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