Is the ans E?

I say B

If 2 different represenatives 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 employees are women.
Out of 10 employees there are at least 6 women. We cannot assume there are more. So let's see the probability with 6 women.
P=6/10*5/9=1/3 <1/2 Not engouh information
2)The probability that both representatives will be men is less than 1/10.
The maximum number of men is 3 because 3/10*2/9=1/15 and 4/10*3/9=2/15>1/10
P=7/10*6/9=7/15 > 1/2

B

Yes, ans is 'E'...can u pls explain ?

I am getting an answer E

Let us see what makes the event space

(two selected are males) or (two selected are females) or (one selected male and one selected female)

stmt 1 : there can be less than half for 7 women in the group or more than 1/2 for 8 women in the group -------- insufficient.

stmt 2 the probabilty that both representative will be female or one male one female > 9/10 again we do not have a way to solve this case.

Together still we do not have a clue to solve what is the prob if one female-one male is chosen

let me know if my approach is wrong.

regards,

The answer is E, also because I have the answer.

I followed the same route as pepeprepa, however, pepeprepa's conclusion is not correct. First of all because 7/15 is not > 1/2 but less than 1/2.

I actually think that the maximum number of men can be three and thus there would be seven women. But there could also be just two men (or maybe even just one), then there are 8 women. This can still be a scenario in the second case, it seems to me. In that case:

With 3 men and 7 women, we get 7/10*6/9 < 1/2
With 2 men and 8 women, we get 8/10*7/9 > 1/2

So that is why I would choose E.