I am studying probability theory and one of the questions that I have faced is this. The problem is that I either don't know where to go about with this question or even if I do do something about it, I have no way of knowing if I'm on the right path. Is it possible to please help explain this?
Given a three-sided biased die. The probability of throwing $2$ or $3$ is $k$ times higher than the probability of throwing $1$. Compute the probability space. Show that the event $A$ = ‘number is greater than $1$’ and $B$ = ‘number is less than $3$’ are not independent.
If you throw the die, then you may get one element from $\Omega = \{1,2,3\}$. You also know that $$\mathbb P(2) = \mathbb P(3) = k \mathbb P(1),$$ where I've omitted the set notation brackets and interpreted "$k$ times more than other" as that (could be $k+1$ instead).
Since $1 = \mathbb P(\Omega) = \mathbb P(1) + \mathbb P(2) + \mathbb P(3) = (2k+1)\mathbb P(1)$, then $\mathbb P(1) = 1/(2k+1)$ and $\mathbb P(2) = \mathbb P(3) = k/(2k+1)$. Therefore, we have
$$\mathbb P(A) = 1 - \mathbb P(1) = \frac{2k}{2k+1},\qquad\mathbb P(B) = 1 - \mathbb P(3) = \frac{k+1}{2k+1}.$$
The events are not indendent because $$\mathbb P(A\cap B) = \mathbb P(2) = \frac{k}{2k+1} \ne \mathbb P(A)\mathbb P(B) = \frac{2k(k+1)}{(2k+1)^2}.$$