An ordinary (fair) die is a cube with the numbers 1 through 6 on the sides (represented by painted spots). Imagine that such a die is rolled twice in succession and that the face values of the two rolls are added together. This sum is recorded as the outcome of a single trial of a random experiment. Compute the probability of each of the following events: Event A : The sum is greater than 6 . Event B : The sum is divisible by 2 or 5 (or both).

Question

An ordinary (fair) die is a cube with the numbers  1  through  6  on the sides (represented by painted spots). Imagine that such a die is rolled twice in succession and that the face values of the two rolls are added together. This sum is recorded as the outcome of a single trial of a random experiment.  Compute the probability of each of the following events: Event  A  : The sum is greater than  6  . Event  B  : The sum is divisible by  2  or  5  (or both).

anonymous · Answer

This would be a Normal Distribution. There are 36 possible outcomes, with numbers that can be {2,...,12}, distributed {1,2,3,4,5,6,5,4,3,2,1}. You can figure this out manually (since there aren't that many possibilities) by counting the number of possibilities that are 6 or less then subtracting 1 from the fraction. In this case, the distribution would be {1, 2, 3, 4, 5} which adds up to 15. Therefore, the total number of possibilites that are greater than 6 is equal to 1 - (15/36) = 21/36.

For part B: repeat the last process, only now counting numbers that are divisible by 2 or 5. These numbers are 2 (1), 4 (3), 5 (4), 6 (5), 8 (5), 10 (3), and 12 (1). The number of outcomes now equals 22, so the probability that the total is divisble by 2 or 5 is 22/36.