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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?
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Verified Answer
The correct answer is:
$\frac{1}{2}$
$\begin{aligned} & \mathrm{P}=\mathrm{P}\left(\mathrm{E}_{1}\right) \mathrm{P}\left(\overline{\mathrm{E}}_{2}\right)+\mathrm{P}\left(\overline{\mathrm{E}}_{1}\right) \mathrm{P}\left(\mathrm{E}_{2}\right) \\ &=\frac{1}{2}\left(1-\frac{1}{3}\right)+\left(1-\frac{1}{2}\right) \frac{1}{3} \\ &=\frac{1}{2} \times \frac{2}{3}+\frac{1}{2} \times \frac{1}{3} \\ &=\frac{1}{3}+\frac{1}{6}=\frac{1}{2} \\ \therefore & \text { Option (a) is correct. } \end{aligned}$
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