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 Your Progress
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
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:
Six friends live in the city of Monrovia. There are four nat
[#permalink]
Show Tags
03 Jul 2013, 06:27
19
00:00
A
B
C
D
E
Difficulty:
85% (hard)
Question Stats:
37% (02:12) correct 63% (01:45) wrong based on 35 sessions
HideShow timer Statistics
Found this question in one of the Veritas Prep blog posts...
Six friends live in the city of Monrovia. There are four natural attractions around Monrovia – a waterfall, a safari, a lake and some caves. The friends decide to take a vacation together at one of these attractions. To select the attraction, each one of them votes for one of the attractions. What is the probability that each attraction gets at least one vote?
The answer is 1560/(4^6). I was wondering if we can get to this answer by removing the cases where [an "attraction" gets zero votes] can be subtracted from the [total number of cases].
Re: Six friends live in the city of Monrovia. There are four nat
[#permalink]
Show Tags
24 Mar 2014, 02:12
2
v1gnesh wrote:
Found this question in one of the Veritas Prep blog posts...
Six friends live in the city of Monrovia. There are four natural attractions around Monrovia – a waterfall, a safari, a lake and some caves. The friends decide to take a vacation together at one of these attractions. To select the attraction, each one of them votes for one of the attractions. What is the probability that each attraction gets at least one vote?
The answer is 1560/(4^6). I was wondering if we can get to this answer by removing the cases where [an "attraction" gets zero votes] can be subtracted from the [total number of cases].
Counting those is harder than a direct approach when we need to count only: {3, 1, 1, 1} --> \(\frac{4!}{3!}*C^3_6*3!=480\). {2, 2, 1, 1} --> \(\frac{4!}{2!2!}*C^2_6*C^2_4*C^1_2=1,080\).
The total number of ways in which the 6 friends can vote = 4*4*4*4*4*4 = 4^6.
Re: Six friends live in the city of Monrovia. There are four nat
[#permalink]
Show Tags
28 Apr 2014, 05:35
Bunuel wrote:
v1gnesh wrote:
Found this question in one of the Veritas Prep blog posts...
Six friends live in the city of Monrovia. There are four natural attractions around Monrovia – a waterfall, a safari, a lake and some caves. The friends decide to take a vacation together at one of these attractions. To select the attraction, each one of them votes for one of the attractions. What is the probability that each attraction gets at least one vote?
The answer is 1560/(4^6). I was wondering if we can get to this answer by removing the cases where [an "attraction" gets zero votes] can be subtracted from the [total number of cases].
Counting those is harder than a direct approach when we need to count only: {3, 1, 1, 1} --> \(\frac{4!}{3!}*C^3_6*3!=480\). {2, 2, 1, 1} --> \(\frac{4!}{2!2!}*C^2_6*C^2_4*C^1_2=1,080\).
The total number of ways in which the 6 friends can vote = 4*4*4*4*4*4 = 4^6.
P = (480 + 1,080)/4^6 = 1560/4^6.
Hope it helps.
Bunuel
Please Elaborate how you got the 3 Equations :
Counting those is harder than a direct approach when we need to count only: {3, 1, 1, 1} --> \(\frac{4!}{3!}*C^3_6*3!=480\). {2, 2, 1, 1} --> \(\frac{4!}{2!2!}*C^2_6*C^2_4*C^1_2=1,080\).
The total number of ways in which the 6 friends can vote = 4*4*4*4*4*4 = 4^6.
Re: Six friends live in the city of Monrovia. There are four nat
[#permalink]
Show Tags
28 Apr 2014, 10:11
3
1
niyantg wrote:
Bunuel wrote:
v1gnesh wrote:
Found this question in one of the Veritas Prep blog posts...
Six friends live in the city of Monrovia. There are four natural attractions around Monrovia – a waterfall, a safari, a lake and some caves. The friends decide to take a vacation together at one of these attractions. To select the attraction, each one of them votes for one of the attractions. What is the probability that each attraction gets at least one vote?
The answer is 1560/(4^6). I was wondering if we can get to this answer by removing the cases where [an "attraction" gets zero votes] can be subtracted from the [total number of cases].
Counting those is harder than a direct approach when we need to count only: {3, 1, 1, 1} --> \(\frac{4!}{3!}*C^3_6*3!=480\). {2, 2, 1, 1} --> \(\frac{4!}{2!2!}*C^2_6*C^2_4*C^1_2=1,080\).
The total number of ways in which the 6 friends can vote = 4*4*4*4*4*4 = 4^6.
P = (480 + 1,080)/4^6 = 1560/4^6.
Hope it helps.
Bunuel
Please Elaborate how you got the 3 Equations :
Counting those is harder than a direct approach when we need to count only: {3, 1, 1, 1} --> \(\frac{4!}{3!}*C^3_6*3!=480\). {2, 2, 1, 1} --> \(\frac{4!}{2!2!}*C^2_6*C^2_4*C^1_2=1,080\).
The total number of ways in which the 6 friends can vote = 4*4*4*4*4*4 = 4^6.
Thankyou!
For each attraction to get at least one vote we should have the following distributions of 6 votes: {3, 1, 1, 1} or {2, 2, 1, 1}.
A - B - C - D (attractions) {3, 1, 1, 1} --> \(\frac{4!}{3!}*C^3_6*3!=480\), where: \(\frac{4!}{3!}\) is the # assignments of {3, 1, 1, 1} to attractions (A gets 3, B gets 1, C gets 1, D gets 1; A gets 1, B gets 3, C gets 1, D gets 1; A gets 1, B gets 1, C gets 3, D gets 1; A gets 1, B gets 1, C gets 1, D gets 3); \(C^3_6\) is selecting the 3 people who will vote for the attraction with 3 votes; 3! is the distribution of other 3 votes among the remaining attractions.
