WAEC 2014 · Paper 2 · Q6

A committee of 3 is formed from a panel of 5 men and 3 women. Find the:

(a)
number of ways of forming the committee;
(b)
probability that at least one woman is on the committee.

Worked solution (try it first)

(a)

The panel has $5 + 3 = 8$ people.
Choose 3 of them: ${}^8C_3 = \dfrac{8 \times 7 \times 6}{3 \times 2 \times 1}$
$= 56$ ways.

(b)

A committee with no woman is chosen from the 5 men: ${}^5C_3 = 10$ ways.
So P(no woman) $= \dfrac{10}{56}$ .
P(at least one woman) $= 1 - \dfrac{10}{56} = \dfrac{46}{56}$ .
Simplify: the probability is $\dfrac{23}{28}$ .

Report a problem with this question