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aritrar4
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Joe is among N people in a group, where N > 3. If 3 people are randoml 05 Aug 2020, 22:10
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VeritasKarishma BrentGMATPrepNow

I was initially inclined to resolve this using the combinatorics approach, but then I tried the below approach. Please let me know if this approach is correct.

Joe can be selected in either of the 3 below slots from a group of N people.

____ ____ ____

Joe ___ ____
___ Joe ____
___ ____ Joe

Probability of selecting Joe from N people is 1/N

Therefore Joe can be selected in 3/N ways.
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Joe is among N people in a group, where N > 3. If 3 people are randoml 10 Aug 2020, 03:34
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aritrar4
VeritasKarishma BrentGMATPrepNow

I was initially inclined to resolve this using the combinatorics approach, but then I tried the below approach. Please let me know if this approach is correct.

Joe can be selected in either of the 3 below slots from a group of N people.

____ ____ ____

Joe ___ ____
___ Joe ____
___ ____ Joe

Probability of selecting Joe from N people is 1/N

Therefore Joe can be selected in 3/N ways.
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The question asks for selection only, not arrangements. So there are no slots to arrange Joe in.

You should use (n-1)C2/nC3 = 3/n

(n-1)C2 - Number of ways of selecting Joe and 2 people from leftover (n-1) people.
nC3 - Number of ways of selecting any 3 people from n people.

Say n = 10,
9C2 / 10C3 = (9*8/2) / (10*9*8)/(3*2) = 3/10
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Joe is among N people in a group, where N > 3. If 3 people are randoml 22 Feb 2022, 07:11
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aritrar4
VeritasKarishma BrentGMATPrepNow

I was initially inclined to resolve this using the combinatorics approach, but then I tried the below approach. Please let me know if this approach is correct.

Joe can be selected in either of the 3 below slots from a group of N people.

____ ____ ____

Joe ___ ____
___ Joe ____
___ ____ Joe

Probability of selecting Joe from N people is 1/N

Therefore Joe can be selected in 3/N ways.
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This works and is the simplest approach.

Similarly:

Probability of not being chosen in any of the three draws

1st: (n-1)/n
2nd: (n-2)/(n-1)
3rd: (n-3)/(n-2)

Probability of not being chosen

(n-1)/n * (n-2)/(n-1) * (n-3)/(n-2) =

(n-3)/n

Probability of being chosen is 1- the above or

1-((n-3)/n) = (n-(n-3)/n)=

3/n


Everyone has an equal chance of being chosen and since there are 3 opportunities the probability is just 3/n

Another way of thinking about it, say n=10.

_ _ _ | _ _ _ _ _ _ _

What's the probability of any given person sitting in the first 3 spots: 3/10

And it doesn't matter which 3 spots you designate

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Joe is among N people in a group, where N > 3. If 3 people are randoml 01 May 2026, 23:03
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total = N people. (N>4)
we need to select 3 people.

the way to select 3 people from N is pretty straightforward.
Its NC3 = N(N-1)(N-2)(N-3)! / (N-3)! * 6
we get, NC3 = N(N-1)(N-2)/6
this is our denominator.

now we need to find probability that joe is selected in those 3 people.
so once joe is selected, we have 2 people left to choose from N-1 people.
for that to find, we will simple use (N-1)C2
we get, (N-1)(N-2)(N-3)! / (N-3)*2
(N-1)C2 = (N-1)(N-2)/2.
this is out numerator.

now when we use both these values, we get= [(N-1)(N-2)/2] * [ 6/N(N-1)(N-2)]
upon solving this, we get 3/N

ans is C
BrentGMATPrepNow
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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Joe is among N people in a group, where N > 3. If 3 people are randoml 01 May 2026, 23:23
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Many people using numbers to solve however this one can simply be done by using N as well.

(N-1)C2/NC3 is our required result
= (N-1)(N-2) * 3! / (2! * N(N-1)(N-2))
= 3/N

Answer: Option C

________________________

PS:- Problem would be pushed into the hard level if an option like "3/(N-2)" existed.
Such an answer would stem from going for NC2/NC3 but Joe is himself part of the group so it gives a wrong result. A very common mistake!
BrentGMATPrepNow
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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Joe is among N people in a group, where N > 3. If 3 people are randoml 18 May 2026, 22:55
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KarishmaB / Bunuel
Is this approach right?
the probability that the joe should be in the group is that, he should be among the first 3 people selected. So it is 3/N

Thanks,
Swetha
BrentGMATPrepNow
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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Joe is among N people in a group, where N > 3. If 3 people are randoml 18 May 2026, 23:19
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SwethaReddyL
KarishmaB / Bunuel
Is this approach right?
the probability that the joe should be in the group is that, he should be among the first 3 people selected. So it is 3/N

Thanks,
Swetha

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Yes, that approach is right.

Imagine the selected committee members as the first 3 people in a random ordering of all N people. Joe is equally likely to be in any of the N positions, and 3 of those positions get selected.

So the probability is 3/N.
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Joe is among N people in a group, where N > 3. If 3 people are randoml 21 May 2026, 04:59
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SwethaReddyL
KarishmaB / Bunuel
Is this approach right?
the probability that the joe should be in the group is that, he should be among the first 3 people selected. So it is 3/N

Thanks,
Swetha

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Yes, the way I like to think about it in this case is:
Say they are standing randomly in a line. Joe can be in any of the N spots. Of those 3 are the favored spots i.e. the selected spots. The probability that he is in one of those 3 spots is simply 3/N.
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Joe is among N people in a group, where N > 3. If 3 people are randoml 20 Jun 2026, 22:34
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Option A also gives the same answer as 3/5 when plugged in. We should try with 6 or solve using N as a variable.
gmatkarma20
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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