Question 10

Let $x$ and $y$ be the numbers rolled on the first and second rolls, respectively. There are $6 \times 6 = 36$ equally likely ordered outcomes $(x, y)$. Counting the pairs where $y$ is an integer multiple of $x$:

• $x = 1$: $y \in \{1, 2, 3, 4, 5, 6\}$ ($6$ pairs)
• $x = 2$: $y \in \{2, 4, 6\}$ ($3$ pairs)
• $x = 3$: $y \in \{3, 6\}$ ($2$ pairs)
• $x = 4$: $y \in \{4\}$ ($1$ pair)
• $x = 5$: $y \in \{5\}$ ($1$ pair)
• $x = 6$: $y \in \{6\}$ ($1$ pair)

Total favorable outcomes $= 6 + 3 + 2 + 1 + 1 + 1 = 14$. Thus the probability is $\frac{14}{36} = \frac{7}{18}$.