WAEC 2019 · Paper 2 · Q5

Probability & distributions

(a)
Three soldiers, $X$ , $Y$ and $Z$ , have probabilities $\frac13$ , $\frac15$ and $\frac14$ respectively of hitting a target. If each of them fires once, find, correct to two decimal places, the probability that only one of them hits the target.

Worked solution (try it first)

The miss chances are $\frac23$ , $\frac45$ and $\frac34$ .
Only $X$ : $\frac13 \cdot \frac45 \cdot \frac34 = \frac{12}{60}$ .
Only $Y$ : $\frac23 \cdot \frac15 \cdot \frac34 = \frac{6}{60}$ .
Only $Z$ : $\frac23 \cdot \frac45 \cdot \frac14 = \frac{8}{60}$ .
Add: $\frac{26}{60} = \frac{13}{30}$
$\approx 0.43$ .

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