SOLUTION: A committee of four is chosen at random from a group of 7 women and 5 men. Find the probability that the committee contains at least one man.
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Question 1137006: A committee of four is chosen at random from a group of 7 women and 5 men. Find the probability that the committee contains at least one man. Answer by ikleyn(52781) (Show Source):
Such problems are always easier to solve by considering the complement events and complement probability.
The complement event is that the committee contains no man, and its probability is .
The denominator is the number of all possible different subsets of 4 elements in the set of 7+5 = 12 elements.
It is the number of all combinations of 12 items taken 4 at a time = = 495.
The numerator is the number of all possible different committee of 4 women selected from 7 women.
It is the number of all possible combinations of 7 items taken 4 at a time = = 35.
495 is the number of all elements in the full space of events.
35 is the number of elements in the "favorable" set of events.
The probability P' of the complementary event thus is the ratio of "favorable" to "all"
P' = = = .
Then the probability under the problem's question is the complement to it
P = 1 - P' = 1 - = = = 0.9292(92). . . = 92.92% (approximately) ANSWER
