WAEC 2009 · Paper 2 · Q13

(a)
Simplify ${}^{n+1}C_3 - {}^{n-1}C_3$ .
(b)
A fair die is thrown five times. Calculate, correct to three decimal places, the probability of obtaining (i) at most two sixes; (ii) exactly three sixes.
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

(a)

Use ${}^nC_3 = \dfrac{n(n - 1)(n - 2)}{6}$ : ${}^{n+1}C_3 = \dfrac{(n + 1)n(n - 1)}{6}$ and ${}^{n-1}C_3 = \dfrac{(n - 1)(n - 2)(n - 3)}{6}$ .
Take out the common factor $\dfrac{n - 1}{6}$ : $\dfrac{n - 1}{6}[(n + 1)n - (n - 2)(n - 3)]$ .
Inside the bracket: $n^2 + n - (n^2 - 5n + 6) = 6n - 6$
$= 6(n - 1)$ .
So ${}^{n+1}C_3 - {}^{n-1}C_3 = (n - 1)^2$ .

(b)

Each throw is a six with $p = \frac16$ and not a six with $q = \frac56$ .
The number of sixes in 5 throws is binomial.

(i)

$P(0) = (\frac56)^5 = \frac{3125}{7776}$ , $P(1) = 5(\frac16)(\frac56)^4$
$= \frac{3125}{7776}$ , $P(2) = 10(\frac16)^2(\frac56)^3$
$= \frac{1250}{7776}$ .
Add: $P(\le 2) = \frac{7500}{7776} = 0.965$ (3 d.p.).

(ii)

$P(3) = 10(\frac16)^3(\frac56)^2$
$= \frac{250}{7776}$
$= 0.032$ (3 d.p.).

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