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 odd number is $\qquad$.
Answer: $\frac{1}{10}$.
Solution: When $a b c+d e f$ is odd, $a b c, d e f$ must be one odd and one even. If $a b c$ is odd, then $a, b, c$ are permutations of $1,3,5$, and $d, e, f$ are permutations of $2,4,6$. This results in $3!\times 3!=36$ cases. By symmetry, the number of cases that satisfy the condition is $36 \times 2=72$. Therefore, the required probability is $\frac{72}{6!}=\frac{72}{720}=\frac{1}{10}$.