WAEC Further Mathematics 2011 Paper

Concept:When only one hits, the other must miss. So add the two separate cases.Explanation:Probability that Kofi hits is $\frac{1}{6}$, so probability that Kofi misses is $1 - \frac{1}{6} = \frac{5}{6}$.Probability that Ama hits is $\frac{1}{9}$, so probability that Ama misses is $1 - \frac{1}{9} = \frac{8}{9}$.Case 1: Kofi hits and Ama misses.$P(K \text{ hits and } A \text{ misses}) = \frac{1}{6} \times \frac{8}{9} = \frac{8}{54}$.Case 2: Ama hits and Kofi misses.$P(A \text{ hits and } K \text{ misses}) = \frac{1}{9} \times \frac{5}{6} = \frac{5}{54}$.Add the two cases because they are mutually exclusive.$P(\text{only one hits}) = \frac{8}{54} + \frac{5}{54} = \frac{13}{54}$.Answer:$\frac{13}{54}$, which is option B.