Question
A box has 6 red, 5 blue, 4 green balls. Two balls drawn
without replacement. If at least one is blue, what is the probability that both are blue?Solution
ATQ,
P(both blue | at least one blue) = P(BB)/P(at least one blue). Total balls = 15. P(BB) = C(5,2)/C(15,2) = 10/105 = 2/21. P(no blue) = C(10,2)/C(15,2) = 45/105 = 3/7. So P(at least one blue) = 1 − 3/7 = 4/7. Thus required = (2/21)/(4/7) = (2/21)·(7/4) = 14/84 = 1/6.
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