The same logic applies to {2, 2, 1, 1} case.
As for 4^6: each friend can vote in 4 ways (for waterfall, safari, lake or caves). So, the total number of ways in which the 6 friends can vote = 4*4*4*4*4*4 = 4^6.
Re: Six friends live in the city of Monrovia. There are four nat
[#permalink]
Show Tags
20 Mar 2019, 15:32
Hi guys ,I dont' understand why you considered them different objects I think the votes and the attractions should be considered identical , let's count the total way of placing 6 votes into 4 rooms with no constraints , so any room can get a zero or any number , (we don't care who voted for which , we car only about the number of votes each room gets )
then let's find how many ways all rooms get at least one vote
not sure if my approach is correct , please correct me :
6 identical votes need to go into 4 identical rooms , first let's find the total number with no constraints : let's make 3 walls , vvvvvv||| ,9 objects , let's choose 3 of them to place our walls : 9C3=84 ways
now , let's give each room a vote at first we are left with 2 votes and 4 rooms , let's make 3 walls , vv||| , ok : 5 objects , let's choose 3 of them to place our walls : 5C3 = 10 ways so p = 10/84
_________________
I am not able to understand here that why does it matter who has voted for which place. why are we multiplying by 6C4 here?? the question simply asks that each place should have atleast one vote. why does it matter from whom that vote has been casted???
Re: Six friends live in the city of Monrovia. There are four nat
[#permalink]
Show Tags
26 May 2019, 15:10
1
v1gnesh wrote:
Found this question in one of the Veritas Prep blog posts...
Six friends live in the city of Monrovia. There are four natural attractions around Monrovia – a waterfall, a safari, a lake and some caves. The friends decide to take a vacation together at one of these attractions. To select the attraction, each one of them votes for one of the attractions. What is the probability that each attraction gets at least one vote?
The answer is 1560/(4^6). I was wondering if we can get to this answer by removing the cases where [an "attraction" gets zero votes] can be subtracted from the [total number of cases].
Re: Six friends live in the city of Monrovia. There are four nat
[#permalink]
Show Tags
15 Jun 2019, 06:18
Bunuel wrote:
v1gnesh wrote:
Found this question in one of the Veritas Prep blog posts...
Six friends live in the city of Monrovia. There are four natural attractions around Monrovia – a waterfall, a safari, a lake and some caves. The friends decide to take a vacation together at one of these attractions. To select the attraction, each one of them votes for one of the attractions. What is the probability that each attraction gets at least one vote?
The answer is 1560/(4^6). I was wondering if we can get to this answer by removing the cases where [an "attraction" gets zero votes] can be subtracted from the [total number of cases].
Counting those is harder than a direct approach when we need to count only: {3, 1, 1, 1} --> \(\frac{4!}{3!}*C^3_6*3!=480\). {2, 2, 1, 1} --> \(\frac{4!}{2!2!}*C^2_6*C^2_4*C^1_2=1,080\).
The total number of ways in which the 6 friends can vote = 4*4*4*4*4*4 = 4^6.
P = (480 + 1,080)/4^6 = 1560/4^6.
Hope it helps.
Hi Bunuel,
Could you please evaluate the below logic?
The ways of voting in which at most 3 attarctions participate: ABC or ABD or ACD or BCD. So overall \(4*3^6\) cases. But those votings double count the cases consisting of only AB, AC, AD, BC, BD, and CD. Thus we need to subtract \(6*2^6\) from \(4*3^6\). Next, AB, AC, AD, BC, BD, and CD themselves triple count the cases consisting of only A, B, C, and D. So, we we need to add \(4\) back to overall computation in order to include 4 cases - AAAAAA, BBBBBB, CCCCCC, DDDDDD, thus:
\(4*3^6 - 6*2^6 + 4 = 2536\)
Number of cases with at least one attraction: \(4^6 - 2536 = 1560\)
Is this logic correct?
Thank you beforehand!
_________________
Bruce Lee: “I fear not the man who has practiced 10,000 kicks once, but I fear the man who has practiced one kick 10,000 times.” GMAC: “I fear not the aspirant who has practiced 10,000 questions, but I fear the aspirant who has learnt the most out of every single question.”
I did it this way: out of 6 friends lets select 4 friends who vote for the 4 different places: 6C4 ways. These votes can be arranged in 4! ways.. For the remaining 2 votes each has 4 ways. So the numerator becomes: 6C4*4!*4*4.... Denominator is obviously 4^6 Please tell me where am i wrong? Thanks
I did it this way: out of 6 friends lets select 4 friends who vote for the 4 different places: 6C4 ways. These votes can be arranged in 4! ways.. For the remaining 2 votes each has 4 ways. So the numerator becomes: 6C4*4!*4*4.... Denominator is obviously 4^6 Please tell me where am i wrong? Thanks
You need the number of distinct voting patterns for the 6 friends: F1, F2, ..., F6 The voting patterns would be like WWWWWW, WLCLCW, SLLLLL, LCLCWS etc
Using your method, you are double counting some favourable cases. Say you select F1, F2, F3 and F5. They vote for WSLC. F4 votes for L and F6 votes for C so you get WSLLCC.
Now imagine the selected friends are F1, F2, F3 and F6. They vote for WSLC. F4 votes for L and F5 votes for C so you get WSLLCC again.
But since you are selecting 4 friends in different ways, you are counting them as different voting patterns which is not correct.
_________________
Karishma Veritas Prep GMAT Instructor
Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >
close
37% (02:12) correct 
