Conditional probability using single die. "A single die is rolled. Find the probability that the die lands on 5, given that the outcome is an odd number."
I'm not sure of the correct way to do this. I think its 1/3 because 5 is one of the three odd numbers, but I'm not sure of the work to get there.
 A: Let $X$ be our random variable representing the dice roll. Clearly, $X$ can take any value in the space $\{1, 2, 3, 4, 5, 6\}$. Now, we're interested in the probability
\begin{equation}
 \mathbb{P}(X = 5~|~ X \text{ is odd})
\end{equation}
So, if $X$ is odd we're restricting our sample space to $\{1,3,5\}$. Therefore, our probability of interest can be written as
\begin{equation}
 \mathbb{P}(X = 5~|~ X \text{ is one of } \{1, 3, 5\})
\end{equation}
and from this the result falls into our hands.
A: "Given that" suggests that you could use conditioning. If we assume this is a fair die, and  I call $B$ the event of rolling an odd number $\{1,3,5\}$, and $A$ the event of rolling a five $\{5\}$, the the problem asks
$$P(A\mid B) = \frac{P(A\cap B)}{P(B)} = \frac{P(A)}{P(B)} =\frac{1/6}{3/6} = \frac{1}{3}$$
where the first equality is true by definition and the intersection of $A$ and $B$ is $A = \{5\}$.
Of course, that's the long way. Your intuitive reasoning is perfectly right.
A: Since this is a problem of conditional probability, I like to check Bayes' theorem:
$$P(5 | odd)= P(odd | 5)\frac{P(5)}{P(odd)} = 1\frac{\frac16}{\frac12} = \frac13$$
A: Your instinct is trustworthy.   You have equally probable outcomes and $1$ of the $3$ odd numbers is five.   Therefore the (conditional)probability is $1/3$.
That is all the work needed to get there.   Everything else is just double checking.
