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) \(\frac{8}{15}\)
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Sandy
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Explanation

Because this is an “at least” question, use the 1 – x shortcut:

(The probability of picking at least one man) + (The probability of picking no men) = 1

The probability of picking no men is an and setup: woman and woman and woman.

For the first choice, there are 8 women out of 10 people: \(\frac{8}{10}=\frac{4}{5}\).

For the second choice, there are \(\frac{7}{9}\)(because one woman has already been chosen).

For the third choice, there are \(\frac{6}{8}=\frac{3}{4}\).

Multiply the three probabilities together to find the probability that the committee will be comprised of woman and woman and woman:

\(\frac{4}{5} \times \frac{7}{9} \times \frac{3}{4}=\frac{7}{15}\).

To determine the probability of picking at least one man, subtract this result from 1:

\(1- \frac{7}{15}=\frac{8}{15}\).
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Sandy
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