I Suck at Probability need desperate help

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I Suck at Probability need desperate help

by Kaunteya » Thu Feb 28, 2008 11:30 am
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

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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!
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by Kaunteya » Thu Feb 28, 2008 1:40 pm
Yeah I appreciate your help Stuart. Also, your reasonning for an educated guess was what I actually did when taking the timed exam. Could you take a look at the question I posted: Permutation and Combination Manhattan. Thanks.

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by ash g » Mon Mar 10, 2008 6:16 pm
Hey Stuart,

Regarding
"(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.) "


I think
[1] these probability problems would normally be in 700-800 range and to be good at strategic guessing to deal with such problems would be a very good skill to have - quickkill.
[2] this usually comes with practice but is there any other way to roughly get an idea of approximate i.e. thought process ?

The reason I ask is that there are lots of such problems requring the same approach...
FOR EXAMPLE, Please dont solve
-------------------
From a group of 3 boys and 3 girls, 4 children are to be randomly selected. What is the probability that equal numbers of boys and girls will be selected?

A. 1/ 10 B. 4/9 C 1/2 D 3/5 E 2/3
--------------------
Richard has 3 green, 2 red, and 3 blue balls in a bag.He randomly picks 5 from the bag without replacement.What is the probability that of the 5 drawn balls, Richard has picked 1red, 2 green, and 2 blue balls?
8/28 9/28 10/28 10/18 11/18
---------------------


Regards,
Ash

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by fleshins » Sat Nov 15, 2008 6:31 pm
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?

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by cramya » Sat Nov 15, 2008 6:34 pm
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
I dont have the answers but thsi is how I wil solve 2 (if this is the question u r referring to)

Richard has 3 green, 2 red, and 3 blue balls in a bag.He randomly picks 5 from the bag without replacement.What is the probability that of the 5 drawn balls, Richard has picked 1red, 2 green, and 2 blue balls?
8/28 9/28 10/28 10/18 11/18


2C1*3C2*3C2 / 8C5 = 9/28

B)

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by fleshins » Sun Nov 16, 2008 12:39 am
Thanks Cramya. What if after each selection, the ball was replaced? How would that change the solution/answer?

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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
Hello !

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)=67/91

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by NikolayZ » Sat Oct 03, 2009 2:57 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 for double posting =)

Second one also may be solved next way :

All possible ways to grab 5 balls out of 8 ( 2+3+3) :

1 2 3 4 5 6 7 8
Y Y Y Y Y N N N =8!/5!*3!= 56

Since there are 2 red balls, then you could grab 1 of 2 , 2!/1!= 2.

2 green and 2 blue balls grabbing ways would be - 3!/2! =3 each.

2*3*3 = 18 ways to organize 5 grabbed things in the way as problem says. then the probability will be equal to 18/56= 9/28.

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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

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by Stuart@KaplanGMAT » Sat Oct 03, 2009 8:43 pm
The "one minus" approach does indeed work very well on GMAT at least/at most questions; questions are often designed to reward those who see non-traditional approaches.
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by sanjana » Sat Oct 03, 2009 10:04 pm
Thanks Stuart! :D

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by pkit » Mon Aug 09, 2010 12:06 pm
It is very easy . I have solved this in less than a minute. 8-)

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!

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by tanviet » Sun Nov 07, 2010 7:58 pm
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

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by kushal.adhia » Mon Nov 08, 2010 6:09 am
@pkit

u might want to have a look at this article:

https://www.beatthegmat.com/mba/2010/11/ ... vernacular