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

Question

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

KarishmaB · Accepted Answer

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.

AKProdigy87 · Answer

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)}

subhashghosh · Answer

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)

Answer - D

ngoctraiden1905 · Answer

But I end up choose the 1/n^y ways

powerka · Answer

winning outcomes = n (and not 1)
total outcomes = n^y
=> n/n^y = 1/n^(y-1)

Mehadihasanshawon · Answer

Total number of travelers= y
number of destinations= n
total number of destinations= (n)^y
probability of having one destination=Nc1/(n)^y

Posted from my mobile device

imaityaroy · Answer

Here is my approach 
there are there are y = 3 travellers and n=5 destination

the 1st traveller has option to go to 5 destination 
the 2nd traveller will only have the option to go 1 destination ( the destination where the 1st traveller has gone)
the 3rd traveller will only have the option to go 1 destination ( the destination where the 1st and 2nd traveller have gone)

so favourable outcomes = 5 *1*1
total outcomes = 5*5*5

Therefore the probability = 5/5^3 
= 1/5^2
= 1/n^(y-1)

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

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions

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

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards