Four boys participated in a competition in which their respective chances of winning prizes are \(\frac{1}{5}, \frac{1}{4}, \frac{1}{3}\) and \(\frac{1}{2}\). What is the probability that at most two of them win prizes?
Take Free Practice Test On 2026 JAMB UTME, Post UTME, WAEC SSCE, GCE, NECO SSCE
Correct Answer: Option n
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) = \(1 - [\frac{1}{120} + \frac{10}{120}]\)
= \(\frac{109}{120}\)
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) = \(1 - [\frac{1}{120} + \frac{10}{120}]\)
= \(\frac{109}{120}\)