3. If $a, b, c$ are three elements chosen from the set $\{1,2,3,4,5\}$ (not necessarily distinct), then the probability that $ab+c$ is even is ( ).
A. $\frac{2}{5}$
B. $\frac{59}{125}$
C. $\frac{1}{2}$
D. $\frac{64}{125}$
E. $\frac{3}{5}$
3. B. Reason: The total number of ways to choose three elements from the set is $5^{3}=125$.

Now consider the case of $c$: There are $C_{3}^{1}=3$ ways to choose one from $\{1,3,5\}$, making $c$ an odd number; there are $C_{2}^{1}=2$ ways to choose one from $\{2,1\}$, making $c$ an even number.
The number of cases where $a b$ is odd is $3^{2}=9$, and the number of cases where $a b$ is even is $5^{2}-3^{2}=16$.
By "odd + odd = even" and "even + even = even", the number of cases where $a b+c$ is even is $3 \times 9 + 2 \times 16 = 59$ (cases),
Thus, the required probability is $\frac{59}{125}$.