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 1/10.

arorag wrote: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 1/10.

I'm confused - Did you omit the word "percent" after "p > 21" and after "21 (percent) of the 10 employees are women"???

the correct question is this - can someone help with the shortest way for this?
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 half of of the 10 employees are women.
(2) The probability that both representatives selected will be men is less than 1/10.

wonder wrote:the correct question is this - can someone help with the shortest way for this?
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) More than of the 10 employees are women.
(2) The probability that both representatives selected will be men is less than .

Were you trying to copy paste the question, for some reason the question is still missing some vital content.

I think Answer choice is E

Explanation -
Lets consider, there are n women in the group of 10.
The probability for selecting both women from 10 people group is nC2/10C2. Now according to this, p>10 only when nos. of women be greater than or equal to 8.

1) lets consider there are 6 women in the group. The probability will be 1/3. Also Lets consider, there are 8 women, probability will be 28/45 which is greater than 1/2.
2) It is clearly insufficient to answer.
Combining both will also not give any concrete answer.

Let me know if this answer is correct.

Thanks

Dude you need to answer the question in less than 2 minutes...
The best approach should be...
when is p>1/2
that is When is no of females more than 7.
So A does not give the answer because no. of females can be either 6 or 8...
B tells us the number of males is less than 4. i.e. they can be 1,2,3...
But for the number of females to be more than 7, the number of males should be either 1 or 2
combining both is also of no help...
hence the answer is E

Can you explain why exactly 7 I don't get it ??

Yeah, neither did I. Could someone please elaborate?

My approach:

Let women be w.
men: (10-w)

Probability of selecting two women = W/10×(w-1)/10-1
(W^2 -w )/90 >1/2
=>w^2 -w >45
So, w > 7

The question actually asked whether the number of women is greater than 7 or not.

Stmt1:
More than ½ of 10 is women.
W may be 6, 7, 8, 9. NOT SUFF.

Stmt2:
(10-w)/10 ×(10-w-1)/9 <1/10. NOT SUFF.

1 and 2:
NOT SUFF>
Answer is E.

Robinmrtha wrote:Dude you need to answer the question in less than 2 minutes...
The best approach should be...
when is p>1/2
that is When is no of females more than 7.
So A does not give the answer because no. of females can be either 6 or 8...
B tells us the number of males is less than 4. i.e. they can be 1,2,3...
But for the number of females to be more than 7, the number of males should be either 1 or 2
combining both is also of no help...
hence the answer is E

I s a question for Robinmrtha.
For statement 2, they said that the probability for getting two men is less than 1/10. How did you get that there is les than 4 mens. Thanks