WAEC 2021 · Paper 2 · Q5

(a)
There are 25 boys and 15 girls in a class, all of them equally likely to be chosen for a contest. If two students are chosen one after the other without replacement for the contest, find the probability that: (i) two boys are chosen; (ii) a boy and a girl are chosen; (iii) two boys or two girls are chosen.
Separate values with commas, e.g. 3, −2
(b)
A box contains 5 red, 3 green and 4 blue identical beads. Calculate the probability that a girl takes away two red beads, one after the other, from the box.

Worked solution (try it first)

Two boys: $\frac{25}{40} \times \frac{24}{39} = \frac{600}{1560}$
$= \frac{5}{13}$ .

A boy and a girl, in either order: $\frac{25}{40} \times \frac{15}{39} + \frac{15}{40} \times \frac{25}{39} = \frac{375 + 375}{1560}$
$= \frac{750}{1560}$
$= \frac{25}{52}$ .

Two girls: $\frac{15}{40} \times \frac{14}{39} = \frac{210}{1560}$ .
Two boys or two girls: $\frac{600 + 210}{1560} = \frac{810}{1560}$
$= \frac{27}{52}$ .
(Check: (ii) and (iii) cover every case, and $\frac{25}{52} + \frac{27}{52} = 1$ .)

12 beads, 5 red.
After one red bead is taken away, 11 remain with 4 red: $\frac{5}{12} \times \frac{4}{11} = \frac{5}{33}$ .