Eight women and two men are available to serve on a

Question

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

GMATGuruNY · Answer

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.

fskilnik@GMATH · Answer

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. Regards, Fabio.

Brent@GMATPrepNow · Answer

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

Cheers,
Brent

Brent Hanneson - Creator of GMATPrepNow.com

deloitte247 · Answer

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
$$=\ 1-probability\ \ of\ committee\ not\ pick\ any\ man$$
$$=1-\frac{7}{15}=\frac{8}{15}$$
$$answer\ is\ Option\ E$$

Scott@TargetTestPrep · Answer

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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