Question
A bag contains 8 blue balls, 7 green balls and 5 yellow
balls. Three balls are drawn at random. Find the probability that at least two balls are green.Solution
Total balls = 8 + 7 + 5 = 20
Required probability = (Two green and one other) + (Three green)
= 3 × [(7/20) × (6/19) × (13/18)] + (7/20 × 6/19 × 5/18)
= (1638/6840) + (210/6840)
= (1848/6840) = (77/285)
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