# Independence of dice rolls

Suppose that a fair dice is rolled and that the number $x$ appears. Let $E_1$ be the event that the number $x$ is even, $E_2$ be the event that the number $x$ is greater than or equal to $3$, $E_3$ be the event that the number $x$ is a $4,5$ or $6$. Are the events $E_1$ and $E_2$ independent? Are the events $E_1$ and $E_3$ independent?

I would say events $E_1$ and $E_2$ are independent because

$\mathbb{P}(E1 \cap E2) = \mathbb{P}(E1) \cdot \mathbb{P}(E2) \Leftrightarrow \mathbb{P}(\{4,6\}) = \mathbb{P}(E1) \cdot \mathbb{P}(E2) \Leftrightarrow \frac{1}{3} = \frac{1}{2} \cdot \frac{4}{6}$

• Well, you still haven't considered the $E_1, E_3$ case. Commented May 3, 2018 at 19:10
• This would be also dependent because $1/2 \neq 1/2 \cdot 1/2$?
– user528814
Commented May 3, 2018 at 19:12
• How many numbers from $1$ to $6$ are both even and at least $3$? Also, how many are both even and equal to $4$, $5$, or $6$? Commented May 3, 2018 at 19:18

You initially got it backwards. Because, for example,

$$\mathbb{P}(E1 \cap E2) \neq \mathbb{P}(E1) \cdot \mathbb{P}(E2),$$

they're dependent, or not independent.

But, if the joint probability equals the product of the individual probabilities, then the events are independent.

• yes thats what ive meant but is my way for the solution correct? I am still not complete safe with the definitions of probability
– user528814
Commented May 3, 2018 at 19:10
• You did the work, but reached the wrong conclusion.
– John
Commented May 3, 2018 at 19:11
• If you change the line above your equation to: "I would say the Events are all dependent because ..." then you're good.
– John
Commented May 3, 2018 at 19:13
• The reasoning is now correct, but I believe the work is wrong, since $P(E_1\cap E_2)=\frac{1}{3}$. Commented May 3, 2018 at 20:39
• But then its is $\frac{1}{3} = \frac{1}{3}$ what would mean they are independent
– user528814
Commented May 4, 2018 at 15:31