\displaystyle \textbf{Question 1: }~\text{If }P(A)=\frac{7}{13},~P(B)=\frac{9}{13}\text{ and }P(A\cap B)=\frac{4}{13},\text{ find }P(A/B).
\displaystyle \text{Answer:}
\displaystyle \text{Given:}
\displaystyle P(A)=\frac{7}{13}
\displaystyle P(B)=\frac{9}{13}
\displaystyle P(A\cap B)=\frac{4}{13}
\displaystyle \text{Now,}
\displaystyle P(A\mid B)=\frac{P(A\cap B)}{P(B)}
\displaystyle \Rightarrow P(A\mid B)=\frac{\frac{4}{13}}{\frac{9}{13}}=\frac{4}{9}

\displaystyle \textbf{Question 2: }~\text{If }A\text{ and }B\text{ are events such that }P(A)=0.6,~P(B)=0.3\text{ and } \\ P(A\cap B)=0.2,\text{ find }P(A/B)\text{ and }P(B/A).
\displaystyle \text{Answer:}
\displaystyle \text{Given:}
\displaystyle P(A)=0.6
\displaystyle P(B)=0.3
\displaystyle P(A\cap B)=0.2
\displaystyle \text{Now,}
\displaystyle P(A\mid B)=\frac{P(A\cap B)}{P(B)}
\displaystyle \Rightarrow P(A\mid B)=\frac{0.2}{0.3}=\frac{2}{3}
\displaystyle P(B\mid A)=\frac{P(A\cap B)}{P(A)}
\displaystyle \Rightarrow P(B\mid A)=\frac{0.2}{0.6}=\frac{1}{3}

\displaystyle \textbf{Question 3: }~\text{If }A\text{ and }B\text{ are two events such that }P(A\cap B)=0.32\text{ and } \\ P(B)=0.5,\text{ find }P(A/B).
\displaystyle \text{Answer:}
\displaystyle \text{Given:}
\displaystyle P(B)=0.5
\displaystyle P(A\cap B)=0.32
\displaystyle \text{Now,}
\displaystyle P(A\mid B)=\frac{P(A\cap B)}{P(B)}
\displaystyle \Rightarrow P(A\mid B)=\frac{0.32}{0.5}=\frac{32}{50}=0.64

\displaystyle \textbf{Question 4: }~\text{If }P(A)=0.4,~P(B)=0.8,~P(B/A)=0.6.\text{ Find }P(A/B)\text{ and } \\ P(A\cup B).
\displaystyle \text{Answer:}
\displaystyle \text{Given:}
\displaystyle P(A)=0.4
\displaystyle P(B)=0.8
\displaystyle P(B\mid A)=0.6
\displaystyle \text{Now,}
\displaystyle P(B\mid A)=\frac{P(A\cap B)}{P(A)}
\displaystyle \Rightarrow 0.6=\frac{P(A\cap B)}{0.4}
\displaystyle \Rightarrow P(A\cap B)=0.24
\displaystyle P(A\mid B)=\frac{P(A\cap B)}{P(B)}=\frac{0.24}{0.8}=0.3
\displaystyle P(A\cup B)=P(A)+P(B)-P(A\cap B)
\displaystyle \Rightarrow P(A\cup B)=0.4+0.8-0.24=0.96

\displaystyle \textbf{Question 5: }~\text{If }A\text{ and }B\text{ are two events such that:} \\ \text{(i) }P(A)=\frac13,~P(B)=\frac14,~P(A\cup B)=\frac{5}{12}; \\ \text{ (ii) }P(A)=\frac6{11},~P(B)=\frac5{11},~P(A\cup B)=\frac7{11}; \\ \text{ (iii) }P(A)=\frac7{13},~P(B)=\frac9{13},~P(A\cap B)=\frac4{13}; \\ \text{ (iv) }P(A)=\frac12,~P(B)=\frac13,~P(A\cap B)=\frac14. \\ \text{ Find the required conditional probabilities.}
\displaystyle \text{Answer:}
\displaystyle \text{(i)  }
\displaystyle \text{We have,}
\displaystyle P(A)=\frac{1}{3},\ P(B)=\frac{1}{4}\text{ and }P(A\cup B)=\frac{5}{12}
\displaystyle \text{As, }P(A\cup B)=P(A)+P(B)-P(A\cap B)
\displaystyle \Rightarrow P(A\cap B)=P(A)+P(B)-P(A\cup B)
\displaystyle \Rightarrow P(A\cap B)=\frac{1}{3}+\frac{1}{4}-\frac{5}{12}
\displaystyle \Rightarrow P(A\cap B)=\frac{2}{12}
\displaystyle \Rightarrow P(A\cap B)=\frac{1}{6}
\displaystyle \text{Now,}
\displaystyle P(A\mid B)=\frac{P(A\cap B)}{P(B)}=\frac{\frac{1}{6}}{\frac{1}{4}}=\frac{4}{6}=\frac{2}{3}
\displaystyle P(B\mid A)=\frac{P(A\cap B)}{P(A)}=\frac{\frac{1}{6}}{\frac{1}{3}}=\frac{3}{6}=\frac{1}{2}

