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If 2 different representatives are to be selected at random

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If 2 different representatives are to be selected at random [#permalink] New post 15 Jan 2006, 09:53
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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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 [#permalink] New post 15 Jan 2006, 10:48
1) We don't know the actual number of women; this is what makes the statement insufficient.

Assume 5 women:

P= (5/10)*(4/9)= 2/9 not greater than 1/2
P=(9/10)*(8/9)= 72/90= 8/10= 4/5 greater than 1/2

2) Alike

Assume 4 men, than the probability that the two women are in would be less than 1/2

etc.

E
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Re: DS probability [#permalink] New post 15 Jan 2006, 10:53
sperumba 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 .


'E'!

St1: More than 1/2 of the 10 employees are women.

For 5 women,
p = 5C2/10C2 = ~0.2
If No. of women > 5, p will increase. YES/NO

St2:The probability that both representatives selected will be men is less than 1/10

We can't get 'p' from this information. YES/NO
since 1-(1/10) is not only 'p' but also prob of 1 man + 1 woman

Together:
YES/NO.

So, it should be 'E'
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 [#permalink] New post 15 Jan 2006, 22:14
Why not B guys..

P(both are women)= 1- P(both are men)
=1-(no<1/10) = >1/2

How am i wron here?
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 [#permalink] New post 15 Jan 2006, 22:19
(1) We know there are 6 and above # of employees that are women.

For w =6, p = 6/10 * 5/9 = 1/3 < 1/2
For w =7, p = 7/10 * 6/9 = 7/15 < 1/2
For w =8, p = 8/10 * 7/9 = 28/45 > 1/2
For w =9, p = 9/10 * 8/9 = 4/5 > 1/2
For w =10, p = 1 > 1/2

Insufficient.

(2) p(both men) = 1/10 -> not useful since we know p(both women) + p(1women + 1 men) = 9/10. Insufficient

Using both we cannot yield any more useful information.

Ans E
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 [#permalink] New post 16 Jan 2006, 00:32
andy_gr8 wrote:
Why not B guys..

P(both are women)= 1- P(both are men)
=1-(no<1/10) = >1/2

How am i wron here?


E its is

Andy you are wrong here:

P(both men) = 1- P(both women) + P(1women + 1 men)

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SAID BUSINESS SCHOOL, OXFORD - MBA CLASS OF 2008

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 [#permalink] New post 16 Jan 2006, 11:42
good job guys OA is E.
  [#permalink] 16 Jan 2006, 11:42
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