<s>[INST] Conditional probability with dependent events We have 2 dice. One is fair. The other one lands by the following probabilities:
6: 1/2
5: 1/10
4: 1/10
3: 1/10
2: 1/10
1: 1/10

We roll both dice.
What is the probability that the sum is 9 given that exactly one die lands on 6?

Here's what I have so far:
Cases where sum is 9:
Case1: Loaded = 6 | Fair = 3 -> 1/2 * 1/6
Case2: Loaded = 3 | Fair = 6 -> 1/10 * 1/6
Case3: Loaded = 5 | Fair = 4 -> 1/10 * 1/6
Case4: Loaded = 4 | Fair = 5 -> 1/10 * 1/6
------------------------------------
                  P(sum is 9) = 2/15

P(exactly one die lands on 6) = P(loaded lands on 6) + P(fair lands on 6) - P(both land on 6)
P(exactly one die lands on 6) = 1/2 + 1/6 - (1/2 * 1/6) = 7/12

P(sum is 9 AND exactly one die lands on 6)
= (P(case1) + P(case2)) / P(sum is 9)
= ((1/2 * 1/6) + (1/10 * 1/6)) / (2/15)
= 3/4

P(sum is 9 | exactly one die lands on 6)
= P(sum is 9 AND exactly one die lands on 6) / P(exactly one die lands on 6)
= (3/4) / (7/12)
> 1

What am I doing wrong?
 [/INST] In short, you erroneously divided by $\Pr\{\text{sum is 9}\}$ when calculating $\Pr\{\text{sum is 9 and exactly one die lands on 6}\}$.
$$
  \begin{align*}
    \Pr\{\text{sum is 9 and exactly one die lands on 6}\}
    & = \Pr\{(6,3),(3,6)\} \\
    & = \Pr\{(6,3)\} + \Pr\{(3,6)\} \\
    & = \left(\frac{1}{2}\right)\left(\frac{1}{6}\right) + \left(\frac{1}{10}\right)\left(\frac{1}{6}\right) \\
    & = \frac{1}{10}
  \end{align*}
$$
</s>