WAEC 2020 · Paper 2 · Q12

Probability & distributions

In a bottle manufacturing company, it was observed that $5\%$ of the bottles manufactured were defective. In a random sample of 150 bottles manufactured, find the probability that: $[\text{Take } e = 2.7183]$

(a)
exactly 3;
(b)
between 3 and 6;
(c)
at most 4, bottles are defective.

Worked solution (try it first)

$n = 150$ is large and $p = 0.05$ is small, so use the Poisson approximation with $\lambda = np = 7.5$ .
$e^{-7.5} \approx 0.000553$ .

(a)

$P(3) = e^{-7.5} \times \dfrac{7.5^3}{3!}$
$\approx 0.0389$ .

(b)

Between 3 and 6 means 4 or 5: $P(4) + P(5) \approx 0.0729 + 0.1094$
$= 0.1823$ .

(c)

At most 4: $P(0) + \ldots + P(4) \approx 0.0006 + 0.0041 + 0.0156 + 0.0389 + 0.0729$
$= 0.1321$ .

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