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

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GMAT Club Legend
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Joined: 07 Jun 2014
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GRE 1: Q167 V156
WE: Business Development (Energy and Utilities)
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Eight women and two men are available to serve on a committe [#permalink] New post 05 Aug 2018, 14:59
Expert's post
00:00

Question Stats:

87% (01:13) correct 12% (01:54) wrong based on 16 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)
GMAT Club Legend
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User avatar
Joined: 07 Jun 2014
Posts: 4749
GRE 1: Q167 V156
WE: Business Development (Energy and Utilities)
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Kudos [?]: 1652 [0], given: 396

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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
Expert's post
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] 23 Aug 2018, 07:47
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