You understood Me

For divisibility of 33 we should check the divisibity of 11 and 3. I think I need not to explain that for divisibility of 3 we should have $(a+b) = 2,5,8,11,14,17$ Also for divisibility of 11 we need $(3+2+b)-(a+5) = 0$ (Since, the given numbers are not so large that we can get a multiple of 11 as difference) So we get, $a+5 = 5+b$ Or, $a=b$ Or, $a+b=2a=2b$ So $a+b$ is even. Hence only $2,8,14$ are required 3 sums and there are only three sets of $a$ & $b$ . The number of ways in which $a$ & $b$ can be chosen are $10\times 10 = 100$ Therefore required probability = $\frac{3}{100} = 0.03$