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If 2 different representatives are to be selected at random

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If 2 different representatives are to be selected at random

by mitzwillrockgmat » Tue Jun 08, 2010 1:45 am
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.

I understand how stat 1 is insufficient:

more than half are women then the no, of women can be 6, 7, 8, 9 or even 10

so prob when there are 6 women is 6^C2 /10^C2 = 15/45 = 1/3 so 1/3 < 1/2

but when prob 10^C2 /10^C2 = 45/45 = 1 so 1 is not < 1/2

we dont get an unique value, so insufficient.


But I don't understand why stat 2 is insufficient. Couldn't statement two also mean that there are no males employees since the probability is less than 1/10???

If there are no male employees then the prob of 2 women is easy to find out.

OR does it mean that prob 2 men is : m/10 * (m-1)/9 < 1/10

& i can simplify it to:

m(m-1)/90 < 1/10

m(m-1) < 9

ans : m<9 & m-1 < 9 so m< 10

so number of men are less than 9. There could be 8,7,6,5,4,3,2,1 and all of these would give different values less than or equal to or greater than half.

does this make sense???

Please help clarify stat 2 for me! thanks! :)
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by jube » Tue Jun 08, 2010 1:53 am
Hi,

I interpreted S2 to mean:

Probability that the 2 people are women < 1- (1/10) = 9/10

Since this can be less than 0.5 or more than 0.5, hence it's insufficient.
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by mitzwillrockgmat » Tue Jun 08, 2010 2:06 am
jube wrote:Hi,

I interpreted S2 to mean:

Probability that the 2 people are women < 1- (1/10) = 9/10

Since this can be less than 0.5 or more than 0.5, hence it's insufficient.
Okay so to reiterate;

you are saying that i should see it the other way round , that prob of women is < 9/10

so there could be 8,7,6,5,4,3,2,1 or 0 woman & all of these give different answers < or > 1/2 ? as stat 1 demonstrated.

Alright so that makes sense!

But I'm still unsure why I cannot see it as prob of 2 men < 1/10 means no women ?????????
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by liferocks » Tue Jun 08, 2010 2:08 am
from question P= 1-(probability that both representatives selected will be men+probability that one of the representative selected will be women and another will be men)

from 2
probability that both representatives selected will be men<1/10
but no information for 'probability that one of the representative selected will be women and another will be men'

so we cannot tell for sure whether p>1/2
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by liferocks » Tue Jun 08, 2010 2:12 am
in another way ,
probability that both representatives selected will be men<1/10
so number of men can be 0,1,2,3
probability that both representatives selected will be men=3/45
if number of men is 3,probability that one of the representative selected will be women and another will be men =
(3*7)/45 or 21/45
so P=1-(21/45 + 3/45)=21/45 <1/2 but if number of men is 0 or 1 P>1/2
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by mitzwillrockgmat » Tue Jun 08, 2010 2:45 am
liferocks wrote:in another way ,
probability that both representatives selected will be men<1/10
so number of men can be 0,1,2,3
probability that both representatives selected will be men=3/45
if number of men is 3,probability that one of the representative selected will be women and another will be men =
(3*7)/45 or 21/45
so P=1-(21/45 + 3/45)=21/45 <1/2 but if number of men is 0 or 1 P>1/2

okay!! got it! so i should see it as;

prob of selecting 2 men is <1/10

so there can be only 4,3,2,1, or 0 men

hence, there can be 6,7,8,9 or 10 women

if i calculate the prob for 6,7,8,9 or 10 women after 7 all values are greater than 1/2. so data is insuff.

i guess, i overlooked the fact that prob of selecting two men from a group of 10 is different prob of selecting 1 from group of 10.

If the question stated that the prob of selecting 1 man out of grp of 10 is <1/10 then clearly there are no men in the grp right?

THANKS!!
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by Simon » Fri May 13, 2011 1:14 pm
The question is P(w,w)>1/2?
Statement I states that women are more than half so number of women/ men is
6/4
7/3
8/2
9/1
10/0
so let's try two extremes to see if the p(w,w) is more than 1/2.
If 6 women then 6/10*5/9 = 30/90 . Now is P(w,w) >1/2 The answer is NO.

If there are 10 women then the P(w,w) is 1 . Now is the P(w,w) >1/2 > The answer is YES.

We have a YES & NO so statement I is Insufficient.

Statement 2: Probability of men is less than 1/10.
Let us try some numbers to see how many men would give us a probability less than 1/10.
We know for sure if there are 10 women the p(m,m) is zero. Therefore, if number of women is 10 then the P(w,w) is I and at the same time P(w,w) is greater than 1/2.

If there are 7 women and 3 men ( I tried 6 women and 4 men and it did not stand up to the second statement) then P(m,m) is 6/90 , which is less than 1/10, and the P(w,w) is less than 1/2.
We have a YES & NO so statement II is Insufficient.