\displaystyle \text{(ii)  }
\displaystyle \text{We have,}
\displaystyle P(A)=\frac{6}{11},\ P(B)=\frac{5}{11}\text{ and }P(A\cup B)=\frac{7}{11}
\displaystyle \text{As, }P(A\cup B)=P(A)+P(B)-P(A\cap B)
\displaystyle \Rightarrow P(A\cap B)=P(A)+P(B)-P(A\cup B)
\displaystyle \Rightarrow P(A\cap B)=\frac{6}{11}+\frac{5}{11}-\frac{7}{11}
\displaystyle \Rightarrow P(A\cap B)=\frac{6+5-7}{11}
\displaystyle \Rightarrow P(A\cap B)=\frac{4}{11}
\displaystyle \text{Now,}
\displaystyle P(A\mid B)=\frac{P(A\cap B)}{P(B)}=\frac{\frac{4}{11}}{\frac{5}{11}}=\frac{4}{5}
\displaystyle P(B\mid A)=\frac{P(A\cap B)}{P(A)}=\frac{\frac{4}{11}}{\frac{6}{11}}=\frac{4}{6}=\frac{2}{3}

\displaystyle \text{(iii)  }
\displaystyle \text{We have,}
\displaystyle P(A)=\frac{7}{13},\ P(B)=\frac{9}{13}\text{ and }P(A\cap B)=\frac{4}{13}
\displaystyle \text{As, }P(\overline{A}\cap B)=P(B)-P(A\cap B)
\displaystyle \Rightarrow P(\overline{A}\cap B)=\frac{9}{13}-\frac{4}{13}
\displaystyle \Rightarrow P(\overline{A}\cap B)=\frac{9-4}{13}
\displaystyle \Rightarrow P(\overline{A}\cap B)=\frac{5}{13}
\displaystyle \text{Now,}
\displaystyle P(\overline{A}\mid B)=\frac{P(\overline{A}\cap B)}{P(B)}=\frac{\frac{5}{13}}{\frac{9}{13}}=\frac{5}{9}

\displaystyle \text{(iv)  }
\displaystyle \text{We have,}
\displaystyle P(A)=\frac{1}{2},\ P(B)=\frac{1}{3}\text{ and }P(A\cap B)=\frac{1}{4}
\displaystyle \text{Also, }P(\overline{B})=1-P(B)=1-\frac{1}{3}=\frac{2}{3}
\displaystyle \text{As, }P(A\cup B)=P(A)+P(B)-P(A\cap B)
\displaystyle =\frac{1}{2}+\frac{1}{3}-\frac{1}{4}
\displaystyle =\frac{6+4-3}{12}
\displaystyle \Rightarrow P(A\cup B)=\frac{7}{12}
\displaystyle \text{Also, }P(\overline{A}\cap B)=P(B)-P(A\cap B)
\displaystyle \Rightarrow P(\overline{A}\cap B)=\frac{1}{3}-\frac{1}{4}
\displaystyle \Rightarrow P(\overline{A}\cap B)=\frac{4-3}{12}
\displaystyle \Rightarrow P(\overline{A}\cap B)=\frac{1}{12}
\displaystyle \text{And, }P(\overline{A}\cap\overline{B})=1-P(A\cup B)
\displaystyle =1-\frac{7}{12}
\displaystyle =\frac{5}{12}
\displaystyle \text{Now,}
\displaystyle P(A\mid B)=\frac{P(A\cap B)}{P(B)}=\frac{\frac{1}{4}}{\frac{1}{3}}=\frac{3}{4}
\displaystyle P(B\mid A)=\frac{P(A\cap B)}{P(A)}=\frac{\frac{1}{4}}{\frac{1}{2}}=\frac{2}{4}=\frac{1}{2}
\displaystyle P(\overline{A}\mid B)=\frac{P(\overline{A}\cap B)}{P(B)}=\frac{\frac{1}{12}}{\frac{1}{3}}=\frac{3}{12}=\frac{1}{4}
\displaystyle P(\overline{A}\mid\overline{B})=\frac{P(\overline{A}\cap\overline{B})}{P(\overline{B})}=\frac{\frac{5}{12}}{\frac{2}{3}}=\frac{15}{24}=\frac{5}{8}

