If a jury of 12 people is to be selected randomly from a

D

\(P=\frac{C^{10}_{8}*C^{5}_{4}+C^{10}_{9}*C^{5}_{3}+C^{10}_{10}*C^{5}_{2}}{C^{15}_{12}}=\frac{\frac{10*9}{2}*5+10*10+1*10}{\frac{15*14*13}{3*2}}=\frac{5*(45+20+2)}{5*91}=\frac{67}{91}\)

You have 10 men and 5 women.

Required probability is at least 8 men out of the 12 jurors.

This means we may have 8, 9 or 10 men in the 12 member jury.

Find the probability for each and add them up.

Probability of choosing 8 men and 4 women is:

(10C8*5C4)/15C12

Similarly for 9 and 10: (10C9*5C3)/15C12 and (10C10*5C2)/15C12.

Add them up and you get 67/91.

Hi,

You have 10 men and 5 women in the initial jury. The question asks that you should have at least 2/3 men which means that in 12 people selected at least 8 are men.

But if you select 12 out of 15 you can do it in (15!)/(12!*3!) or 455 ways. Note that the worst option i.e. with least men is 7men 5 women which can be done in 120 ways.

All other options are OK meaning 8m 4W, 9M 3W etc.

Solution is 1- 120/455=335/455=67/91

Regards

There can be below scenarios.
a). 8M + 4W = 10C8 5C4 = 225 ways.
b). 9M + 3W = 10C9 5C3 = 100
c). 10M + 2W = 10C10 5C2 = 10
So total selections possible = a+b+c = 335
Total ways to select 12 = 15C12 = 455

Probability = 335/455 = 67/91.

I would apppreciate kudos

It is very easy . I have solved this in less than a minute.

there are total 10 men and 5 women
in 12 jury members, could be:

M W T
10+2=12
9+3=12
8+4=12
7+5=12
M-men , W-women T - total

So out of 4 possible outcomes 3 are favorbale, 1 is unfavorable, it is 3/4=0.75

67/91~73%., which is the closest to 3/4.

Bingo!

There are 10 Men & 5 Females

Possible choices :
10M & 2F
9M & 3F
8M & 4F
7M & 5F

so :
P(at least 2/3 males = 8 males at least)
= 1 - P(7males and 5 females)
= 1 - [C(10,7)*C(5,5)/C(15,12)]
= 1 - 120*1/455
= 335/455
= 67/91

OA : D

A guessing strategy from Magoosh:

When you see that the problem can be solved using the 1-(P of opposite), in this case 1-(P of 5 women being selected), find two choices among the answers that added together yield 1. (no calculations what so ever, just the addition of any two answers must have 1 as result)

In this case its A and D. No other answers combo gives you 1 as result of addition.

When you subtract 24/91 from 1 you get 67/97, so the final answer. And we are looking for these "pairs" in the answer choices.

The P(5 women being selected) is a trap answer for those who forget to subtract that from 1. So the existence of these trap-answers that are supposed to confuse you allows you to use this strategy on a well-designed, typical GMAT question.

Then you have a 50% guess or:

see that you have a +- 33% chance (A) and a +-66% chance (D) choices. You can reason that the nº of men is quite large so its likely that the jury will be comprised mostly of men. so D looks like a good guess.

Thanks to Magoosh

Approach 1:-



Another way to solve this is "1-x" prob shortcut.
We need to know what is the probability of 7 men in jury, since it is the only way there are fewer than 8 men in jury ( maximum number of women is 5 ==> so 12-5=7 must be men). Then you have to 1-prob.of.7men.

1) All possible ways to assemble jury are : 15!/12!3!=455.
2) 7 of 10 men : 10!/7!3!=120
3) and the only way to include all women in jury is 5!/5!=1.

All possible ways to assemble 7men-5women jury will be 120*1=120.
Then, 120/455=24/91 - is the probability of 7 men and 5 woman in jury.
Hence, the probability that there will be more than 7 men in the jury is (1 - 24/91)=67/


Approach 2:-


there are total 10 men and 5 women
in 12 jury members, could be:

M W T
10+2=12
9+3=12
8+4=12
7+5=12
M-men , W-women T - total

So out of 4 possible outcomes 3 are favourbale, 1 is unfavorable, it is 3/4=0.75

67/91~73%., which is the closest to 3/4.

Hi All,

Yes, you CAN answer the question by solving the individual calculations that you listed. Here's how….

Since we're selecting 12 people from a group of 15, and the order doesn't matter, we can use the combination formula:

N!/[K!(N-K)!]

15c12 = 15!/[12!(3!)] = 455 possible groups of 12

Now we calculate each of the possible options that fits what we're looking for:

8 men (from 10) and 4 women (from 5) = (10c8)(5c4) = (45)(5) = 225
9 men (from 10) and 3 women (from 5) = (10c9)(5c3) = (10)(10) = 100
10 men (from 10) and 2 women (from 5) = (10c10)(5c2) = (1)(10) = 10

Total options with at least 8 men = 335

Probability = 335/455 = 67/91

Final Answer:

GMAT assassins aren't born, they're made,
Rich

I fell into the atleast trap. But the main trick really is understanding that P of ATLEAST 8 men = 1 - P(7 Men)

so, 1 - (10c7 * 5c5)/15c12 will give us 67/91

Asked: If a jury of 12 people is to be selected randomly from a pool of 15 potential jurors, and the jury pool consists of 2/3 men and 1/3 women, what is the probability that the jury will comprise at least 2/3 men?

Total men = (2/3)*15 = 10
Total women = (1/3)*15 = 5

Total jury members = 12
Men jury members > (2/3)*12 = 8

Ways that the jury will comprise at least 2/3 men = 10C8*5C4+ 10C9*5C3 + 10C10*5C2 = 45*5 + 10*10 + 1*10 = 225 + 100 + 10 = 335

Total ways = 10C7*5C5 + 10C8*5C4 + 10C9*5C3 + 10C10*5C2 = 120*1 + 45*5 + 10*10 + 1*10 = 120 + 225 + 100 + 10 = 455

The probability that the jury will comprise at least 2/3 men = 335/455 = 67/91

IMO D

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