I am Confused. So I took Guess
ANSWER: D
Its the wildest possible guess for me even if it is correct.
I Suck at Probability need desperate help
- Deependra1
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thanks stuart i was not able able to understand and solve the question but once i looked at your both the methods to approach this was able to solve it infact the second method is less time consuming to find 1-P(m<8)
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i chose the answer option D.my way of solving was as follows:i found out the probability for selecting 8men 4women,9men 3 women,10 men 2 women and then add them up which is 335 dividing by the probability for selecting 12 jury people from 15 which is 455 getting 67/91
- jmckenna14
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@pkit...I liked the simplicity behind your solution.
But, how did you arrive with 7 + 4 = 12 as a possible outcome. If 2/3 of the jurors are supposed to men, then that would equate to 8 or more? What am I missing here?
But, how did you arrive with 7 + 4 = 12 as a possible outcome. If 2/3 of the jurors are supposed to men, then that would equate to 8 or more? What am I missing here?
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Total no of ways of selecting 12 people from 15 jurors = (15)C(12) = 15*14*13/1*2*3 = 35*13 = 455
no of men in jury = 2/3 * 15 = 10
no of women in jury = 1/3 * 15 = 5
No of ways of choosing atleast 2/3 men (8 men)
8 men and 4 women = (10)C(8) * 5C4 = 45*5 = 225
9 men and 3 women = 10C9 * 5C3 = 10*10 = 100
10 men and 2 women = 10C10 * 5C2 = 1*10 = 10
No of ways of selecting atleast 2/3 men = 225+100+10 = 335
Probability = 335/455 = 67/91
Option D is the correct option.
no of men in jury = 2/3 * 15 = 10
no of women in jury = 1/3 * 15 = 5
No of ways of choosing atleast 2/3 men (8 men)
8 men and 4 women = (10)C(8) * 5C4 = 45*5 = 225
9 men and 3 women = 10C9 * 5C3 = 10*10 = 100
10 men and 2 women = 10C10 * 5C2 = 1*10 = 10
No of ways of selecting atleast 2/3 men = 225+100+10 = 335
Probability = 335/455 = 67/91
Option D is the correct option.
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This is a great question. You can take the (1-x) approach or just do it the normal way. I am just going to go ahead and solve it normally.
Total number of ways jury of 12 can be selected from 15= 15C12 = 455
If atleast 2/3 of the jury has to be male, then the combinations can be:
1. 8male and 4 female
2. 9male and 3 female
3. 10male and 2 female
So our overall combination for all possible cases is = 10C8X5C4 + 10C9 X 5C3 + 10C10 X 5C2
= 335
Hence our probability is 335/455= 67/91
D Wins !!!
Total number of ways jury of 12 can be selected from 15= 15C12 = 455
If atleast 2/3 of the jury has to be male, then the combinations can be:
1. 8male and 4 female
2. 9male and 3 female
3. 10male and 2 female
So our overall combination for all possible cases is = 10C8X5C4 + 10C9 X 5C3 + 10C10 X 5C2
= 335
Hence our probability is 335/455= 67/91
D Wins !!!
- bpdulog
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I got it right with a lucky guess.
I knew it wasn't 2/3 because the selection pool isn't the same as the number in the chosen group (12, 15) and E seemed too close to absolute certainty.
I knew it wasn't 2/3 because the selection pool isn't the same as the number in the chosen group (12, 15) and E seemed too close to absolute certainty.
NO EXCUSES
"Winston tastes good like a cigarette should."
"Winston tastes good like a cigarette should."
- tuanquang269
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The number of way to choose jury in which have 12 people from 15 people are: 15C12
There are 10 Men and 5 Women
In jury of 12 people, we need AT LEAST 2/3 is men. So, we have some cases:
8 Man 4 Women: 10C8x5C4
9 Men 3 Women: 10C9x5C3
10 Men 2 Women: 10C10x5C2
So the probability is: (10C8x5C4 + 10C9x5C3 + 10C10x5C2)/15C12 = 67/91
There are 10 Men and 5 Women
In jury of 12 people, we need AT LEAST 2/3 is men. So, we have some cases:
8 Man 4 Women: 10C8x5C4
9 Men 3 Women: 10C9x5C3
10 Men 2 Women: 10C10x5C2
So the probability is: (10C8x5C4 + 10C9x5C3 + 10C10x5C2)/15C12 = 67/91
- nisagl750
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I am always stuck at probability. Can anyone tell me how to prepare for it?
