WAEC 2023 · Paper 2 · Q6

Probability & distributions

In an examination, $80\%$ of the candidates passed. If 8 of the candidates are selected at random, calculate, correct to three decimal places, the probability that:

(a)
exactly seven passed;
(b)
more than three-fourth passed.

Worked solution (try it first)

$X \sim B(8, 0.8)$ .

(a)

$P(7) = \binom87(0.8)^7(0.2)$
$= 0.33554$
$\approx 0.336$ .

(b)

More than three-fourths of 8 is more than 6, so 7 or 8: $0.33554 + 0.8^8 = 0.33554 + 0.16777$
$= 0.50332$
$\approx 0.503$ .

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