Mutually Exclusive Events Are Not Independent

Mutual Exclusivity and Independence Answer Different Questions

Two events can both describe outcomes in the same probability model, but their relationship may be described in two different ways. Mutual exclusivity asks whether the events can happen together. Independence asks whether knowing that one event occurred changes the probability of the other, expressed precisely by a probability equation. Confusing the two leads to a common error: disjoint events with positive probabilities are not independent.

The distinction is aligned with the SOA VEE Mathematical Statistics syllabus. The reliable test is the factorization definition of independence; descriptions such as “unrelated” or “cannot happen together” are not substitutes for checking the probabilities.

Definition: Events $A$ and $B$ are mutually exclusive (or disjoint) if $A\cap B=\varnothing$, so they have no outcomes in common. They are independent if $$ P(A\cap B)=P(A)P(B). $$ Thus, mutual exclusivity concerns which outcomes the events contain, while independence concerns whether a probability factorization holds.

For disjoint events, $P(A\cap B)=0$. That fact alone does not establish independence: the product $P(A)P(B)$ must also equal zero. If both events have positive probability, their product is positive, so the independence equation fails. There is one important exception: if at least one event has probability zero, the product is zero, and disjoint events do satisfy the independence equation.

Why Disjoint Events With Positive Probability Are Dependent

The result follows directly from the definition, and its two directions are useful when checking a problem.

Use disjointness.
If $A\cap B=\varnothing$, then $P(A\cap B)=0$.

Apply the independence criterion.
Independence would require $0=P(A)P(B)$.

Interpret the product.
Since probabilities are nonnegative, $P(A)P(B)=0$ exactly when $P(A)=0$ or $P(B)=0$.

Theorem: If $A$ and $B$ are mutually exclusive, then they are independent if and only if at least one of $P(A)$ and $P(B)$ is zero. In particular, two mutually exclusive events that both have positive probability are dependent.

This conclusion can also be understood conditionally when $P(B)>0$. Disjointness gives $P(A\mid B)=0$. If $P(A)>0$, learning that $B$ occurred rules out $A$, so the probability of $A$ has changed from $P(A)$ to zero. This conditional comparison is not available when $P(B)=0$: the usual definition $P(A\mid B)=P(A\cap B)/P(B)$ has a zero denominator. In that case, use the factorization definition instead.

Worked Examples

Worked Example: Disjoint Events in a Uniform Model

A number is selected uniformly from the integers $1$ through $12$, so each integer has probability $1/12$. Let $A$ be the event that the selected number is at most $4$, and let $B$ be the event that it is at least $9$. Determine whether $A$ and $B$ are independent. Also compare $P(A\mid B)$ with $P(A)$.

Method. List the outcomes in each event, calculate their probabilities under the stated uniform model, and test the factorization equation.

The outcome sets are $A=\{1,2,3,4\}$ and $B=\{9,10,11,12\}$. They have no outcomes in common, so $A\cap B=\varnothing$ and $P(A\cap B)=0$.

There are $12$ equally likely outcomes. Each event contains $4$ outcomes, so

P(A)=\frac{4}{12}=\frac{1}{3}, \qquad P(B)=\frac{4}{12}=\frac{1}{3}.

If they were independent, the intersection probability would be

P(A)P(B)=\frac{1}{3}\cdot\frac{1}{3}=\frac{1}{9}.

But $P(A\cap B)=0\ne 1/9$, so the events are not independent. Since $P(B)>0$, the conditional probability is defined and equals

P(A\mid B)=\frac{P(A\cap B)}{P(B)} =\frac{0}{1/3}=0, \qquad P(A)=\frac{1}{3}.

The probabilities differ, confirming that occurrence of $B$ changes the probability of $A$. The common mistake this calculation avoids is to treat disjointness as a reason for independence; here it is exactly why the intersection probability is zero while the product is positive. Check: $P(A\cap B)=0$, as required for disjoint events, and the factorization test gives the same “not independent” conclusion as the conditional comparison.

Worked Example: A Nonempty Event of Probability Zero

Let $X$ have a continuous uniform distribution on the interval $(0,1)$, with density $f_X(x)=1$ for $0<x<1$. Let $A=\{X=0.4\}$ and $B=\{X<0.4\}$. Are these events mutually exclusive and independent?

Method. Check the event sets first, then use the factorization criterion. The distribution is continuous, so a single point has probability zero.

The events have no outcomes in common: $X$ cannot equal $0.4$ and also be strictly less than $0.4$. Thus $A\cap B=\varnothing$. The point event is nonempty as a set, but its probability is

P(A)=P(X=0.4)=\int_{0.4}^{0.4}1\,dx=0.

The probability of $B$ is the length of the interval $(0,0.4)$, because the density is $1$ throughout that interval:

P(B)=\int_{0}^{0.4}1\,dx=0.4.

