WAEC 2010 · Paper 2 · Q5

1. Choosing 4 from $12$ people: ${}^{12}C_4 = 495$ ways.

(a)

1. 2 of the 5 boys and 2 of the 7 girls: ${}^5C_2 \times {}^7C_2 = 10 \times 21$
   $= 210$ ways.
2. Probability $= \dfrac{210}{495} \approx 0.42$.

(b)

1. At least 3 boys means 3 boys and 1 girl, or 4 boys.
2. 3 boys and 1 girl: ${}^5C_3 \times {}^7C_1 = 10 \times 7 = 70$ ways. 4 boys: ${}^5C_4 = 5$ ways.
3. Probability $= \dfrac{70 + 5}{495}$
   $= \dfrac{75}{495}$
   $\approx 0.15$.