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# There are y different travelers who each have a choice of va

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Manager
Joined: 22 Jul 2009
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There are y different travelers who each have a choice of va  [#permalink]

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22 Sep 2009, 16:59
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Difficulty:

85% (hard)

Question Stats:

44% (01:42) correct 56% (01:27) wrong based on 244 sessions

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There are y different travelers who each have a choice of vacationing at one of n different destinations. What is the probability that all y travelers will end up vacationing at the same destination?

A. 1/n!
B. n/n!
C. 1/n^y
D. 1/n^(y-1)
E. n/y^n

Source: Manhattan GMAT Archive (tough problems set).doc

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Re: PS, Probability - There are y different travelers ...  [#permalink]

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10 May 2011, 06:52
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powerka wrote:
There are y different travelers who each have a choice of vacationing at one of n different destinations. What is the probability that all y travelers will end up vacationing at the same destination?
a) 1/n!
b) n/n!
c) 1/n^y
d) 1/n^(y-1)
e) n/y^n

Source: Manhattan GMAT Archive (tough problems set).doc

There are 2 ways to handle a question with variables. Using logic which I endorse and plugging in numbers which I discuss for those situations where you run out of time or are thoroughly confused or are exhausted. I will take both though the logic has pretty much been discussed above.

The first traveler has n options to choose from (n destinations). The moment he chooses one of those n, every one else has to take the same destination. So number of favorable outcomes = n
If there were no constraints, each of the y travelers could choose any one of the n destinations. So total number of combinations = n*n*... (y times)
$$Probability = \frac{n}{n^y} = \frac{1}{n^{y-1}}$$

Plugging in numbers:
Say y = 1 and n = 2 (1 traveler, 2 places)
What is the probability that all travelers will go to the same place? 1 of course since there is only one traveler. Where ever he goes is the place where all travelers are! Plug it in options. Only options b and d give 1 when you plug in y = 1 and n = 2.
Say y = 2 and n = 2 (2 travelers, 2 places)
They could either be together at a place or at two different places so probability of being together is 1/2. Plug n = 2, y = 2 in options b and d. Only option d gives you 1/2.
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Manager
Joined: 11 Sep 2009
Posts: 127
Re: PS, Probability - There are y different travelers ...  [#permalink]

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22 Sep 2009, 17:15
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The answer is D: 1/n^(y-1).

The simplest way to look at it is to assume that the first traveler picks a place. All the remaining travelers (given as y-1) now need to pick the same place for vacation.

As a result the probability is equivalent to (1/n) for each traveler to pick that same spot as the original traveler, which has to happen (y-1) times.

$$P = \frac{1}{n}*\frac{1}{n}*...*\frac{1}{n}$$ (y-1) times

$$P = {\frac{1}{n}}^{(y-1)}$$
##### General Discussion
Manager
Joined: 11 Aug 2008
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Re: PS, Probability - There are y different travelers ...  [#permalink]

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Updated on: 25 Oct 2009, 09:32
But I end up choose the 1/n^y ways

Originally posted by ngoctraiden1905 on 21 Oct 2009, 21:08.
Last edited by ngoctraiden1905 on 25 Oct 2009, 09:32, edited 1 time in total.
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Re: PS, Probability - There are y different travelers ...  [#permalink]

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22 Oct 2009, 06:25
4
1
winning outcomes = n (and not 1)
total outcomes = n^y
=> n/n^y = 1/n^(y-1)
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Re: PS, Probability - There are y different travelers ...  [#permalink]

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09 May 2011, 01:12
Cool solution.
1/n^(y-1).
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Re: PS, Probability - There are y different travelers ...  [#permalink]

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10 May 2011, 06:34
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Prob that traveler 1 will choose 1st place(any particular place) out n places = 1/n

Prob that traveler 2 will choose same place out n places = 1/n
......

Total = 1/n * ... 1/n (y times)

= 1/n^y

But this can happen for all of the n places.

So Prob = n/(n^y) = 1/n^(y-1)

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Re: There are y different travelers who each have a choice of va  [#permalink]

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22 Sep 2014, 05:34
powerka wrote:
There are y different travelers who each have a choice of vacationing at one of n different destinations. What is the probability that all y travelers will end up vacationing at the same destination?

A. 1/n!
B. n/n!
C. 1/n^y
D. 1/n^(y-1)
E. n/y^n

Source: Manhattan GMAT Archive (tough problems set).doc

Excellent question. Totally Stumped.
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Re: There are y different travelers who each have a choice of va  [#permalink]

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07 Apr 2017, 07:48
AKProdigy87 wrote:
The answer is D: 1/n^(y-1).

The simplest way to look at it is to assume that the first traveler picks a place. All the remaining travelers (given as y-1) now need to pick the same place for vacation.

As a result the probability is equivalent to (1/n) for each traveler to pick that same spot as the original traveler, which has to happen (y-1) times.

$$P = \frac{1}{n}*\frac{1}{n}*...*\frac{1}{n}$$ (y-1) times

$$P = {\frac{1}{n}}^{(y-1)}$$

thank you for sharing nice explanation
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Joined: 23 Jul 2015
Posts: 38
Re: There are y different travelers who each have a choice of va  [#permalink]

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24 May 2019, 14:50
This is so confusing. Can someone explain with hypothetical numbers?
Re: There are y different travelers who each have a choice of va   [#permalink] 24 May 2019, 14:50
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