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# Eight women and two men are available to serve on a committee. If thre

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Math Expert
Joined: 02 Sep 2009
Posts: 56268
Eight women and two men are available to serve on a committee. If thre  [#permalink]

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07 Aug 2018, 04:11
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Difficulty:

35% (medium)

Question Stats:

74% (01:55) correct 26% (02:24) wrong based on 171 sessions

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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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Joined: 13 Feb 2018
Posts: 401
GMAT 1: 640 Q48 V28
Re: Eight women and two men are available to serve on a committee. If thre  [#permalink]

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07 Aug 2018, 05:24
1
We have to apply 1-p approach

What is the probability that the committee doesn't include man?
$$\frac{8}{10}$$*$$\frac{7}{9}$$*$$\frac{6}{8}$$=$$\frac{7}{15}$$

What is the probability that the committee includes at least one man? 1-$$\frac{7}{15}$$=$$\frac{8}{15}$$

IMO
Ans: E
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Re: Eight women and two men are available to serve on a committee. If thre  [#permalink]

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07 Aug 2018, 06:30
1
Bunuel 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?

2 men 1 women = 2C2 * 8C1 = 8
1 men 2 women = 2C1 * 8C2 = 56
Combing both 8 + 56 = 64
Total = 10C3 = 120

Required probabiity = 64 / 120 = 8/15

Hence, E.
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Re: Eight women and two men are available to serve on a committee. If thre  [#permalink]

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11 Nov 2018, 09:32
Top Contributor
Bunuel 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) $$\frac{8}{15}$$

When it comes to probability questions involving "at least," it's best to try using the complement.
That is, P(Event A happening) = 1 - P(Event A not happening)
So, here we get: P(getting at least 1 man) = 1 - P(not getting at least 1 man)
What does it mean to not get at least 1 man? It means getting ZERO men.
So, we can write: P(getting at least 1 man) = 1 - P(getting ZERO men)

P(getting ZERO men)
P(getting ZERO men) = P(all 3 selections are women)
= P(1st selection is a woman AND 2nd selection is a woman AND 3rd selection is a woman)
= P(1st selection is a woman) x P(2nd selection is a woman) x P(3rd selection is a woman)
= 8/10 x 7/9 x 6/8
= 7/15

So we get: P(getting at least 1 man) = 1 - 7/15
= 8/15

Cheers,
Brent
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Re: Eight women and two men are available to serve on a committee. If thre  [#permalink]

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22 Dec 2018, 14:52
Bunuel 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) $$\frac{8}{15}$$

P at least = 1-(none/total)>>1-( 8c3/10c3) =56/120=8/15
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Re: Eight women and two men are available to serve on a committee. If thre   [#permalink] 22 Dec 2018, 14:52
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