Conditional probability - sum of dice is even given that at least one is a five

A = event when one of the tosses gives a five. (Sample space for the conditional probability) Let (n1, n2) represent the outcomes of die1 and die2 A = { (1,5), (2,5), (3,5), (4,5), (5,5), (6,5), (5,1), (5,2), (5,3), (5,4), (5,6) } // (5,5) must be counted once only

Thus n(A) = 11

B = sum of two dice tosses is even n(B|A) = { (1,5), (3,5), (5,5), (5,1), (5,3)|

P(B|A) = n(B|A)/n(A) = 5/11

By the definition of conditional probability, $P(A|B) = \frac{P(A \cap B)}{P(B)}$.

Let $A$ be the event of the two rolls adding to an even number and let $B$ be the event of rolling at least one five.

Note that $A \cap B$ = the event of the two rolls adding to an even number AND rolling at least one five. Now we need to find $P(A \cap B)$ and $P(B)$:

There are three ways we can roll at least one five.

1) You roll a five on the first roll but not the second. The probability of this is $(1/6)*(5/6) = 5/36.$

2) Not rolling a five on the first roll, but rolling a five on the second: Probability of this is $(5/6)*(1/6) = 5/36.$

3) Or you can roll two fives: Probability $= (1/6)*(1/6) = 1/36.$

These three events are disjoint, so to find the probability of getting $\textit{at least}$ one five, we add them: $5/36 + 5/36 + 1/36 = 11/36$.

Now to calculate $P(A \cap B)$: If you roll two fives, the sum is even. Otherwise, the roll that is not a five must be an odd number (ODD + ODD = EVEN). In each of cases 2) and 3), the only other rolls that give you an even sum are 1 and 3. That's a total of 5 scenarios out of a possible 36. So $P(A \cap B) = 5/36.$ Finally,

$$P(A|B) = \frac{5/36}{11/36} = \frac{5}{11}.$$

As suggested above, the Bayes th. can help too:

Let $S$ be the event that the sum is even, and $A5$ the event that at least one $5$ shows up, and $S^c$ is the event that the sum is odd.

Then, $$P(S|A5) = \frac{P(A5|S)P(S)}{P(A5|S)P(S)+P(A5|S^c)P(S^c)} $$

$P(S)=P(S^c)=0.5$, $P(A5|S)= 5/18$, as there are $5$ in $A5$ for even sum. Also, $P(A5|S^c)= 6/18$ when the sum is odd.

Thus, $P(S|A5) = 5/11$.