WAEC 2013 · Paper 2 · Q5

1. There are $\,{}^9C_3 = 84$ ways to choose 3 prefects from 9.

(a)

1. Only boys: $\,{}^5C_3 = 10$, so $P = \dfrac{10}{84}$
   $= \dfrac{5}{42}$
   $\approx 0.119$.

(b)

1. More girls than boys means 3 girls, or 2 girls and 1 boy.
2. 3 girls: $\,{}^4C_3 = 4$. 2 girls and 1 boy: $\,{}^4C_2 \times {}^5C_1 = 6 \times 5$
   $= 30$.
3. So $P = \dfrac{4 + 30}{84}$
   $= \dfrac{34}{84}$
   $= \dfrac{17}{42}$
   $\approx 0.405$.