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Probability & Combination Question

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by rickyishere » Sat Mar 13, 2010 9:32 am
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?

a)1/14
b)1/7
c) 2/7
d)3/7
e)1/2

OA D

I have a question about the solution given, but I will post my query after you guys reply so that I don't give any hints whatsoever.

Thanks as always..
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by outreach » Sat Mar 13, 2010 10:04 am
prob of choosing 2 women out of 5 is 5 C2= 10
prob of choosing 2 men out of 3 is 3 C2 = 3

total prob of choosing 4 out of 8 is 8 C4= 70


=>(10*3)/70=3/7

answer is D
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by rickyishere » Sat Mar 13, 2010 10:16 am
outreach wrote:prob of choosing 2 women out of 5 is 5 C2= 10
prob of choosing 2 men out of 3 is 3 C2 = 3

total prob of choosing 4 out of 8 is 8 C4= 70


=>(10*3)/70=3/7

answer is D
Thank you for your reply. My question : why do we need to select the case of men when we have already selected 2 women for a committee of 4 to be formed? If 2 women have been selected then it automatically means that 2 men need to be selected in order for a committee of 4 to be formed, so I dont understand why case of choosing 2 men had to be included.

Thanks
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by kstv » Sat Mar 13, 2010 10:55 am
Got to select 2 men from 4. Which two men ? Suppose the men are A B C D can't just select A&B and not give a chance to poor C&D.
Atleast be fair and tell them that there are 3C2 ways of selecting either two of them.
AB or AC or AD
BC or BD
CD
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by rickyishere » Sat Mar 13, 2010 11:32 am
kstv wrote:Got to select 2 men from 4. Which two men ? Suppose the men are A B C D can't just select A&B and not give a chance to poor C&D.
Atleast be fair and tell them that there are 3C2 ways of selecting either two of them.
AB or AC or AD
BC or BD
CD
Thanks :), it makes sense
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