Two fair dice are rolled simultaneously. What is the probability that the sum of the numbers shown on the dice is divisible by 4 and at least one of the dice shows a 6?

Step-by-step solution

1. Sums divisible by $4$ up to $12$ are $4,;8,;12$.
2. List ordered pairs for each: - Sum $4$: $(1,3),(2,2),(3,1)$ – none contain a $6$. - Sum $8$: $(2,6),(3,5),(4,4),(5,3),(6,2)$ – pairs with a $6$ are $(2,6)$ and $(6,2)$. - Sum $12$: $(6,6)$ – contains a $6$.
3. Favourable ordered pairs: $(2,6),(6,2),(6,6)$ – $3$ outcomes.
4. There are $6\times 6=36$ equally likely ordered pairs.
5. Probability $=3/36=1/12$.

Concept tested: Combinatorial probability with multiple constraints

Published 14 September 2026 · Last updated 16 September 2026 · How this solution was prepared and checked