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

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Senior Manager
Joined: 08 Dec 2015
Posts: 287
GMAT 1: 600 Q44 V27
If a jury of 12 people is to be selected randomly from a  [#permalink]

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08 Jun 2016, 09:41
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
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Re: If a jury of 12 people is to be selected randomly from a  [#permalink]

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07 Jan 2017, 09:30
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.
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Re: If a jury of 12 people is to be selected randomly from a  [#permalink]

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05 Mar 2018, 12:31
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

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Rich
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Re: If a jury of 12 people is to be selected randomly from a  [#permalink]

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16 Mar 2019, 19:04
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Re: If a jury of 12 people is to be selected randomly from a   [#permalink] 16 Mar 2019, 19:04

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