Question
A box has 5 red and 3 blue balls. Two balls drawn
without replacement. Probability of at least one blue?Solution
ATQ,
P(no blue) = C5,2 / C8,2 = 10/28 = 5/14 So P(at least one blue) = 1 – 5/14 = 9/14
Compute P(AǀB), if P(B) = 1.0 and P(A∩B) = 0.64
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