fair die probability

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 9

.

Event B

: The sum is divisible by 3

.

Answer 1

Roll 1 → 1 2 3 4 5 6

Roll

2

↓ -----------------------------------

1 | 2 3 4 5 6 7

|

2 | 3 4 5 6 7 8

|

3 | 4 5 6 7 8 9

|

4 | 5 6 7 8 9 10

|

5 | 6 7 8 9 10 11

|

6 | 7 8 9 10 11 12

Above is the sample space (all possible rolls). Counting them all, you see 36 possibilities. A sum greater than 9 is 10 (there are 3 possible ways to get 10), 11 (there are 2 possible ways to get 11) and 12 (there is one possible way to get 12). Since 3 + 2 + 1 = 6 the probability of getting a sum greater than 9 is 6/36 (or 1/6).

Do the same analysis for numbers divisible by 3.