JAMB 1992 · UME · Q48

Probability

The probability of an event $P$ is $\frac34$ while that of another event $Q$ is $\frac16$ . If the probability of both $P$ and $Q$ is $\frac1{12}$ , what is the probability of either $P$ or $Q$ ?

Worked solution (try it first)

For "either or", add the two probabilities and take away the overlap, which is counted twice: $P(P \cup Q) = P(P) + P(Q) - P(P \cap Q)$ .
Over 12: $\frac{9}{12} + \frac{2}{12} - \frac{1}{12} = \frac{10}{12}$ .
So the probability is $\frac56$ , option C.

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