jmc

First we find the number of $4$ digit palindromes. There are ten palindromes for every distinct thousandth digit from $1$ to $9$ because there are $10$ numbers from $0$ to $9$ we could pick for the second and third digit. This gives us a total of $9\cdot 10$ palindromes.

Next, we can get that all palindromes are multiples of $11$. The divisibility rule for $11$ tells us that for a number $abcd$ to be divisible by $11$, then $a-b+c-d$ is divisible by $11$. Since $a=d$ and $b=c$, $a-b+c-d$ is always divisible by $11$ so all four digit palindromes are divisible by $11$.

Now we want to find now many of these palindromes are divisible by $9$. For a number to be divisible by $9$, the sum of the digits must be divisible by $9.$ It's impossible for the sum of the digits to be equal to $9$ or $27$ because it must be an even number (the sum is $a+b+c+d=2(a+b)$). We find the number of palindromes whose digits add up to $18.$ Since $a+b+c+d=2(a+b)=18,$ we get that $a+b=9.$ There are $9$ possible answers, where $a$ goes from $1$ to $9$ and $b=9-a$. We then find the number of palindromes whose digit add up to $36.$ There is only one four-digit number that does so, $9999.$

Therefore, we have that there are $9+1=10$ four-digit palindromes that are divisible by $99.$

Since there is a total of $90$ palindromes, the probability that it is divisible by $99$ is $\frac{10}{90}=\boxed{\frac{1}{9}}$.