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Six friends live in the city of Monrovia. There are four nat
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03 Jul 2013, 06:27
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Difficulty:
95% (hard)
Question Stats:
38% (02:30) correct 62% (01:45) wrong based on 39 sessions
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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
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24 Mar 2014, 02:12
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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
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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
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28 Apr 2014, 10:11
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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
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26 May 2019, 15:10
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].
Re: Six friends live in the city of Monrovia. There are four nat
[#permalink]
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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!
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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.
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38% (02:30) correct 
