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

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Joined: 17 Oct 2017
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For one roll of a certain number cube with six faces, numbered 1 throu  [#permalink]

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Updated on: 22 Oct 2017, 20:37
4
00:00

Difficulty:

55% (hard)

Question Stats:

57% (02:17) correct 43% (02:28) wrong based on 63 sessions

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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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Posts: 7981
Re: For one roll of a certain number cube with six faces, numbered 1 throu  [#permalink]

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22 Oct 2017, 20:34
1
1
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 $$\frac{1}{6}^4+4*\frac{1}{6}^3\frac{5}{6}$$

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

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22 Oct 2017, 21:13
1
"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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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)

$$(1/6)^4$$

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

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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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