WAEC 2008 · Paper 2 · Q15

(a)
Simplify $^nC_r + {^nC_{r-1}}$ .
Model answer
Write both terms with the common denominator $r!(n - r + 1)!$ and add: the top becomes $n!(n - r + 1) + n!\,r = n!(n + 1) = (n + 1)!$ , so the sum is $\dfrac{(n + 1)!}{r!(n + 1 - r)!} = {^{n+1}C_r}$ .
(b)(i)
In an examination $4\%$ of the candidates passed with distinction. If 6 of the candidates are selected at random, what is the probability that 2 of them obtained distinction?
(b)(ii)
What is the probability that at most 3 of them obtained distinction?

Worked solution (try it first)

$^nC_r = \dfrac{n!}{r!(n - r)!}$ and $^nC_{r-1} = \dfrac{n!}{(r - 1)!(n - r + 1)!}$ .
Common denominator $r!(n - r + 1)!$ : multiply the first fraction by $\frac{n - r + 1}{n - r + 1}$ and the second by $\frac{r}{r}$ .
Add the tops: $n!(n - r + 1) + n!\,r = n!(n + 1)$
$= (n + 1)!$ .
So $^nC_r + {^nC_{r-1}} = \dfrac{(n + 1)!}{r!(n + 1 - r)!}$
$= {^{n+1}C_r}$ .

Binomial with $n = 6$ , $p = 0.04$ , $q = 0.96$ : $P(2) = {^6C_2}(0.04)^2(0.96)^4$ .
$15 \times 0.0016 \times 0.84935 \approx 0.0204$ .

"At most 3" is easier as $1 - P(4, 5 \text{ or } 6)$ .
$P(4) = 15(0.04)^4(0.96)^2$
$\approx 0.0000354$ .
$P(5) = 6(0.04)^5(0.96) \approx 0.0000006$ .
$P(6) = (0.04)^6 \approx 0$ .
So $P(\text{at most } 3) \approx 1 - 0.0000360$
$= 0.99996$ .