5. In $1,2,3, \cdots, 10$ a number $a$ is randomly selected, and in $-1,-2,-3, \cdots,-10$ a number $b$ is randomly selected, then the probability that $a^{2}+b$ is divisible by 3 is $\qquad$ .
Answer: $\frac{37}{100}$.
Solution: The number of pairs $(a, b)$ is $10^{2}=100$ with equal probability.
Consider the number of pairs $N$ that make $a^{2}+b$ divisible by 3.
If $a$ is divisible by 3, then $b$ is also divisible by 3. In this case, $a$ and $b$ each have 3 choices, resulting in $3^{2}=9$ pairs of $(a, b)$. If $a$ is not divisible by 3, then $a^{2} \equiv 1(\bmod 3)$, thus $b \equiv-1(\bmod 3)$. In this case, $a$ has 7 choices, and $b$ has 4 choices, resulting in $7 \times 4=28$ pairs of $(a, b)$.
Therefore, $N=9+28=37$. Hence, the required probability is $\frac{37}{100}$.