The probability that $A$ hit a target is $1 / 4$ and the probability that $B$ hits the target is $1 / 3$ . If each of them fired once, what is the probability that the target will be hit atleast once?

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Solution

$P$ = $A$ hits $B$ missed+ $B$ hits $A$ missed+Both hits
$\begin{matrix} = \frac{1}{4} \times \frac{2}{3} + \frac{1}{3} \times \frac{3}{4} + \frac{1}{3} \times \frac{1}{4} = \frac{2}{12} \times \frac{3}{12} \times \frac{1}{12} = \frac{1}{2} \end{matrix}$

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The probability that A hit a target is 1/4 and the probability that B hits the target is 1/3. If each of them fired once, what is the probability that the target will be hit atleast once?

P=A hits B missed+B hits A missed+Both hits=14×23+13×34+13×14=212×312×112=12

The probability that $A$ hit a target is $1 / 4$ and the probability that $B$ hits the target is $1 / 3$ . If each of them fired once, what is the probability that the target will be hit atleast once?

$P$ = $A$ hits $B$ missed+ $B$ hits $A$ missed+Both hits
$\begin{matrix} = \frac{1}{4} \times \frac{2}{3} + \frac{1}{3} \times \frac{3}{4} + \frac{1}{3} \times \frac{1}{4} = \frac{2}{12} \times \frac{3}{12} \times \frac{1}{12} = \frac{1}{2} \end{matrix}$