is the answer A?

prob will be atleat 6c2/10,

alsergi wrote:Please, help me with this, problem, I am a little bit confused ...


If 2 different representatives are about to be selected from a group of 10 employees and if x is the probability that both representatives are women , is x > 1/2 ?

1. More than 1 / 2 of the employees are women.
2. The probability that both representatives are men is less than 1/10.

Thanks in advance !!!

I go with B

heres how...

Statement 1.
more than half are women .. Assume women to be 6 and men to be 4
selecting 2 women out of 10 will be
6c2/10c2 = 1/3 ................. Insufficient

statement 2.

probability of 2 men is < 1/10

thus probability of 2 women will be 1-(<)1/10

thus it will be >9/10..... sufficient

hence B

hope it helps...

Thank you very much sudhir3127, but I dont agree with you ...

I agree with the first statement but not with the second

1 - 1/10 would be the probability of AT LEAST one man, not the probability of 2 women so, you cant answer B ...

The right answer is E, but I dont understand exactly why ...

Thank you very much!!! I am still waiting for the right explanation, thinking it too ... :roll:

This trouble me during some minutes, I thought it was B first...

2)The probability that both representatives are men is less than 1/10

You fastly find that this is possible when men are either 2 or 3.
When 2 men: P=2/10*1/9=1/45
When 3 men: P=3/10*2/9=1/15
They are both less than 1/10 and when 4men the probability is above 1/10

In these cases the probability to get 2 women are:
When 2 men: W=8/10*7/9=28/45>1/2
When 3 men: W=7/10*6/9=21/45<1/2

So we can't really now, and given 2) gives us more information than 1) that's E...

It has been discussed before.

in B commonntrait is to ignore total sample space

which can be selecting two women (or) selecting one woman and one man (or) selecting two men

As selecting one woman and one man can not be determined , hence E