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Six friends live in the city of Monrovia. There are four nat [#permalink]

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03 Jul 2013, 06:27

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A

B

C

D

E

Difficulty:

85% (hard)

Question Stats:

48% (03:12) correct
53% (02:12) wrong based on 40 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].

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]

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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.

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]

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03 Jan 2016, 04:39

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