jmc

Let $\ell$ be the perpendicular bisector of segment $AC$. We note that the points that are closer to $A$ than they are to $C$ are the points that are on the same side of $\ell$ as $A$.

Since $ABC$ is a 3-4-5 right triangle with a right angle at $C$, $\ell$ is parallel to line $BC$. Since it passes through the midpoint of $AC$, it also passes through the midpoint of $AB$, which we'll call $D$.

Let $m$ be the perpendicular bisector of segment $BC$. As before, the points that are closer to $C$ than they are to $B$ are those that lie on the same side of $m$ as $A$, and $m$ also passes through $D$.

Therefore the points that are closer to $C$ than they are to $A$ or $B$ are the points in the shaded rectangle below. The probability we want is then this rectangle's area divided by triangle $ABC$'s area. There are a few different ways to see that this ratio is $\boxed{\frac{1}{2}}$. One way is to note that we can divide $ABC$ into 4 congruent triangles, 2 of which are shaded: Another way is to notice that the rectangle's sides have length $\frac{3}{2}$ and $\frac{4}{2}$, so that the rectangle's area is $\frac{3\cdot 4}{2\cdot 2}$. Since triangle $ABC$'s area is $\frac{3\cdot 4}{2}$, it follows that the probability we seek is $\boxed{\frac{1}{2}}$, as before.