jmc

A three-digit palindrome must be of the form $1\square 1,2\square 2,\cdots 9\square 9$, where $\square$ is any digit from 0 to 9. So there are $9\cdot 10=90$ three-digit palindromes. Now we look at which ones are multiples of 3. Recall that a positive integer is a multiple of 3 if and only if the sum of its digits is a multiple of 3. If we look at $1\square 1$, we want $1+1+\square$ to be a multiple of 3, so $\square$ could be 1, 4, or 7. With $2\square 2$, $2+2+\square$ should be a multiple of 3, so $\square$ could be 2, 5, or 8. With $3\square 3$, $\square$ could be 0, 3, 6, or 9. The possible $\square$ values then repeat, $4\square 4$ giving 1, 4, or 7, $5\square 5$ giving 2, 5, or 8, etc. So the number of multiples of 3 is $3\times (3+3+4)=30$. Since there are 90 three-digit palindromes in all, we get a probability of $\frac{30}{90}=\boxed{\frac{1}{3}}$.