The first die has faces $1$ through $6$ and the second die has faces $1$ through $8$.

Total number of outcomes:

$6 \times 8 = 48$

Find all pairs $(a, b)$ where $a + b = 12$, with $1 \le a \le 6$ and $1 \le b \le 8$:

• $(4, 8)$: $4 + 8 = 12$ $\checkmark$
• $(5, 7)$: $5 + 7 = 12$ $\checkmark$
• $(6, 6)$: $6 + 6 = 12$ $\checkmark$

There are $3$ favorable outcomes, so:

$P(\text{sum} = 12) = \frac{3}{48}$