Mathematics/Lessons/Probability/Mutually Exclusive & Independent Events
Statistics · Probability · Year 1

Mutually Exclusive & Independent Events

Two special relationships between events — never happening together, and never affecting each other — each collapse the general rules into something much simpler.

9 min read 4 worked examples

Mutually exclusive events

Two events are mutually exclusive if they can never happen at the same time — on a Venn diagram, their circles don't overlap at all, so $P(A\cap B)=0$. The addition rule then simplifies:

$P(A\cup B) = P(A)+P(B) \quad\text{(mutually exclusive events)}.$

This is just the general addition rule with the overlap term removed, since there's nothing to subtract.

A useful special case: exhaustive events

A set of mutually exclusive events is called exhaustive if, together, they cover the entire sample space — every outcome belongs to exactly one of them. Their probabilities then sum to exactly $1$: $P(A)+P(B)+\cdots=1$. Rolling a die and defining events "even" and "odd" gives a simple exhaustive, mutually exclusive pair.

Independent events

Two events are independent if one occurring has no effect on the probability of the other. This gives the multiplication rule:

$P(A\cap B) = P(A)\times P(B) \quad\text{(independent events)}.$

This formula also serves as a test: if $P(A\cap B) = P(A)\times P(B)$ holds for given data, the events are independent; if it doesn't, they aren't.

Mutually exclusive and independent are opposite, not synonyms

These terms are easy to confuse but describe very different (in fact, almost opposite) relationships. Mutually exclusive events with nonzero probabilities can never be independent — if $A$ happening rules $B$ out completely, $A$ occurring is about as much "effect" on $B$ as possible. Genuinely independent events, by contrast, will generally overlap somewhat, since neither restricts the other.

Two Venn diagrams side by side: on the left, two non-overlapping circles A and B representing mutually exclusive events; on the right, two overlapping circles A and B with the overlap shaded, representing independent events.
Side by side: two non-overlapping circles for mutually exclusive events ($P(A\cap B)=0$), next to two overlapping circles for independent events, where the overlap satisfies $P(A\cap B)=P(A)P(B)$.

Worked example 1

Mutually exclusive events
A card is drawn from a standard deck. Let $A$ be "the card is a King" and $B$ be "the card is a Queen". Explain why $A$ and $B$ are mutually exclusive, and find $P(A\cup B)$.
MethodConfirm no overlap, then use the simplified addition rule
1
No single card can be both a King and a Queen, so $A\cap B=\varnothing$ — mutually exclusive.
The two events share no outcomes.
2
$P(A) = \dfrac{4}{52}$, $P(B) = \dfrac{4}{52}$.
4 Kings and 4 Queens in a 52-card deck.
3
$P(A\cup B) = P(A)+P(B) = \dfrac{4}{52}+\dfrac{4}{52} = \dfrac{8}{52} = \dfrac{2}{13}$.
No overlap to subtract.
▪ solved
Examiner's eye

Always state why events are mutually exclusive in context ("a card can't be both a King and a Queen"), not just assert it — this is usually a method mark in its own right, separate from the probability calculation.

Worked example 2

Independent events
A fair coin is flipped and a fair six-sided die is rolled. Find the probability of getting heads and rolling a number greater than $4$.
MethodConfirm independence, then multiply
1
The coin and die are physically separate, so the outcome of one has no effect on the other — the events are independent.
Independence from the context of the experiment.
2
$P(\text{heads}) = \dfrac12$. $P(\text{die}>4) = P(5\text{ or }6) = \dfrac{2}{6}=\dfrac13$.
Each event's individual probability.
3
$P(\text{heads and die}>4) = \dfrac12\times\dfrac13 = \dfrac16$.
Multiplication rule for independent events.
▪ solved
The key insight

Independence here comes from the physical setup, not from a calculation — two unconnected devices simply can't influence each other. Once independence is justified, "and" translates directly into multiplication, exactly as "or" (for mutually exclusive events) translates into addition.

Worked example 3

Testing for independence
For two events, $P(A)=0.4$, $P(B)=0.3$, $P(A\cap B)=0.12$. Determine whether $A$ and $B$ are independent.
MethodCompare P(A)×P(B) against the given P(A∩B)
1
$P(A)\times P(B) = 0.4\times0.3=0.12$.
Compute what independence would require.
2
This matches the given $P(A\cap B)=0.12$ exactly.
Compare to the actual value.
3
Since $P(A\cap B) = P(A)\times P(B)$, the events are independent.
The independence test is satisfied.
▪ tested
Common mistake

The independence test only checks whether the multiplication rule happens to hold for the given numbers — it doesn't require any physical reason for independence. Even seemingly unrelated events must be checked numerically; the test can't be assumed just because a scenario "seems" unconnected.

Worked example 4

Combining both ideas · Challenge
Events $A$ and $B$ are independent, with $P(A)=0.5$ and $P(B)=0.6$. Event $C$ is mutually exclusive with $A\cap B$, and $P(C)=0.1$. Find $P((A\cap B)\cup C)$.
MethodFind P(A∩B) using independence, then add using mutual exclusivity
1
Since $A$ and $B$ are independent: $P(A\cap B) = 0.5\times0.6=0.3$.
Multiplication rule.
2
Since $C$ is mutually exclusive with $A\cap B$: $P((A\cap B)\cup C) = P(A\cap B)+P(C)$.
No overlap to subtract, by the simplified addition rule.
3
$= 0.3+0.1 = 0.4$.
Add the two probabilities directly.
▪ solved
Look deeper — treating a combined event as a single block

Once $A\cap B$ is found as a number, it can be treated exactly like any single event for the remaining calculation — its internal structure (that it's itself an intersection) stops mattering once its probability is known. Breaking a compound question into "find this piece, then combine" is the standard strategy whenever multiple rules apply in sequence.

Common mistake

Applying $P(A\cap B)=P(A)\times P(B)$ to events that aren't independent gives a wrong answer with no warning — the formula doesn't know whether it's valid. Always confirm independence (from context or a stated test) before multiplying; otherwise the general rules, or a tree diagram, are needed instead.

Check yourself

A spinner has three mutually exclusive, exhaustive outcomes: red, blue, green, with $P(\text{red})=0.35$ and $P(\text{blue})=0.4$. Find $P(\text{green})$, and find $P(\text{red or blue})$.
(Hint: exhaustive probabilities sum to 1; mutually exclusive events just add.)

Method
1
$P(\text{green}) = 1-0.35-0.4=0.25$.
Exhaustive: all three sum to 1.
2
$P(\text{red or blue}) = 0.35+0.4=0.75$.
Mutually exclusive: simple addition.
▪ solved