Existing Understanding:
I understand that mutually exclusive events cannot be independent.
If A and B are independent, the probability of A and B both happening would be:
P(A and B) = P(A)P(B)
If A and B are mutually exclusive, then the probability of A and B happening at the same time is zero.
P(A and B) = 0, which does not agree with the first equation stated above.
Question:
Can there be a scenario where two mutually inclusive events can be dependent? Right now, I do not think so because how can two events affect each other when they are occurring at the same time?
$\endgroup$ 32 Answers
$\begingroup$Suppose that $A$ and $B$ are mutually exclusive and also independent.
Then you have $P(A\cap B)=P(\emptyset)=0$ by the mutual exclusivity while at the same time $P(A\cap B)=P(A)\times P(B)$ by the independence.
You get then that if $A$ and $B$ are simultaneously mutually exclusive and independent then $P(A)\times P(B)=0$ which implies that either $Pr(A)=0$, that $Pr(B)=0$, or both equal zero.
That is possible, for example when $A=B=\emptyset$. It is possible for two events to simultaneously be mutually exclusive and independent, but that occurs only when one or both of the events are "almost impossible" events, meaning occurs with probability zero.
For $A$ and $B$ which are both possible events, meaning occurs with positive probability, then it is impossible for them to simultaneously be independent and mutually exclusive. If you know they are independent then they cannot be mutually exclusive and vice versa.
$\endgroup$ 1 $\begingroup$Example 1. Roll a 6-sided die (numbered 1-6). Let $A=\{7\}$, $B=\{8\}$. Then:$$P(A\cap B)=P(A)\cdot P(B|A)=P(A)\cdot P(B)=0\cdot 0=0.$$Example 2. Roll a 6-sided die (numbered 1-6). Let $A=\{1\}$, $B=\{8\}$. Then:$$P(A\cap B)=P(A)\cdot P(B|A)=P(A)\cdot P(B)=\frac16\cdot 0=0.$$
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