\displaystyle \textbf{Question 6: }~\text{If }A\text{ and }B\text{ are two events such that }2P(A)=P(B)=\frac5{13} \\ \text{ and }P(A/B)=\frac25,\text{ find }P(A\cup B).
\displaystyle \text{Answer:}
\displaystyle \text{Given:}
\displaystyle 2P(A)=P(B)=\frac{5}{13}
\displaystyle P(A\mid B)=\frac{2}{5}
\displaystyle \therefore P(A)=\frac{5}{26}
\displaystyle P(B)=\frac{5}{13}
\displaystyle \text{Now,}
\displaystyle P(A\mid B)=\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}{5}\times\frac{5}{13}=\frac{2}{13}
\displaystyle P(A\cup B)=P(A)+P(B)-P(A\cap B)
\displaystyle =\frac{5}{26}+\frac{5}{13}-\frac{2}{13}
\displaystyle \Rightarrow P(A\cup B)=\frac{11}{26}

\displaystyle \textbf{Question 7: }~\text{If }P(A)=\frac6{11},~P(B)=\frac5{11}\text{ and }P(A\cup B)=\frac7{11}, \\ \text{ find (i) }P(A\cap B),\text{ (ii) }P(A/B),\text{ (iii) }P(B/A).
\displaystyle \text{Answer:}
\displaystyle \text{Given:}
\displaystyle P(A)=\frac{6}{11}
\displaystyle P(B)=\frac{5}{11}
\displaystyle P(A\cup B)=\frac{7}{11}
\displaystyle \text{(i) }P(A\cup B)=P(A)+P(B)-P(A\cap B)
\displaystyle \Rightarrow \frac{7}{11}=\frac{6}{11}+\frac{5}{11}-P(A\cap B)
\displaystyle \Rightarrow P(A\cap B)=\frac{6}{11}+\frac{5}{11}-\frac{7}{11}=\frac{4}{11}
\displaystyle \text{(ii) }P(A\mid B)=\frac{P(A\cap B)}{P(B)}=\frac{\frac{4}{11}}{\frac{5}{11}}=\frac{4}{5}
\displaystyle \text{(iii) }P(B\mid A)=\frac{P(A\cap B)}{P(A)}=\frac{\frac{4}{11}}{\frac{6}{11}}=\frac{4}{6}=\frac{2}{3}

\displaystyle \textbf{Question 8: }~\text{A coin is tossed three times. Find }P(A/B)\text{ where: (i) }A=\text{Head on third toss, } \\ B=\text{Head on first two tosses; (ii) }A=\text{At least two heads, }B=\text{At most two heads; (iii) }A=\text{At most two tails, }B=\text{At least one tail.}
\displaystyle \text{Answer:}
\displaystyle \text{(i)  }
\displaystyle \text{Consider the given events.}
\displaystyle A=\text{Heads on third toss}
\displaystyle B=\text{Heads on first two tosses}
\displaystyle \text{Clearly,}
\displaystyle A=\{(H,H,H),(H,T,H),(T,H,H),(T,T,H)\}
\displaystyle B=\{(H,H,H),(H,H,T)\}
\displaystyle \text{Now,}
\displaystyle A\cap B=\{(H,H,H)\}
\displaystyle \therefore\ \text{Required probability}=P(A\mid B)=\frac{n(A\cap B)}{n(B)}=\frac{1}{2}