For me this was a tough question, though i somehow managed to solve it, but it took me almost 15 mins to solve
For me this was a tough question, though i somehow managed to solve it, but it took me almost 15 mins to solve
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could someone please help clarify why wouldn't solve these 2 problems? and also if we wanted to solve, how it would go about?ash g wrote: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
I solved for both. and they all took time. I'm wondering how my methods compare.
thx
ray
My approach would be:-
The Number of ways in which this can happen i mean selection a jury in of 12 members from a group of 10 men and 5 women are
5 women and 7 men
4 women and 8 men
3 women and 9 men
2 women and 10 men
We need a minimum of 8 men..so 3 out of 4 events.3/4 approx =0.75 and the answer 69/91=approx 0.75.
Please correct me if my method is wrong
The Number of ways in which this can happen i mean selection a jury in of 12 members from a group of 10 men and 5 women are
5 women and 7 men
4 women and 8 men
3 women and 9 men
2 women and 10 men
We need a minimum of 8 men..so 3 out of 4 events.3/4 approx =0.75 and the answer 69/91=approx 0.75.
Please correct me if my method is wrong
Hi Stuart,
I tried solving this problem using the following approach:
at least 2/3 men => we can have 8M (12-8 = 4W) or 9M (12-9=3W) or 10M (12-10=2W) in the jury of 12.
So, probability of having 8 men and 4 women = 8/15 * 4/7 --A
probability of having 9 men and 3 women = 9/15 * 3/6 --B
probability of having 10 men and 2 women = 10/15 * 2/5 --C
required probability = A + B + C = 61/70 (neither of the listed options)
Please guide me where am I going wrong in this approach.
I tried solving this problem using the following approach:
at least 2/3 men => we can have 8M (12-8 = 4W) or 9M (12-9=3W) or 10M (12-10=2W) in the jury of 12.
So, probability of having 8 men and 4 women = 8/15 * 4/7 --A
probability of having 9 men and 3 women = 9/15 * 3/6 --B
probability of having 10 men and 2 women = 10/15 * 2/5 --C
required probability = A + B + C = 61/70 (neither of the listed options)
Please guide me where am I going wrong in this approach.
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!
- Shubhu@MBA
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No. of men in Jury pool = 2/3*15=10
No. of women in Jury pool = 1/3*15=5
Now we need to create a jury of 12 people, out of which at least 2/3rd(8 out of 12) are men.Thus at max we can select 4 women to form this jury.
But,if we select all the 5 women, above stated condition will fail(because to create the jury with at least 2/3 men, we need at least 8 people out of 12 to be men. Thus at max we can select 4 women).
12 people out of a pool of 15 people can be selected in 15C12 ways.
We can select 5 women by 5C5 ways
Remaining 7 will men which can be selected in 10C7 ways
Thus a pool of 5 women & 7 men can be selected in
(10C7*5C5)/15C12 ways
But, we don't want this condition. Hence our answer will be 1 - (10C7*5C5)/15C12.
1- (120/455) = 1-(24/91) = 67/91
Thus option D is the answer.
Got the answer in 1 min or so.
No. of women in Jury pool = 1/3*15=5
Now we need to create a jury of 12 people, out of which at least 2/3rd(8 out of 12) are men.Thus at max we can select 4 women to form this jury.
But,if we select all the 5 women, above stated condition will fail(because to create the jury with at least 2/3 men, we need at least 8 people out of 12 to be men. Thus at max we can select 4 women).
12 people out of a pool of 15 people can be selected in 15C12 ways.
We can select 5 women by 5C5 ways
Remaining 7 will men which can be selected in 10C7 ways
Thus a pool of 5 women & 7 men can be selected in
(10C7*5C5)/15C12 ways
But, we don't want this condition. Hence our answer will be 1 - (10C7*5C5)/15C12.
1- (120/455) = 1-(24/91) = 67/91
Thus option D is the answer.
Got the answer in 1 min or so.
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
Hi,
But P(M=6),P(M=5),....,P(M=0) are all equal to 0 because women number is limited to 5. we cannot have (6 men,6 women), ( 5 men, 7 w), etc. Then in this case, P(M>=8) = 1- P(M=7).