Question

Two men hit at a target with probabilities $\frac{1}{2}$ and $\frac{1}{3}$ respectively. What is the probability that exactly one of them hits the target?

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

Answer 1

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

$E_1$ denotes the event when first man hits the target

$E_2$ denotes the event when second man hits the target

$P\left(E_1\right)=\frac{1}{2},P\left(E_2\right)=\frac{1}{3}$

Probability that exactly one of them hits the target, $P=P\left(E_1\right)P\left({\stackrel{\_}{E}}_2\right)+P\left(E_2\right)P\left({\stackrel{\_}{E}}_1\right)$

$=\frac{1}{2}\times \frac{2}{3}+\frac{1}{3}\times \frac{1}{2}=\frac{1}{2}$