Question

$A$ can hit a target $4$ times out of $5$ shots, $B$ thrice in $4$ shots and $C$ twice in $3$ shots, independent of each other. They fire a volley. Two shots hit the target. Then, the probability that it is $C$ who has missed the target, is

• A. $\frac{1}{5}$
• B. $\frac{13}{30}$
• C. $\frac{6}{13}$
• D. $\frac{2}{5}$

Answer 1

The correct option is C $\frac{6}{13}$

Given that
The chance of hitting by $A,P\left(A\right)=\frac{4}{5}$ and $P\left(\overline{A}\right)=\frac{1}{5}$
The chance of hitting by $B,P\left(B\right)=\frac{3}{4}$ and $P\left(\overline{B}\right)=\frac{1}{4}$
The chance of hitting by $C,P\left(C\right)=\frac{2}{3}$ and $P\left(\overline{C}\right)=\frac{1}{3}$

$P\left(\frac{A\text{and}B\text{hit the target}}{2\text{shots}\text{hit the target}}\right)$
$=\frac{P\left(A\right)\cdot P\left(B\right)\cdot P\left(\overline{C}\right)}{P\left(A\right)\cdot P\left(B\right)\cdot P\left(\overline{C}\right)+P\left(A\right)\cdot P\left(\overline{B}\right)\cdot P\left(C\right)+P\left(\overline{A}\right)\cdot P\left(B\right)\cdot P\left(C\right)}$
$=\frac{\frac{4}{5}\times \frac{3}{4}\times \frac{1}{3}}{\frac{4}{5}\times \frac{3}{4}\times \frac{1}{3}+\frac{4}{5}\times \frac{1}{4}\times \frac{2}{3}+\frac{1}{5}\times \frac{3}{4}\times \frac{2}{3}}$
$=\frac{12}{12+8+6}=\frac{6}{13}$