Question

100 points!!!!

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 4: The sum is greater than 6.
Event B: The sum is not divisible by 3 and not divisible by 6.
Round your answers to two decimal places.

Answer 1

There are 36 possible outcomes from rolling a die twice (or two dice once) and these result in sums from 2 to 12.

Possible ways to make a sum of 2: 1

Possible ways to make a sum of 3: 2

Possible ways to make a sum of 4: 3

Possible ways to make a sum of 5: 4

Possible ways to make a sum of 6: 5

Possible ways to make a sum of 7: 6

Possible ways to make a sum of 8: 5

Possible ways to make a sum of 9: 4

Possible ways to make a sum of 10: 3

Possible ways to make a sum of 11: 2

Possible ways to make a sum of 12: 1

This means there are 21 ways to make a sum greater than 6.

21/36 ≈ 58.33%

There are 24 ways to make sums that are not divisible by 3 and not divisible by 6. (Note that if a number isn't divisible by 3, it's also not divisible by 6, so that's a weird thing to include.)

24/36 ≈ 66.67%