Rolling dice

What we are wanting to calculate is the conditional probability that both dice show a $6$ given that at least one die shows a $6$ on the last roll. Letting $A$ be the event that at least one die shows a $6$ and $B$ the event that both dice show a $6$ , the desired probability is $P(B|A) = \dfrac{P(B \cap A)}{P(A)}$ .

Now $B$ occurs whenever $A$ does, so $P(B \cap A) = P(B) = \left(\dfrac{1}{6}\right)^{2} = \dfrac{1}{36}.$

Next, it is easiest to calculate $P(A)$ by way of its complement, i.e.,

$P(A) = 1 - P(\bar{A}) = 1 - \left(\dfrac{5}{6}\right)^{2} = \dfrac{11}{36}.$

Thus $P(B|A) = \dfrac{\dfrac{1}{36}}{\dfrac{11}{36}} = \dfrac{1}{11}$ , and so $a + b = 1 + 11 = \boxed{12}.$