If the events A and B are mutually exclusive (incompatible)…
If the events $A$ and $B$ are mutually exclusive (incompatible), does it follow that they are independent?
Explanation of Method
Compare the two definitions. Mutually exclusive: $AB=\varnothing$, so $P(AB)=0$. Independent: $P(AB)=P(A)P(B)$. Both hold together only if $P(A)P(B)=0$. So, apart from the trivial case where one event has probability $0$, mutually exclusive events are dependent — in fact strongly so, since the occurrence of one excludes the other.
Step 1: The two definitions
| Notion | Definition | Meaning |
|---|---|---|
| mutually exclusive | $AB=\varnothing$ | they cannot occur together |
| independent | $P(AB)=P(A)P(B)$ | knowing one does not change the chance of the other |
Step 2: Combine them
If $A$ and $B$ are mutually exclusive, $P(AB)=0$. They are independent iff also $P(A)P(B)=0$, i.e. iff $P(A)=0$ or $P(B)=0$.
\[ \boxed{\text{No: mutually exclusive events with } P(A)\gt0,\ P(B)\gt0 \text{ are always dependent.}} \]
Step 3: In terms of conditional probability
If $A$ and $B$ are mutually exclusive and $P(B)\gt0$, then $P(A\mid B)=\frac{P(AB)}{P(B)}=0$. Knowing that $B$ occurred tells us for sure that $A$ did not — the strongest possible dependence. For independence we would need $P(A\mid B)=P(A)\gt0$.
Step 4: Example
Roll a die: $A$ = "a $1$", $B$ = "a $2$". They are mutually exclusive, $P(AB)=0$, but $P(A)P(B)=\frac1{36}\ne0$ — dependent.
By contrast, $C$ = "an even number" and $D$ = "at most $2$" are independent: $P(CD)=P(\{2\})=\frac16=\frac12\cdot\frac13$, but not mutually exclusive.
Step 5: The trivial exception
If $P(A)=0$ (e.g. $A=\varnothing$), then $A$ is both mutually exclusive with and independent of any event $B$.
Result
| Answer | |
|---|---|
| Mutually exclusive $\Rightarrow$ independent? | No, unless $P(A)=0$ or $P(B)=0$ |
| Reason | $P(AB)=0\ne P(A)P(B)$ when both are positive; $P(A\mid B)=0\ne P(A)$ |