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26 May 2019, 14:09
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Difficulty:
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85% (01:02) correct
15%
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wrong
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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 one of them votes for the safari?
A) (1/6)^4
B) (1/4)^6
C) (1/24)
D) (1/6)
E) (1/4)
A) (1/6)^4
B) (1/4)^6
C) (1/24)
D) (1/6)
E) (1/4)
26 May 2019, 14:10
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dabaobao
Veritas Prep Official Solution
Here, A, the event for which we want to find the probability is ‘all six friends vote for the safari’
P(A) = No of ways in which all six can vote for the safari/Total no. of ways in which they can vote.
What is the no. of ways in which all six vote for the safari? Only one way. They all vote for the safari!
What is the no. of ways in which the friends can vote? Say, the friends are A, B, C, D, E and F. A can vote in 4 ways. B can vote in 4 ways. C can vote in 4 ways and so on… Total no of ways in which the 6 friends can vote = 4*4*4*4*4*4 = 4^6 (Using our old friend, the basic counting principle). We discussed this concept in our post on Unfair Distributions.
Therefore, P(A) = (1/4)^6
ANSWER: B
24 Mar 2021, 04:04
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dabaobao
I don't understand why we do 4^6 to count the number of possible outcomes. This means that we consider that if A votes for safari first and then B for the lake (S,L, X, X), it is not the same as if A votes for the lake first and then B votes for the Safari (L, S, X, X).
But shouldn't this be counted only once ? Hence the number of outcomes would be 9C3 because you assign 6 identical votes to 4 choices.
Please tell me where I am wrong.
