3. Arrange $1,2,3,4,5,6$ randomly in a row, denoted as $a, b, c, d, e, f$, then the probability that $a b c+\operatorname{def}$ is an even number is $\qquad$.
Answer: $\frac{9}{10}$.
Solution: First consider the case where $a b c+d e f$ is odd, at this point $a b c, d e f$ 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 $1-\frac{72}{6!}=1-\frac{72}{720}=\frac{9}{10}$.