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

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Bunuel
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Eight women and two men are available to serve on a committee. If thre [#permalink] New post   07 Aug 2018, 04:11
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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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E
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LevanKhukhunashvili
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Re: Eight women and two men are available to serve on a committee. If thre [#permalink] New post   07 Aug 2018, 05:24
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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] New post   07 Aug 2018, 06:30
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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] New post   29 Feb 2020, 08:47
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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 one man) = 1 - P(no men) = 1 - P(all women)

P(all women) = 8/10 x 7/9 x 6/8 = 4/5 x 7/9 x 3/4 = 3/5 x 7/9 = 21/45 = 7/15

P(at least one man) = 1 - 7/15 = 8/15

Answer: E
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BrentGMATPrepNow
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Re: Eight women and two men are available to serve on a committee. If thre [#permalink] New post   11 Nov 2018, 09:32
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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] New post   14 Feb 2020, 05:28
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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) \(\frac{8}{15}\)


Let's find how many committees we can make with no men. We can subtract that out of the total number of committees to get how many committees will have at least 1 man.

8 women and 2 men

Select a committee with all women = 8C3 = 8*7*6/3*2 = 56

Select a committee in all possible ways = 10C3 = 10*9*8/3*2 = 120

No of committees with at least 1 man = 120 - 56 = 64

Req Probability = 64/120 = 8/15

Answer (E)
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Re: Eight women and two men are available to serve on a committee. If thre [#permalink] New post   11 Sep 2023, 12:10
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?

Probability(committee includes at least one man) = 1 - Probability(committee includes no man) = 1 - 8C3/10C3 = 1 - 7/15 = 8/15

Hence E
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Re: Eight women and two men are available to serve on a committee. If thre [#permalink]
11 Sep 2023, 12:10
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