# Two dice rolls, intersection of two events

Let's say we throw two dice and event $$A$$ is at least one die is 5, event $$B$$ is sum of two numbers is even.

So we have a set of outcomes for:

$$A$$ = $$\{(5,1),(5,2),(5,3),(5,4),(5,5),(5,6),(1,5),(2,5),(3,5),(4,5),(6,5)\}, P(A) = 11/36$$

$$B$$ = $$\{(1,1),(1,3),(1,5),(2,2),(2,4),(2,6),(3,3),(3,5),(3,1),(5,1),(4,2),(6,2),(5,3),(4,4),(4,6),(5,5),(6,6)\}, P(B) = 18/36 = 1/2$$

I am trying to calculate $$P(A∩B)$$. I know that there are five outcomes so it would be $$5/36$$, but I want to use the formula for intersection which is $$P(A∩B) = P(A) P(A|B)$$.

This is wrong formula. Right formula should be $$\mathbb P(A\cap B) = \mathbb P(A)\cdot \mathbb P(B\mid A).$$ Here $$\mathbb P(B\mid A) = \frac5{11}$$ since only $$5$$ outcomes from $$11$$ outcomes of $$A$$ also belong to $$B$$.
As NCh pointed out in another answer, you're not using the right formula. Indeed, the probability that events $$A$$ and $$B$$ both happen equals:
$$P(A, B) = P(A) P(B | A) = P(B) P (A | B)$$
It is worth noting that often, for problems like these, it is easier to calculate the event space $$A, B$$ directly. For instance, consider the first die. If its value is $$5$$, the second die can equal $$1$$, $$3$$ or $$5$$ in order to achieve an even sum. If it is not $$5$$, the second die should equal $$5$$, and the first die should equal $$1$$ or $$3$$. Indeed, there are only five valid options and the probability thus becomes:
$$P(A, B) = \frac{5}{36}$$
• @GeonguAidenPark I have calculated the probability $P(A, B)$ of the events $A$ and $B$ happening, which is the same as $P(A \cap B)$. In the other answer, $P(B | A)$ is calculated, which is the conditional probability that the event $B$ happens if $A$ happens. – jvdhooft Feb 10 at 9:14