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I Suck at Probability need desperate help

This topic has 6 expert replies and 63 member replies
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pesfunk Master | Next Rank: 500 Posts
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Mon Nov 08, 2010 7:00 am
Nice Sanjana...One Minus is probably the best approach

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

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?

Thanks,
Sanjana

anshumishra Legendary Member
Joined
15 Jun 2010
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Tue Nov 30, 2010 7:47 am
duongthang 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

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?

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
I would agree with you.
However, luckily here, P(M=6), P(M=5).....P(M=0) is 0, as the number of women is 5 only. [So, P(M=6) which should have meant P(M=6,W=6) is an impossible event and so on...]
That is why, the answer still matched.

tomada Master | Next Rank: 500 Posts
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Sat Dec 25, 2010 3:12 pm
Amen!

_________________
I'm really old, but I'll never be too old to become more educated.

amitpundir007 Newbie | Next Rank: 10 Posts
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Tue Mar 01, 2011 8:19 pm
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
Kaunteya
Let us see probability of not happening of 2/3 of man in selected group.
7 Men 5 women + 1 women 11 men (not possible)

so 7 m 5 w = 10c7 X 5 c5 = 120

Total no of outcome = 15C12 = 455

probaility of not happening = 120/455

Probability of heppening the event = 1- 120/455 = 335/455 = 67/91 choice D

coolly01 Junior | Next Rank: 30 Posts
Joined
20 Feb 2011
Posted:
29 messages
Wed Mar 02, 2011 8:36 pm
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!
Hi steward,
Thanks for your explanation, but i always keep in my mind another reasoning that i don't know what's wrong
with it. Could you please point out where does it go wrong. I really much appreciate that

All the probalibilities should occour like below:
10 M : 2 W (1/4)
9 M : 3 W (1/4)
8 M : 4 W (1/4)
7 M : 5 W (1/4)
So the probability that at least 8 men were chosen should be 3/4. I really can't figure out why we should need factorial

GMAT/MBA Expert

Stuart Kovinsky GMAT Instructor
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Thu Mar 03, 2011 1:55 pm
coolly01 wrote:
Thanks for your explanation, but i always keep in my mind another reasoning that i don't know what's wrong
with it. Could you please point out where does it go wrong. I really much appreciate that

All the probalibilities should occour like below:
10 M : 2 W (1/4)
9 M : 3 W (1/4)
8 M : 4 W (1/4)
7 M : 5 W (1/4)
So the probability that at least 8 men were chosen should be 3/4. I really can't figure out why we should need factorial
Hi,

there definitely isn't a 1/4 probability of each event occurring.

Let's look at another example: if we flip a fair coin 5 times, there are 6 things that can happen:

1H, 4T
2H, 4T
3H, 2T
4H, 1T
5H, 0T

According to your reasoning, there's a 1/6 chance of each event happening. However, it's definitely more likely that you'll get 4H and 1T than 5H.

Here's why: probability = # of desired outcomes / total # of possibilities. When we flip a coin 5 times, there's only 1 way to get all heads - HHHHH. However, there are 5 ways to get 4 heads - HHHHT, HHHTH, HHTHH, HTHHH and THHHH. Since there are 5 ways to get 4 heads and only 1 way to get 5 heads, it's 5 times as likely to get 4 heads as 5.

Similarly, in the question here, there are considerably more ways to choose 8 men and 4 women than 10 men and 2 women, so the probability of each scenario is skewed.

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pankajks2010 Master | Next Rank: 500 Posts
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Tue Apr 05, 2011 9:25 am
duongthang 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

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?

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
I beg to differ. In this case Sanjana's approach is correct as the maximum number of females in this case is 5. Thus, one cannot consider the option of P(M<=6).

