3. Arrange $1,2,3,4,5,6$ randomly in a row, denoted as $a, b, c, d, e, f$. Then the probability that $a b c+d e f$ is an even number is $\qquad$
3. $\frac{9}{10}$.

Consider the case where $a b c+d e f$ is odd. In this case, $a b c$ and $d e f$ are one odd and one even. If $a b c$ is odd, then $a, b, c$ are permutations of 1, 3, 5, and thus, $d, e, f$ are permutations of 2, 4, 6, which gives $3! \times 3! = 36$ cases.

By symmetry, the number of cases where $a b c+d e f$ is odd is $36 \times 2 = 72$.
Therefore, the probability that $a b c+d e f$ is even is
$$
1-\frac{72}{6!}=1-\frac{72}{720}=\frac{9}{10} .
$$