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 or (or both).
Write your answers as fractions

Answer 1

I need to confirm what numbers "divisible by or (or both)" means. A common interpretation is "divisible by 2 or 3 (or both)"; I'll give that solution. If you meant different divisors, tell me which and I'll redo it.

Sample space: 36 equally likely outcomes (6×6).

Event A: sum > 6.
Sums greater than 6 are 7–12. Their frequencies: 7 (6), 8 (5), 9 (4), 10 (3), 11 (2), 12 (1). Total = 6+5+4+3+2+1 = 21.
P(A) = 21/36 = 7/12.

Event B (assuming "divisible by 2 or 3"):
- Divisible by 2 (even sums 2,4,6,8,10,12): frequencies 1+3+5+5+3+1 = 18 → 18/36 = 1/2.
- Divisible by 3 (sums 3,6,9,12): frequencies 2+5+4+1 = 12 → 12/36 = 1/3.
- Divisible by both 2 and 3 (i.e. by 6: sums 6,12): frequencies 5+1 = 6 → 6/36 = 1/6.
By inclusion–exclusion: P(B) = 1/2 + 1/3 − 1/6 = 2/3 (i.e. 24/36).

Answers:
- P(A) = 7/12
- P(B) = 2/3 (if "2 or 3" was intended).