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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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37% (02:12) correct 63% (01:45) wrong based on 35 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].
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Re: Six friends live in the city of Monrovia. There are four nat
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24 Mar 2014, 02:12
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]. The cases when attractions get zero votes are: {6, 0, 0, 0} {5, 1, 0, 0} {4, 2, 0, 0} {4, 1, 1, 0} {3, 2, 1, 0} {3, 1, 1, 1} {2, 2, 2, 0} {2, 2, 1, 1} 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.
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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]. The cases when attractions get zero votes are: {6, 0, 0, 0} {5, 1, 0, 0} {4, 2, 0, 0} {4, 1, 1, 0} {3, 2, 1, 0} {3, 1, 1, 1} {2, 2, 2, 0} {2, 2, 1, 1} 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!



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Re: Six friends live in the city of Monrovia. There are four nat
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28 Apr 2014, 10:11
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]. The cases when attractions get zero votes are: {6, 0, 0, 0} {5, 1, 0, 0} {4, 2, 0, 0} {4, 1, 1, 0} {3, 2, 1, 0} {3, 1, 1, 1} {2, 2, 2, 0} {2, 2, 1, 1} 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. Hope it's clear.
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Re: Six friends live in the city of Monrovia. There are four nat
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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
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Re: Six friends live in the city of Monrovia. There are four nat
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23 Mar 2019, 13:00
BunuelI 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???



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Re: Six friends live in the city of Monrovia. There are four nat
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26 May 2019, 15:10
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]. Veritas Prep Official Solution Here, A, the event for which we want to find the probability is ‘each attraction gets at least one vote’. P(A) = No of ways in which each attraction gets at least one vote /Total no. of ways in which the friends can vote. Each attraction should get at least one vote. 6 votes can be divided among 4 attractions in the following ways: (1, 1, 1, 3) and (1, 1, 2, 2) Case 1: (1, 1, 1, 3) First, we select the attraction that will get 3 votes in 4 ways (= 4C1) Now, we can select the 3 people who will vote for this attraction in 6*5*4/3! = 20 ways (= 6C3 ) The other 3 votes will be distributed among the other 3 attractions in 3! = 6 ways The 6 people could vote for the 4 attractions in this case in 4*20*6 = 480 ways Case 2: (1, 1, 2, 2) Let’s select the two attractions that will get 2 votes each in 4*3/2! = 6 ways (= 4C2). Say we select caves and waterfall. Now, we can select the 2 people who will vote for one of the selected attractions in 6*5/2! = 15 ways (= 6C2) We can select the other 2 people who will vote for the other selected attraction in 4*3/2! = 6 ways (= 4C2) The other 2 votes will be distributed among the other 2 attractions in 2! = 2 ways The 6 people could vote for the 4 attractions in this case in 6*15*6*2 = 1080 ways Total number of ways in which 6 votes can be distributed among 4 attractions such that each attraction gets at least one vote = 480 + 1080 = 1560 ways As we saw in the questions above, the total no. of ways in which the friends can vote = 4^6 Therefore, P(A) = 1560/(4^6)
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Re: Six friends live in the city of Monrovia. There are four nat
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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]. The cases when attractions get zero votes are: {6, 0, 0, 0} {5, 1, 0, 0} {4, 2, 0, 0} {4, 1, 1, 0} {3, 2, 1, 0} {3, 1, 1, 1} {2, 2, 2, 0} {2, 2, 1, 1} 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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Re: Six friends live in the city of Monrovia. There are four nat
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15 Jun 2019, 07:14
VeritasKarishma chetan2u Bunuel generisI 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



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Re: Six friends live in the city of Monrovia. There are four nat
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16 Jun 2019, 23:25
Debashis Roy wrote: VeritasKarishma chetan2u Bunuel generisI 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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Re: Six friends live in the city of Monrovia. There are four nat
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