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
I Suck at Probability need desperate help
- pesfunk
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Nice Sanjana...One Minus is probably the best approach
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I would agree with you.duongthang 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
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.
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Amen!
kushal.adhia wrote:@pkit
u might want to have a look at this article:
https://www.beatthegmat.com/mba/2010/11/ ... vernacular
I'm really old, but I'll never be too old to become more educated.
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Let us see probability of not happening of 2/3 of man in selected group.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
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
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!
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
Please help, I really makes me headached
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Hi,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
Please help, I really makes me headached
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:
0 heads, 5 tails
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.
Stuart Kovinsky | Kaplan GMAT Faculty | Toronto
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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).duongthang 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
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
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
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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
Lasted about 30 seconds..
My advice: Think about potential shortcuts for an additional 5-10 seconds in the beginning and you can save several minutes!
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
Lasted about 30 seconds..
My advice: Think about potential shortcuts for an additional 5-10 seconds in the beginning and you can save several minutes!
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
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
Lasted about 30 seconds..
My advice: Think about potential shortcuts for an additional 5-10 seconds in the beginning and you can save several minutes!
- olegpoi
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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)
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)
Stuart, but why exactly 3 scenarios? Why not 11 + 1, 7 + 5 and etc. ?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!
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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.
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
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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.
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.