WAEC 2012 · Paper 2 · Q13

Probability & distributions

A hunter hits the target 3 times out of every five trials made. If 4 hunters aim at the target, calculate:

(a)
Calculate, correct to four significant figures, the probability that: (i) none of them hit the target; (ii) between 1 and 3 hunters inclusive hit the target; (iii) at least 2 hunters hit the target.
Separate values with commas, e.g. 3, −2
(b)
Given that the target is hit, find the probability that at most 3 hunters hit the target.

Worked solution (try it first)

The number of hunters who hit is binomial with $n = 4$ , $p = \frac35 = 0.6$ , $q = 0.4$ .
$P(0) = 0.4^4 = 0.0256$ , $P(1) = 4(0.6)(0.4)^3 = 0.1536$ , $P(4) = 0.6^4 = 0.1296$ .

(a)(i)

$P(\text{none}) = 0.02560$ .

(ii)

$P(1 \le X \le 3) = 1 - P(0) - P(4)$
$= 1 - 0.0256 - 0.1296$
$= 0.8448$ .

(iii)

$P(X \ge 2) = 1 - P(0) - P(1)$
$= 1 - 0.0256 - 0.1536$
$= 0.8208$ .

(b)

The target is hit means $X \ge 1$ : $P(X \ge 1) = 1 - 0.0256 = 0.9744$ .
At most 3 and at least 1 is $1 \le X \le 3$ , with probability $0.8448$ from (a)(ii).
Conditional probability: $\dfrac{0.8448}{0.9744} \approx 0.8670$ .

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