Data Sufficiency

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 > 21?

(1) More than 21 of the 10 employees are women.

(2) The probability that both representatives selected will be men is less than 101.

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.

How do I resolve this question fastly?

Am pretty sure the first statement is wrong.

How can "more than 21 of employees are women" be even remotely possible??

Second by the definition of Probability we know that for an even E

0<= P(E) <= 1

So p cannot> 1
therefore p cannot be greater than 21

Sorry, but i made a mistake in the text.

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.

(1) More than 1/2 of the 10 employees are women.

This implies that number of Women is between 6 and 10.

When # women = 6
Probability of selecting both Women representatives = 6C2/10C2 = 1/3

When # women = 10
Probability of selecting both Women representatives = 10C2/10C2 = 1

1/3 <= Probability of Selecting Women representatives <= 1

NOT SUFFICIENT.

(2) The probability that both representatives selected will be men is less than 1/10.

Let number of men be M. Then, probabilty of selecting both Men representatives = MC2/10C2

MC2 = (M)(M-1)/2; 10C2=45

Prob of both Men = MC2/10C2 = (M)(M-1)/90 < 1/10

M(M-1) < 9

M = 1,2,3 Satisfy this criteria. Women therefore can be 9,8,7

When # Women = 9
Probability of BOTH Women = 9C2/10C2 = 12/15 > 1/2

When # Women = 8
Probability of BOTH Women = 8C2/10C2 = 28/45 > 1/2

When # Women = 7
Probability of BOTH Women = 7C2/10C2 = 7/15 < 1/2

As probabilty can be greater and less than 1/2. NOT SUFFICIENT.

1) consider stmt#1.
Number of women = 5+n, where n>0
probability two women are chosen:
((5+n)/10)*((4+n)/9)
n=1 => p = 1/3 which is less than 1/2
n=4 => p = 0.8 which is greater than 1/2
Therefore this statement is not sufficient.

2) consider stmt# 2
Probability that two men will be chosen is less than 1/10
if number of men = 2, probability is 1/45 which is less than 1/10
if number of men = 3, probability is 1/15 which is less than 1/10
if number of men = 4, probability is 2/15 which is greater 1/10.
Therefore, number of women >= 7. if number of women = 7, probability of women being chosen is 7/45 (<1/2), if number = 8, probability is 28/45 (>1/2)
- So stmt# 2is not sufficient.

It is fairly obvious that combining 1 and 2 won't be sufficient.

Therefore answer is E.

What is OA?

The OA is E.

jay2007 wrote:1) consider stmt#1.
Number of women = 5+n, where n>0
probability two women are chosen:
((5+n)/10)*((4+n)/9)
n=1 => p = 1/3 which is less than 1/2
n=4 => p = 0.8 which is greater than 1/2
Therefore this statement is not sufficient.

2) consider stmt# 2
Probability that two men will be chosen is less than 1/10
if number of men = 2, probability is 1/45 which is less than 1/10
if number of men = 3, probability is 1/15 which is less than 1/10
if number of men = 4, probability is 2/15 which is greater 1/10.
Therefore, number of women >= 7. if number of women = 7, probability of women being chosen is 7/45 (<1/2), if number = 8, probability is 28/45 (>1/2)
- So stmt# 2is not sufficient.

It is fairly obvious that combining 1 and 2 won't be sufficient.

Therefore answer is E.

What is OA?

Where do you get probability of 1/45 ???? Thanks