Theory question 106
ProbabilityFour boys participated in a competition in which their respective chances of winning prizes are 1/5, 1/4, 1/3 and 1/2. What is the probability that at most two of them win prizes?
Show the model answer
Model answer
P(win): 1/5, 1/4, 1/3, 1/2; P(not win): 4/5, 3/4, 2/3, 1/2
P(all four win) = 1/5 × 1/4 × 1/3 × 1/2 = 1/120
P(exactly three win) = (4/5)(1/4)(1/3)(1/2) + (1/5)(3/4)(1/3)(1/2) + (1/5)(1/4)(2/3)(1/2) + (1/5)(1/4)(1/3)(1/2) = (4 + 3 + 2 + 1)/120 = 10/120
P(at most two) = 1 − 1/120 − 10/120 = 109/120 ≈ 0.908
Explanation
It is quicker to subtract P(three win) and P(four win) from 1.