E ?

10 employees, 2 women have to be selected. i.e. p
is p>1/2 ?

Statement I
More than half employees are women.
if women=6
probability = 6/10 * 5/10 = 1/3 < 1/2

If women=9
probability = 9/10*8/9 = 4/5 > 1/2

Insufficient.


Statement II
Probability that both employees are men is less 1/10

lets take 1/10

probability of both women = 1-1/10 = 9/10 > 1/2

if probability of men can be anything less than 1/10 the probability of women will increase. 1/10 is satisfying our question stem, therefore every number less than 1/10 will satisfy the question stem. Sufficient.

Hence B

parallel_chase wrote:10 employees, 2 women have to be selected. i.e. p
is p>1/2 ?

Statement I
More than half employees are women.
if women=6
probability = 6/10 * 5/10 = 1/3 < 1/2

If women=9
probability = 9/10*8/9 = 4/5 > 1/2

Insufficient.


Statement II
Probability that both employees are men is less 1/10

lets take 1/10

probability of both women = 1-1/10 = 9/10 > 1/2

if probability of men can be anything less than 1/10 the probability of women will increase. 1/10 is satisfying our question stem, therefore every number less than 1/10 will satisfy the question stem. Sufficient.

Hence B

Let me know if you still have any doubts.

You da man!!

Joined: 20 Jun 2007
Posts: 912

Will you ever take the test or you just like to hang out here ?

logitech wrote: You da man!!

Joined: 20 Jun 2007
Posts: 912

Will you ever take the test or you just like to hang out here ?

My Test is scheduled for tomorrow 1:15 PM third attempt.

Yeah I jonied this beautiful forum on 20th June 2007 during my first attempt. After my first two attempts i dint study for more than a year because of work.

Lets hope this will be my last time. When are you taking the test.

parallel_chase wrote:

logitech wrote: You da man!!

Joined: 20 Jun 2007
Posts: 912

Will you ever take the test or you just like to hang out here ?

My Test is scheduled for tomorrow 1:15 PM third attempt.

Yeah I jonied this beautiful forum on 20th June 2007 during my first attempt. After my first two attempts i dint study for more than a year because of work.

Lets hope this will be my last time. When are you taking the test.

TOMORROW!!!!!!!!!!!!!!!

Well I live in California, I am not sure what time it is over where you live but...FUCK IT MAN!! GO GET SOME REST!!

KICK SOME ASS TOMORROW!!!

I want to see you for the last time in this forum in couple of days when you are telling us how you BEAT it this time.

Good luck comrade!

parallel_chase wrote:10 employees, 2 women have to be selected. i.e.

Statement II
Probability that both employees are men is less 1/10

lets take 1/10

probability of both women = 1-1/10 = 9/10 > 1/2

if probability of men can be anything less than 1/10 the probability of women will increase. 1/10 is satisfying our question stem, therefore every number less than 1/10 will satisfy the question stem. Sufficient.

Hence B

parallel,
have a doubt: if p(both men)=1/10 then would 9/10 not represent p(not both men? which will include 1 man, 1 woman or both women. if P(one woman, 1 man )=9/10, can we infer p(both women)>1/2?

scoobydooby wrote: parallel,
have a doubt: if p(both men)=1/10 then would 9/10 not represent p(not both men? which will include 1 man, 1 woman or both women. if P(one woman, 1 man )=9/10, can we infer p(both women)>1/2?

You are absolutely right. This is exactly what I hate about myself.

Nice Spot.

answer cannot be B.

Thanks.

parallel_chase wrote:

scoobydooby wrote: parallel,
have a doubt: if p(both men)=1/10 then would 9/10 not represent p(not both men? which will include 1 man, 1 woman or both women. if P(one woman, 1 man )=9/10, can we infer p(both women)>1/2?

You are absolutely right. This is exactly what I hate about myself.

Nice Spot.

answer cannot be B.

Thanks.

you will do very well, all the best for tomorrow

Interesting question.

let's say there are x women and y men.

x+ y =10 -------------------(1)

Now 2 people out of 10 can be selected 2C10 ways.
2 women from x women can be selected 2Cx ways.

Therefore, the probability of selecting both women
P(W)=(2Cx)/(2C10)=x*(x-1)/90 ------------------(2)

x =6, 7, 8, 9 Therefore, P(W) can be .33, .46, .62, .8

(A) is NOT SUFFICIENT

Probability of selecting both men P(M)= y*(y-1)/90 ------(3)



y*(y-1)/90 < 1/10

OR y(y-1) <9 ------------------- (3)

Therefore, y = either 2 or 3 (there should be at least 2 men)
Therefore there are either 7 or 8 women.

We have seen that when for 7 or 8 women, P(W) is can be less than (.46) or greater than 0.5 (.62)

Therefore, (B) is NOT SUFFICIENT.

Together we get no additional info.

Therefore (E)