jmc

The quantity $ab+c$ is even if and only if $ab$ and $c$ are both odd or both even. The probability that $c$ is odd is $\frac{3}{5},$ and the probability that $ab$ is odd is ${\left(\frac{3}{5}\right)}^2=\frac{9}{25}$ (because both $a$ and $b$ must be odd). Therefore, the probability that $ab+c$ is even is $$\frac{3}{5}\cdot \frac{9}{25}+\left(1-\frac{3}{5}\right)\left(1-\frac{9}{25}\right)=\boxed{\frac{59}{125}}.$$