<s>[INST] Conditional probability - sum of dice is even given that at least one is a five Question:

Calculate the conditional probability that the sum of two dice tosses is even given that at least one of the tosses gives a five.

I'm a bit confused by this. Shouldn't the probability just be 1/2, since we know that at least one of the dice tosses gave us a five, thus the other must give us an odd number?
 [/INST] If the dies were colored, say red and blue, we asked the question, what's chance that sum is even, given that the red die rolls a five, that's $1/2$. But, you see, either red or the blue die could roll a five. So, your sample space is slightly larger. 
Fix $X_{i}$ to be the random variable representing the number on die $i$, $i=1,2$. You could draw a nice tree to think of the possible outcomes of this experiment and count them.
(1) When the first die is rolled, there are three possibilities - $X_{1}=5$, $X_{1}=1 \cup X_{1}=3$, $X_{1}\text{ is even}$. 
$P(X_{1}=5)=1/6$. 
$P(X_{1}=1 \cup X_{1}=3)=1/3$. 
$P(X_{1} \text{ is even})=1/2$.
(2) The second die is now rolled, again having three possibilities each - $X_{2}=5$, $X_{2}=1 \cup X_{2}=3$, $X_{2}\text{ is even}$. So, we have $3 \times 3=9$ branches.
$P(X_{1}=5 \cap X_{2}=5)=1/6 \times 1/6=1/36$. 
$P(X_{1}=5 \cap (X_{2}=1 \cup X_{2}=3))=1/6 \times 1/3=2/36$.
$P(X_{1}=5 \cap X_{2}\text{ is even})=1/6 \times 1/2=3/36$.
$P((X_{1}=1 \cup X_{1}=3) \cap X_{2}=5)=1/3 \times 1/6 = 2/36$. 
$P((X_{1}=1 \cup X_{1}=3) \cap (X_{2}=1 \cup X_{2}=3))=1/3 \times 1/3 = 4/36$. 
$P((X_{1}=1 \cup X_{1}=3) \cap X_{2}\text{ is even})=1/3 \times 1/2 = 6/36$. 
$P(X_{1} \text{ is even} \cap X_{2}=5)=1/2 \times 1/6 = 3/36$.
$P(X_{1} \text{ is even} \cap (X_{2}=1 \cup X_{2}=3))=1/2 \times 1/3 = 6/36$.
$P(X_{1} \text{ is even} \cap X_{2}\text{ is even})=1/2 \times 1/2 = 9/36$.
Counting the outcomes where the sum is even and atleast one roll is 5, the numerator $1/36+2/36+2/36=5/36$.
Restricting our attention to events, where atleast one roll is 5, the denominator = $1/36+2/36+3/36+2/36+3/36=11/36$.
So, the chance is $5/11 < 1/2$.
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