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) = \(1 - \frac{1}{75}\)

= \(\frac{74}{75}\)

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