GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 19 Jun 2018, 12:48

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

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

Author Message
TAGS:

### Hide Tags

Manager
Joined: 22 Jul 2009
Posts: 188
There are y different travelers who each have a choice of va [#permalink]

### Show Tags

22 Sep 2009, 16:59
19
00:00

Difficulty:

85% (hard)

Question Stats:

43% (01:01) correct 57% (00:47) wrong based on 300 sessions

### HideShow timer Statistics

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

_________________

Please kudos if my post helps.

 Manhattan GMAT Discount Codes Kaplan GMAT Prep Discount Codes Magoosh Discount Codes
Manager
Joined: 11 Sep 2009
Posts: 129
Re: PS, Probability - There are y different travelers ... [#permalink]

### Show Tags

22 Sep 2009, 17:15
7
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

$$P = {\frac{1}{n}}^{(y-1)}$$
Manager
Joined: 11 Aug 2008
Posts: 133
Re: PS, Probability - There are y different travelers ... [#permalink]

### Show Tags

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.
Manager
Joined: 22 Jul 2009
Posts: 188
Re: PS, Probability - There are y different travelers ... [#permalink]

### Show Tags

22 Oct 2009, 06:25
4
winning outcomes = n (and not 1)
total outcomes = n^y
=> n/n^y = 1/n^(y-1)
_________________

Please kudos if my post helps.

VP
Status: There is always something new !!
Affiliations: PMI,QAI Global,eXampleCG
Joined: 08 May 2009
Posts: 1140
Re: PS, Probability - There are y different travelers ... [#permalink]

### Show Tags

09 May 2011, 01:12
Cool solution.
1/n^(y-1).
_________________

Visit -- http://www.sustainable-sphere.com/
Promote Green Business,Sustainable Living and Green Earth !!

Retired Moderator
Joined: 16 Nov 2010
Posts: 1472
Location: United States (IN)
Concentration: Strategy, Technology
Re: PS, Probability - There are y different travelers ... [#permalink]

### Show Tags

10 May 2011, 06:34
6
2
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)

_________________

Formula of Life -> Achievement/Potential = k * Happiness (where k is a constant)

GMAT Club Premium Membership - big benefits and savings

Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8100
Location: Pune, India
Re: PS, Probability - There are y different travelers ... [#permalink]

### Show Tags

10 May 2011, 06:52
5
6
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.
_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Get started with Veritas Prep GMAT On Demand for \$199

Veritas Prep Reviews

Manager
Joined: 20 Jan 2014
Posts: 166
Location: India
Concentration: Technology, Marketing
Re: There are y different travelers who each have a choice of va [#permalink]

### Show Tags

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.
_________________

Manager
Joined: 28 Jul 2016
Posts: 126
Re: There are y different travelers who each have a choice of va [#permalink]

### Show Tags

07 Apr 2017, 07:48
AKProdigy87 wrote:

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
Non-Human User
Joined: 09 Sep 2013
Posts: 7004
Re: There are y different travelers who each have a choice of va [#permalink]

### Show Tags

22 Apr 2018, 04:59
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
Re: There are y different travelers who each have a choice of va   [#permalink] 22 Apr 2018, 04:59
Display posts from previous: Sort by