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WAEC 2011 · Paper 2 · Q5

XX and YY are two events such that P(X∪Y)=1115P(X \cup Y) = \frac{11}{15} and P(X)=13P(X) = \frac13. Find P(Y)P(Y) if the events are:

  1. (a)

    mutually exclusive;

  2. (b)

    independent.

Worked solution (try it first)

(a)

  1. Mutually exclusive: P(X∪Y)=P(X)+P(Y)P(X \cup Y) = P(X) + P(Y), so P(Y)=1115−13P(Y) = \frac{11}{15} - \frac13
    =615= \frac{6}{15}
    =25= \frac25.

(b)

  1. Independent: P(X∪Y)=P(X)+P(Y)−P(X)P(Y)P(X \cup Y) = P(X) + P(Y) - P(X)P(Y), so 1115=13+P(Y)−13P(Y)\frac{11}{15} = \frac13 + P(Y) - \frac13P(Y).
  2. Collect: 23P(Y)=1115−515\frac23P(Y) = \frac{11}{15} - \frac{5}{15}
    =615= \frac{6}{15}.
  3. So P(Y)=615×32P(Y) = \frac{6}{15} \times \frac32
    =35= \frac35.

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