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

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