WAEC 2011 · Paper 2 · Q13

(a)
A committee of five is to be formed among 6 Ghanaians, 8 Nigerians and 5 Gambians. In how many ways can the committee be formed if: (i) there is no restriction; (ii) at most 2 Ghanaians are on the committee; (iii) 1 Nigerian is on the committee?
Separate values with commas, e.g. 3, −2
(b)
Five out of 12 articles are known to be defective. If three articles are picked, one after the other without replacement, find the probability that all the three articles are non-defective.

Worked solution (try it first)

At most 2 Ghanaians means 0, 1 or 2.
The rest come from the 13 others.
$\,{}^6C_0 \times {}^{13}C_5 = 1287$ , $\,{}^6C_1 \times {}^{13}C_4 = 6 \times 715$
$= 4290$ and $\,{}^6C_2 \times {}^{13}C_3 = 15 \times 286$
$= 4290$ .
Add: $1287 + 4290 + 4290 = 9867$ .

1 Nigerian from 8, and the other 4 from the 11 non-Nigerians: $8 \times 330 = 2640$ .

There are 7 good articles.
Without replacement: $\dfrac{7}{12} \times \dfrac{6}{11} \times \dfrac{5}{10} = \dfrac{210}{1320}$
$= \dfrac{7}{44}$
$\approx 0.159$ .