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This question already has an answer here:

So, two events are mutually exclusive if they have an empty set, and two events are independent if they do not affect each other.

However, is it possible for two events to be mutually exclusive but not independent?

For example, the p(A) = {card < 5 is drawn} and p(B) = {face card is drawn}.

p(A&B) = 0, p(A) = 12/52, p(B) = 12/52

Test for independence fails since p(A&B) does not equal p(A)*p(B). Therefore, this must mean that the events are not independent but are mutually exclusive.

Is this logically possible?

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marked as duplicate by José Carlos Santos, Xander Henderson, Cave Johnson, mechanodroid, Namaste logic Oct 13 '17 at 13:27

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ What does “if they have an empty set” mean? $\endgroup$ – José Carlos Santos Oct 12 '17 at 15:07
  • $\begingroup$ By that, I mean the two events A and B = 0 $\endgroup$ – bssm Oct 12 '17 at 15:08
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Suppose mutually exclusive events $A,B$ are such that $P(A) > 0$ and $P(B) > 0$.

Then $A,B$ can't be independent, since $P(A \cap B) = 0$, but $P(A)P(B) > 0$.

Alternatively, $P(A|B) = 0$, so $P(A|B) \ne P(A)$. Thus, knowledge that $B$ has occurred affects the probability of the simultaneous occurrence of $A$, reducing it from what it was originally, namely $P(A)$, down to zero.

As an example, lets say we roll a die.

Let $A$ be the event of getting a $3$, and let $B$ be the event that the die comes up an even number.

Then $A,B$ are mutually exclusive, but definitely not independent. After all, if for a given roll, you are told that $B$ occurred, doesn't that affect your view of the likelihood that for the same roll, $A$ occurred?

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  • $\begingroup$ Therefore, two events can be mutually exclusive and not independent, right? $\endgroup$ – bssm Oct 12 '17 at 15:09
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    $\begingroup$ Stronger than that. If they are mutually exclusive, and have positive probabilities, they are can't be independent. $\endgroup$ – quasi Oct 12 '17 at 15:11

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