I Suck at Probability need desperate help

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by Jim@StratusPrep » Wed May 28, 2014 6:07 am
Just something to add here: This is NOT an easy problem. You must have a firm grasp of several concepts and pick out some tough steps. It's unique - while these concepts are good to know, you shouldn't think of this as a problem that you must be able to solve, but rather a top notch problem. Use the small pieces to identify areas you can work on, but don't get distraught by having difficulty with this.
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by ujjwal.dheer » Sun Aug 10, 2014 1:50 am
tanviet wrote:
sanjana wrote:
Kaunteya 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 2/3 men and 1/3 women, what is the probability that the jury will comprise at least 2/3 men?
a. 24/91
b. 5/91
c. 2/3
d. 67/91
e. 84/91

Very similiar to the last problem that I posted but still haven't figured out how to solve these. If someone can please guide that I would really
appreciate it. If you could solve this one, could you please solve the other one I posted titled: Combination and Permutation Manhattan. Thanks again guys.

Kaunteya
To this problem I came up with the below solution,

As found out by Stuart, there are 10 men and 5 Women from which the committee of 12 members needs to be picked.
Now the committee needs to be have atleast 2/3 Men,i.e atleast 8 men.
Thats P(M>=8) = 1-P(M<8)
=1-P(M=7)
=1-(10C7*5C5)15C12
=67/91,which is the correct Answer.

Stuart,

I have a question for you here..
For all problems that Involve Atleast or Atmost I almost always use the Rule P(A) = 1-P(A') and so far I have never got the wrong answer.
Is this the right approach or am I missing something?

Look forward to your valuable advice!

Thanks,
Sanjana
you are wrong. P(M>=8) is not equal to 1-P(M=7)

but P(M>=8) = 1- P(M=7)-P(M=6)-P(M=5)-P(M=4)-P(M=3)-P(M=2)-P(M=1)-P(M=0)

so the approach "one minus"here is not good
To your reply , I would like to say this approach perfectly works for this question because
M cannot be less than 7 as it would mean having 6 women in the jury which is not possible as per the data in the question. This approach is a great time saver. Kudos Sanjana

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by mlevit89 » Thu May 14, 2015 9:20 am
This problem is easier if you just find the probability that there will be all 5 women in the group and subtract that probability from 1. This will give you the same answer in a shorter amount of time.

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by nikhilgmat31 » Sun May 31, 2015 10:25 pm
I like pkit's way to solve jury question.
Bingo...

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by umbe » Mon Jul 13, 2015 2:49 am
fleshins wrote:Does anyone know the answers / solutions to the questions above? I *think* I'm solving the 1st correctly - but have no idea how to do the 2nd.

Solution for the first is:

1. 6C4 total possible combinations of boys and girls
2. only successful combination is 2 girls and 2 boys
3. number of ways we can chose 2 girls and 2 boys is 3C2 boys x 3C2 girls = 9
4. F/T rule is 9/6C4 = 9/15 = 3/5 (D)

Is that right?

Anyone know how to do the 2nd one?
sorry to bother, i'm new in the forum
isn't the solution to the first 1/10(A)?
we want to end up with the same number of girls and boys so 1/2 x 2/5 x 3/4 x 2/3, no matter with which category we start first having the two the same number

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by lotrgandalf » Wed Aug 02, 2017 10:46 am
Awesome strategy.

Now we have a big split among the remaining choices. (a) and (b) are both very small (less than 1/3) and (d) and (e) are both big (more than 2/3). Since 2/3 of the jury pool are men, does it make any sense that there would be a small probability that 2/3 of the actual jury will be men too? Of course not, so (a) and (b) don't really make sense. So, if we're guessing, choose (d) or (e). (Further, we predict that the answer should be a bit more than 2/3 - (e) really seems too big, so (d) looks like the best guess.)

Never discount the power of strategic guessing, especially on really time consuming questions![/quote]

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by Matt@VeritasPrep » Sun Aug 06, 2017 11:04 pm
Yeah, this one is all about strategic guessing - it almost punishes you for doing it "properly".

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by Matteo_A » Fri Oct 06, 2017 12:41 am
Stuart Kovinsky wrote:
Kaunteya 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 2/3 men and 1/3 women, what is the probability that the jury will comprise at least 2/3 men?
a. 24/91
b. 5/91
c. 2/3
d. 67/91
e. 84/91

Very similiar to the last problem that I posted but still haven't figured out how to solve these. If someone can please guide that I would really appreciate it. If you could solve this one, could you please solve the other one I posted titled: Combination and Permutation Manhattan. Thanks again guys.

Kaunteya
If 2/3 are men, we have 10 men and 5 women.

We want to know the probability that at LEAST 2/3 of the people actually selected will be men. In other words, that at least 8 out of the 12 jury members will be men.

There are three scenarios in which this could happen:

8 men and 4 women;

9 men and 3 women; and

10 men and 2 women.

Let's see how many different ways we can make each of these occur.

There are 10 men total, so there are 10C8 different groups of 8 men. There are 5 women, so there are 5C4 different groups of 4 women.

Therefore, scenario 1 has 10C8 * 5C4 = 10!/8!2! * 5!/4!1! = 45 * 5 = 225 possible juries.

For scenario 2, we have 10C9 * 5C3 = 10!/9!1! * 5!/3!2! = 10 * 10 = 100 possible juries.

For scenario 3, we have 10C10 * 5C2 = 10!/10!0! * 5!/2!3! = 1 * 10 = 10 possible juries.

[remember, 0!=1]

Now, since this is a probability question, we we want to use the probability formula.

Probability = #desired outcomes / total # of possible outcomes.

We've already calculated the # of desired outcomes: 225 + 100 + 10 = 335 juries with at least 8 men on them.

The total # of possible outcomes is the total # of possible juries, which is simply 15C12 = 15!/12!3! = 15*14*13/3*2*1 = 5*7*13 = lots, so let's reduce instead!

So:

335/5*7*13 = 67/7*13 = 67/91

choose (d).

Let's also look at this question from a strategic guessing point of view.

2/3 of the jury pool are men. Let's eliminate (c) 2/3, because that's way too easy.

Now we have a big split among the remaining choices. (a) and (b) are both very small (less than 1/3) and (d) and (e) are both big (more than 2/3). Since 2/3 of the jury pool are men, does it make any sense that there would be a small probability that 2/3 of the actual jury will be men too? Of course not, so (a) and (b) don't really make sense. So, if we're guessing, choose (d) or (e). (Further, we predict that the answer should be a bit more than 2/3 - (e) really seems too big, so (d) looks like the best guess.)

Never discount the power of strategic guessing, especially on really time consuming questions!
I've tried to do in a different way, could you please tell me why it is wrong?
p(at least 4 women) = 1 - p(5 women, 7 men) = 1 - [(5x4x3x2x1x10x9x8x7x6x5x4)/(15x14x13x12x11x10x9x8x7x6x5x4)] = 1 - (1/3003) = 0.99
So I would choose E!
I can't see why it is wrong...

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by deepak4mba » Sun Feb 25, 2018 4:03 am

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