I Suck at Probability need desperate help
- Jim@StratusPrep
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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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- ujjwal.dheer
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To your reply , I would like to say this approach perfectly works for this question becausetanviet wrote:you are wrong. P(M>=8) is not equal to 1-P(M=7)sanjana wrote:To this problem I came up with the below solution,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
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
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
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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sorry to bother, i'm new in the forumfleshins 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?
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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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]
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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Yeah, this one is all about strategic guessing - it almost punishes you for doing it "properly".
I've tried to do in a different way, could you please tell me why it is wrong?Stuart Kovinsky wrote:If 2/3 are men, we have 10 men and 5 women.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
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!
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...
- deepak4mba
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- deepak4mba
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