Suppose we have two events, $A$ and $B$. I want to find out if $A$ and $B^c$ are which one of mutually exclusive or independent. I have figured out that they are not mutually exclusive.. does that mean they are definitely independent? If so, how would I justify this?

  • $\begingroup$ Isn't this a duplicate of this? $\endgroup$ Commented Oct 21, 2015 at 8:49

4 Answers 4


No. You can have dependent events that are not mutually exclusive.

Consider the events:

$A$: The radio traffic report says that traffic is "heavy".

$B$: I am late for work
$B^c$: I am not late for work

Neither $B$ nor $B^c$ are independent of $A$ (since I am more likely to be late when the radio says traffic is heavy than otherwise), but neither is mutually exclusive of it since I would generally still be there in sufficient time, but will not always be there on time irrespective of $A$ happening.

[To be independent, two events with non-zero probability cannot be mutually exclusive, but the converse doesn't hold.]


Suppose for your two events $A$ and $B$, $P(A)\ne 0$ and $P(B)\ne 0$.

Now suppose $A$ and $B$ are mutually exclusive

which means $A\cap B =\varnothing$ then $P(A\cap B)=0 \tag{1}$

If $A$ and $B$ are independent then

$P(A\cap B)=P(A)P(B)\ne0 \tag{2}$

since neither $P(A)$ nor $P(B)$ are $0$.

You can see $(1)$ and $(2)$ are contradictory.

So you can see if $A$ and $B$ are mutually exclusive then, under the additional condition that $P(A)$ and $P(B)$ are both non-zero, they are in fact dependent i.e $P(A\cap B)\ne P(A)P(B)$.

One example is a disease and its vaccine.

Suppose you have a 100% effective vaccine such that you cannot contract the disease if you were vaccinated. Then being vaccinated and contracting the disease are mutually exclusive and they are dependent.

  • 2
    $\begingroup$ You don't need empty events, just events of probability zero, such as $A= \{X=2\}$ where $X$ is a continuous random variable. Then, regardless of the choice of nonempty event $B$, $P(A)$ and $P(A\cap B)$ both have probability zero, and $P(A\cap B)=P(A)P(B)$ holds (both sides equaling $0$). Thus, an event of probabilty $0$ is independent of all events $B$. Heck, it is even independent of itself!! $\endgroup$ Commented Oct 20, 2015 at 21:17
  • $\begingroup$ Thank you very much. and your last sentence make me laugh and it is true. $\endgroup$
    – Deep North
    Commented Oct 20, 2015 at 21:34
  • 1
    $\begingroup$ Consider further that if $A$ is an event of probability $0$, then it is independent of $A^c$ too!! An intuitive notion of independence is that if $A$ and $B$ are independent events, then knowing that $A$ has occurred should not tell you anything that you didn't already know about the occurrence or non-occurrence of $B$. But, knowing that $A$ has occurred does tell me that $A^c$ has not occurred! Oh well, fortunately, this occurs so infrequently that I can dismiss it as a nightmare and continue to sleep well through the night. $\endgroup$ Commented Oct 21, 2015 at 14:55
  1. Event that can't occur simultaneously are mutually exclusive. Eg. If you toss a coin you will get head or tail, so event of getting head and event of getting tail are mutually exclusive.
  2. If events are independent outcome of one does not affect other event.Eg. 2 different events of tossing two coins are independent as outcome of one does not affect other.

So mutual dependence is prerequisite for mutual exclusivity therefore mutually exclusive events can't be independent.

PS. These are layman definitions might not be mathematically correct.


Consider this example:

Rolling the dice only once:

  1. Event A: The result is 1 or 6.
  2. Event B: The result is 1 or 3.
  • Does Event A and Event B mutually Exclusive? No, two events occur simultaneously if the result is 1.
  • Does Event A and Event B independent? No, because if event A occurs, it increases the chance that event B occurs.

So does "not mutual exclusive" implies independence? NO!

  • $\begingroup$ Your example tells us nothing about this implication. To demonstrate that a proposition P does not imply Q, you need to exhibit a case where P is true but Q is false--and you haven't done that. $\endgroup$
    – whuber
    Commented Apr 25, 2021 at 18:54

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