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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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\frac{1}{2}

\frac{1}{3}

\frac{1}{6}

\frac{2}{3}

answer is A.

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Detailed Solution

Given that, the probability of hitting a target by two men is

\frac{1}{2}

and

\frac{1}{3}

respectively.
Let,

E_{1}

denotes the event when first man hits the target

E_{2}

denotes the event when second man hits the target, so,

P (E_{1}) = \frac{1}{2}, P (E_{2}) = \frac{1}{3}

{E^{_}}_{1}

is the probability that first man does not hit the target,

P ({E^{_}}_{1}) = 1 - P (E_{1})

P ({E^{_}}_{1}) = 1 - \frac{1}{2}

P ({E^{_}}_{1}) = \frac{1}{2}

{E^{_}}_{2}

is the probability that first man does not hit the target,

P ({E^{_}}_{2}) = 1 - P (E_{2})

P ({E^{_}}_{2}) = 1 - \frac{1}{3}

P ({E^{_}}_{2}) = \frac{2}{3}

Probability that exactly one of them hits the target,

P = P (E_{1}) P ({E^{_}}_{2}) + P (E_{2}) P ({E^{_}}_{1})

\Rightarrow \frac{1}{2} \times \frac{2}{3} + \frac{1}{3} \times \frac{1}{2}

\Rightarrow \frac{1}{2}

Therefore, option (1) is correct.

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