Step 1: Set up the sample space.
Rolling a fair die twice gives $6\times6=36$ equally likely pairs $(a,b)$, where $a$ is the first roll and $b$ is the second.
Step 2: Work out how many pairs give each sum.
The number of pairs adding to a sum $s$ rises from $1$ at $s=2$ up to $6$ at $s=7$ and then falls back to $1$ at $s=12$, moving by $1$ each step. So the counts for sums $2$ through $12$ are $1,2,3,4,5,6,5,4,3,2,1$.
Step 3: Pick out the prime sums.
Among $2$ to $12$, the primes are $2,3,5,7,11$. Reading off their counts from the list above: sum $2$ gives $1$, sum $3$ gives $2$, sum $5$ gives $4$, sum $7$ gives $6$, and sum $11$ gives $2$, using the symmetric count for $11$, which mirrors sum $3$.
Step 4: Add these counts.
\[ 1+2+4+6+2=15 \]
Step 5: Divide by the total outcomes.
\[ P=\frac{15}{36} \]
So out of every $36$ equally likely rolls, $15$ give a prime sum.
\[ \boxed{\dfrac{15}{36}} \]