Rolling dice and summing up to 9 or greater

A pair of honest dice are rolled. Find the probability of the sum of the values being equal to 9 or greater if a 6 occurs in at least one of them.

So I made a table of the possible sums. I called Event A the event of the sum being 9 or greater with at least one of the dice being 6, and Event B the event of the sum being 9 or greater with no die being 6. My reasoning was that the sum of the probabilities of both events would be equal to the probability of the sum being equal to 9 or greater, period. There are 10 ways out of 36 that sum up to 9 or greater, and 3 ways out of 36 that sum up to 9 or greater without any of the dice being 6. However, that's not the answer (7/11). What am I getting wrong?

What you have actually calculated here is the probability that a 6 occurs given the sum of the values being equal to 9 or greater. Note how both cases you mentioned involved the sum being 9 or greater.

By your method, the real way to do it is by calling Event A the event of the sum being 9 or greater and at least one of the dice being 6 (as you did before), and by calling Event B the event of the sum having a 6 with the sum not being above 9.

Since Event A is a 7-in-36 case and event B is a 4-in-36 case, the probability is $\frac{\frac{7}{36}}{\frac{7}{36}+\frac{4}{36}}=\boxed{\frac{7}{11}},$ as the answer key says.

This is an example of a conditional probability question. You are given that at least one of the die is a $6$. That means this is not accounted into your probability.

Now consider, given one die is a $6$, the probability of summing to a $9$ or greater. The possibilities are $6+3, 6+4, 6+5, 6+6$. Now, either the first die rolls a $6$ or the second die rolls a $6$. So, each possibility is counted twice except $6+6$ because in this case both of the dice roll a $6$.

Then, there are $7$ ways, given at least one die is a $6$, to roll a $9$ or greater. In total, however, you also have $6+1, 6+2, 2+6, 1+6$. Therefore there are $11$ total possibilities.

Then, the probability is $\frac{7}{11}$. Hope this helps!