2. When two dice are rolled simultaneously, if $P(a)$ represents the probability that the sum of the numbers facing up on the two dice is $a$, then
$$
\begin{aligned}
& P(1)+P(2)+P(3)+P(4)+P(5) \\
= &
\end{aligned}
$$
2. $\frac{5}{18}$.

Let the two dice be A and B. Throwing two dice has a total of $6 \times 6=36$ possible outcomes.
Obviously, $P(1)=0$.
The sum of the numbers facing up on the two dice being 2 only has one possibility: both A and B show 1, so $P(2)=\frac{1}{36}$.

The sum of the numbers facing up on the two dice being 3 has two possibilities: A shows 1 and B shows 2, or A shows 2 and B shows 1. Thus, $P(3)=\frac{2}{36}$.
Similarly, $P(4)=\frac{3}{36}, P(5)=\frac{4}{36}$.
Therefore, $P(1)+P(2)+\cdots+P(5)$
$$
=\frac{1+2+3+4}{36}=\frac{5}{18} \text {. }
$$