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?
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})