WAEC 2019 · Paper 2 · Q12

Probability & distributions

In an examination, $60\%$ of the candidates passed. If 10 candidates are selected at random, find, correct to four decimal places, the probability that:

(a)
at least two of them failed;
(b)
exactly half of them passed;
(c)
at most two of them failed.

Worked solution (try it first)

Let $F$ be the number who failed: $F \sim B(10, 0.4)$ .

(a)

At least two failed $= 1 - [0.6^{10} + 10(0.4)(0.6)^9]$
$= 1 - 0.0464$
$= 0.9536$ .

(b)

Half passed means 5 passed: $\binom{10}{5}(0.6)^5(0.4)^5 = 252 \times 0.07776 \times 0.01024$
$\approx 0.2007$ .

(c)

At most two failed: $0.0060 + 0.0403 + 45(0.4)^2(0.6)^8 = 0.0060 + 0.0403 + 0.1209$
$= 0.1673$ .

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