e. }52\text{ full weeks }+2\text{ extra days}
\displaystyle \text{For these two days, there are seven possible combinations:}
\displaystyle \text{(i) Mon and Tue \quad (ii) Tue and Wed \quad (iii) Wed and Thu}
\displaystyle \text{(iv) Thu and Fri \quad (v) Fri and Sat \quad (vi) Sat and Sun}
\displaystyle \text{(vii) Sun and Mon}
\displaystyle \text{The last two combinations favour the event that there will be }53\text{ Sundays}
\displaystyle \therefore \text{Required probability}=\frac{2}{7}
\\

\displaystyle \textbf{Question 71: } \text{A bag has } 4 \text{ red and } 5 \text{ black, a second bag has } 3 \text{ red and 7 black balls. }
\displaystyle \text{One ball is drawn from first bag and two balls are drawn from second bag. Find }
\displaystyle \text{the probability } \text{that two balls are black and one is red. (ISC 2004)}
\displaystyle \text{Answer:}
\displaystyle \text{Let } B_{1}:\text{ first bag has }4\text{ red and }5\text{ black balls}
\displaystyle \text{and } B_{2}:\text{ second bag has }3\text{ red and }7\text{ black balls}
\displaystyle \text{Case I: 1 red ball from }B_{1}\text{ and 2 black balls from }B_{2}\text{ are drawn}
\displaystyle P_{1}=\frac{{}^{4}C_{1}}{{}^{9}C_{1}}\times\frac{{}^{7}C_{2}}{{}^{10}C_{2}}
\displaystyle =\frac{4}{9}\times\frac{21}{45}=\frac{84}{405}=\frac{168}{810}
\displaystyle \text{Case II: 1 black ball from }B_{1}\text{, and 1 red and 1 black ball from }B_{2}\text{ are drawn}
\displaystyle P_{2}=\frac{{}^{5}C_{1}}{{}^{9}C_{1}}\times\frac{{}^{3}C_{1}\times{}^{7}C_{1}}{{}^{10}C_{2}}
\displaystyle =\frac{5}{9}\times\frac{21}{45}=\frac{105}{405}=\frac{210}{810}
\displaystyle \text{Thus, cases I and II are mutually exclusive events}
\displaystyle \therefore \text{Probability of drawing 2 black balls and 1 red ball}
\displaystyle =P_{1}+P_{2}=\frac{168}{810}+\frac{210}{810}=\frac{378}{810}=\frac{7}{15}
\\

\displaystyle \textbf{Question 72: } \text{One number is chosen at random from the numbers } 1 \text{ to } 21. \text{ Find the}
\displaystyle \text{probability that it may be a prime } \text{number. (ISC 2003)}
\displaystyle \text{Answer:}
\displaystyle \text{The prime numbers from }1\text{ to }21\text{ are }\{2,3,5,7,11,13,17,19\}
\displaystyle \therefore \text{Probability of choosing a prime number}=\frac{8}{21}
\\

\displaystyle \textbf{Question 73: } \text{The probability that a contractor will get a plumbing } \text{contract is } \frac{2}{3}
\displaystyle \text{ and electric contract is } \frac{4}{9}. \text{If the probability of getting atleast one contract is } \frac{4}{5},
\displaystyle \text{ find the probability that he will get both the }  \text{contracts. (ISC 2003)}
\displaystyle \text{Answer:}
\displaystyle \text{Let } A:\text{ contractor gets plumbing contract, } B:\text{ contractor gets electric contract}
\displaystyle \text{Given, } P(A)=\frac{2}{3},\ P(B)=\frac{4}{9},\ P(A\cup B)=\frac{4}{5}
\displaystyle P(A\cup B)=P(A)+P(B)-P(A\cap B)
\displaystyle \frac{4}{5}=\frac{2}{3}+\frac{4}{9}-P(A\cap B)
\displaystyle P(A\cap B)=\frac{2}{3}+\frac{4}{9}-\frac{4}{5}
\displaystyle =\frac{30+20-36}{45}=\frac{14}{45}
\\

\displaystyle \textbf{Question 74: } \text{There are } 10 \text{ persons who are to be seated together around a circular table.}
\displaystyle \text{Find the probability that two } \text{particular person will always sit together. (ISC 2002)}
\displaystyle \text{Answer:}
\displaystyle \text{If }n\text{ objects are arranged in a circle, total arrangements }=(n-1)!
\displaystyle \text{Here }n=10\Rightarrow \text{total arrangements}=9!
\displaystyle \text{If two persons always sit together, treat them as one unit}
\displaystyle \Rightarrow \text{arrangements of }9\text{ units }=8!
\displaystyle \text{These two persons can interchange in }2\text{ ways}
\displaystyle \text{Favourable cases}=2\times 8!
\displaystyle \text{Required probability}=\frac{2\times 8!}{9!}=\frac{2}{9}
\\

\displaystyle \textbf{Question 75: } \text{A bag contains } 20 \text{ balls marked } 1 \text{ to } 20. \text{ One ball is drawn at random from}
\displaystyle \text{the bag. What is the probability } \text{that the ball drawn is marked with a number which is a}
\displaystyle \text{multiple of } 5 \text{ or } 7\text{? (ISC 2001)}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }20\text{ balls numbered }1\text{ to }20
\displaystyle A=\text{multiples of }5=\{5,10,15,20\},\ B=\text{multiples of }7=\{7,14\}
\displaystyle A\cap B=\phi
\displaystyle P(A\cup B)=P(A)+P(B)-P(A\cap B)
\displaystyle =\frac{4}{20}+\frac{2}{20}-0=\frac{6}{20}=\frac{3}{10}
\\

 

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