\displaystyle \text{(ii)  }
\displaystyle \text{Consider the given events.}
\displaystyle A=\text{At least two heads}
\displaystyle B=\text{At most two heads}
\displaystyle \text{Clearly,}
\displaystyle A=\{ (H,H,H),(H,T,H),(T,H,H),(H,H,T) \}
\displaystyle B=\{ (T,T,T),(H,T,H),(T,H,H),(H,H,T),(T,H,T),(H,T,T) \}
\displaystyle \text{Now,}
\displaystyle A\cap B=\{(H,T,H),(T,H,H),(H,H,T)\}
\displaystyle \therefore\ \text{Required probability}=P(A\mid B)=\frac{n(A\cap B)}{n(B)}=\frac{3}{6}=\frac{1}{2}

\displaystyle \text{(iii)  }
\displaystyle \text{Consider the given events.}
\displaystyle A=\text{At most two tails}
\displaystyle B=\text{At least one tail}
\displaystyle \text{Clearly,}
\displaystyle A=\{(T,T,H),(T,H,H),(H,H,T),(T,H,T),(H,T,T),(H,H,H),(T,T,T)\}\setminus\{(T,T,T) \}
\displaystyle A=\{(T,T,H),(T,H,H),(H,H,T),(T,H,T),(H,T,T),(H,H,H)\}
\displaystyle B=\{(T,T,T),(T,T,H),(T,H,H),(H,H,T),(T,H,T),(H,T,T),(H,H,H)\}\setminus\{(H,H,H) \}
\displaystyle B=\{(T,T,T),(T,T,H),(T,H,H),(H,H,T),(T,H,T),(H,T,T)\}
\displaystyle \text{Now,}
\displaystyle A\cap B=\{(T,T,H),(T,H,H),(H,H,T),(T,H,T),(H,T,T)\}
\displaystyle \therefore\ \text{Required probability}=P(A\mid B)=\frac{n(A\cap B)}{n(B)}=\frac{6}{7}

\displaystyle \textbf{Question 9: }~\text{Two coins are tossed once. Find }P(A/B)\text{ where: (i) }A=\text{Tail on one coin, } \\ B=\text{One coin shows head; (ii) }A=\text{No tail appears, }B=\text{No head appears.}
\displaystyle \text{Answer:}
\displaystyle \text{(i)  }
\displaystyle \text{Consider the given events.}
\displaystyle A=\text{Tail appears on one coin}
\displaystyle B=\text{One coin shows head}
\displaystyle \text{Clearly,}
\displaystyle A=\{(H,T),(T,H)\}
\displaystyle B=\{(H,T),(T,H)\}
\displaystyle \text{Now,}
\displaystyle A\cap B=\{(H,T),(T,H)\}
\displaystyle \therefore\ \text{Required probability}=P(A\mid B)=\frac{n(A\cap B)}{n(B)}=\frac{2}{2}=1

\displaystyle \text{(ii)  }
\displaystyle \text{Consider the given events.}
\displaystyle A=\text{No tail appears}
\displaystyle B=\text{No head appears}
\displaystyle \text{Clearly,}
\displaystyle A=\{(H,H)\}
\displaystyle B=\{(T,T)\}
\displaystyle \text{Now,}
\displaystyle A\cap B=\varnothing
\displaystyle \therefore\ \text{Required probability}=P(A\mid B)=\frac{n(A\cap B)}{n(B)}=\frac{0}{1}=0

\displaystyle \textbf{Question 10: }~\text{A die is thrown three times. Find }P(A/B)\text{ and }P(B/A)\text{ if }A=\text{4 on third toss, }B=\text{6 and 5 on first two tosses.}
\displaystyle \text{Answer:}
\displaystyle \text{Consider the given events.}
\displaystyle A=\text{Getting }4\text{ on third throw}
\displaystyle B=\text{Getting }6\text{ on first throw and }5\text{ on second throw}
\displaystyle \text{Clearly,}
\displaystyle A=\{(1,1,4),(1,2,4),(1,3,4),(1,4,4),(1,5,4),(1,6,4),(2,1,4),(2,2,4),(2,3,4),(2,4,4), \\ (2,5,4),(2,6,4),(3,1,4),(3,2,4),(3,3,4),(3,4,4),(3,5,4),(3,6,4),(4,1,4),(4,2,4), \\ (4,3,4),(4,4,4),(4,5,4),(4,6,4),(5,1,4),(5,2,4),(5,3,4),(5,4,4),(5,5,4),(5,6,4), \\ (6,1,4),(6,2,4),(6,3,4),(6,4,4),(6,5,4),(6,6,4)\}
\displaystyle B=\{(6,5,1),(6,5,2),(6,5,3),(6,5,4),(6,5,5),(6,5,6)\}
\displaystyle \text{Now,}
\displaystyle A\cap B=\{(6,5,4)\}
\displaystyle \therefore\ \text{Required probability}=P(A\mid B)=\frac{n(A\cap B)}{n(B)}=\frac{1}{6}
\displaystyle \therefore\ \text{Required probability}=P(B\mid A)=\frac{n(A\cap B)}{n(A)}=\frac{1}{36}

