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Joe is among N people in a group, where N > 3. If 3 people are randoml

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New post 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?

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

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New post 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

Thus, Probability is:

(N-1)*(N-2)/2*1
————————
N*(N-1)*(N-2)/3*2*1

= 3/N

Option C

GMATPrepNow, Is this approach better?

Kudos are always appreciated

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Originally posted by Vinit800HBS on 15 May 2019, 09:03.
Last edited by Vinit800HBS on 15 May 2019, 09:06, edited 1 time in total.
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Re: Joe is among N people in a group, where N > 3. If 3 people are randoml  [#permalink]

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New post 15 May 2019, 08:45
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Just plug in a number. Choose N=5, for instance.

# of ways you can form a group of 3: 5C3 = 10

# of ways you can form a group with Joe excluded: 4C3 = 4

Prob(Joe is not selected in the group of 3) = 4/10 = 2/5

Prob(Joe is selected) = 1 - 2/5 = 3/5


C jumps out at you when you plug in N=5 in the options.

So, C.
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New post 15 May 2019, 08:59
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gmatkarma20 wrote:
Just plug in a number. Choose N=5, for instance.

# of ways you can form a group of 3: 5C3 = 10

# of ways you can form a group with Joe excluded: 4C3 = 4

Prob(Joe is not selected in the group of 3) = 4/10 = 2/5

Prob(Joe is selected) = 1 - 2/5 = 3/5


C jumps out at you when you plug in N=5 in the options.

So, C.


Good idea, but when you plug N = 5 into answer choice A, we get 3/5 as well.

A) \(\frac{5^2-2(5)-6}{(5-2)^2+6}=\frac{25-10-6}{3^2+6}=\frac{9}{15}=\frac{3}{5}\)
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Re: Joe is among N people in a group, where N > 3. If 3 people are randoml  [#permalink]

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New post 15 May 2019, 09:04
GMATPrepNow wrote:
gmatkarma20 wrote:
Just plug in a number. Choose N=5, for instance.

# of ways you can form a group of 3: 5C3 = 10

# of ways you can form a group with Joe excluded: 4C3 = 4

Prob(Joe is not selected in the group of 3) = 4/10 = 2/5

Prob(Joe is selected) = 1 - 2/5 = 3/5


C jumps out at you when you plug in N=5 in the options.

So, C.


Good idea, but when you plug N = 5 into answer choice A, we get 3/5 as well.

A) \(\frac{5^2-2(5)-6}{(5-2)^2+6}=\frac{25-10-6}{3^2+6}=\frac{9}{15}=\frac{3}{5}\)


Wonderful; down goes my gmat score. Wonderfully deceptive option C.
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Re: Joe is among N people in a group, where N > 3. If 3 people are randoml  [#permalink]

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New post 15 May 2019, 09:14
gmatkarma20

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.

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Re: Joe is among N people in a group, where N > 3. If 3 people are randoml  [#permalink]

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New post 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

Thus, Probability is:

(N-1)*(N-2)/2*1
————————
N*(N-1)*(N-2)/3*2*1

= 3/N

Option C

GMATPrepNow, Is this approach better?

Kudos are always appreciated

Posted from my mobile device


That's a great approach!
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New post 15 May 2019, 09:23
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gmatkarma20 wrote:
Wonderful; down goes my gmat score. Wonderfully deceptive option C.


The correct answer is still C.
I was just reminding you that, if we use the INPUT-OUTPUT approach, then we need to check all of the answer choices.

Cheers,
Brent
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Re: Joe is among N people in a group, where N > 3. If 3 people are randoml  [#permalink]

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New post 15 May 2019, 09:26
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gmatkarma20 wrote:
GMATPrepNow wrote:
gmatkarma20 wrote:
Just plug in a number. Choose N=5, for instance.

# of ways you can form a group of 3: 5C3 = 10

# of ways you can form a group with Joe excluded: 4C3 = 4

Prob(Joe is not selected in the group of 3) = 4/10 = 2/5

Prob(Joe is selected) = 1 - 2/5 = 3/5


C jumps out at you when you plug in N=5 in the options.

So, C.


Good idea, but when you plug N = 5 into answer choice A, we get 3/5 as well.

A) \(\frac{5^2-2(5)-6}{(5-2)^2+6}=\frac{25-10-6}{3^2+6}=\frac{9}{15}=\frac{3}{5}\)


Wonderful; down goes my gmat score. Wonderfully deceptive option C.


Good, but if you plug N = 10, you get a probability greater than 1.

