How does $\Omega$ space look for independent and mutually exclusive events?

Question

Given we roll an unbiased dice. We define events $A$ and $B$ as follows:

• $A$ <- even number, i.e. $\{2, 4, 6 \}$,
• $B$ <- odd number, i.e. $\{1, 3, 5\}$

Now, if we have rolled a dice once - we can say that $A$ and $B$ are mutually exclusive.

However, if we roll a dice twice - we can say that getting $A$ in the first roll and $B$ in the second roll are independent?

What's the general rule to differentiate between the above?
How does $\Omega$ space look in both scenarios above?

Answer 1

Mutually exclusive: $P(A\cap B)=0$.

Sample space of throwing one dice once: $\Omega = \{ 1,2,3,4,5,6 \}$.

Since we are throwing the dice once we cannot get an outcome from $A$ and $B$ at the same time. It can be also seen using conditional probabilities.

Independent: $P(A\cap B)=P(A)P(B)$.

Sample space of throwing one dice twice: $\Omega = \{ (1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)\}$

where in the second case $(x,y)$ means getting $x$ in the first throw and $y$ in the second.

Answer 2

For the first situation, where we consider a single roll of a 6-sided die, the state space $\Omega_1$ is simply 1-6, where A and B form a partition of the state space. Now, because A and B partition $\Omega_1$, by definition, $P(A \cap B) = 0$

For the second experiment, we are considering the rolling of a die twice, which we will denote as the experiment X. Hence, we now have the state space $\Omega_2$ = {(1,1),(1,2),(1,3),...,(2,1),...,(6,6)}. Notice here that order is important. Now, it is clear that there are 36 outcomes in $\Omega_2$, and since we are rolling a fair dice, each is as likely as the other, such that the probability of the experiment X giving an output x is$ P(X = x) = \frac{1}{36}$, where x could be, for example, (5,6).

Alternatively, we can see this as $P(A = a, B=b)$, where A is the roll of the first dice, with a its output, and similar for B.

Is is clear that $P(A = a, B=b) = P(A = a)P(B = b) = P(X = x) = \frac{1}{36}$,

and since factorisation in probability implies independence, we have shown that events A and B are independent.