Explanation

P(at most two winning prizes) = 1 - [P(all winning) + P(three winning)]

P(three winning) = \(P(ABCD') + P(ABC'D) + P(AB'CD) + P(A'BCD)

P(A) = \(\frac{1}{5}\), P(A') = \(\frac{4}{5}\)

P(B) = \(\frac{1}{4}\), P(B') = \(\frac{3}{4}\)

P(C) = \(\frac{1}{3}\), P(C') = \(\frac{2}{3}\)

P(D) = \(\frac{1}{2}\), P(D') = \(\frac{1}{2}\)

P(three winning) = \((\frac{1}{5} \times \frac{1}{4} \times \frac{1}{3} \times \frac{1}{2}) + (\frac{1}{5} \times \frac{1}{4} \times \frac{2}{3} \times \frac{1}{2}) + (\frac{1}{5} \times \frac{3}{4} \times \frac{1}{3} \times \frac{1}{2}) + (\frac{4}{5} \times \frac{1}{4} \times \frac{1}{3} \times \frac{1}{2})\)

= \(\frac{1}{120} + \frac{2}{120} + \frac{3}{120} + \frac{4}{120}\)

= \(\frac{10}{120}\)

P(all winning) = \(\frac{1}{5} \times \frac{1}{4} \times \frac{1}{3} \times \frac{1}{2}\)

= \(\frac{1}{120}\)

P(at most 2 winning) = \(1 - [\frac{1}{120} + \frac{10}{120}]\)

= \(\frac{109}{120}\)

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