Two dice were rolled simultaneously. Find the probability that the sum of the numbers on them was a prime number.

When two dice are rolled, total outcomes $= 6 \times 6 = 36$

The minimum sum $= 1 + 1 = 2$ and maximum sum $= 6 + 6 = 12$

Prime numbers between 2 and 12 are: $2, 3, 5, 7, 11$

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Favorable outcomes for each prime sum:

Sum $= 2$: $(1,1)$ $\Rightarrow$ 1 outcome

Sum $= 3$: $(1,2), (2,1)$ $\Rightarrow$ 2 outcomes

Sum $= 5$: $(1,4), (4,1), (2,3), (3,2)$ $\Rightarrow$ 4 outcomes

Sum $= 7$: $(1,6), (6,1), (2,5), (5,2), (3,4), (4,3)$ $\Rightarrow$ 6 outcomes

Sum $= 11$: $(5,6), (6,5)$ $\Rightarrow$ 2 outcomes

Total favorable outcomes $= 1 + 2 + 4 + 6 + 2 = 15$

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$P(\text{prime sum}) = \dfrac{15}{36}$

$= \dfrac{5}{12}$

Since $\dfrac{5}{12}$ does not match any of the given specific options, the answer is None of these.