\displaystyle \textbf{Question 11: }~\text{Mother, father and son line up at random. If }A=\text{Son on one end, }B=\text{Father in the middle, find }P(A/B)\text{ and }P(B/A).
\displaystyle \text{Answer:}
\displaystyle \text{Consider the given events.}
\displaystyle A=\text{Son standing on one end}
\displaystyle B=\text{Father standing in the middle}
\displaystyle \text{Clearly,}
\displaystyle S=\{MFS,MSF,FSM,FMS,SMF,SFM\}
\displaystyle A=\{MFS,FMS,SMF,SFM\}
\displaystyle B=\{MFS,SFM\}
\displaystyle \text{Now,}
\displaystyle A\cap B=\{MFS,SFM\}
\displaystyle (i)\ \text{Required probability}=P(A\mid B)=\frac{n(A\cap B)}{n(B)}=\frac{2}{2}=1
\displaystyle (ii)\ \text{Required probability}=P(B\mid A)=\frac{n(A\cap B)}{n(A)}=\frac{2}{4}=\frac{1}{2}

\displaystyle \textbf{Question 12: }~\text{Two dice are thrown and the sum is }6.\text{ Find the conditional probability} \\ \text{that }4\text{ has appeared at least once.}
\displaystyle \text{Answer:}
\displaystyle \text{Consider the given events.}
\displaystyle A=\text{4 appears on one die}
\displaystyle B=\text{The sum of the numbers on two dice is }6
\displaystyle \text{Clearly,}
\displaystyle A=\{(1,4),(2,4),(3,4),(4,1),(4,2),(4,3),(4,4),(4,5),(4,6),(5,4),(6,4)\}
\displaystyle B=\{(1,5),(5,1),(2,4),(4,2),(3,3)\}
\displaystyle \text{Now,}
\displaystyle A\cap B=\{(2,4),(4,2)\}
\displaystyle \therefore\ \text{Required probability}=P(A\mid B)=\frac{n(A\cap B)}{n(B)}=\frac{2}{5}

\displaystyle \textbf{Question 13: }~\text{Two dice are thrown. Find the probability that the sum is }8\text{ if the} \\ \text{second die always shows }4.
\displaystyle \text{Answer:}
\displaystyle \text{Consider the given events.}
\displaystyle A=\text{4 appears on second die}
\displaystyle B=\text{The sum of the numbers on two dice is }8
\displaystyle \text{Clearly,}
\displaystyle A=\{(1,4),(2,4),(3,4),(4,4),(5,4),(6,4)\}
\displaystyle B=\{(4,4),(3,5),(5,3),(2,6),(6,2)\}
\displaystyle \text{Now,}
\displaystyle A\cap B=\{(4,4)\}
\displaystyle \therefore\ \text{Required probability}=P(B\mid A)=\frac{n(A\cap B)}{n(A)}=\frac{1}{6}

\displaystyle \textbf{Question 14: }~\text{A pair of dice is thrown. Find the probability of getting }7\text{ as the} \\ \text{sum if the second die always shows an odd number.}
\displaystyle \text{Answer:}
\displaystyle \text{Consider the given events.}
\displaystyle A=\text{Number appearing on second die is odd}
\displaystyle B=\text{The sum of the numbers on two dice is }7
\displaystyle \text{Clearly,}
\displaystyle A=\{(1,1),(1,3),(1,5),(2,1),(2,3),(2,5),(3,1),(3,3),(3,5),(4,1),(4,3),(4,5),(5,1),(5,3), \\ (5,5),(6,1),(6,3),(6,5)\}
\displaystyle B=\{(2,5),(5,2),(3,4),(4,3),(1,6),(6,1)\}
\displaystyle \text{Now,}
\displaystyle A\cap B=\{(2,5),(4,3),(6,1)\}
\displaystyle \therefore\ \text{Required probability}=P(B\mid A)=\frac{n(A\cap B)}{n(A)}=\frac{3}{18}=\frac{1}{6}

