5. From the set $\{1,2,3, \cdots, 99,100\}$, arbitrarily select $a$, and then from the same set arbitrarily select $b$, the probability that the last digit of $3^{a}+7^{b}$ is 8 equals $\qquad$.
5. $\frac{3}{16}$ Detailed Explanation: The last digits of $3^{n}$ and $7^{n}$ change periodically as $3,9,7,1,7,9,3,1$, respectively. In the numbers from $1$ to $100$, the numbers of the form $4k, 4k+1, 4k+2, 4k+3$ are equal in number. Thus, the number of ways to choose pairs $(a, b)$ from these four types of numbers is $C_{4}^{1} \cdot C_{4}^{1}=16$. Among these, only the cases $7+1, 1+7, 9+9$ result in a last digit of 8. Therefore, the required probability is $\frac{3}{16}$.