A jury pool consists of 6 men and w women. If 2 jurors

Question

A jury pool consists of 6 men and w women. If 2 jurors are selected from the pool at random, is the probability that 2 men will be selected higher than the probability that 1 man and 1 woman will be selected?

(1) w ≥ 3
(2) w < 6

(1) w ≥ 3
(2) w < 6

OE
 Rephrase the question as Is 30 / (6 + w)(5 + w) > 12w / (6 + w)(5 + w)?
(1): Always No
Sufficient
(2): w might be less than 3 but could be higher
Insufficient

Hi, Although I didn't put whole OE, I want to know how we can solve this question, please.

Bunuel · Accepted Answer

A jury pool consists of 6 men and w women. If 2 jurors are selected from the pool at random, is the probability that 2 men will be selected higher than the probability that 1 man and 1 woman will be selected? The total number of people in the jury = 6+w . The probability of selecting 2 men when selecting 2 jurors = \frac{6}{6+w}*\frac{5}{(6+w)-1}=\frac{30}{(6+w)(5+w)} ; The probability of selecting 1 man and 1 woman when selecting 2 jurors = 2*\frac{6}{6+w}*\frac{w}{(6+w)-1}=\frac{12w}{(6+w)(5+w)} : multiplying by 2 as MW can occur in two ways MW or WM; The question asks: is \frac{30}{(6+w)(5+w)}>\frac{12w}{(6+w)(5+w)} ? --> is 30>12w ? --> is w<2.5 ? So, the question basically asks whether there are 2, 1, or 0 women in the jury. (1) w ≥ 3. Directly gives a NO answer to the question. Sufficient. (2) w < 6. Not sufficient. Answer: A. Similar questions to practice: a-jar-contains-8-red-marbles-and-y-white-marbles-if-joan-101748.html (almost the same question from MGMAT) a-certain-jar-contains-only-b-black-marbles-w-white-marbles-104924.html if-2-different-representatives-are-to-be-selected-at-random-128233.html a-certain-box-contains-only-blue-b-green-g-and-red-r-153384.html Hope this helps.

Bunuel · Answer

Similar questions to practice: a-jar-contains-8-red-marbles-and-y-white-marbles-if-joan-101748.html (almost the same question from MGMAT) a-certain-jar-contains-only-b-black-marbles-w-white-marbles-104924.html if-2-different-representatives-are-to-be-selected-at-random-128233.html a-certain-box-contains-only-blue-b-green-g-and-red-r-153384.html Hope this helps.

christianbze · Answer

I guess you did not understand the solution.

First, you have to find the probability for 2 men being picked and for 1 man and 1 woman being picked.

2 men:
- combinatorics:  6 x 5 
- all possibilities:  (6+w)(6+w-1)

1 man and 1 woman:
- combinatorics:  6 x w x 2 
- all possibilities:  (6+w)(6+w-1) 
You can choose out of 6 men, w woman and you have two different places to set them.

So your inequality is:
 \frac{6x5}{(6+w)(6+w-1)} > \frac{6 x w x 2}{(6+w)(6+w-1)}

\frac{30}{(6+w)(5+w)} > \frac{12 w}{(6+w)(5+w)}

as both denominators are the same, we just check the numerators.
So how long is  30 > 12w ?
As long as w (which has to be an integer!) is smaller than 3. So 0, 1, and 2.

So look at your options.

joshlevin90 · Answer

i dont understand how you are getting your answer is there another way to explain it or if you can add in more steps as you go along solving it thank you :-D

joshlevin90 · Answer

could you pls explain in detail how you got that equation?

sanjoo · Answer

I tried...

2/6+w>1/6*1/w

so i got 12w>6+W
11w>6

w>6/11??
so question becums is w>6/11?

ans is A..S>=3

B is not sufficient ..because we cant say its greater than 6/11 or not..

My ans is correct.. Bt i think m not doing it rite !

oss198 · Answer

Can someone tell me where i'm wrong?

