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Probability

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Senior Manager
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Probability [#permalink]

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New post 13 Oct 2008, 08:23
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Kudos [?]: 39 [0], given: 1

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Re: Probability [#permalink]

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New post 13 Oct 2008, 10:01
Hi,

I am going to take a shot at this. Hope I am correct and my methodology works. Please let me know.

(1) If more than 1/2 of the employees are women, then there could be 6, 7, 8, or 9 women. Using 6 women - 6/10 x 5/9 = 30/90. Therefore p < 1/2.
Using 9 women - 9/10 x 8/9 = 72/90. Therefore p > 1/2.

therefore statement 1 is not good.

(2) Since the probability for men will be 1/10 then probability of no women is 1/10. then probability of women (p) will be > 1/2.

My ans is B.

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Re: Probability [#permalink]

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New post 13 Oct 2008, 17:08
Hi,

I am also going to take a shot at this problem

(1)
as it is given that number of women is greaten than 1/2 therefore, number of women is eitehr 6 , 7 , 8 , 9

therefore,
6/10 * 7/10= .42

or
9/10 * 8/10 = .72

therefore, i agree statement 1 is not sufficient

2)

(2) Since the probability for men will be less then 1/10 then it is safe to assume that probably of women being selected is 9/10

therefore, b is my answer. Someone please verify.
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Re: Probability [#permalink]

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New post 13 Oct 2008, 21:03
spiridon wrote:
...


(E)

(1) For 6 women, probability of both selections being women = (6/10) * (5/9) = 1/3
For 9 women, probability = (9/10) * (8/9) = 4/5
1/3 < 1/2 < 4/5 --- insufficient

(2) Probability of all men getting selected is below 1/10 when there are 3 or lesser men
With 3 men (and hence 7 women), p = 7/10 * 6/9 = 7/15
With 2 men (and hence 8 women), p = 8/10 * 7/9 = 28/45
7/15 < 1/2 < 28/45 --- insufficient

(1) and (2) do not provide any additional information. Insufficient.
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Re: Probability [#permalink]

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New post 05 Dec 2008, 04:27
i AM STRONGLY CONVINCED THAT ANSWER IS B. SOMEONE PLEASE GET THE OA
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Re: Probability   [#permalink] 05 Dec 2008, 04:27
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