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Joe is among N people in a group, where N > 3. If 3 people are randoml
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14 May 2019, 13:01

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Joe is in a group consisting of N people (N > 4). If 3 people from the group are randomly selected to be on a committee, what is the probability that Joe is selected?

Joe is among N people in a group, where N > 3. If 3 people are randoml
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Updated on: 15 May 2019, 09:06

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There are two easy ways to solve this question.

First is what gmatkarma20 explained in the first post. The benefit of this method is that you can solve by taking any value of N and get the same answer.

The other is to solve in terms of N. First, total number of cases in which we can select 3 people from N people is: C(n,3) or nC3 N*(N-1)*(N-2)/3*2*1

If Joe has to be in the team, we need to select only 2 more people from the remaining ‘N-1’ people. This can be done in: (n-1)C2 or C(n-1,2) (N-1)*(N-2)/2*1

The extra effort that you have to take when substituting values is to check all options and eliminate the ones that will not end with the desired answer.

Re: Joe is among N people in a group, where N > 3. If 3 people are randoml
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15 May 2019, 09:22

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Vinit800HBS wrote:

There are two easy ways to solve this question.

First is what gmatkarma20 explained in the first post. The benefit of this method is that you can solve by taking any value of N and get the same answer.

The other is to solve in terms of N. First, total number of cases in which we can select 3 people from N people is: C(n,3) or nC3 N*(N-1)*(N-2)/3*2*1

If Joe has to be in the team, we need to select only 2 more people from the remaining ‘N-1’ people. This can be done in: (n-1)C2 or C(n-1,2) (N-1)*(N-2)/2*1

Joe is among N people in a group, where N > 3. If 3 people are randoml
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Updated on: 16 May 2019, 06:13

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GMATPrepNow wrote:

Joe is in a group consisting of N people (N > 4). If 3 people from the group are randomly selected to be on a committee, what is the probability that Joe is selected?

A) \(\frac{N^2-2N-6}{(N-2)^2+6}\)

B) \(\frac{2N+3}{N^2-N}\)

C) \(\frac{3}{N}\)

D) \(\frac{N}{(N-1)(N-2)}\)

E) \(\frac{N^2-1}{5N}\)

My solution:

If there are N people, then we can select 3 people in NC3 (N choose 3) different ways. NC3 = \(\frac{(N)(N-1)(N-2)}{3!}\) = \(\frac{(N)(N-1)(N-2)}{6}\) So, this is our denominator

In how many of the (N)(N-1)(N-2)/6 possible outcomes is Joe chosen? Well, once we make Joe one of the selected people, there are N-1 people remaining. We can select 2 people from the remaining N-1 people in (N-1)C2 ways (N-1)C2 = \(\frac{(N-1)(N-2)}{2!}\) = \(\frac{(N-1)(N-2)}{2}\) This is our numerator

So, P(Joe is selected) = \(\frac{(N-1)(N-2)}{2}\) ÷ \(\frac{(N)(N-1)(N-2)}{6}\)

= \(\frac{(N-1)(N-2)}{2}\) x \(\frac{6}{(N)(N-1)(N-2)}\)

That's sound advice. However, I believe Render was saying that, once we know the correct answer is A or C, we can quickly eliminate A by testing N = 10 Although it's ideal to eliminate 4 of the 5 answer choices in the first round, Render's approach helps us quickly arrive at the correct answer.

I'm not able to understand the concept behind your approach. I think in the problem as follows:

Probability to choose Joe = (choose Joe from from N persons) * (choose 2nd person from N-1 persons) * (choose 3rd person from N-2 persons) + (choose 1st person from N persons) *(choose Joe from from N-1 persons) * (choose 3rd person from N-2 persons) + (choose 1st person from N persons) *(choose 2nd person from N-1 persons) * (choose Joe from from N-2 persons)

I'm not able to understand the concept behind your approach. I think in the problem as follows:

Probability to choose Joe = (choose Joe from from N persons) * (choose 2nd person from N-1 persons) * (choose 3rd person from N-2 persons) + (choose 1st person from N persons) *(choose Joe from from N-1 persons) * (choose 3rd person from N-2 persons) + (choose 1st person from N persons) *(choose 2nd person from N-1 persons) * (choose Joe from from N-2 persons)

Is my approach incorrect?

Thanks in advance

In your approach, you basically saying that order matters That is, you're saying selecting Joe then Al then Bea is different from Al then Joe then Bea

That approach approach will work as long as you treat the denominator the same way. That is, the TOTAL number of ways to select 3 people from N people = (N)(N-1)(N-2)

If you do that, then you should arrive at the correct answer.

Re: Joe is among N people in a group, where N > 3. If 3 people are randoml
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16 May 2019, 13:05

GMATPrepNow wrote:

GMATPrepNow wrote:

Joe is in a group consisting of N people (N > 4). If 3 people from the group are randomly selected to be on a committee, what is the probability that Joe is selected?

A) \(\frac{N^2-2N-6}{(N-2)^2+6}\)

B) \(\frac{2N+3}{N^2-N}\)

C) \(\frac{3}{N}\)

D) \(\frac{N}{(N-1)(N-2)}\)

E) \(\frac{N^2-1}{5N}\)

My solution:

If there are N people, then we can select 3 people in NC3 (N choose 3) different ways. NC3 = \(\frac{(N)(N-1)(N-2)}{3!}\) = \(\frac{(N)(N-1)(N-2)}{6}\) So, this is our denominator

In how many of the (N)(N-1)(N-2)/6 possible outcomes is Joe chosen? Well, once we make Joe one of the selected people, there are N-1 people remaining. We can select 2 people from the remaining N-1 people in (N-1)C2 ways (N-1)C2 = \(\frac{(N-1)(N-2)}{2!}\) = \(\frac{(N-1)(N-2)}{2}\) This is our numerator

So, P(Joe is selected) = \(\frac{(N-1)(N-2)}{2}\) ÷ \(\frac{(N)(N-1)(N-2)}{6}\)

= \(\frac{(N-1)(N-2)}{2}\) x \(\frac{6}{(N)(N-1)(N-2)}\)

= \(\frac{6(N-1)(N-2)}{2(N)(N-1)(N-2)}\)

= \(\frac{6}{2N}\)

= \(\frac{3}{N}\)

Answer: C

Cheers, Brent

Dear Brent,

Thanks a lot for your reply. I followed your advice and it yielded the same result.

But I want to understand 2 things in your solution as it is more easier:

1- I used permutation as I think order matters. However, you used combination. why? Does not order matter?

2- In you solution you choose only 2 out of N-1. I do not understand the logic itself. How Joe is included in your solution. I hope you can elaborate.

Re: Joe is among N people in a group, where N > 3. If 3 people are randoml
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17 May 2019, 05:19

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Mo2men wrote:

Dear Brent,

Thanks a lot for your reply. I followed your advice and it yielded the same result.

But I want to understand 2 things in your solution as it is more easier:

1- I used permutation as I think order matters. However, you used combination. why? Does not order matter?

2- In you solution you choose only 2 out of N-1. I do not understand the logic itself. How Joe is included in your solution. I hope you can elaborate.

Thanks in advance

1) We can go either way with this (i.e., order matters or order does not matter), as long as you use the same construct for numerator and denominator. In my approach, I'm assuming order does not matter For example, selecting Joe then Al then Bea is the SAME as selecting Al then Joe then Bea

2) To count the number of outcomes where Joe is selected to be on the 3-person committee, I first placed Joe on the committee (which means there are N-1 people left to choose from), and then I selected 2 more people to join Joe on the committee. So, we are selecting 2 people (from N-1 people) to be on the committee with Joe.