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For one roll of a certain number cube with six faces, numbered 1 throu
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24 Aug 2021, 05:15
24 Aug 2021, 05:15
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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
(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
Re: For one roll of a certain number cube with six faces, numbered 1 throu
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27 Aug 2021, 10:20
27 Aug 2021, 10:20
1
Expert Reply
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.
- 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
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16 Oct 2022, 00:02
16 Oct 2022, 00:02
Given that For one roll of a certain number cube with six faces, numbered 1 through 6, the probability of rolling a two is 1/6 and the cube is rolled 4 times and We need to find which of the following is the probability that the outcome will be a two at least 3 times?
As we are rolling a cube 4 times => Number of cases = \(6^4\) = 1296
Now, P(getting a two at least 3 times) = P(Getting 2 exactly 3 times) + P(Getting 2 all four times)
P(Getting 2 exactly 3 times)
Now, lets find the three places out of 4 where we will get a 2. This can be done in 4C3 ways
=> \(\frac{4!}{3!*1!}\) = 4 ways
P(getting a 2) = \(\frac{1}{6}\) (as there is one way out of 6 in which we can get a 2)
P(getting any number except 2) = \(\frac{5}{6}\) (as we can get any of the 5 numbers, out of 6, except 2)
P(Getting 2 exactly 3 times) = P(Getting exactly three 2's and getting any number except 2 in 1 roll) = Number of ways * P(Getting 2) * P(Getting 2) * P(Getting 2) * P(Getting any number except 2) = 4 * \(\frac{1}{6}\) * \(\frac{1}{6}\) * \(\frac{1}{6}\) * \(\frac{5}{6}\) = 4*(\(\frac{1}{6}\))^3 * \(\frac{5}{6}\)
P(Getting 2 all four times)
P(Getting 2 all four times) = \(\frac{1}{6}\) * \(\frac{1}{6}\) * \(\frac{1}{6}\) * \(\frac{1}{6}\) = (\(\frac{1}{6}\))^4
=> P(getting a two at least 3 times) = P(Getting 2 exactly 3 times) + P(Getting 2 all four times) = 4*(\(\frac{1}{6}\))^3 * \(\frac{5}{6}\) + (\(\frac{1}{6}\))^4
So, Answer will be D
Hope it helps!
Watch the following video to learn How to Solve Dice Rolling Probability Problems
As we are rolling a cube 4 times => Number of cases = \(6^4\) = 1296
Now, P(getting a two at least 3 times) = P(Getting 2 exactly 3 times) + P(Getting 2 all four times)
P(Getting 2 exactly 3 times)
Now, lets find the three places out of 4 where we will get a 2. This can be done in 4C3 ways
=> \(\frac{4!}{3!*1!}\) = 4 ways
P(getting a 2) = \(\frac{1}{6}\) (as there is one way out of 6 in which we can get a 2)
P(getting any number except 2) = \(\frac{5}{6}\) (as we can get any of the 5 numbers, out of 6, except 2)
P(Getting 2 exactly 3 times) = P(Getting exactly three 2's and getting any number except 2 in 1 roll) = Number of ways * P(Getting 2) * P(Getting 2) * P(Getting 2) * P(Getting any number except 2) = 4 * \(\frac{1}{6}\) * \(\frac{1}{6}\) * \(\frac{1}{6}\) * \(\frac{5}{6}\) = 4*(\(\frac{1}{6}\))^3 * \(\frac{5}{6}\)
P(Getting 2 all four times)
P(Getting 2 all four times) = \(\frac{1}{6}\) * \(\frac{1}{6}\) * \(\frac{1}{6}\) * \(\frac{1}{6}\) = (\(\frac{1}{6}\))^4
=> P(getting a two at least 3 times) = P(Getting 2 exactly 3 times) + P(Getting 2 all four times) = 4*(\(\frac{1}{6}\))^3 * \(\frac{5}{6}\) + (\(\frac{1}{6}\))^4
So, Answer will be D
Hope it helps!
Watch the following video to learn How to Solve Dice Rolling Probability Problems
