Himalayan wrote:

vikramjit_01 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.

Stay tuned for the OA

10c2 = 45

we need total 23 pairs of women to have p>1/2

to have 23 pair, we need 8 women.

from 1:

if w = 6, 6c2 = 15, p = 15/45 <1> 1/2

so nsf.

from 2:

if p < 1/10 for both men, no of pair of both men is less than 4.5. since 4.5 pair can not be possible therefore it must be 4 or 3 or 2 or 1. what ever be the pair, no of men can not be more than 3 because if no of men is 4, then the pair of both men becomes 6, which is not possible.. therefore the no of men is 3 or less than 3.

if n (men) = 3......p (both women) <1> 1/2

so still not suff.....

togather is also nsf.

this is not what i have posted. i tried reposting what i did but it is not posting what exactly i wrote.

10c2 = 45

we need total 23 pairs of women to have p > 1/2

to have 23 pair, we need 8 women.

from 1:

if w = 6, 6c2 = 15, p = 15/45 is less 1/2

if w = 7, 7c2 = 21, p = 21/45 is less 1/2

if w = 8, 8c2 = 28, p = 28/45 is grater 1/2

so nsf.

from 2:

if p < 1/10 for both men, no of pair of both men is less than 4.5. since 4.5 pair can not be possible therefore it must be 4 or 3 or 2 or 1. what ever be the pair, no of men can not be more than 3 because if no of men is 4, then the pair of both men becomes 6, which is not possible.. therefore the no of men is 3 or less than 3.

if m = 3, w = 7, 7c2 = 21, p = 21/45 is less 1/2

if m = 2, w = 8, 8c2 = 28, p = 28/45 is grater 1/2

so still not suff.....

togather is also nsf.

EEE