WAEC Mathematics 2011 Paper

Concept:When events overlap, the probability of "A or B" must subtract the overlap once.Use $P(A \cup B)=P(A)+P(B)-P(A\cap B)$.Explanation:The set is $\{4,5,6,7,8,9,10,11,12,13,14,15\}$, so there are $12$ equally likely numbers.Multiples of 3 in this set: $6,9,12,15$; number of favorable outcomes $=4$.Multiples of 4 in this set: $4,8,12$; number of favorable outcomes $=3$.The number $12$ is counted in both groups, so it must be subtracted once.Favorable outcomes for "multiple of 3 or multiple of 4" $=4+3-1=6$.Therefore, $P=\frac{6}{12}=\frac{1}{2}$.The value $\frac{7}{12}$ ignores the common element $12$ and is therefore incorrect.Answer:$\frac{1}{2}$Since $\frac{1}{2}$ is not among the options, none of the provided options is correct.