Conditional probability (two dice) Two dice are rolled, what is the probability that die 1 shows a 5 given that a 10 has been rolled?
I'm a little confused as to how to solve this exercise.
I get two different results depending on how I go about solving it.
Method 1 (logic):
The probability of having die 1 show a 5 when two dice are rolled is 1/36.
The probability that a 10 is rolled is 3/36.
Now I put this in the basic probability format and it gives me (1/36)/(3/36)=1/3.
Method 2 (baye's theorem):
P(A/B) = P(A)xP(B/A) / P(B) = 1/36 x 1/6 (bc if die 1 shows a five and I need to get a 10, then die 2 needs to show a five also, hence 1/6) / (3/36) = 1/18
 A: Presumably $B$ is the event a $10$ has been rolled, and $A$ is the event Die 1 shows a $5$. Then $\Pr(A)=1/6$, or, if you prefer, $6/36$,  $\Pr(B\mid A)=1/6$,  and $\Pr(B)=3/36$.
Substituting and simplifying, we get that $\Pr(A\mid B)=1/3$. The two methods give the same answer.
A: Observe that our probability space is $(\Omega, \mathbb P)$ where
$$\text{sample space: }\Omega = \{\{1,1\}, \{1,2\}, ..., \{6,5\}, \{6,6\}\}$$
$$\text{probability measure: }\forall \omega \in \Omega, \mathbb P(\omega) = \frac{1}{36}$$
We want to compute
$$P(\text{5 is rolled} | \text{sum is 10}) \tag{*}$$
Observe that the event $\text{5 is rolled}$ corresponds to the sample outcomes $\{5,1\}, ..., \{5,6\}$ and the event $\text{sum is 10}$ corresponds to the sample outcomes $\{5,5\}, \{4,6\}, \{6,4\}$.
Hence,
$$(*) = P((\{5,1\} \cup ... \cup \{5,6\}) | \{5,5\} \cup \{4,6\} \cup \{6,4\})$$
$$ = \frac{P((\{5,1\} \cup ... \cup \{5,6\}) \cap (\{5,5\} \cup \{4,6\} \cup \{6,4\}))}{P(\{5,5\} \cup \{4,6\} \cup \{6,4\})}$$
$$ = \frac{P(\{5,5\})}{P(\{5,5\} \cup \{4,6\} \cup \{6,4\})}$$
$$ = \frac{P(\{5,5\})}{P(\{5,5\}) + P(\{4,6\}) + P(\{6,4\})}$$
$$ = \frac{1/36}{3/36} = 1/3$$

Another way to look at it is to change our probability space
$$\text{sample space: }\Omega = \{\{5,5\}, \{4,6\}, \{6,4\}\}$$
$$\text{probability measure: }\forall \omega \in \Omega, \mathbb P(\omega) = \frac{1}{3}$$
So if we want to compute $P(\{5,5\})$, by definition of our probability measure, we have 1/3.
