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# If 2 different representatives are to be selected at random

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VP
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If 2 different representatives are to be selected at random [#permalink]  23 May 2006, 17:54
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If 2 different representatives are to be selected at random from a group of 10 employees and if p is the probability that both representatives selected will be women, is p > 1/2?

(1) More than 1/2 of the 10 employees are women
(2) The probability that both representatives selected will be men is less than 1/10

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Re: DS: Employees [#permalink]  23 May 2006, 18:53
TeHCM wrote:
If 2 different representatives are to be selected at random from a group of 10 employees and if p is the probability that both representatives selected will be women, is p > 1/2?

(1) More than 1/2 of the 10 employees are women
(2) The probability that both representatives selected will be men is less than 1/10

seems a good question.

1. no of women could be 6 or 10. if 6, p = 15/45 not suff. if 10, suff.
2. no of men is less than 4.
if men = 4, the prob of getting both men = 6/45 which is > 1/10.
if men = 3, the prob(2 men) = 3/45 = 1/15
if men = 2, the prob(2 men) = 1/45

so we are npt sure about the no of women that could be 7 or 8 or so on.
E.

Last edited by HIMALAYA on 23 May 2006, 19:36, edited 1 time in total.
Senior Manager
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E it is.

(1) More than 1/2 of the 10 employees are women

6 women and 4 men.
6/10*5/9=1/3 < 1/2,
9 women and 1 man
9/10*8/9= 72/90 > 1/2, so not suff

(2) The probability that both representatives selected will be men is less than 1/10

We can not find P from this info, so 2) insuff also.
Senior Manager
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HIMALAYA you forgot about that 1-(1/10) is not only P but also the probability of 1man + 1woman.
VP
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Me too E.

Is wC2/10C2 > 1/2
or Is w*(w-1) > 50
Only possible if w >= 8.
Effectively IS w >= 8?

1. w > 5 --> INSUFF
2. mC2/10C2 < 10
=> m*(m-1) < 10
Only possible values are m <= 3
If m>= 3 --> w >=7 ... STILL INSUFF.
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# of men in the company = m
# of women in the company = 10-m

St1:
If # of women = 6, # of men = 4, then P = 6/10 * 5/9 = 1/3 (< 1/2)
If # of women = 9, # of men = 1, then P = 9/10 * 8/9 = 4/5 (> 1/2)

Insufficient.

St2:
If m = 5, P = 5/10 * 4/9 = 2/9 (> 1/10) --> out
If m = 4, P = 4/10 * 3/9 = 2/15 (> 1/10) --> out
If m = 3, P = 3/10 * 2/9 = 1/15 (< 1/10)

So m is 3 and below.

If m = 3, w = 7, then P = 7/10 * 6/9 = 7/15 (< 1/2)
If m = 2, w = 8, then P = 8/10 * 7/9 = 28/45 (> 1/2)

Insufficient.

Using St1 and St2:
We will still have two cases presented in the analysis of St2. Insufficient.

Ans E
VP
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Great job as always, OA is E
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