Question
A bag contains 4 red, 5 blue, and 3 green balls. Two
balls are drawn one by one without replacement. If it is given that at least one ball drawn is blue, what is the probability that both balls are blue?Solution
We need: P(both blue | at least one blue) = P(both blue) / P(at least one blue) P(both blue) = C(5,2) / C(12,2) = 10 / 66 = 5/33 P(at least one blue) = 1 - P(no blue) = 1 - C(7,2) / C(12,2) = 1 - 21/66 = 45/66 = 15/22 So, Probability = (5/33) ÷ (15/22) = (5/33) × (22/15) = 2/9
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