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by GMATGuruNY » Sun Nov 11, 2018 3:39 am
BTGmoderatorLU wrote:Source: Manhattan GMAT

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. 1/32
B. 1/4
C. 2/5
D. 7/15
E. 8/15
P(at least 1 man) = 1 - P(no men).

P(no men):
P(1st person selected is a woman) = 8/10. (Of the 10 people, 8 are women.)
P(2nd person selected is a woman) = 7/9. (Of the 9 remaining people, 7 are women.)
P(3rd person selected is a woman) = 6/8. (Of the 8 remaining people, 6 are women.)
To combine these probabilities, we multiply:
8/10 * 7/9 * 6/8 = 7/15.

Thus:
P(at least 1 man) = 1 - 7/15 = 8/15.

The corrected answer is E.
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by fskilnik@GMATH » Sun Nov 11, 2018 5:59 am
BTGmoderatorLU wrote:Source: Manhattan GMAT

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. 1/32
B. 1/4
C. 2/5
D. 7/15
E. 8/15
\[\left\{ \begin{gathered}
\,8\,\,{\text{women}} \hfill \\
\,2\,\,{\text{men}} \hfill \\
\end{gathered} \right.\,\,\,\,\,\,\,;\,\,\,\,\,\,\,?\,\, = \,\,P\left( {\, \geqslant 1\,\,{\text{men}}\,\,{\text{in}}\,\,3\,\,{\text{people}}\,} \right)\]
\[?\,\,\, = \,\,\,1 - P\left( {3\,\,{\text{women}}\,\,{\text{in}}\,\,3\,\,{\text{people}}} \right)\,\,\, = \,\,\,1 - \frac{{C\left( {8,3} \right)}}{{\,C\left( {10,3} \right)\,}}\,\,\, = \,\,\,1 - \frac{7}{{15}}\,\,\, = \,\,\,\frac{8}{{15}}\]

This solution follows the notations and rationale taught in the GMATH method.

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by Brent@GMATPrepNow » Sun Nov 11, 2018 8:26 am
BTGmoderatorLU 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. 1/32
B. 1/4
C. 2/5
D. 7/15
E. 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

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by deloitte247 » Sun Nov 11, 2018 2:14 pm
There are 8 women and 2 men
What is the probability of picking at least one man if 3 people are picked
Total = 8 + 2
Probability that the committee doesn't include a man.
$$\frac{8}{10}\cdot\frac{7}{9}\cdot\frac{6}{8}=\frac{336}{720}=\frac{7}{15}$$
Probability that the committee includes at least one man = 1 - p
$$=\ 1-probability\ \ of\ committee\ not\ pick\ any\ man$$
$$=1-\frac{7}{15}=\frac{8}{15}$$
$$answer\ is\ Option\ E$$
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by Scott@TargetTestPrep » Mon Jan 21, 2019 6:10 pm
BTGmoderatorLU wrote:Source: Manhattan GMAT

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. 1/32
B. 1/4
C. 2/5
D. 7/15
E. 8/15

The OA is E
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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