WAEC - Further Mathematics (2011 - No. 25)
The probability that Kofi and Ama hit a target in a shooting competition are \(\frac{1}{6}\) and \(\frac{1}{9}\) respectively. What is the probability that only one of them hit the target?
\(\frac{1}{54}\)
\(\frac{13}{54}\)
\(\frac{20}{27}\)
\(\frac{41}{54}\)
Explanation
P(only one hit target) = P(Kofi not Ama) + P(Ama not Kofi)
P(Kofi not Ama) = P(Kofi and Ama') = \(\frac{1}{6} \times \frac{8}{9} = \frac{8}{54}\)
P(Ama not Kofi) = P(Ama and Kofi') = \(\frac{1}{9} \times \frac{5}{6} = \frac{5}{54}\)
P(only one hit target) = \(\frac{8}{54} + \frac{5}{54} = \frac{13}{54}\)
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