Question

a pair of dice is rolled 3x. what is the probability that a sum of 8 on the 2 dice will occur at least once. round your answer to three decimal places

Answer 1

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Now then, to your question. :) I believe this question can be solved by using a permutation. Try looking that up and seeing what happens! :)

Answer 2

thank you! i still need help though with this problem

Answer 3

This is my take on it.

The first dice roll needs to be at LEAST a 2. if you roll a 1 then there is no way you can get a sum of 8.

so we start with the probability of getting at least a 2, which is 5/6.

then the second dice roll MUST be the difference of 8 - the first roll. and the probability of that happening is 1/6.

so if you roll the dice once you get a probability of \[(5*1)/(6 * 6) = 5/36\]
however you roll the dice 3 times, so the probability should be 3 times the old probability

\[(3*5)/36 = 5/12\]

Answer 4

i also feel like this COULD be solved using binomial Distribution but its been a while since i took a stats course

Answer 5

i know you need to multiply 6*6*6 to give you the amount of times each dice could be rolled

Answer 6

The sample space for each roll of the pair has 36 possible combinations of numbers. These can be set out in column form as follows:
6,6 5,6 4,6 3,6 2,6 1,6
6,5 5,5 4,5 3,5 2,5 1,5
6,4 5,4 4,4 3,4 2,4 1,4
6,3 5,3 4,3 3,3 2,3 1,3
6,2 5,2 4,2 3,2 2,2 1,2
6,1 5,1 4,1 3,1 2,1 1,1
The probability of getting a sum of 8 on each roll of the pair is 5/36.
The question asks for the probability that a sum of 8 will be obtained at least once in the 3 rolls, meaning that the probability of getting a sum of eight on one, two or three rolls is required.
The required probability can be found by subtracting the probability of zero sums of 8 from 1.
P(1, 2 or 3 sums of 8) = 1.0000 - P(0 sums of 8)
\[P(0\ sums\ of\ 8)=(\frac{31}{36})^{3}\]
\[P(1, 2\ or\ 3\ sums\ of\ 8)=1-(\frac{31}{36})^{3}=you\ can\ calculate\]