WAEC 2016 · Paper 2 · Q5

(a)
If ${}^nP_5 \div {}^nC_4 = 24$ , find the value of $n$ .
(b)
A fair die is thrown five times. Find, correct to three decimal places, the probability of obtaining a number less than 6 on three of the throws.

Worked solution (try it first)

(a)

$\dfrac{{}^nP_5}{{}^nC_4} = \dfrac{n!}{(n - 5)!} \times \dfrac{4!\,(n - 4)!}{n!}$ .
Cancel $n!$ , and write $(n - 4)! = (n - 4)(n - 5)!$ : the quotient is $24(n - 4)$ .
So $24(n - 4) = 24$ , which gives $n - 4 = 1$ and $n = 5$ .

(b)

A number less than 6 has probability $p = \frac56$ .
With $n = 5$ throws and exactly 3 successes:
$P = {}^5C_3\left(\frac56\right)^3\left(\frac16\right)^2$
$= 10 \times \dfrac{125}{216} \times \dfrac{1}{36}$
$= \dfrac{1250}{7776}$
$\approx 0.161$ .

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