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# A small company employs 3 men and 5 women. If a team of 4

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Intern
Joined: 11 Mar 2007
Posts: 39
Followers: 0

Kudos [?]: 3 [0], given: 1

A small company employs 3 men and 5 women. If a team of 4 [#permalink]  11 Aug 2007, 04:04
A small company employs 3 men and 5 women. If a team of 4 employees is to be randomly selected to organize the company retreat, what is the probability that the team will have exactly 2 women?
1/14
1/7
2/7
3/7
1/2

Why isn't it straight 5C2/8C4??
Senior Manager
Joined: 04 Jun 2007
Posts: 348
Followers: 1

Kudos [?]: 15 [0], given: 0

Re: Probability [#permalink]  11 Aug 2007, 07:50
gkslko101 wrote:
A small company employs 3 men and 5 women. If a team of 4 employees is to be randomly selected to organize the company retreat, what is the probability that the team will have exactly 2 women?
1/14
1/7
2/7
3/7
1/2

Why isn't it straight 5C2/8C4??

I get D (3/7) for this.

You have not considered the ways to select the 2 men in the group. So the probability should be (5C2/8C4) * 3C2.
Director
Joined: 11 Jun 2007
Posts: 932
Followers: 1

Kudos [?]: 67 [0], given: 0

Re: Probability [#permalink]  11 Aug 2007, 22:33
gkslko101 wrote:
A small company employs 3 men and 5 women. If a team of 4 employees is to be randomly selected to organize the company retreat, what is the probability that the team will have exactly 2 women?
1/14
1/7
2/7
3/7
1/2

Why isn't it straight 5C2/8C4??

i get 3/7

probability = M*W / total

M (2 men): 3C2 = 3
W (2 women): 5C2 = 10
total = 8C4 = 70

(3*10) / 70 = 30/70 = 3/7
Manager
Joined: 08 Jan 2007
Posts: 67
Location: D.C.
Followers: 1

Kudos [?]: 1 [0], given: 0

Can someone explain the setup a little more please... The math looks good, but can you elaborate on how you set up the equations.
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