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

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?

• One should note that the sample spaces suggested in the answers below are only one choice among a wealth of possible others. For example, the space $\Omega=\{1,2,\ldots,6\}^2$, which was suggested for two dice rolls, may also be used for one dice roll, defining $X(i,j)=i$ for every $(i,j)$ in $\Omega$. Which is one reason why asking the question "What is $\Omega$?" is almost always useless. – Did Oct 7 '16 at 8:05
• Thanks. Does it all depend on how we choose to model these scenarios using probability? Do we need to decide whether events can be assumed mutually exclusive or independent based on logic and domain we're modeling? – ThamP Oct 8 '16 at 10:18
• ?? What is "it" in "Does it all depend"? – Did Oct 8 '16 at 11:18

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.

• Could you clarify how factorisation in probability implies independence? I initially thought that it's independence that implies factorisation, not the other way around. – ThamP Oct 8 '16 at 10:41
• The mathematical definition of independence is that the functions can be factorised. That is, we can state that two events are independent, but the only way to formally show that is to verify that the joint distribution of the events can be factorised into its marginals. If one demonstrated that the joint can be factorised into the marginals, then that immediately implies independence. That is to say, its not that one leads to the other, but that they both must be present. – Pablo Oct 8 '16 at 13:25

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.