probability DS

[sochatte]

Q2:
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.

A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
C. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
D. EACH statement ALONE is sufficient.
E. Statements (1) and (2) TOGETHER are NOT sufficient.

Please let me know if I am making any mistake

1. More than 1/2 of the 10 are women.
So the number will be between 6 and 9.
Considering 6, the probability will be p = 10c2 / 6c2 = 1/3 <1> 1/2
so 1 is not sufficient.

2. probability of both men is less than 1/10 = .1
so probability(both women) + probability(men+women) > .9
since we don't know probability of (men+women), 2 is not suff as well.

so ans is E.

[sankruth]

St 1: More than 1/2 of 10 are women.

So atleast 6 women in the group.

P(Selecting 2 women for the comittee) = (6/10).(5/9) = 30/90 = 1/3 less than 1/2

So INSUFF

St 2: P(Both are men) less than 1/10

If there are 3 men in the group then P(Both men) = (3/10).(2/9)=6/90
which is less than 1/ 10

If there are 4 men in the group then P(Both men) = (4/10).(3/9)=12/90
which is greater than 1/ 10

So men less than equal to 3, women greater than equal to 7

If 7 women in group then P(Both Women) = (7/10).(6/9)=63/90 greater than 1/2

SUFF

So answer is B (Is this correct?)

[ri2007]

How is B sufficient

If u have 3 men in the group u will have 7 women.

So P(both women) = 7/10 * 6/9 = 7/15 which is less than half.

Pls explain

[sankruth]

Whoops! Apologies, my mistake! B is INSUFF!

Answer must be E!