Such Relations, Much Wow!

Let $p$ be a prime greater than $3$ . Clearly, $p\not\equiv0, 2, 4 \pmod{6}$ because it would be even, and $p\not\equiv 3\pmod{6}$ because it would be divisible by $3$ . Therefore, $p\equiv 1,5\pmod{6}$ , meaning it can be expressed in the form $6k\pm 1$ for some $k\in\mathbb{N}$ .

Hence the set $A$ is simply $\{2,3\}$ . There are $2\cdot 2=4$ mappings in total, but only two of these satisfy the condition: $\{2,3\}\rightarrow\{2,3\}$ and $\{2,3\}\rightarrow\{3,2\}$ . Hence the answer is $\frac{2}{4}=\frac{1}{2}\implies a+b=\boxed{3}$ .