Two dice are thrown together. Find the probability that the sum of the points…

Concept — For a random experiment in which every outcome is equally likely, the probability of an event E is a ratio of counts: P(E) = (number of outcomes favourable to E) / (total number of possible outcomes).

When an event is described as "A or B" and the two cases can never happen on the same trial, they are mutually exclusive, so their favourable counts simply add: n(A or B) = n(A) + n(B). A divisibility condition on a sum must first be converted into the list of multiples that actually lie inside the range of possible sums.

Applying it here

1. Two dice thrown together give ordered outcomes of the form (first die, second die). Each die has 6 faces, so the total number of equally likely outcomes is 6 x 6 = 36.

2. The smallest possible sum is 1 + 1 = 2 and the largest is 6 + 6 = 12, so every sum lies between 2 and 12.

3. Multiples of 7 inside that range: only 7, because 14 is already greater than 12. Multiples of 9 inside that range: only 9, because 18 is greater than 12. So the condition "divisible by 7 or 9" means the sum equals 7 or the sum equals 9.

4. Outcomes whose sum is 7: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1). That is 6 outcomes.

5. Outcomes whose sum is 9: (3,6), (4,5), (5,4), (6,3). That is 4 outcomes.

6. A single roll cannot total 7 and 9 at the same time, so the two cases are mutually exclusive and their counts add: 6 + 4 = 10 favourable outcomes.

7. Apply the ratio: P = 10 / 36. Dividing numerator and denominator by 2 gives P = 5/18.

Cross-check — Cross-check with the standard tally for two dice: the number of ordered outcomes with sum s is (s - 1) for sums from 2 to 7, and (13 - s) for sums from 7 to 12. This gives 7 - 1 = 6 outcomes for sum 7 and 13 - 9 = 4 outcomes for sum 9, confirming 10 favourable outcomes out of 36.

Result — Probability that the sum of the two faces is divisible by 7 or 9 = 5/18.