Question

A & B are sharp shooters whose probabilities of hitting a target are $\frac{9}{10}$ & $\frac{14}{15}$ respectively. If it is knownthat exactly one of them has hit the target, then the probability that it was hit by A is equal to

• A. $\frac{24}{55}$
• B. $\frac{27}{55}$
• C. $\frac{9}{23}$
• D. $\frac{10}{23}$

Answer 1

Answer: (C)

$\frac{9}{23}$

$E_1$ : only A hits the target
$E_2$ : only B hits the target
$E$ : exactly one hits the target.
$\therefore P\left(E_1/E\right)=\frac{P\left(E_1\right).P\left(E/E_1\right)}{P\left(E_1\right).P\left(E/E_1\right)+P\left(E_2\right).P\left(E/E_2\right)}$
$=\frac{\frac{9}{10}\times \frac{1}{15}}{\frac{9}{10}\times \frac{1}{15}+\frac{14}{15}\times \frac{1}{10}}=\frac{9}{23}$