WAEC 2020 · Paper 2 · Q5✱

Probability & distributions

Thirty percent ( $30\%$ ) of the bulbs in a large box are defective. If $12$ bulbs are selected randomly from the box, calculate the probability that:

(a)
exactly 6 are defective;
(b)
more than 2 but fewer than 5 are defective.

Worked solution (try it first)

$X \sim B(12, 0.3)$ , with $q = 0.7$ .

(a)

$P(X = 6)$
$= \binom{12}{6}(0.3)^6(0.7)^6$
$= 924 \times 0.000729 \times 0.117649$
$\approx 0.0792$ .

(b)

More than 2 but fewer than 5 means 3 or 4: $\binom{12}{3}(0.3)^3(0.7)^9 + \binom{12}{4}(0.3)^4(0.7)^8$ .
$\approx 0.2397 + 0.2311 = 0.4708$ .

Report a problem with this question