So I have a question:
Two dice are rolled. Find the probability that at least one die is 6 if it is known that at least one die had an odd number.
Is this the right way to calculate conditional probability?
If I understood correctly, then I need to find this
\begin{align} P(A \mid B) &= \frac{AB}{B} \end{align}
There is an event A -> at least one die is 6. First, I will find probability, that both dice don't have 6.
\begin{align} P(\neg A) &= \frac{5}{6} \cdot \frac{5}{6} = \frac{25}{36} \\\\ \end{align} So \begin{align} P(A) &= 1 - \frac{25}{36} = \frac{11}{36} \\\\ \end{align} Next, we have event B -> came odd number
\begin{align} P(B) &= \frac{3}{6} = \frac{1}{2} \\\\ \end{align}
In the end we have: \begin{align} P(A|B) &= \frac{\frac{11}{36} \cdot \frac{1}{2}}{\frac{1}{2}} = \frac{11}{36} = 0.305 \\\\ \end{align}
Does this look fine?
That's not correct. Also the question is not stated clearly. It says at least one of the dice is $6$. But if at least one of the dice is odd then either exactly one die is $6$ or none of them is $6$.
Event $A$ - one of the dice is $6$
Event $B$ - at least one of the dice is odd
Probability that at least one of the dice is odd
$ \displaystyle P(B) = 1 - \frac 12 \cdot \frac 12 = \frac 34 ~$, we take complement of the probability that none of the dice is odd.
$ \displaystyle P(A \cap B) = 2 \cdot \frac 12 \cdot \frac 16~$, as one of them is $6$ and the other is odd.
Now, $ \displaystyle P(A \mid B) = \frac{P(A \cap B)}{P(B)}$