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If 2 different representatives are to be selected at random [#permalink]

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05 Sep 2008, 11:03

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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 than1/10 .

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

Last edited by Nihit on 05 Sep 2008, 11:15, edited 1 time in total.

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 than1/10 .

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

(1) More than 1/2 of the 10 employees are women. Women count >6 assume 6W 4 M

P = 6C2/10C2 = 60/90=2/3 >1/2 so for W>=6 p>1/2 so for W=9 p= 9C2/10C2= 9*8/10*9 = 72/90 >1/2

not suffcient (2) The probability that both representatives selected will be men is less than1/10 M 10-M women.

MC2/10C2 <1/10 --> M(M-1)/10*9<1/10 M(M-1)<9 this is possible only when M=4 or less. So W>=6 same as (1) not suffcient.

combined same result E
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Your attitude determines your altitude Smiling wins more friends than frowning

(1) More than 1/2 of the 10 employees are women. Women count >6 assume 6W 4 M

P = 6C2/10C2 = 3/5 >1/2 so for W>=6 p>1/2 suffcient (2) The probability that both representatives selected will be men is less than1/10 M 10-M women.

I DONT FOLLOW YOU HERE, PLEASE FORGIVE ME. ASSUMING 6 OF 10 ARE WOMEN, WOULDNT THAT PROBABILITY BE 6/10*5/9 OR 30/90 WHICH IS 3/9 (SO NOT ABOVE 1/2). ALTERNATIVELY, IT COULD BE ANOTHER EXTREME, 9 OF 10, WHICH WOULD MAKE PROBABILITY 9/10*8/9 WHICH IS 72/90 AND CLEARLY OVER 1/2. IT SEEMS TO ME THAT THIS IS NOT SUFFICIENT. PERHAPS I AM READING QUESTION WRONG? THANKS.

MC2/10C2 <1/10 --> M(M-1)/10<1/10 M(M-1)<1 this is possible only when M=1 or M=0 So W=9 or 10 p= 9C2/10C2 > 1/2 SUFFCIENT I AGREE WITH YOU HERE, MAKES PERFECT SENSE. D

(1) More than 1/2 of the 10 employees are women. Women count >6 assume 6W 4 M

P = 6C2/10C2 = 3/5 >1/2 so for W>=6 p>1/2 suffcient (2) The probability that both representatives selected will be men is less than1/10 M 10-M women.

I DONT FOLLOW YOU HERE, PLEASE FORGIVE ME. ASSUMING 6 OF 10 ARE WOMEN, WOULDNT THAT PROBABILITY BE 6/10*5/9 OR 30/90 WHICH IS 3/9 (SO NOT ABOVE 1/2). ALTERNATIVELY, IT COULD BE ANOTHER EXTREME, 9 OF 10, WHICH WOULD MAKE PROBABILITY 9/10*8/9 WHICH IS 72/90 AND CLEARLY OVER 1/2. IT SEEMS TO ME THAT THIS IS NOT SUFFICIENT. PERHAPS I AM READING QUESTION WRONG? THANKS.

MC2/10C2 <1/10 --> M(M-1)/10<1/10 M(M-1)<1 this is possible only when M=1 or M=0 So W=9 or 10 p= 9C2/10C2 > 1/2 SUFFCIENT I AGREE WITH YOU HERE, MAKES PERFECT SENSE. D

Thanks for pointing out.. thats silly mistake.. I updated the original post.
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Your attitude determines your altitude Smiling wins more friends than frowning