Hillel Newbie | Next Rank: 10 Posts
Joined
04 Apr 2011
Posted:
3 messages
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Sun Apr 17, 2011 1:51 pm
All options are choose 12 out of 15, which is (n!/(k!(n-k)!))
15
12
= 15*14*13/3! = 455

by one minus:
only option to choose not 2/3 men is to choose all women:
5 out of 5 = 1 option multiple by choosing 7 men out of 10:
10
7
= 10!/7!3! = 120

---------
1-120/455 = 335/455 = 67/91

ranjithreddy.k9 Senior | Next Rank: 100 Posts
Joined
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Posted:
71 messages
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Sat Apr 23, 2011 6:27 pm
ans is 67/91

(10c8*5c4)+(10c9*5c3)+(10c10*5c2)/15c12 which equals 67/91

pizza2equity Newbie | Next Rank: 10 Posts
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Wed Jul 13, 2011 6:54 am
The answer is D, but I would suggest a much faster way to solve this problem.
I think the key is to think a little bit before starting to crunch the numbers.

10 out of the 15 potential jurors are men, 5 women. The question asks for the probability of jury that will comprise at least 2/3 men.

We can MUCH easier and faster answer the question What is the probability that the jury will comprise LESS THAN 2/3 men? (*)

It is only possible when each of the persons WHO ARE NOT CHOSEN are men.

All the computations you have to make:

Probability of (*): 10/15 * 9/14 * 8/13
After fast simplification: 3*8 / 7*13 = 24/91

Of course, the correct answer will be 1-(24/91)=67/91

My advice: Think about potential shortcuts for an additional 5-10 seconds in the beginning and you can save several minutes!

krishp84 Junior | Next Rank: 30 Posts
Joined
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Posted:
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Sat Jul 16, 2011 10:30 am
pizza2equity - Can you explain the below solution ?
I feel this should be the best solution, if you can explain the same.

More specifically - the highlighted part

pizza2equity wrote:
The answer is D, but I would suggest a much faster way to solve this problem.
I think the key is to think a little bit before starting to crunch the numbers.

10 out of the 15 potential jurors are men, 5 women. The question asks for the probability of jury that will comprise at least 2/3 men.

We can MUCH easier and faster answer the question What is the probability that the jury will comprise LESS THAN 2/3 men? (*)

It is only possible when each of the persons WHO ARE NOT CHOSEN are men.

All the computations you have to make:

Probability of (*): 10/15 * 9/14 * 8/13
After fast simplification: 3*8 / 7*13 = 24/91

Of course, the correct answer will be 1-(24/91)=67/91

My advice: Think about potential shortcuts for an additional 5-10 seconds in the beginning and you can save several minutes!

olegpoi Senior | Next Rank: 100 Posts
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Wed Aug 03, 2011 12:28 am
IMO D
1 - unwanted(with 5 female jurors) = wanted outcome(at least 8 male jurors)
Total combinations of 12-person jury is (15x14x13x.....x4)/12! = 5x7x13;
unwanted combinations = (10x9x8x...x4)/7!=5x3x8

5x7x13/5x7x13 - 5x3x8/5x7x13 = 67/91 (D)

fr743 Junior | Next Rank: 30 Posts
Joined
22 Dec 2010
Posted:
18 messages
Wed Aug 03, 2011 1:24 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!
Stuart, but why exactly 3 scenarios? Why not 11 + 1, 7 + 5 and etc. ?

saketk Legendary Member
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Sat Aug 20, 2011 10:24 am
10 men and 5 women
12 people

atleast 8 men should be there.
10C8*5C4
10C9*5C3
10C10*5C2

do the total ... and divide it by 15C12. you will get the answer.

Or the shorter version.. find for 7 men and 5 women.. 1- the result of this equation will be answer.

olegpoi Senior | Next Rank: 100 Posts
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Thu Aug 25, 2011 11:11 am
IMO: 67/91
First calculate total combinations - 91

Then 2\3 of male members of jury of 12 is 8 men => 8 or more men in jury are wanted

The following scenarios of combinaions divided by total 91 to add:
1) 4w + 8m = 5*5*9
2) 3w + 9m = 5*2*10
3) 2w + 10m = 10

Result in 67/91 - probability of selecting 12 person jury where at least 2/3 are men.

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