sandy wrote:
Eight women and two men are available to serve on a committee. If three people are picked, what is the probability that the committee includes at least one man?
(A) \(\frac{1}{32}\)
(B) \(\frac{1}{4}\)
(C) \(\frac{2}{5}\)
(D) \(\frac{7}{15}\)
(E) \(\frac{8}{15}\)
We can use the equation:
The number of committees with at least one man = (the total number of committees) - (the number of committees without a man)
The total number of committees = 10C3 = 10!/(3! x 7!) = (10 x 9 x 8)/3! = 720/6 = 120
The number of committees without a man = 8C3 x 2C0 = (8 x 7 x 6)/3! x 1 = 56
Therefore, the number of committees with at least one man = 120 - 56 = 64 and the probability of selecting such a committee is 64/120 = 8/15.
Answer: E