5. In $1,2, \cdots, 10$ a number $a$ is randomly selected, and in $-1,-2, \cdots,-10$ a number $b$ is randomly selected. The probability that $a^{2}+b$ is divisible by 3 is $\qquad$.
5. $\frac{37}{100}$.

The array $(a, b)$ has $10^{2}=100$ equally probable selection methods.
Consider the number of selection methods $N$ that make $a^{2}+b$ divisible by 3.
If $3 \mid a$, then $3 \mid b$, in this case, $a$ and $b$ each have three selection methods, such $(a, b)$ has $3^{2}=9$ groups; If $3+a$, then $a^{2} \equiv 1(\bmod 3)$, thus, $b \equiv-1(\bmod 3)$, at this time, $a$ has seven selection methods, $b$ has four selection methods, such $(a, b)$ has $7 \times 4=28$ groups.
Therefore, $N=9+28=37$.
Thus, the required probability is $\frac{37}{100}$.