crackinggmat 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.

[/spoiler]E

If w were the number of women and 10 - w the number of men in the group of 10 employees, then the probability that both representatives selected will be women

= wC2/10C2; and wC2/10C2 > ½ is possible only when the women in group are in nearly an absolute majority, which we won't get as right when they are 6 or 7 in number, but we cannot deny their majority once they are 6 in number.

Statement is insufficient for the reasons cited above.

Statement (2) pledges that there is a chance that two men could also be selected from the group of 10, hence the number of women could not exceed 8, or may be something less; but the given range of probability (less than 1/10) only establishes that there could be at most 3 men in the group. If there were 8 women, the required probability would have exceeded ½, but if there were 7 women, the required probability, 7C2/10C2 = 21/45, remains less than ½. Insufficient

Even if taken together, we are still left with two possibilities, one agrees and the other denies the same thing.


[spoiler]E[/spoiler]

sanju09......i did not get ur explanation for statement B ....how did u say that number of men can be atmost 3 ?????

can u pls take some time to explain further

crackinggmat wrote:sanju09......i did not get ur explanation for statement B ....how did u say that number of men can be atmost 3 ?????

can u pls take some time to explain further

If there are 3 men in the group, then the probability that both representatives selected will be men

= 3C2/10C2 = 1/15 < 1/10, but as if it's 4, the same probability

= 4C2/10C2 = 2/15 > 1/10.