Two events $A$ and $B$ can't be mutually exclusive and independent at the same time if $P(A)\neq0$ and $P(B)\neq0$. Suppose there is a group of $100$ students. No student has failed in physics. $30$ students have failed in chemistry. Let $A$ be the event of a student to fail in physics and $B$ be the event of a student to fail in chemistry.enter image description here Clearly $P(A)=0$. Knowing that a student has failed in chemistry surely tells us that he has passed in physics because there is no common part of these two events, which means a student cannot fail in physics and chemistry at the same time. (So they are mutually exclusive.) But we don't actually need to go that far. We know no student has failed in physics. So a student failing or passing in chemistry does not effect the event $A$. So they are both independent and mutually exclusive.

Now my question is when I knew that $P(A)=0$, then without thinking anything else could have I said that they are both mutually exclusive and independent events? More specifically, If the probability of any event $A$ is zero, that automatically makes it both independent and mutually exclusive with any other event $B$? Or they maybe both independent and mutually exclusive or they may not be?

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    In the discrete case , $P(A)=0$ means that $A$ is impossible, so $A$ and $B$ cannot occur both because $A$ cannot occur. Moreover $P(A\cap B)=0=P(A)\cdot P(B)$, so $A$ and $B$ are both mutually exclusive and independent. – Peter Oct 05 '17 at 12:07
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    Does this answer your question? [What is the difference between independent and mutually exclusive events?](https://math.stackexchange.com/questions/941150/what-is-the-difference-between-independent-and-mutually-exclusive-events) – NNOX Apps Dec 24 '21 at 09:01

1 Answers1


Independent: the occurrence of one does not affect the occurrence of the other; $$P(A\cap B)= P(A) \, P(B)\tag{1}$$

Mutually exclusive: there is no way for the events to occur simultaneously; $$P(A\cup B) = P(A)+P(B) \iff P(A\cap B)=0\tag{2}$$

This comes from the general formula for combined events:

$$P(A\cup B) + P(A\cap B) = P(A)+P(B)$$

As you indicated, the data in your Venn diagram imply $$P(A)=0\tag{3}$$ $$P(A\cap B)=0\tag{4}$$

Equation $(4)$ immediately satisfies criterion $(2)$. If you look at equations $(3)$ and $(4)$ together, you can also see that criterion $(1)$ is satisfied:

$$\begin{align} P(A\cap B) &= P(A)\,P(B) \\ 0 &= 0\,P(B) \\ 0&=0\\ \end{align}$$

You should remember, however, that these data are observed (aka “experimental” aka “empirical”), and these conclusions might need a larger sample to be corroborated. It seems fairly fishy that it is impossible to fail physics and that it is impossible to fail both classes.

gen-ℤ ready to perish
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