\displaystyle \textbf{Question 15: }~\text{A pair of dice is thrown. Find the probability of getting }7\text{ as the} \\ \text{sum if the second die always shows a prime number.}
\displaystyle \text{Answer:}
\displaystyle \text{Consider the given events}
\displaystyle A=\text{A prime number appears on second die}
\displaystyle B=\text{The sum of the numbers on two dice is }7
\displaystyle \text{Clearly,}
\displaystyle A=\{(1,2),(1,3),(1,5),(2,2),(2,3),(2,5),(3,2),(3,3),(3,5),(4,2),(4,3),(4,5), \\ (5,2),(5,3),(5,5),(6,2),(6,3),(6,5)\}
\displaystyle B=\{(2,5),(5,2),(3,4),(4,3),(1,6),(6,1)\}
\displaystyle \text{Now,}
\displaystyle A\cap B=\{(2,5),(5,2),(4,3)\}
\displaystyle \therefore\ \text{Required probability}=P(B\mid A)=\frac{n(A\cap B)}{n(A)}=\frac{3}{18}=\frac{1}{6}

\displaystyle \textbf{Question 16: }~\text{A die is rolled. If the outcome is odd, what is the probability that} \\ \text{it is prime?}
\displaystyle \text{Answer:}
\displaystyle \text{Consider the given events.}
\displaystyle A=\text{The number is odd}
\displaystyle B=\text{The number is prime}
\displaystyle \text{Clearly,}
\displaystyle A=\{1,3,5\}
\displaystyle B=\{2,3,5\}
\displaystyle \text{Now,}
\displaystyle A\cap B=\{3,5\}
\displaystyle \therefore\ \text{Required probability}=P(B\mid A)=\frac{n(A\cap B)}{n(A)}=\frac{2}{3}

\displaystyle \textbf{Question 17: }~\text{A pair of dice is thrown. Find the probability that the sum is } \\ 8\text{ or more if }4\text{ appears on the first die.}
\displaystyle \text{Answer:}
\displaystyle \text{Consider the given events.}
\displaystyle A=\text{4 appears on first die}
\displaystyle B=\text{The sum of the numbers on two dice is }8\text{ or more}
\displaystyle \text{Clearly,}
\displaystyle A=\{(4,1),(4,2),(4,3),(4,4),(4,5),(4,6)\}
\displaystyle n(A)=6
\displaystyle B=\{(2,6),(3,5),(3,6),(4,4),(4,5),(4,6),(5,3),(5,4),(5,5),(5,6),(6,2),(6,3), \\ (6,4),(6,5),(6,6)\}
\displaystyle n(B)=15
\displaystyle \text{Now,}
\displaystyle A\cap B=\{(4,4),(4,5),(4,6)\}
\displaystyle \therefore\ \text{Required probability}=P(B\mid A)=\frac{n(A\cap B)}{n(A)}=\frac{3}{6}=\frac{1}{2}

\displaystyle \textbf{Question 18: }~\text{Find the probability that the sum on two dice is }8\text{ given that at least} \\ \text{one die does not show }5.
\displaystyle \text{Answer:}
\displaystyle \text{Consider the given events.}
\displaystyle A=\text{At least one die does not show }5
\displaystyle B=\text{The sum of the numbers on two dice is }8
\displaystyle \text{Clearly,}
\displaystyle A=\{(1,1),(1,2),(1,3),(1,4),(1,6),(2,1),(2,2),(2,3),(2,4),(2,6),(3,1),(3,2),(3,3), \\ (3,4),(3,6),(4,1),(4,2),(4,3),(4,4),(4,6),(6,1),(6,2),(6,3),(6,4),(6,6)\}
\displaystyle B=\{(2,6),(6,2),(3,5),(5,3),(4,4)\}
\displaystyle \text{Now,}
\displaystyle A\cap B=\{(4,4),(6,2),(2,6)\}
\displaystyle \therefore\ \text{Required probability}=P(B\mid A)=\frac{n(A\cap B)}{n(A)}=\frac{3}{25}

