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A division of a company consists of seven men and five women

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A division of a company consists of seven men and five women [#permalink] New post 08 Jan 2013, 10:42
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A division of a company consists of seven men and five women. If two of these twelve employees are randomly selected as representatives of the division, what is the probability that both representatives will be female?
(A) 1/6
(B) 2/5
(C) 2/9
(D) 5/12
(E) 5/33

For a full discussion of probability questions that involve counting techniques, and for a full solution to this particular question, see this post:
http://magoosh.com/gmat/2013/gmat-proba ... echniques/
For some shortcuts for doing these calculations without a calculator, see this post:
http://magoosh.com/gmat/2012/gmat-math- ... binations/
[Reveal] Spoiler: OA

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Re: A division of a company consists of seven men and five women [#permalink] New post 09 Jan 2013, 07:44
Total number of ways to select 2 members from 12 members = 12C2 - denominator
Total number of ways to select 2 females from 5 total females = 5C2 - numerator

Therefore probability of having 2 females reps is 5C2 / 12C2 = 5/33 (E)
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Re: A division of a company consists of seven men and five women [#permalink] New post 09 Jan 2013, 19:56
A different way.

There are a total of 12 employees (7 men and 5 women)

Probability of 1st person selected being a woman is 5/12
Probability of the 2nd person selected being a woman is 4/11 (4 women left in the 11 remaining employees)

Probability of 1st and 2nd selected both being women
=5/12 X 4/11
= 20/132
which reduces to
= 5/33

Answer E
Re: A division of a company consists of seven men and five women   [#permalink] 09 Jan 2013, 19:56
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