\(\frac{10^2-2(10)-6}{(10-2)^2+6}=\frac{100-20-6}{8^2+6}=\frac{74}{70}\)
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Re: Joe is among N people in a group, where N > 3. If 3 people are randoml  [#permalink]

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New post 15 May 2019, 09:27
GMATPrepNow wrote:
gmatkarma20 wrote:
Wonderful; down goes my gmat score. Wonderfully deceptive option C.


The correct answer is still C.
I was just reminding you that, if we use the INPUT-OUTPUT approach, then we need to check all of the answer choices.

Cheers,
Brent


Precisely what I stated to gmatkarma20

Kindly give kudos GMATPrepNow if you liked my approach

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Render wrote:

Good, but if you plug N = 10, you get a probability greater than 1.

\(\frac{10^2-2(10)-6}{(10-2)^2+6}=\frac{100-20-6}{8^2+6}=\frac{74}{70}\)


Great idea! That approach quickly eliminates answer choice A
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New post 15 May 2019, 09:30
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Vinit800HBS wrote:

Precisely what I stated to gmatkarma20

Kindly give kudos GMATPrepNow if you liked my approach

Posted from my mobile device


Here comes a Kudos . . . . done!!
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New post 15 May 2019, 09:33
Render

But you still need to know that the probability that you are looking for is 3/5.

Let me give you a good idea for substituting values: NEVER PUT THE EXTREME VALUES.

In this, question says N>4. DON’T TAKE N=5 and solve.

Mostly the options are created from the view that student will take N=5 to avoid nasty calculations and will get stuck on multiple options.

Take values that are further away from N= 5. That will avoid falling into these small traps.

Hope that helps.

GMATPrepNow, does this approach sound good?

Posted from my mobile device
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New post 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 \(\frac{6}{(N)(N-1)(N-2)}\)

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

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

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

Answer: C

Cheers,
Brent

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Originally posted by GMATPrepNow on 15 May 2019, 09:42.
Last edited by GMATPrepNow on 16 May 2019, 06:13, edited 1 time in total.
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Re: Joe is among N people in a group, where N > 3. If 3 people are randoml  [#permalink]

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New post 15 May 2019, 09:45
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Vinit800HBS wrote:
Render

But you still need to know that the probability that you are looking for is 3/5.

Let me give you a good idea for substituting values: NEVER PUT THE EXTREME VALUES.

In this, question says N>4. DON’T TAKE N=5 and solve.

Mostly the options are created from the view that student will take N=5 to avoid nasty calculations and will get stuck on multiple options.

Take values that are further away from N= 5. That will avoid falling into these small traps.

Hope that helps.

GMATPrepNow, does this approach sound good?

Posted from my mobile device


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.

Cheers,
Brent
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New post 16 May 2019, 03:32
GMATPrepNow wrote:
Render wrote:

Good, but if you plug N = 10, you get a probability greater than 1.

\(\frac{10^2-2(10)-6}{(10-2)^2+6}=\frac{100-20-6}{8^2+6}=\frac{74}{70}\)


Great idea! That approach quickly eliminates answer choice A


Dear GMATPrepNow Brent


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
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New post 16 May 2019, 06:05
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Mo2men wrote:
GMATPrepNow wrote:
Render wrote:

Good, but if you plug N = 10, you get a probability greater than 1.

\(\frac{10^2-2(10)-6}{(10-2)^2+6}=\frac{100-20-6}{8^2+6}=\frac{74}{70}\)


Great idea! That approach quickly eliminates answer choice A


Dear GMATPrepNow Brent


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.

Cheers,
Brent
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Re: Joe is among N people in a group, where N > 3. If 3 people are randoml  [#permalink]

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New post 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.

Thanks in advance
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Re: Joe is among N people in a group, where N > 3. If 3 people are randoml  [#permalink]

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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.

Does that help?

Cheers,
Brent
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Re: Joe is among N people in a group, where N > 3. If 3 people are randoml  [#permalink]

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New post 03 Jun 2019, 17:14
I don't understand the approach that GMATkarma20 used.

I did it this way:

Probability of picking Joe:

Probability of picking him 1st:

(1/N) * (N-2/N-1) * (N-3/N-2)

since we can have him picked on the 2nd slot or 3rd slot as well we multiple this probability by 3.

we get 3((1/N) * (N-2/N-1) * (N-3/N-2)) = 3(N-1)/N(N-1) =3/N
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Re: Joe is among N people in a group, where N > 3. If 3 people are randoml   [#permalink] 03 Jun 2019, 17:14
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