Probability

[crackgmat007]

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.

Can someone help me understand the logic pls. Answer seems to be D

[bluementor]

Are you sure of the OA. I'm getting a different answer.

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is p>1/2?

Statement 1: More than half are women.

This means there could be 6, 7, 8, 9 or 10 women.

if number of women = 6, then p = (6/10)(5/9) = 1/3, i.e. < 1/2
if number of women = 9, then p = (9/10)(8/9) = 4/5, i.e. > 1/2

Insufficient.

Statement 2: Prob(both men) < 1/10

if number of men = 2, Prob(both men) = (2/10)(1/9) = 1/45
if number of men = 3, Prob(both men) = (3/10)(2/9) = 1/15
if number of men = 4, Prob(both men) = (4/10)(3/9) = 2/15 (i.e. > 1/10)

so the number of men must be less than 4. this means, there must be at least 7 women.

if number of women = 7, then p = (7/10)(6/9) = 14/30 (i.e. < 1/2)
if number of women = 8, then p = (8/10)(7/9) = 28/45 (i.e. > 1/2).

Insufficient.

Both statements together:

we will still arrive the same conclusion as statement 2. Insufficient.

I believe its E.

-BM-

[crackgmat007]

thanks for the detailed explanation. there may be error in the answer.

[HumbleMe]

I Think answer is B...

According to the question:

p = 1/2
let's say w is number of women in group, w/10x(w-1)/9 > 1/2
w^2-1 > 45
w^2 > 46
w > 6.5 so in order for probability to be larger than 1/w, w should be 7,8,9 or 10.

1. More than half of women involves 6 women... Which doesn't fit or conclusion above hence insufficient...

2. m/10 x (m-1)/9 < 1/10
m^2-1 < 9
m^2 < 10
m < 3.3 so men should be either 3,2,1,0 which is essentially samething as mentioned in the question.

Hence B alone is sufficient...

[HumbleMe]

OK. I have calculation issues in my answer... Blue mentor is right...