In a firing contest, the probabilities that Kojo and Kwame hit the target are \(\frac{2}{5}\) and \(\frac{1}{3}\) respectively. What is the probability that...

Assessment: WAEC SSCE - Further Mathematics - 2006 (Objective) Subject: Further Mathematics

Question 1 Report

In a firing contest, the probabilities that Kojo and Kwame hit the target are \(\frac{2}{5}\) and \(\frac{1}{3}\) respectively. What is the probability that none of them hit the target?

Answer Details
The probability that Kojo hits the target is \(\frac{2}{5}\), which means that the probability that he misses the target is \(1-\frac{2}{5}=\frac{3}{5}\). Similarly, the probability that Kwame misses the target is \(1-\frac{1}{3}=\frac{2}{3}\). Now, we want to find the probability that none of them hit the target. Since they are shooting independently, we can multiply their probabilities of missing the target: Probability that both miss the target = Probability that Kojo misses the target AND Probability that Kwame misses the target = \(\frac{3}{5} \times \frac{2}{3} = \frac{6}{15} = \frac{2}{5}\) Therefore, the probability that none of them hit the target is \(\frac{2}{5}\).

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