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WAEC 2012 · Paper 2 · Q15

A class consists of 6 girls and 10 boys. If a committee of 3 is chosen at random from the class, find the probability that:

  1. (a)

    all the members are boys;

  2. (b)

    exactly 2 of them are girls;

  3. (c)

    at least one is a boy.

Worked solution (try it first)
  1. Choosing 3 from 16: 16C3=16×15×146{}^{16}C_3 = \dfrac{16 \times 15 \times 14}{6}
    =560= 560 ways.

(a)

  1. All boys: 10C3=120{}^{10}C_3 = 120 ways, so the probability is 120560=314\dfrac{120}{560} = \dfrac{3}{14}.

(b)

  1. 2 girls and 1 boy: 6C2×10C1=15×10{}^6C_2 \times {}^{10}C_1 = 15 \times 10
    =150= 150 ways, so 150560=1556\dfrac{150}{560} = \dfrac{15}{56}.

(c)

  1. The only committee with no boy is 3 girls: 6C3=20{}^6C_3 = 20 ways.
  2. P(at least one boy)=1−20560P(\text{at least one boy}) = 1 - \dfrac{20}{560}
    =540560= \dfrac{540}{560}
    =2728= \dfrac{27}{28}.

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