How to check if two events are independent

The independence test in one line

Two events $A$ and $B$ are independent exactly when

where $P(A \cap B)$, read "the probability of A and B", is the chance that both happen. Compute each side separately, then compare. Equal means independent. Not equal means dependent, with nothing in between.

There is a second form of the same test:

read "the probability of A given B equals the probability of A". In words: learning that $B$ happened does not change the chance of $A$. Both forms appear in the AP course and never disagree, so use whichever one fits the numbers you were handed.

Why the two checks always agree

Start from the definition of conditional probability:

Set the left side equal to $P(A)$, which is the conditional form of the test, then multiply both sides by $P(B)$:

That is the product form: the two statements are one equation written two ways, so every table gives the same verdict either way. The conditional form needs $P(B) > 0$ to be defined, while the product form still works when a probability is 0, which makes it the safer default.

The same algebra run the other way gives $P(B \mid A) = P(B)$. Independence is symmetric: if knowing $B$ tells you nothing about $A$, then knowing $A$ tells you nothing about $B$. What is not symmetric is which conditional a problem is asking for, sorted out in P(A|B) vs P(B|A).

Both checks on one two-way table

A science museum recorded 200 Saturday visitors, noting whether each bought a ticket online in advance or at the door, and whether each went to the planetarium show.

Let $A$ be "bought online" and $B$ be "saw the show". Pull the three probabilities straight off the table, using the denominators in how to read a two-way table.

Product check. $P(A) \cdot P(B) = 0.40 \times 0.25 = 0.10$, and $P(A \cap B) = 0.10$. The two sides match, so $A$ and $B$ are independent.

Conditional check. $P(A \mid B) = \frac{20}{50} = 0.40$, and $P(A) = 0.40$. They match as well. Running it the other direction, $P(B \mid A) = \frac{20}{80} = 0.25$, which equals $P(B) = 0.25$.

Both checks return the same verdict, as the algebra promised. Independence is visible across the whole table too: the show rate is $\frac{20}{80} = 0.25$ among online buyers and $\frac{30}{120} = 0.25$ among door buyers. Identical conditional distributions in every row is what independence looks like in a table.

Change the inside of the table and the verdict flips

Margins do not decide independence. Cells do. Here is a second table with exactly the same margins as the first, 80 and 120 down the side, 50 and 150 along the bottom. Only the interior has moved.

$P(A)$ is still $0.40$ and $P(B)$ is still $0.25$, so the product $P(A) \cdot P(B)$ is still $0.10$. The joint probability is not: $P(A \cap B) = \frac{32}{200} = 0.16$.

Since $0.16 \neq 0.10$, the events are dependent. The conditional check agrees: $P(A \mid B) = \frac{32}{50} = 0.64$, well above $P(A) = 0.40$, so knowing that a visitor saw the show makes an online purchase much more likely.

The rows disagree now as well. The show rate is $\frac{32}{80} = 0.40$ among online buyers and $\frac{18}{120} = 0.15$ among door buyers. There is also a count version of the product check hiding in this table: independence would put $\frac{80 \times 50}{200} = 20$ visitors in the top-left cell, and the table shows 32. That row total times column total over table total is the expected count formula used in a chi-square test.

Independent is not the same as mutually exclusive

These two words describe different relationships, and trading one for the other is the most common error in this part of Unit 2.

Mutually exclusive is about whether the two events overlap. Independent is about whether one carries information about the other. In the first museum table, "bought online" and "saw the show" overlap in 20 visitors, so they are not mutually exclusive, and separately they turned out to be independent. Disjoint vs independent events runs through more pairs.

A two-way table settles half of this automatically. Any two events from the same margin, two rows or two columns, are mutually exclusive, because each visitor sits in one row and one column. The pair worth testing for independence is one row event against one column event.

Mutually exclusive events with nonzero probability are never independent

Suppose $A$ and $B$ are mutually exclusive and both have probability above 0. Then $P(A \cap B) = 0$ by definition, while $P(A) \cdot P(B)$ is a product of two positive numbers and so is above 0. The two sides cannot match, so the events fail the independence test. That is a proof, not a tendency.

Try it on the first museum table with $B$ as "saw the show" and $C$ as "skipped the show". Those cannot both happen, so $P(B \cap C) = 0$, while $P(B) \cdot P(C) = 0.25 \times 0.75 = 0.1875$. Since $0 \neq 0.1875$, the two events are dependent.

The conditional form says it more bluntly. $P(B \mid C) = 0$ while $P(B) = 0.25$, so learning that a visitor skipped the show moves the chance they saw it from $0.25$ all the way to $0$. That is not "no information". It is complete information, the strongest dependence available.

So mutually exclusive events with nonzero probabilities are not merely non-independent. They are as dependent as two events can be, and the only way to be both disjoint and independent is for one of them to have probability 0.

When the two sides are close but not equal

Real data almost never produces exact equality, which raises a fair question about what "check" means.

Read what is being asked. If a question hands you a table and asks whether two events are independent, the check is exact: $0.16$ is not $0.10$, so the answer is no for these data. Do not round until the two sides look alike.

If the question is whether two variables are independent in the population the data came from, that is inference rather than arithmetic. Sampling variation alone pushes $P(A \cap B)$ off $P(A) \cdot P(B)$ even when the population is genuinely independent. The tool is a chi-square test for independence, which compares each observed count with its expected count and asks whether the gaps exceed what chance explains. Chi-square tests explained covers the procedure.

One more distinction worth keeping straight. Independent events is the property on this page, while independent trials means that one repetition of a process does not affect the next, which is what the binomial setting requires.

Independent samples means two groups were selected separately, which is a condition for two-sample inference. Same word, three different settings.

AP note: where this shows up on the exam

Independence is topic 2.7 of Unit 2 in the Fall 2026 AP Statistics course. The framework states the test in three forms at once: $A$ and $B$ are independent when $P(A \mid B) = P(A)$, when $P(B \mid A) = P(B)$, and when $P(A \cap B) = P(A) \cdot P(B)$. Mutually exclusive events are the separate topic 2.5, defined by $P(A \cap B) = 0$, and conditional probability is topic 2.6.

Unit 2 is 15 to 25 percent of the multiple-choice section. Two item types recur:

• A two-way table with a question asking whether two events are independent. Show both sides of the product, or both sides of the conditional, and then write a sentence giving the verdict. A comparison with no conclusion does not earn the point.
• A pair of mutually exclusive events with a question asking whether they are independent. The answer is no whenever both probabilities are above 0.

The chi-square test for independence in Unit 3, topic 3.15, is the inference version of the same idea. Practice the reading and the arithmetic on the conditional probability set. The official framework is at AP Central.