1) Let's take the case when W=3. So there are 6 men and 3 women.

probability of taking 2 men 
(6C1*5C1)/9C2 = 30/72

probability of taking 1 man and 1 woman 
(6C1*3C1)/9C2 = 18/72

30/72 > 18/72, for 3 women. But as the number of women will rise, the probabilty will reverse. So insufficient.

Do i have to multiply 6C1*3C1 by two as we can take 1 man first, or 1 woman first?

Bunuel · Answer

If there are 3 women, then:

The probability of selecting 2 men is (6C2)/(9C2) = 15/36.

The probability of selecting 1 men and woman is (6C1*3C1)/9C2 = 18/36.

15/36 < 18/36. If you increase the number of women the probability of the first case will always be less than the probability of the second.

Hope it's clear.

oss198 · Answer

Thank you for the answer but you mean  2* (6C1*3C1) don't you?

we have to multiply by 2 because it can be WM or MW?

Bunuel · Answer

No. When you do with combinations method you don't need to account for different ways of occurring because this is already taken care of with this method.

oss198 · Answer

Thank you very much for your patience

gayam · Answer

The no. of ways in which you can select 2 men from 6 men = 6c2 = 6!/(2! * 4!) = 15
no. of ways in which you can select a man from 6 men & a women from w women = 6C1 * WC1 = 6W

As per statement 1, W>=3 =>6W >=18 so always choosing a men & women has more chances than 2 men being selected from 6
As per statement 2, W<6 =>if w = 1 no. of ways you can choose a man & women is 6 which is less than 15(no. of ways for choosing 2 men from 6)
 but if W = 5 no. of ways you can choose a man & women is 6 * 5 = 30 which is greater than 15. Hence statement 2 alone cannot be used for solving the problem.

So the answer is A.

gooner · Answer

Fantastic Bunuel!! I was struggling with the "multiply by 2" issue and trying to figure out when to do so and when not to.

Problem solved! Thanks Much.

iPen · Answer

I did it a slightly different way...

Using Combinations....

The probability of 2 random jurors from the entire juror pool is (6+w)!/ .

Looking at the first condition, (1) w≥3, just use 3 to represent women - 2 men have the greatest odds of selection with the fewest number of women. If the probability of selecting 2 men is  worse  than the probability of 1 man and 1 woman when there are only 3 women, then selecting 2 men has even worse odds when we increase the number of women (e.g. w=4,5,6, etc.).

So, total combination of all in the juror pool is 9!/(2!*7!) = 9*8/2 = 36

The combination of 2 men is 6!/(4!*2!) = 6*5/2 = 15

The combination of 1 man and 1 woman is  *  = 6*3 = 18

2 men < 1 man, 1 woman
15/36 < 18/36

This answers as a  no  to condition (1) and thus it's sufficient.

Using a similar logic to condition (1), we test the fewest possible number of women for condition (2), which is 1 (or 0...). We already know that having 3 women will mean that selecting 2 men has a smaller probability. Then, will the min. number of women still result in the same outcome (less chance for two men...)? If so, then it satisfies. If not, then two men have a greater probability of selection and condition (2) is insufficient.

The total combination of all in the juror pool is 7!/(2!*5!) = 7*6/2 = 21

Combination of 1 man and 1 women with only 1 female juror in the pool is  *  = 6*1 = 6

Combination of 2 men is 6!/(4!*2!) = 6*5/2 = 15

1 man, 1 woman < 2 men
6/21 < 15/21

So 2 men have a greater probability. We already know that 3 women in the juror pool means a greater probability for the 1 man, 1 woman probability, so this condition is insufficient.

