Question

In a combat, A targets B, and both $B$ and $C$ target $A$. The probabilities of A, B, C hitting their targets are $2 / 3,1 / 2$ and $1 / 3$ respectively. They shoot simultaneously and $A$ is hit. the probability that $B$ hits his target whereas $C$ does not

• A. $ \frac{1}{2}$
• B. $\frac{1}{3}$
• C. $\frac{1}{4}$
• D. $\frac{1}{6}$

Answer 1

Answer: (A)

$E_1: A \text { targets } B ; P\left(E_1\right)=\frac{2}{3}$
$E _2: B \operatorname{targets} A ; P \left( E _2\right)=\frac{1}{2}$
$E _3: C \text { targets } A ; P \left( E _3\right)=\frac{1}{3}$
Given $A$ is hit. Hence $H _1: B$ hits $A$ and $C$ does not $H _2$ : $C$ hits $A$ and $B$ does not $H _3: B$ and $C$ both hit $A$
$\therefore P \left( H _1\right)=\frac{1}{2} \cdot \frac{2}{3}=\frac{1}{3}$
$P \left( H _2\right)=\frac{1}{3} \cdot \frac{1}{2}=\frac{1}{6}$
$P \left( H _3\right)=\frac{1}{3} \cdot \frac{1}{2}=\frac{1}{3} \cdot \frac{1}{2}$
now, $P \left( H _1 / H _1 \cup H _2 \cup H _3\right)=\frac{\frac{1}{3}}{\frac{1}{3}+\frac{1}{6}+\frac{1}{6}}=\frac{1}{2}$