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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Karishma
Owner of Angles and Arguments at https://anglesandarguments.com/
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