Problem Solving

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)45/91
C)2/3
D)67/91
E)84/91

This is how I solve it

Possible combinations =15*14*13*12*11*10*9*8*7*6*5*4 =455
12!

To get atleast 2/3 males in the panel there are 3possibilities

1)8M and 4F
2)9M and 3F
3)10M and 2F

For the first case let the 8 males be already chosen.We only need to find the number of ways in which the females can be chosen. Since we got to select 4 females out of 5 we have 5*4*3*2=5
4!

For the second case let the 9 males be already chosen.We only need to find the number of ways in which the 3 females can be chosen. Since we got to select 3 females out of 5 we have 5*4*3 =10
3!

For the third case let the 10 males be already chosen.We only need to find the number of ways in which the females can be chosen. Since we got to select 2 females out of 5 we have 5*4 =10
2!

Thus 10+10+5=25

Favourable combinations=25/455
Possible combinations

Why am I wrong?

Look at the following problem


An engagement team consists of a project manager, team leader, and four consultants. There are 2 candidates
for the position of project manager, 3 candidates for the position of team leader, and 7 candidates for the 4
consultant slots. If 2 out of 7 consultants refuse to be on the same team, how many different teams are possible?

a.25
b.35
c.150
d.210
e.300


First, recognize that you need to solve for project manager (PM), team leader (TL), and consultants (C) separately:

PM: one slot, two people = 2

TL: one slot, three people = 3

C: four slots, seven people = 7 * 6 * 5 * 4. With the slot method, you need to then stop and realize that there are 4! ways to rearrange the four people you have chosen. So you need to divide 840 by 24 to get 35 unique teams that can be created. Finally, since 2 of the consultants can't work together, we need to remove these specific teams from the 35 possibilities. If these two people are selected, we have two slots remaining with five other people to fill those slots, so there are 5 * 4 = 20, different ways to select the two other people, divided by 2, because there are 2! ways to choose those two other people. This means there are 35 - 10 = 25 consultant teams that can be selected.

Overall: 2 * 3 * 25 = 150

Hi!

Here's another thread on this question in which I looked at a both algebra and strategic guessing.

https://www.beatthegmat.com/i-suck-at-pr ... t8927.html

Stuart

I liked that method,but im still not able to reason out why Im wrong.Ive also attached another problem there and bold'd it.There we directly select the 2 people and only have to find the number of ways in which the other 2 are selected.Why cant I use the same approach in the jury problem too?


thanks

Dan

Stuart Kovinsky wrote:Hi!

Here's another thread on this question in which I looked at a both algebra and strategic guessing.

https://www.beatthegmat.com/i-suck-at-pr ... t8927.html

Stuart

Speaking specifically to your solution, the problem in each case is when you begin with:

For the first case let the 8 males be already chosen

There are 10 males total, so there are 10C8 ways to select those 8 males. You need to factor all those possibilities into your solution as well. (10C8 = 10*9/2*1 = 45)

In the second case, you have to multiply by 10C9 = 10.

In the third case, 10C10 only leaves us with 1 choice, so no need to mess with that part of your solution.

Also, a quick note on quick calculations of combinations. You keep factoring out the smaller factorial on the bottom - but it's much quicker to factor out the big one!

For example:

15C12 = 15!/12!3!

If we cancel out the 12!, we get:

15*14*13/3*2*1

which is much quicker to calculate (and write out!) than:

15*14*13*12*11*10*9*8*7*6*5*4/12!

dddanny2006 wrote:I liked that method,but im still not able to reason out why Im wrong.Ive also attached another problem there and bold'd it.There we directly select the 2 people and only have to find the number of ways in which the other 2 are selected.Why cant I use the same approach in the jury problem too?


thanks

Dan

Your final solution for that problem was 2*3*25.

So, in that problem you multiplied by 2 (possible choices for PM) and 3 (possible choices for TL) - you didn't ignore those parts of the puzzle at all.

Thanks Stuart,but why is it different in the engagement team question?Here we directly select the 2 unwanted people first and then find combos for the remaining 2,why the difference?

Stuart Kovinsky wrote:Speaking specifically to your solution, the problem in each case is when you begin with:

For the first case let the 8 males be already chosen

There are 10 males total, so there are 10C8 ways to select those 8 males. You need to factor all those possibilities into your solution as well. (10C8 = 10*9/2*1 = 45)

In the second case, you have to multiply by 10C9 = 10.

In the third case, 10C10 only leaves us with 1 choice, so no need to mess with that part of your solution.

Also, a quick note on quick calculations of combinations. You keep factoring out the smaller factorial on the bottom - but it's much quicker to factor out the big one!

For example:

15C12 = 15!/12!3!

If we cancel out the 12!, we get:

15*14*13/3*2*1

which is much quicker to calculate (and write out!) than:

15*14*13*12*11*10*9*8*7*6*5*4/12!

Finally, since 2 of the consultants can't work together, we need to remove these specific teams from the 35 possibilities. If these two people are selected, we have two slots remaining with five other people to fill those slots, so there are 5 * 4 = 20, different ways to select the two other people, divided by 2, because there are 2! ways to choose those two other people. This means there are 35 - 10 = 25 consultant teams that can be selected.

In this part of the solution, you're ONLY looking at the unacceptable teams. In other words, you're counting how many teams are disqualified due to inclusion of the 2 arch-enemies.

Since we only care about disqualifying teams including the 2 troublemakers, we automatically include both of them on each of the problem teams. When counting those teams, we simply have to fill the 2 remaining spots from the 5 remaining candidates. So, we simply calculate 5C2 = 10 to count the disqualified teams.

Im talking about the part where we got to select 4 consultants out of the 7 consultants where 2 of them refuse to be a part of the same team.

First we assume the 2 consultants who dont want to be together are in the team.After that we find the number of combi's to accomodate the other 2 people.

So its 2(unhappy consultant) and 2(Unhappy consultants)

7*6 * 5*4=210 combi's
2 2

this should have been the way.But they didnt do that.I cant spot the difference.

Stuart Kovinsky wrote:

dddanny2006 wrote:I liked that method,but im still not able to reason out why Im wrong.Ive also attached another problem there and bold'd it.There we directly select the 2 people and only have to find the number of ways in which the other 2 are selected.Why cant I use the same approach in the jury problem too?


thanks

Dan

Your final solution for that problem was 2*3*25.

So, in that problem you multiplied by 2 (possible choices for PM) and 3 (possible choices for TL) - you didn't ignore those parts of the puzzle at all.

So if I can disqualify teams by including the trouble makers and not calculating their combi's at all(Combi's are calculated only for the remaining 2 slots) then I sure can include the 8men directly and only calculate combi's for the remaining 4 women for the first case,include 9men directly and find combi's only for the 3women in the 2nd case and so on for the 3rd case------ relating to the jury problem.Why dont we do that?

Stuart Kovinsky wrote:

Finally, since 2 of the consultants can't work together, we need to remove these specific teams from the 35 possibilities. If these two people are selected, we have two slots remaining with five other people to fill those slots, so there are 5 * 4 = 20, different ways to select the two other people, divided by 2, because there are 2! ways to choose those two other people. This means there are 35 - 10 = 25 consultant teams that can be selected.

In this part of the solution, you're ONLY looking at the unacceptable teams. In other words, you're counting how many teams are disqualified due to inclusion of the 2 arch-enemies.

Since we only care about disqualifying teams including the 2 troublemakers, we automatically include both of them on each of the problem teams. When counting those teams, we simply have to fill the 2 remaining spots from the 5 remaining candidates. So, we simply calculate 5C2 = 10 to count the disqualified teams.

dddanny2006 wrote:So if I can disqualify teams by including the trouble makers and not calculating their combi's at all(Combi's are calculated only for the remaining 2 slots) then I sure can include the 8men directly and only calculate combi's for the remaining 4 women for the first case,include 9men directly and find combi's only for the 3women in the 2nd case and so on for the 3rd case------ relating to the jury problem.Why dont we do that?

Here's the key difference: there's only one way to choose the 2 consultants who can't be together. There are multiple ways to choose the 8 male members of the jury.

Technically, when calculating the number of disqualified teams of consultants, we're calculating:

2C2 * 5C2 = 1*10 = 10

However, since we know that 2C2 = 1 (nCn=1 for all values on n), we're ignoring that part of the math.

Let's change the consultant problem a bit:

Of the 7, there are 3 who refuse to work together. How many teams of 4 are disqualified because they contain a conflicting pair or trio?

