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

OE


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

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.
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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

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 !
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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.
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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.
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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.

I did it a slightly different way...

Using Combinations....

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

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!/(5!*1!)]*[3!/(1!*2!)] = 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!/(5!*1!)]*[1!] = 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.

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
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