\displaystyle \textbf{Question 1: }~\text{Ten cards numbered }1\text{ through }10\text{ are placed in a box and mixed} \\ \text{thoroughly. One card is drawn at random. If it is known that the number on the} \\ \text{drawn card is more than }3,\text{ find the probability that it is an even number.}
\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{Even number appears on the card}
\displaystyle B=\text{A number which is more than }3\text{ appears on the card}
\displaystyle \text{Here,}
\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 2: }~\text{Assume that each child born is equally likely to be a boy or a girl.} \\ \text{If a family has two children, find the conditional probability that both are girls given} \\ \text{that (i) the youngest is a girl, (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}

\displaystyle \textbf{Question 3: }~\text{Given that the two numbers appearing on throwing two dice are } \\ \text{different, find the probability that the sum of the numbers on the dice is }4.
\displaystyle \text{Answer:}
\displaystyle \text{Consider the given events.}
\displaystyle A=\text{Numbers appearing on two dice are different}
\displaystyle B=\text{The sum of the numbers on two dice is }4
\displaystyle \text{Clearly,}
\displaystyle A=\{(1,2),(1,3),(1,4),(1,5),(1,6),(2,1),(2,3),(2,4),(2,5),(2,6),(3,1), (3,2),(3,4), \\ (3,5),(3,6),(4,1),(4,2),(4,3),(4,5),(4,6),(5,1),(5,2), (5,3),(5,4),(5,6),(6,1),(6,2), \\ (6,3), (6,4),(6,5)\}
\displaystyle B=\{(1,3),(3,1),(2,2)\}
\displaystyle \text{Now,}
\displaystyle A\cap B=\{(1,3),(3,1)\}
\displaystyle \therefore\ \text{Required probability}=P(B\mid A)=\frac{n(A\cap B)}{n(A)}=\frac{2}{30}=\frac{1}{15}

\displaystyle \textbf{Question 4: }~\text{A coin is tossed three times. If head occurs on the first two tosses, } \\ \text{find the probability of getting head on the third toss.}
\displaystyle \text{Answer:}
\displaystyle \text{Consider the given events.}
\displaystyle A=\text{Getting head on third toss}
\displaystyle B=\text{Getting head 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 \textbf{Question 5: }~\text{A die is thrown three times. Find the probability that }4\text{ appears} \\ \text{on the third toss if it is given that }6\text{ and }5\text{ appear respectively on the 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 \textbf{Question 6: }~\text{Compute }P(A/B),\text{ if }P(B)=0.5\text{ and }P(A\cap B)=0.32.
\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}=\frac{16}{25}

\displaystyle \textbf{Question 7: }~\text{If }P(A)=0.4,\ P(B)=0.3\text{ and }P(B/A)=0.5,\text{ find }P(A\cap B)\text{ and }P(A/B).
\displaystyle \text{Answer:}
\displaystyle \text{Given:}
\displaystyle P(A)=0.4
\displaystyle P(B)=0.3
\displaystyle P(B\mid A)=0.5
\displaystyle \text{Now,}
\displaystyle P(B\mid A)=\frac{P(A\cap B)}{P(A)}
\displaystyle \Rightarrow 0.5=\frac{P(A\cap B)}{0.4}
\displaystyle \Rightarrow P(A\cap B)=0.2
\displaystyle P(A\mid B)=\frac{P(A\cap B)}{P(B)}=\frac{0.2}{0.3}=\frac{2}{3}

\displaystyle \textbf{Question 8: }~\text{If }A\text{ and }B\text{ are two events such that }P(A)=\frac{1}{3},\ P(B)=\frac{1}{5}\text{ and }P(A\cup B)=\frac{11}{30},\text{ find }P(A/B)\text{ and }P(B/A).
\displaystyle \text{Answer:}
\displaystyle \text{Given:}
\displaystyle P(A)=\frac{1}{3}
\displaystyle P(B)=\frac{1}{5}
\displaystyle P(A\cup B)=\frac{11}{30}
\displaystyle \text{Now,}
\displaystyle P(A\cup B)=P(A)+P(B)-P(A\cap B)
\displaystyle \Rightarrow \frac{11}{30}=\frac{1}{3}+\frac{1}{5}-P(A\cap B)
\displaystyle \Rightarrow P(A\cap B)=\frac{1}{3}+\frac{1}{5}-\frac{11}{30}=\frac{10+6-11}{30}=\frac{5}{30}=\frac{1}{6}
\displaystyle P(A\mid B)=\frac{P(A\cap B)}{P(B)}=\frac{\frac{1}{6}}{\frac{1}{5}}=\frac{5}{6}
\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 \textbf{Question 9: }~\text{A couple has two children. Find the probability that both the} \\ \text{children are (i) males, if it is known that at least one of the children is male,} \\ \text{(ii) females, if it isknown that the elder child is a female.}
\displaystyle \text{Answer:}
\displaystyle \text{Consider the given events.}
\displaystyle A=\text{Both the children are female.}
\displaystyle B=\text{The elder child is a female.}
\displaystyle C=\text{At least one child is a male.}
\displaystyle D=\text{Both children are male.}
\displaystyle \text{Clearly,}
\displaystyle S=\{M_1M_2,M_1F_2,F_1M_2,F_1F_2\}
\displaystyle A=\{F_1F_2\}
\displaystyle B=\{F_1M_2,F_1F_2\}
\displaystyle C=\{M_1M_2,M_1F_2,F_1M_2\}
\displaystyle D=\{M_1M_2\}
\displaystyle \text{Here, first child is elder and second is younger.}
\displaystyle D\cap C=\{M_1M_2\}\text{ and }A\cap B=\{F_1F_2\}
\displaystyle (i)\ \text{Required probability}=P(D\mid C)=\frac{n(D\cap C)}{n(C)}=\frac{1}{3}
\displaystyle (ii)\ \text{Required probability}=P(A\mid B)=\frac{n(A\cap B)}{n(B)}=\frac{1}{2}

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