For the conflicting pairs, we're choosing a pair who can't work together out of the 3 troublemakers and the 2 remaining slots will be filled by the 4 friendlies. So:

3C2 * 4C2 = 3*6 = 18

For the conflicting trio, we're choosing all 3 who can't work together and the remaining team member out of the 4 friendlies. So:

3C3 * 4C1 = 1*4 = 4

Accordingly, there would be 22 disqualified teams of consultants.

Why instead of the 2C2 we cant have a 7C2?The question says 2 people refuse to be a part of the same team,so there could be multiple ways of choosing those 2 people.

Let A,B,C,D,E,F,G be the people.Now AC,or AB or AD..............there could be so many trouble makers right?

Stuart Kovinsky wrote:

dddanny2006 wrote:
Technically, when calculating the number of disqualified teams of consultants, we're calculating:

2C2 * 5C2 = 1*10 = 10

dddanny2006 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)45/91
C)2/3
D)67/91
E)84/91

Question rephrased: What is the probability that at least 8 men will be selected to serve on the 12-member jury?

P(good outcome) = 1 - P(bad outcome).

Here, a BAD outcome means selecting a jury with FEWER than 8 men.
Of the 10 men and 5 women in the jury pool, 3 people must be selected NOT to serve on the jury.
There is only ONE WAY to select fewer than 8 men FOR the jury:
The 3 people selected NOT to serve on the jury must ALL be men, leaving 7 men and all 5 women to serve on the jury.

P(1st non-juror is a man) = 10/15. (Of the 15 people in the jury pool, 10 are men.)
P(2nd non-juror is a man) = 9/14. (Of the 14 remaining people in the jury pool, 9 are men.)
P(3rd non-juror is a man) = 8/13. (Of the 13 remaining people in the jury pool, 8 are men.)
Since a bad outcome requires that all 3 events happen, we MULTIPLY the fractions:
10/15 * 9/14 * 8/13 = 24/91.

Thus:
P(good outcome) = 1 - 24/91 = 67/91.

The correct answer is D.

Why instead of the 2C2 we cant have a 7C2?The question says 2 people refuse to be a part of the same team,so there could be multiple ways of choosing those 2 people.

Let A,B,C,D,E,F,G be the people.Now AC,or AB or AD..............there could be so many trouble makers right?

No - the question is written in such a way that those 2 people are fixed (even if they're not named). For the purposes of our calculation, it doesn't matter which 2 people don't get along.

Now I feel like you're arguing just for the sake of arguing!

Thanks Stuart.Well actually Im not arguing just for the sake of it.I was really confused and got it wrong because of the 7c2 and couldnt convince myself of the 2c2.So what looked like arguing for the sake of it was how confused my mind was because of this problem.Is the stuff in Kaplan enough for the GMAT in the combinatorics and probability areas?I seem to have a problem with these topics,and get a lot of problems wrong.

Thanks and cheers!!

Dan

Stuart Kovinsky wrote:

Why instead of the 2C2 we cant have a 7C2?The question says 2 people refuse to be a part of the same team,so there could be multiple ways of choosing those 2 people.

Let A,B,C,D,E,F,G be the people.Now AC,or AB or AD..............there could be so many trouble makers right?

No - the question is written in such a way that those 2 people are fixed (even if they're not named). For the purposes of our calculation, it doesn't matter which 2 people don't get along.

Now I feel like you're arguing just for the sake of arguing!

dddanny2006 wrote:Thanks Stuart.Well actually Im not arguing just for the sake of it.I was really confused and got it wrong because of the 7c2 and couldnt convince myself of the 2c2.So what looked like arguing for the sake of it was how confused my mind was because of this problem.Is the stuff in Kaplan enough for the GMAT in the combinatorics and probability areas?I seem to have a problem with these topics,and get a lot of problems wrong.

Thanks and cheers!!

Dan

Hey Dan!

I'm glad that the question makes sense now.

Kaplan's Premier Guide has a chapter devoted to statistics, which includes perms & combs (and of course those questions show up on our CATs). Whether that's "enough" to get you through to test day depends on how much work you want/need in those areas. If you're doing well on the test, you're likely to see 1-3 P&C questions.

Our course covers the material in a lot more detail (there's a full lesson on Stats plus numerous quizzes and workshops - the permutations and combinations quiz is one of the most challenging in the entire curriculum).