Since the events are disjoint, $P(A\cap B)=0$. The product is also

P(A)P(B)=0(0.4)=0=P(A\cap B).

Therefore, $A$ and $B$ are independent according to the factorization definition, despite being mutually exclusive. This is the probability-zero exception in the theorem. The common mistake to avoid is claiming that every pair of disjoint events must be dependent. Also, $P(B\mid A)$ is not defined by the usual conditional-probability formula because $P(A)=0$; the valid check here is factorization. Check: the nonempty event $A$ has probability zero, and both sides of the independence equation equal zero.

Worked Example: Disjoint Events With Unequal Outcome Probabilities

A random experiment has outcomes $\omega_1,\omega_2,\omega_3,\omega_4$, with probabilities $0.10, 0.20, 0.30, 0.40$, respectively. Let $C=\{\omega_1,\omega_2\}$ and $D=\{\omega_3\}$. Determine whether $C$ and $D$ are independent, and calculate $P(C\mid D)$.

Method. Add the specified outcome probabilities; do not count outcomes as equally likely when the model assigns them different probabilities.

The probabilities sum to $0.10+0.20+0.30+0.40=1.00$, so the stated outcome probabilities form a probability model. The events $C$ and $D$ have no common outcome, hence $P(C\cap D)=0$. Their individual probabilities are

P(C)=P(\omega_1)+P(\omega_2)=0.10+0.20=0.30, \qquad P(D)=P(\omega_3)=0.30.

The product required by independence is

$$ P(C)P(D)=0.30(0.30)=0.09. $$

Since $0\ne0.09$, the events are not independent. Because $P(D)=0.30>0$, their conditional probability is

P(C\mid D)=\frac{P(C\cap D)}{P(D)} =\frac{0}{0.30}=0,

which differs from $P(C)=0.30$. The common mistake avoided here is counting the two outcomes in $C$ as though they had equal probability; their assigned probabilities must be added. Check: the conditional comparison and the factorization test both show dependence.

Common Exam Traps

A frequent incorrect calculation is to reason, “The events cannot occur together, so they are independent.” For disjoint events, the correct calculation is

P(A\cap B)=0, \qquad P(A)P(B)=P(A)P(B).

If both event probabilities are positive, then $P(A)P(B)>0$, so the two sides are unequal. For example, if $P(A)=0.2$ and $P(B)=0.3$ for disjoint events, the wrong conclusion would label them independent merely because their intersection has probability zero. The correction is to calculate the product: $0.2(0.3)=0.06$, which is not zero. They are dependent.

A second trap is assuming that a probability-zero event must be the empty event. In continuous models, a single possible value can form a nonempty event with probability zero. For disjoint events, that distinction matters: a probability-zero event and a positive-probability event do satisfy the factorization equation. Conversely, do not attempt to use a conditional probability given a probability-zero event; check independence by comparing $P(A\cap B)$ and $P(A)P(B)$.

Key takeaway: Mutual exclusivity makes the intersection probability zero. Independence requires that this intersection probability equal the product of the event probabilities. Thus, disjoint events are independent exactly when at least one has probability zero.

Check Your Understanding

For each problem, use the stated probability model and distinguish disjointness from independence.

A fair ten-sided die has outcomes $1,\ldots,10$, each with probability $1/10$. Let $A=\{1,2\}$ and $B=\{3,4,5\}$. Are the events independent? (Answer: No; $P(A\cap B)=0$, while $P(A)P(B)=(2/10)(3/10)=0.06$.)
One record is selected uniformly from $20$ records. Event $C$ means the selected record is among the first $5$; event $D$ means it is among the last $4$. Are $C$ and $D$ independent? (Answer: No; they are disjoint, and $P(C)P(D)=(5/20)(4/20)=0.05\ne0=P(C\cap D)$.)
A model has outcomes $a,b,c,d$ with probabilities $0.10,0.20,0.30,0.40$. Let $E=\{a\}$ and $F=\{b,c,d\}$. Determine whether they are mutually exclusive and independent. (Answer: They are mutually exclusive but not independent; $P(E\cap F)=0$, whereas $P(E)P(F)=0.10(0.90)=0.09$.)
Let $X$ be uniform on $(0,4)$, and define $G=\{X=2\}$ and $H=\{X>1\}$. Are these events independent? Are they mutually exclusive as sets? (Answer: They are independent because $P(G)=0$ and $P(G\cap H)=P(G)P(H)=0$; they are not mutually exclusive as sets because $X=2$ belongs to both.)
Events $J$ and $K$ are disjoint, with $P(J)=0.25$. What must $P(K)$ be for them to be independent? Explain why, and state whether a conditional probability given $J$ is needed. (Answer: $P(K)=0$; independence requires $0=P(J)P(K)=0.25P(K)$. No conditional probability is needed.)

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