\displaystyle \textbf{Question 19: }~\text{Two numbers are selected from }1\text{ to }9.\text{ If the sum is even, find the} \\ \text{probability that both are odd.}
\displaystyle \text{Answer:}
\displaystyle \text{Suppose }O\text{ represents the event of getting two odd numbers and }S\text{ represents the event} \\ \text{of getting their sum as an even number.}
\displaystyle \text{Now,}
\displaystyle P(O\mid S)=\frac{P(O\cap S)}{P(S)}
\displaystyle =\frac{\binom{5}{2}}{\binom{4}{2}+\binom{5}{2}}
\displaystyle =\frac{10}{16}=\frac{5}{8}

\displaystyle \textbf{Question 20: }~\text{A die is thrown twice and the sum is }8.\text{ Find the conditional probability} \\ \text{that }5\text{ has appeared at least once.}
\displaystyle \text{Answer:}
\displaystyle \text{Consider the given events.}
\displaystyle A=\text{5 appears on the die at least once}
\displaystyle B=\text{The sum of the numbers on two dice is }8
\displaystyle \text{Clearly,}
\displaystyle A=\{(1,5),(2,5),(3,5),(4,5),(5,5),(6,5),(5,1),(5,2),(5,3),(5,4),(5,6)\}
\displaystyle B=\{(2,6),(3,5),(4,4),(5,3),(6,2)\}
\displaystyle \text{Now,}
\displaystyle A\cap B=\{(3,5),(5,3)\}
\displaystyle \therefore\ \text{Required probability}=P(A\mid B)=\frac{n(A\cap B)}{n(B)}=\frac{2}{5}

\displaystyle \textbf{Question 21: }~\text{Two dice are thrown and the first shows }6.\text{ Find the probability} \\ \text{that the sum is }7.
\displaystyle \text{Answer:}
\displaystyle \text{Consider the given events.}
\displaystyle A=\text{First die shows }6
\displaystyle B=\text{The sum of the numbers on two dice is }7
\displaystyle \text{Clearly,}
\displaystyle A=\{(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)\}
\displaystyle B=\{(2,5),(5,2),(4,3),(3,4),(1,6),(6,1)\}
\displaystyle \text{Now,}
\displaystyle A\cap B=\{(6,1)\}
\displaystyle \therefore\ \text{Required probability}=P(B\mid A)=\frac{n(A\cap B)}{n(A)}=\frac{1}{6}

\displaystyle \textbf{Question 22: }~\text{Let }E=\{\text{sum}\ge10\}\text{ and }F=\{\text{5 on first die}\}.\text{ Find }P(E/F).\text{ If }F=\{\text{5 on at least one die}\},\text{ find }P(E/F).
\displaystyle \text{Answer:}
\displaystyle \text{Consider the given events.}
\displaystyle E=\text{The sum of the numbers on two dice is }10\text{ or more}
\displaystyle F=\text{5 appears on first die}
\displaystyle \text{Clearly,}
\displaystyle E=\{(4,6),(5,5),(5,6),(6,4),(6,5),(6,6)\}
\displaystyle F=\{(5,1),(5,2),(5,3),(5,4),(5,5),(5,6)\}
\displaystyle \text{Now,}
\displaystyle E\cap F=\{(5,5),(5,6)\}
\displaystyle \therefore\ \text{Required probability}=P(E\mid F)=\frac{n(E\cap F)}{n(F)}=\frac{2}{6}=\frac{1}{3}
\displaystyle \text{Second case:}
\displaystyle \text{Consider the given events.}
\displaystyle E=\text{The sum of the numbers on two dice is }10\text{ or more}
\displaystyle F=\text{5 appears on a die at least once}
\displaystyle \text{Clearly,}
\displaystyle E=\{(4,6),(5,5),(5,6),(6,4),(6,5),(6,6)\}
\displaystyle F=\{(1,5),(2,5),(3,5),(4,5),(5,5),(6,5),(5,1),(5,2),(5,3),(5,4),(5,6)\}
\displaystyle \text{Now,}
\displaystyle E\cap F=\{(5,5),(5,6),(6,5)\}
\displaystyle \therefore\ \text{Required probability}=P(E\mid F)=\frac{n(E\cap F)}{n(F)}=\frac{3}{11}

