# How to check independence of events

This is my question The fair die is 6-sided and it is rolled twice. We define three events:

E1 = The first roll lies in the set {2, 3, 4}
E2 = The first roll lies in the set {4, 5, 6}
E3 = Sum of two rolls is 5

Are the events E1, E2, E3 independent of each other? In order to find independance, following formula has to be satisfied:

$P (E1 \cap E2) = P(E1)P(E2)$

I already know the probabilities of both events i.e. ee1 and E2 is $\frac 3 6$. And $P (E1 \cap E2) = \frac 1 6$
Putting values in formula gives us : $\frac 1 6 = \frac 3 6 *\frac 3 6$, which is false. So both events are not independent. However, I am confused about finding independence between first and last event.

E3 looks like this = { (1,4), (4,1), (3,2), (2,3)}.

P(E3) = $\frac 1 6 * \frac 1 6 *4$(since there are 4 events).

But how do I find $P(E1 \cap E3)$?

E1={2,3,4}
E2={4,5,6}
E3={(1,4),(2,3),(3,2),(4,1)}

For intersection of E1 and E3, you need outcomes in E3 starting with 2 or 3 or 4.
So, it will consist of {(2,3),(3,2),(4,1)}, having probability $\frac{3}{36}$
So, $P(E_1\cap E_3$)=$\frac{3}{36}$ = $\frac{1}{12}$
and P(E1)*P(E3)= $\frac{3}{6}$ * $\frac{4}{36}$=$\frac{1}{18}$.

So, E1 and E3 are not independent.
Similarly, go for E2 and E3.

• Perfectly explained. Thanks. One last question. My answer stating E1 and E2 are not independent is correct, right? – puffles Sep 17 '18 at 5:42
• yes...perfectly – pooja somani Sep 17 '18 at 5:44

As you worked out, we have $E_3=\{(1,4),(2,3),(3,2),(4,1)\}$. Now observe that the first entry of $(1,4)$ is not in $\{2,3,4\}$, and the first entries of all other members of $E_3$ lie in $\{2,3,4\}$. Therefore, $E_1\cap E_3=\{(2,3),(3,2),(4,1)\}$. Hence, $$P(E_1\cap E_3)=\frac{3}{36}=\frac1{12}.$$

• So, for $P(E2\cap E3)$ = $\frac 1/ 36$. Right? – puffles Sep 17 '18 at 5:40
• Yes, $P(E_2\cap E_3)=1/36$. – eloiprime Sep 17 '18 at 5:42

Use conditional probability. Given: $$E_1=\{2,3,4\}, E_2=\{4,5,6\}, E_3=\{(1,4),(2,3),(3,2),(4,1)\}$$ we get: $$P(E_1\cap E_3)=P(E_1)\cdot P(E_3|E_1)=\frac{3}{6}\cdot \frac{1}{6}=\frac{1}{12},$$ because for each number from $E_1$, there is only one number out of $6$ to get the pair (sum is $5$).