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 $\qquad$
When $a \neq 3 k$, $a^{2} \equiv 1(\bmod 3)$, then $b \equiv -1(\bmod 3)$.
At this time, the number of ways to choose $(a, b)$ is $7 \times 4=28$;
When $a=3 k$, $a^{2} \equiv 0(\bmod 3)$, then $b \equiv 0(\bmod 3)$.
At this time, the number of ways to choose $(a, b)$ is $3 \times 3=9$.
Therefore, the probability that $a^{2}+b$ is divisible by 3 is $P=\frac{28+9}{100}=\frac{37}{100}$.