smc

View the palindrome as some number with form (decimal representation): $a_3\cdot 10^3+a_2\cdot 10^2+a_1\cdot 10+a_0$. But because the number is a palindrome, $a_3=a_0,a_2=a_1$. Recombining this yields $1001a_3+110a_2$. 1001 is divisible by 7, which means that as long as $a_2=0$, the palindrome will be divisible by 7. This yields 9 palindromes out of 90 ($9\cdot 10$) possibilities for palindromes. However, if $a_2=7$, then this gives another case in which the palindrome is divisible by 7. This adds another 9 palindromes to the list, bringing our total to $18/90=\boxed{\frac{1}{5}}=\boxed{\frac{1}{5}}$