Question

If $A$ is targeting to $B,B$ and $C$ are targeting to $A.$ The probability of hitting the target by $A,B$ and $C$ are $\frac{2}{3},\frac{1}{2},\frac{1}{3}$ respectively. Then the probability that $B$ hits the target but $C$ does not is

• A. $\frac{4}{5}$
• B. $\frac{1}{6}$
• C. $\frac{1}{2}$
• D. $\frac{2}{3}$

Answer 1

The correct option is C $\frac{1}{2}$

Let probability of $A$ being hit $=P\left(E\right)$
$\Rightarrow P\left(E\right)=1-$ (Probability of $B$ missing) $\times$ (Probability of $C$ missing)
$\Rightarrow P\left(E\right)=1-\frac{1}{2}\cdot \frac{2}{3}=\frac{2}{3}$
Probability that $A$ is hit by $B$ but not by $C=P\left(\right(B\cap \overline{C}\left)|E\right)=\frac{P\left(B\right)\cdot P\left(\overline{C}\right)}{P\left(E\right)}$
$\Rightarrow$ Probability that $A$ is hit by $B$ but not by $C=\frac{\frac{1}{2}\cdot \frac{2}{3}}{\frac{2}{3}}=\frac{1}{2}$