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 $P(A∩B) = P(A) P(A|B)$.


2 Answers 2


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:

$$P(A, B) = \frac{5}{36}$$

  • $\begingroup$ what is correct? NCh says its 5/11 and u are saying its 5/36... im confused $\endgroup$ Commented Feb 10, 2019 at 8:49
  • 1
    $\begingroup$ @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. $\endgroup$
    – jvdhooft
    Commented Feb 10, 2019 at 9:14

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