\displaystyle \textbf{Question 23: }~\text{A student passes Mathematics with probability }\frac45\text{ and both subjects} \\ \text{with probability }\frac12.\text{ Find }P(\text{Computer Science}\mid\text{Mathematics}).
\displaystyle \text{Answer:}
\displaystyle \text{Consider the given events.}
\displaystyle M=\text{Students pass Mathematics}
\displaystyle C=\text{Students pass Computer Science}
\displaystyle \text{We have,}
\displaystyle P(M)=\frac{4}{5}
\displaystyle P(M\cap C)=\frac{1}{2}
\displaystyle \text{Now,}
\displaystyle P(C\mid M)=\frac{P(M\cap C)}{P(M)}
\displaystyle =\frac{\frac{1}{2}}{\frac{4}{5}}=\frac{5}{8}

\displaystyle \textbf{Question 24: }~\text{A person buys a shirt with probability }0.2,\text{ trousers }0.3,\text{ and }P(\text{shirt}\mid\text{trousers})=0.4.\text{ Find the probabilities required.}
\displaystyle \text{Answer:}
\displaystyle \text{Suppose }S\text{ represents the event of buying a shirt and }T\text{ represents the event of buying a trouser.}
\displaystyle \text{We have,}
\displaystyle P(S)=0.2
\displaystyle P(T)=0.3
\displaystyle P(S\mid T)=0.4
\displaystyle \text{Now,}
\displaystyle P(S\mid T)=\frac{P(S\cap T)}{P(T)}
\displaystyle \Rightarrow P(S\cap T)=P(S\mid T)\times P(T)=0.4\times0.3=0.12
\displaystyle P(T\mid S)=\frac{P(S\cap T)}{P(S)}=\frac{0.12}{0.2}=0.6

\displaystyle \textbf{Question 25: }~\text{Out of }1000\text{ students, }430\text{ are girls and }10\%\text{ of them study in class} \\ \text{XII. Find the conditional probability.}
\displaystyle \text{Answer:}
\displaystyle \text{Suppose }S\text{ represents a student chosen randomly studying in class XII and }G\text{ represents a female student chosen randomly.}
\displaystyle \text{We have,}
\displaystyle P(G)=\frac{430}{1000}
\displaystyle P(S\cap G)=\frac{43}{1000}
\displaystyle \text{Now,}
\displaystyle P(S\mid G)=\frac{P(S\cap G)}{P(G)}
\displaystyle =\frac{\frac{43}{1000}}{\frac{430}{1000}}=\frac{1}{10}

\displaystyle \textbf{Question 26: }~\text{Ten cards numbered }1\text{ to }10\text{ are mixed and one is drawn. Given} \\ \text{it is greater than }3,\text{ find the probability that it is even.}
\displaystyle \text{Answer:}
\displaystyle \text{Sample space},\ S=\{1,2,3,4,5,6,7,8,9,10\}
\displaystyle \text{Consider the given events.}
\displaystyle A=\text{An even number on the card}
\displaystyle B=\text{A number more than }3\text{ on the card}
\displaystyle \text{Clearly,}
\displaystyle A=\{2,4,6,8,10\}
\displaystyle B=\{4,5,6,7,8,9,10\}
\displaystyle \text{Now,}
\displaystyle A\cap B=\{4,6,8,10\}
\displaystyle \therefore\ \text{Required probability}=P(A\mid B)=\frac{n(A\cap B)}{n(B)}=\frac{4}{7}

\displaystyle \textbf{Question 27: }~\text{Each child is equally likely to be a boy or girl. If a family has} \\ \text{two children, find the probability that both are girls given (i) the youngest is a girl,} \\ \text{(ii) at least one is a girl.}
\displaystyle \text{Answer:}
\displaystyle \text{Consider the given events.}
\displaystyle A=\text{Both the children are girls.}
\displaystyle B=\text{The youngest child is a girl.}
\displaystyle C=\text{At least one child is a girl.}
\displaystyle \text{Clearly,}
\displaystyle S=\{B_1B_2,B_1G_2,G_1B_2,G_1G_2\}
\displaystyle A=\{G_1G_2\}
\displaystyle B=\{B_1G_2,G_1G_2\}
\displaystyle C=\{B_1G_2,G_1B_2,G_1G_2\}
\displaystyle A\cap B=\{G_1G_2\}\text{ and }A\cap C=\{G_1G_2\}
\displaystyle (i)\ \text{Required probability}=P(A\mid B)=\frac{n(A\cap B)}{n(B)}=\frac{1}{2}
\displaystyle (ii)\ \text{Required probability}=P(A\mid C)=\frac{n(A\cap C)}{n(C)}=\frac{1}{3}

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