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Eight women and two men are available to serve on a committe

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Eight women and two men are available to serve on a committe [#permalink] New post 05 Aug 2018, 14:59
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73% (01:15) correct 26% (01:34) wrong based on 23 sessions
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) \(\frac{8}{15}\)
[Reveal] Spoiler: OA

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Re: Eight women and two men are available to serve on a committe [#permalink] New post 09 Aug 2018, 01:39
We can use the rule of probability:

The sum of the probabilities of all possible outcomes is 1( ex: p(a)+p)(not a) =1)

P(picking NO MAN) + P(picking 1 MAN) + P(picking 2 MAN) = 1

The probability of at least one man = P(1 MAN) + P(2 MAN)= 1- P(NO MAN)

P(NO MAN)= P( ALL WOMAN) = 8/10 * 7/9 * 6/8

= 7/15
The probability of at least one man = 1- P(NO MAN) = 1-7/15
=8/15.

Answer choice (E)
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Re: Eight women and two men are available to serve on a committe [#permalink] New post 23 Aug 2018, 07:47
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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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Re: Eight women and two men are available to serve on a committe [#permalink] New post 21 Jan 2019, 18:21
Expert's post
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
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Re: Eight women and two men are available to serve on a committe   [#permalink] 21 Jan 2019, 18:21
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