The probabilities that Kofi, Kwasi and Ama will pass a certain examination are \(\frac{9}{10}, \frac{4}{5}\) and x respectively. If the probability that only one of them will pass the examination is \(\frac{9}{50}\), find the :
(a) value of x ;
(b) probability that at least one of them will pass the examination.
Explanation
(a) p(Kofi passes) = \(\frac{9}{10}\), p(Kofi fails) = \(\frac{1}{10}\)
p(Kwasi passes) = \(\frac{4}{5}\), p(Kwasi fails) = \(\frac{1}{5}\)
p(Ama passes) = x, p(Ama fails) = 1 - x.
p(only one passes) = p(only Kofi passes) + p(only Kwasi passes) + p(only Ama passes)
= \((\frac{9}{10} \times \frac{1}{5} \times (1 - x)) + (\frac{4}{5} \times \frac{1}{10} \times (1 - x)) + (x \times \frac{1}{10} \times \frac{1}{5})\)
= \(\frac{9(1 - x)}{50} + \frac{4(1 - x)}{50} + \frac{x}{50}\)
\(\implies \frac{9}{50} = \frac{9 - 9x + 4 - 4x + x}{50} = \frac{13 - 12x}{50}\)
\(\therefore 13 - 12x = 9 \implies 12x = 4\)
\(x = \frac{4}{12} = \frac{1}{3}\).
(b) p(at least one passes) = 1 - p(none passes).
p(none passes) = \(\frac{1}{10} \times \frac{1}{5} \times \frac{2}{3}\)
= \(\frac{1}{75}\)
p(at least one passes) = \(1 - \frac{1}{75}\)
= \(\frac{74}{75}\)