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The probability of getting a '6' on a role of die is 1/6. If this die

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The probability of getting a '6' on a role of die is 1/6. If this die  [#permalink]

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24 Nov 2019, 10:45
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74% (02:08) correct 26% (01:51) wrong based on 19 sessions

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The probability of getting a '6' on a role of die is 1/6. If this die is rolled
4 times, what is the probability of getting a '6' at least 2 times?

(A) $$(1/6)^4$$
(B) $$2(1 /6)^3(5/6) + (1/6)^4$$
(C) $$6(1/6)^2(5/6)^2 + 4(1/6)^3(5/6) + (1/6)^4$$
(D) $$12(1/6)^2(5/6)^2 + 4(1/6)^3(5/6) + (1/6)^4$$
(E) $$12(1/6)^2(5/6)^2 + 6(1/6)^3(5/6) + (1/6)^4$$
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Re: The probability of getting a '6' on a role of die is 1/6. If this die  [#permalink]

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01 Dec 2019, 02:05
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Re: The probability of getting a '6' on a role of die is 1/6. If this die  [#permalink]

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01 Dec 2019, 02:30
fauji wrote:
The probability of getting a '6' on a role of die is 1/6. If this die is rolled
4 times, what is the probability of getting a '6' at least 2 times?

(A) $$(1/6)^4$$
(B) $$2(1 /6)^3(5/6) + (1/6)^4$$
(C) $$6(1/6)^2(5/6)^2 + 4(1/6)^3(5/6) + (1/6)^4$$
(D) $$12(1/6)^2(5/6)^2 + 4(1/6)^3(5/6) + (1/6)^4$$
(E) $$12(1/6)^2(5/6)^2 + 6(1/6)^3(5/6) + (1/6)^4$$

total ways ; 4c2 * ( 5/6)^2*(1/6)^2 + 4c1* (1/6)^3*(5/6) + (1/6)6
IMO C $$6(1/6)^2(5/6)^2 + 4(1/6)^3(5/6) + (1/6)^4$$
Re: The probability of getting a '6' on a role of die is 1/6. If this die   [#permalink] 01 Dec 2019, 02:30
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