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08 Nov 2021, 06:45
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D
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Difficulty:
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49% (01:22) correct
51%
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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 same attraction?
(A) (1/6)^4
(B) (1/6)^3
(C) (1/4)^6
(D) (1/4)^5
(E) (1/4)
(A) (1/6)^4
(B) (1/6)^3
(C) (1/4)^6
(D) (1/4)^5
(E) (1/4)
08 Nov 2021, 06:52
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Bunuel
P(all 6 friends choose the same attraction) = P(1st person selects any attraction AND 2nd person selects same attraction as 1st person AND 3rd person selects same attraction as 1st person AND 4th person selects same attraction as 1st person AND 5th person selects same attraction as 1st person AND 6th person selects same attraction as 1st person)
= P(1st person selects any attraction) x P(2nd person selects same attraction as 1st person) x P(3rd person selects same attraction as 1st person) x P(4th person selects same attraction as 1st person) x P(5th person selects same attraction as 1st person) x P(6th person selects same attraction as 1st person)
= \(\frac{4}{4}\) x \(\frac{1}{4}\) x \(\frac{1}{4}\) x \(\frac{1}{4}\) x \(\frac{1}{4}\) x \(\frac{1}{4}\)
= \(\frac{1}{4^5}\)
Answer: D
06 Jun 2024, 20:54
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