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

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

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New post 07 Aug 2018, 03:11
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71% (01:40) correct 29% (01:45) wrong based on 148 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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Re: Eight women and two men are available to serve on a committee. If thre  [#permalink]

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New post 07 Aug 2018, 04:24
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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New post 07 Aug 2018, 05:30
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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New post 11 Nov 2018, 08:32
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

Answer: E

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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New post 22 Dec 2018, 13:52
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 &nbs [#permalink] 22 Dec 2018, 13:52
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