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Question:

Three numbers are chosen at random without replacement from 1, 2, 3, ..., 8. The probability that their minimum is 3, given that their maximum is 6, is:

15

38

25

14

Solution:

A → Maximum is 6
B → Minimum is 3
P(A) = 5C2 / 8C3
P(B) = 5C2 / 8C3
P(A∩B) = 2C1 / 8C3
P(B|A) = P(A∩B) / P(A) = (2C1 / 5C2) = 2 / 10 = 1/5 = 15