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For one roll of a certain number cube with six faces, numbered 1 through 6, the probability of rolling a two is 1/6. If

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by BTGmoderatorDC » Tue May 12, 2020 9:26 pm

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A

B

C

D

E

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For one roll of a certain number cube with six faces, numbered 1 through 6, the probability of rolling a two is 1/6. If this number cube is rolled 4 times, which of the following is the probability that the outcome will be a two at least 3 times?

(A) (1/6)^4
(B) 2(1/6)^3 + (1/6)^4
(C) 3(1/6)^3 (5/6) + (1/6)^4
(D) 4(1/6)^3 (5/6) + (1/6)^4
(E) 6(1/6)^3 (5/6) + (1/6)^4


OA D

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Source: — Problem Solving |

BTGmoderatorDC wrote:
Tue May 12, 2020 9:26 pm
For one roll of a certain number cube with six faces, numbered 1 through 6, the probability of rolling a two is 1/6. If this number cube is rolled 4 times, which of the following is the probability that the outcome will be a two at least 3 times?

(A) (1/6)^4
(B) 2(1/6)^3 + (1/6)^4
(C) 3(1/6)^3 (5/6) + (1/6)^4
(D) 4(1/6)^3 (5/6) + (1/6)^4
(E) 6(1/6)^3 (5/6) + (1/6)^4


OA D
“At least 3 twos in 4 rolls” means we could have 3 twos and 1 non-two number OR all 4 twos.

P(3 twos and 1 non-two number) = 4C3 x (1/6)^3 x (5/6) = 4(1/6)^3 (5/6)

P(all 4 twos) = 4C4 x (1/6)^4 = (1/6)^4

Therefore, P(at least 3 twos) = 4(1/6)^3 (5/6) + (1/6)^4

Answer: D

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