Combining together we have YES, NO and YES, NO for both statements with same values. Therefore, E.
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by worldmovers » Sat Aug 27, 2011 3:47 am
mitzwillrockgmat 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 > 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.

I understand how stat 1 is insufficient:

more than half are women then the no, of women can be 6, 7, 8, 9 or even 10

so prob when there are 6 women is 6^C2 /10^C2 = 15/45 = 1/3 so 1/3 < 1/2

but when prob 10^C2 /10^C2 = 45/45 = 1 so 1 is not < 1/2

we dont get an unique value, so insufficient.


But I don't understand why stat 2 is insufficient. Couldn't statement two also mean that there are no males employees since the probability is less than 1/10???

If there are no male employees then the prob of 2 women is easy to find out.

OR does it mean that prob 2 men is : m/10 * (m-1)/9 < 1/10

& i can simplify it to:

m(m-1)/90 < 1/10

m(m-1) < 9

ans : m<9 & m-1 < 9 so m< 10

so number of men are less than 9. There could be 8,7,6,5,4,3,2,1 and all of these would give different values less than or equal to or greater than half.

does this make sense???

Please help clarify stat 2 for me! thanks! :)

OR

m(m-1) < 9

means that m is either 2 or 3
=> w = 8 or 7
prob of getting ww = 8*7/10*9 = 56/90 > 1/2

or ww= 7*6/10*9 = 42/90 < 1/2
Hence B is insufficient
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by navami » Sat Sep 03, 2011 12:41 pm
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.

-------------------------------------------------

P(Men) ,1/10

this is possible if number of men = 3 or less
2* 3 / 10* 9


in this case minimum probability of women = 7*6 / 10*9 = 1/2.14 or more

So B is not sufficient
This time no looking back!!!
Navami
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by jbivins » Sat Sep 03, 2011 2:34 pm
I did not read all the replies but they all seem really long and complicated.

I thought this line if reasoning might be easier to follow (if someone else has said it first!)

the sum of the probabilities equals one

probability all women +probability of a man and woman + probability of all men =1

(2) tells us probability of all men =1/10 but it tells us nothing about the other terms so we don't know what the probability of the other two terms is.
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by ilovemgmat » Sun Sep 04, 2011 3:05 am
I think E (had typed 'B' by mistake before, sorry for that!) is the answer. I don't know if I have worked the right way, but here is how I got it:

Probability of 2 women being chosen= [w!/[2!*(w-2)!]]/[10!/(2!*8!)]
= [w!/[2!*(w-2)!]]/45

So, to know if the probability is greater than 1/2, we need to know the range of w.

1)More than 1/2 are women:
So, there are 6, 7, 8, or 9 women.
-If there are 6 women, p<1/2
-If there are 7, 8, or 9 women, p>1/2
So, 1) alone is insufficient!

2) Probability that both are men< 1/10 can be written as
[m!/2!(m-2)!]/45<1/10
= [m(m-1)/2]*(1/45)<1/10
= m(m-1)/90<1/10
=(m^2-m-9)<0
Applying the formula to find the root of quadratic eqation, we get m<=3 (got very messed up here).

From 2), we know that w= 7, 8 or 9
When w=7, p<1/2
and when w=8 or 9, probability>1/2

Therefore, 2 alone is also not sufficient

From 1) and 2), we just know that w= 7, 8, or 9, which is not sufficient

So, IMO E is the answer.

Please comment what you guys feel. I am really not feeling confident :(
Last edited by ilovemgmat on Tue Sep 06, 2011 8:07 am, edited 2 times in total.
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by GMATGuruNY » Sun Sep 04, 2011 2:52 pm
mitzwillrockgmat 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 > 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.
7 women, 3 men:
Statement 1 is satisfied: more than 1/2 the employees are women.
Statement 2 is satisfied: P(MM) = 3/10 * 2/9 = 1/15, which is less than 1/10.
P(WW) = 7/10 * 6/9 = 7/15, so p<1/2.

10 women, 0 men:
Statement 1 is satisfied: more than 1/2 the employees are women.
Statement 2 is satisfied: P(MM) = 0.
P(WW) = 1, so p>1/2.

Since both statements are satisfied, and in the first case p< 1/2 and in the second case p>1/2, the two statements combined are insufficient.

The correct answer is E.
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by bubbliiiiiiii » Mon Sep 05, 2011 10:49 pm
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 .
I used the number plugin method, as stated by the poster of this question and expert, Mitch.

I concluded B to be insufficient in the following way,

We are given the probability that both are men. Thus, we cannot conclude anything on probability of selecting two women. However, if we subtract the given probability from 1, we get the probability that atleast one of the member is men. Thus, inconclusive.

If, there was an additional premise given that there are only two men, then B would have been sufficient. Please correct incase wrong.
Regards,

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