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

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If a jury of 12 people is to be selected randomly from a [#permalink] New post 22 Jan 2007, 18:17
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If a jury of 12 people is to be selected randomly from a pool of 15 potential jurors, and the jury pool consists of 10 men and 5 women, what is the probability that the jury will comprise at least 8 men?

a) 24/91
b) 45/91
c) 2/3
d) 67/91
e) 84/91

I can't get my mind around this one and don't have the OA. Would like to see you probability expert how to handle a question like this.
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Re: PS: Jury (probability) [#permalink] New post 22 Jan 2007, 18:33
jylo wrote:
If a jury of 12 people is to be selected randomly from a pool of 15 potential jurors, and the jury pool consists of 10 men and 5 women, what is the probability that the jury will comprise at least 8 men?

a) 24/91
b) 45/91
c) 2/3
d) 67/91
e) 84/91

I can't get my mind around this one and don't have the OA. Would like to see you probability expert how to handle a question like this.


Total number of ways to select 12 people from 15 is 15C12 = 455

number of ways in which jury has at least 8 men:

= 10C8*5C4+10C9*5C3+10C10*5C2
= 225+ 100+10 = 335

=> probability = 335/455 = 67/91
Ans: D
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Re: PS: Jury (probability) [#permalink] New post 22 Jan 2007, 23:05
hsampath wrote:
jylo wrote:
If a jury of 12 people is to be selected randomly from a pool of 15 potential jurors, and the jury pool consists of 10 men and 5 women, what is the probability that the jury will comprise at least 8 men?

a) 24/91
b) 45/91
c) 2/3
d) 67/91
e) 84/91

I can't get my mind around this one and don't have the OA. Would like to see you probability expert how to handle a question like this.


Total number of ways to select 12 people from 15 is 15C12 = 455

number of ways in which jury has at least 8 men:

= 10C8*5C4+10C9*5C3+10C10*5C2
= 225+ 100+10 = 335

=> probability = 335/455 = 67/91
Ans: D


Just to elucidate:
A jury of 12 with ATLEAST 8 men =
1. 8 men and 4 women
2. 9 men and 3 women
3. 10 men and 2 women

Total number of such juries possible = 1 + 2 + 3
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Re: PS: Jury (probability) [#permalink] New post 09 Dec 2007, 12:21
kripalkavi wrote:
hsampath wrote:
jylo wrote:
If a jury of 12 people is to be selected randomly from a pool of 15 potential jurors, and the jury pool consists of 10 men and 5 women, what is the probability that the jury will comprise at least 8 men?

a) 24/91
b) 45/91
c) 2/3
d) 67/91
e) 84/91

I can't get my mind around this one and don't have the OA. Would like to see you probability expert how to handle a question like this.


Total number of ways to select 12 people from 15 is 15C12 = 455

number of ways in which jury has at least 8 men:

= 10C8*5C4+10C9*5C3+10C10*5C2
= 225+ 100+10 = 335

=> probability = 335/455 = 67/91
Ans: D


Just to elucidate:
A jury of 12 with ATLEAST 8 men =
1. 8 men and 4 women
2. 9 men and 3 women
3. 10 men and 2 women

Total number of such juries possible = 1 + 2 + 3


You can flip this around to figure out the probability of a jury of 12 with at least 5 women, then subtract the probability from 1.
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 [#permalink] New post 09 Dec 2007, 12:47
Having trouble understanding this part:

3. 10 men and 2 women

Wouldn't that be 12!/(10!)(2!). I'm sure i'm doing it wrong. Can someone correct my math?
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 [#permalink] New post 09 Dec 2007, 12:51
Can you elaborate on that? Please explain what I'm seeing here:

10!/10! * 5!/2!3! = 10

Nevermind see it.
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 [#permalink] New post 09 Dec 2007, 20:09
i took approach of calculating probability of each case on its own, and added them all up and ended with D.

is that the OA ?
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 [#permalink] New post 09 Dec 2007, 21:22
Ans is D - I recall it from MGMAT test:

15C12 = 2730 total combinations how set of 12 could be found from 15

"what is the probability that the jury will comprise at least 8 men" = 1 - prob that jury can contain less that 8 men ==> There is only 1 scenario - when m = 7 and w = 5.

So, how we can find total combinations for set of 7 men from 10 men available? ==> 10C7

So, how we can find total combinations for set of 5 women from 5 women available? ==> 5C5

Finally: 10C7 + 5C5 = 720 combinations when m = 7 and w = 5

The probability of such scenario is: 720/2730 = 24/91.

1 - prob that jury can contain less that 8 men = 1 - 24/91 = 67/91 ==> Ans D
  [#permalink] 09 Dec 2007, 21:22
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