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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.
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.
10 Aug 2020, 03:34
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aritrar4
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
22 Feb 2022, 07:11
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aritrar4
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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