AlexIV · Answer

Dear Bunuel, would you please confirm if this method is correct:

If there are 3 woman:
1 woman and 1 man: 6/9*3/8+3/9*6/8=36/72
2 man: 6/9*5/8=30/36
Probability for 1 woman and 1 man is higher than for 2 man

4 woman: 
1 woman and 1 man: 6/10*4/9+4/10*6/9=48/90
2 man: 6/10*5/9=30/90
Probability for 1 woman and 1 man is higher than for 2 man

And so on, the probability of 1w to 1m will be higher as the number of woman increases.

Thank you!

cledgard · Answer

P = Success/ Total possibilities

P(2 men) = P (1st man) * P (2nd man) ={ 6/(6+w) } * { 5/(5+w) } = 30 / {(6+w) (5+w)}

P(1 man + 1 woman ) = P (1st man) * P (2nd woman) + P (1st woman) * P (2nd man) 
= { 6/(6+w) } * { w/(5+w) } + { w/(6+w) } * { 6/(5+w) } = 12w / {(6+w) (5+w)}

Question: is P(2 men) > P(1 man + 1 woman ) ? = is 30 / {(6+w) (5+w)} > = 12w / {(6+w) (5+w)} ?

Divide both sides by 12 / {(6+w) (5+w)} => Question: is 2.5 > w ?

(1) w ≥ 3 SUFFICIENT
(2) w < 6 ; 2 = NO, 3 = YES ; INSUFFICIENT

Answer is A

BrentGMATPrepNow · Answer

Target question :  Is P(2 men) greater than P(1 man and 1 woman)?

Statement 1 : w >  3 
Let's see what happens if w = 3 (note: this is the best chance that P(2 men) will be greater than P(1 man and 1 woman)

P(2 men) = P(man selected 1st and man selected 2nd)
= (6/9)(5/8)
= 30/72

P(1 man and 1 woman) = P(man selected 1st and woman selected 2nd  OR  woman selected 1st and man selected 2nd)
= P(man selected 1st and woman selected 2nd)  +  P(woman selected 1st and man selected 2nd)
= (6/9)(3/8)  +  (3/9)(6/8)
= 18/72  +  18/72
= 36/72

So, when w = 3,  P(2 men) is  not  greater than P(1 man and 1 woman)

IMPORTANT: Now that we've shown that P(2 men) is  not  greater than P(1 man and 1 woman) when w = 3, we can see that, as the value of w increases, the answer to the  target question  will always remain the same.

As such, statement 1 is SUFFICIENT

Statement 2 : w < 6  
Consider these two conflicting cases:

Case a : w = 1
P(2 men) = (6/7)(5/6)
= 30/42

P(1 man and 1 woman) = P(man selected 1st and woman selected 2nd  OR  woman selected 1st and man selected 2nd)
= P(man selected 1st and woman selected 2nd)  +  P(woman selected 1st and man selected 2nd)
= (6/7)(1/6)  +  (1/6)(6/7)
= 6/42  +  6/42
= 12/42

So, when w = 1,  P(2 men)  is  greater than P(1 man and 1 woman)

Case b : w = 3
In statement 1, we already showed that, when w = 3,   P(2 men) is  not  greater than P(1 man and 1 woman)

Since we cannot answer the  target question  with certainty, statement 2 is NOT SUFFICIENT

Answer = A

Cheers,
Brent

Shobhit7 · Answer

Ways of selecting 2 men: 6C2
Ways of selecting 1 man and 1 woman: 6C1*wC1
Total Ways: (6+w)C2

Rephrase Q stem:
Is 6C2 / (6+w)C2 > (6C1*wC1) / (6+w)C2 ?
Is 6C2 > 6C1*wC1 ?
Is 15>6w ?
Is w<15/6?
Is w<2.5?

St.1: No, w>=3
St.2: May be. w<6

Ans A

ankur_1988 · Answer

Hi,

In case of selecting 1 man and 1 women, can anyone please explain why are we multiplying 2 as it is immaterial whether we select a women followed by a man or the other way round.

A jury pool consists of 6 men and w women. If 2 jurors

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A jury pool consists of 6 men and w women. If 2 jurors

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