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For one roll of a certain number cube with six faces, numbered 1 throu

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For one roll of a certain number cube with six faces, numbered 1 throu  [#permalink]

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New post Updated on: 22 Oct 2017, 20:37
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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

Originally posted by Rorschach1337 on 22 Oct 2017, 20:11.
Last edited by Bunuel on 22 Oct 2017, 20:37, edited 1 time in total.
Renamed the topic.
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Re: For one roll of a certain number cube with six faces, numbered 1 throu  [#permalink]

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New post 22 Oct 2017, 20:34
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Rorschach1337 wrote:
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



ATLEAST 3 times means 3 times and 4 times..
1) 3 times :-
3 times "2" will mean a probability of \(\frac{1}{6}^3\) and prob of any other number in 4 throw is 5/6..
but this ANY other number can come in any of 4 throws ..
so prob of 3times = \(\frac{1}{6}^3*4*\frac{5}{6}\)
2) 4 times :-
\(\frac{1}{6}^4\)

total \(\frac{1}{6}^4+4*\frac{1}{6}^3\frac{5}{6}\)

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For one roll of a certain number cube with six faces, numbered 1 throu  [#permalink]

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New post 22 Oct 2017, 21:13
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"The outcome will be a two at least 3 times" means the outcome will be a two 3 times or 4 times.

Case 1: 3 times = 4C3 * (1/6)^3 * 5/6 = 4*(1/6)^3 * 5/6

- select 3 cubes whose the outcome will be two = 4C3
- The prob that 3 cubes will be 2 = (1/6)^3
- The prob of the remaining cube that can be any other value except "2" = 5/6

Case 2: 4 times = (1/6)^4

Hence, total = case 1 + case 2 = 4*(1/6)^3 * 5/6 + (1/6)^4 => Answer D.

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Re: For one roll of a certain number cube with six faces, numbered 1 throu  [#permalink]

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New post 25 Mar 2019, 02:34
Rorschach1337 wrote:
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


Total rolls = 4 times

Dice = 6 sided

Probability of a "2" on every roll = 1/6

Probability of not a "2" but any of the other 5 numbers = 1 - 1/6 = 5/6

Probability of "2" at least 3 times means :

a. 2 three times

b. 2 four times

Case 1 : T T T N (T = two and N = Not two)

\(\frac{1}{6}*\frac{1}{6}*\frac{1}{6}*\frac{1}{6}*\frac{5}{6} * \frac{4!}{3!}\)

We multiply by 4! because permutations/total arrangements of 4 different items can be done in 4! ways

We divide by 3! because we have 3 identical items.

Therefore, 4*(1/6)^4 * 5/6

Case 2 : T T T T (All twos)

\((1/6)^4\)


Add Case I and II

\(4*(1/6)^4 * 5/6\) + \((1/6)^4\)

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Re: For one roll of a certain number cube with six faces, numbered 1 throu   [#permalink] 25 Mar 2019, 02:34
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