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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 30 Oct 2007, 10:13
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A
B
C
D
E

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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.
A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
C. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
D. EACH statement ALONE is sufficient.
E. Statements (1) and (2) TOGETHER are NOT sufficient.

Pls explain.
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 [#permalink] New post 31 Oct 2007, 07:33
the answer should be E

and i hv attached the explanation to the file pls check
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Re: DS - Probability [#permalink] New post 01 Nov 2007, 12:58
rags 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.
A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
C. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
D. EACH statement ALONE is sufficient.
E. Statements (1) and (2) TOGETHER are NOT sufficient.

Pls explain.



S1: Don't bother w/ factorials, least for me, b/c it confused me at first.

Since there are 10 ppl: lets take the two extremes. Women>5.

So Case1: W=6
Case 2: W=9

6/10*5/9=30/90=1/3 P is not greater than 1/2

9/10*8/9 --> 72/90. P is greater than 1/2

Insuff.

S2:

let M equal the number of men.

M/10*(M-1)/9 < 1/10

(M^2-M)/90 <1> 10M^2-10M< 90 divide by 10 and subtract the 9.

M^2-M-9<0> (M - X ) (M + Y ) We know X must be greater than Y to produce -M. Just make some estimate values for X and Y b/c trying to solve exactly will take too long. We know X and Y must be near 3. So lets try -3.5 and 2.5 This will sum to -M and the product of these numbers is close to 9.

So (M-3.5)(M+2.5) M<3> 42/90. Close but not greater than 1/2.

M=1 then W=9. 9/10 *8/9 --> 72/90

Insuff.


taking these statements together adds nothing new and you shouldn't have to test further to realize this.


Ans E.
Re: DS - Probability   [#permalink] 01 Nov 2007, 12:58
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