Testing Independence With Conditional Probabilities

Why Compare Conditional and Unconditional Probabilities?

Suppose a nurse wants to know whether appointment attendance is related to receiving a reminder. One useful question is: does knowing that a patient received a reminder change the probability that the patient attends? If the probability of attendance is the same with or without that information, the two events are independent in the probability model. If the probability changes, they are not independent.

Conditional probability makes this comparison precise. The quantity $P(A)$ describes the chance of event $A$ before restricting attention to another event. The quantity $P(A\mid B)$ describes the chance of $A$ among outcomes where $B$ has occurred. Comparing them asks whether learning $B$ provides information about $A$.

Definition: Events $A$ and $B$ are independent if knowing whether $B$ occurred does not change the probability of $A$. When $P(B)>0$, this is equivalent to $$ P(A\mid B)=P(A). $$ Using the definition of conditional probability, this equality is also equivalent to $P(A\cap B)=P(A)P(B)$. If $P(B)=0$, $P(A\mid B)$ is undefined, so use the product definition $P(A\cap B)=P(A)P(B)$ instead.

For the conditional comparison, the requirement $P(B)>0$ matters because a conditional probability given an event of probability zero has a zero denominator. If both $P(A)>0$ and $P(B)>0$, independence can also be checked in the reverse direction by comparing $P(B\mid A)$ with $P(B)$. Either comparison gives the same conclusion under these conditions.

How to Make the Comparison

First identify the two events and the probability model. Then calculate the overall probability $P(A)$ and the conditional probability $P(A\mid B)$, using only outcomes in $B$ as the conditional denominator. Equality means the probability of $A$ is unchanged when we learn that $B$ occurred; inequality means it changed.

In a two-way table, $P(A)$ is usually found using the total number of outcomes in the full table, while $P(A\mid B)$ uses only the total for the $B$ group. These are different denominators. A conditional probability is not calculated by dividing by the number in $A$, unless the event being conditioned on is $A$.

Worked Example: Independence on a Fair Die

Roll one fair six-sided die. Let $A$ be the event “the result is even,” and let $B$ be the event “the result is greater than 3.” Determine whether $A$ and $B$ are independent.

The six outcomes are equally likely. Event $A$ contains $\{2,4,6\}$, so $$ P(A)=\frac{3}{6}=\frac{1}{2}. $$ Event $B$ contains $\{4,5,6\}$. Among these three outcomes, the even results are 4 and 6, so $$ P(A\mid B)=\frac{2}{3}. $$

Since $\frac{2}{3}\ne\frac{1}{2}$, the events are not independent. In this model, learning that the result is greater than 3 raises the probability that it is even, from one-half to two-thirds. The comparison is valid because $P(B)=3/6>0$.

Worked Example: Comparing Rates in a Two-Way Table

In a hypothetical roster of 80 students, a school records whether each student had breakfast and whether the student biked to school. Let $A$ be “the student biked” and $B$ be “the student had breakfast.” The table summarizes all 80 students.

	Biked	Did not bike	Total
Had breakfast	18	22	40
Did not have breakfast	12	28	40
Total	30	50	80

The overall probability of biking, based on the roster, is $$ P(A)=\frac{30}{80}=0.375. $$ Among students who had breakfast, 18 out of 40 biked. Therefore, $$ P(A\mid B)=\frac{18}{40}=0.450. $$

The probabilities differ: $0.450\ne0.375$. Thus, for a student selected at random from this roster, the events “biked” and “had breakfast” are not independent. The biking rate is 45.0% among students who had breakfast, compared with 37.5% across the full roster. This describes the roster; by itself, it does not establish why the rates differ or show that breakfast causes biking.

Why the Conditional Criterion Works

The comparison follows directly from the definition of conditional probability. When $P(B)>0$, $$ P(A\mid B)=\frac{P(A\cap B)}{P(B)}. $$ If $P(A\mid B)=P(A)$, multiplying both sides by $P(B)$ gives $$ P(A\cap B)=P(A)P(B). $$ This is the product definition of independence. Conversely, if $P(A\cap B)=P(A)P(B)$, divide both sides by $P(B)>0$ to obtain $P(A\mid B)=P(A)$. Thus, whenever the conditioning event has positive probability, the two definitions agree.

The reasoning also explains the meaning of the comparison: the conditional probability measures the share of $B$ outcomes that are also in $A$, while the unconditional probability measures the share of all outcomes that are in $A$. If those shares are equal, restricting attention to $B$ has not changed the probability of $A$.

Worked Example: Two Draws With Replacement

A bag contains 3 green tokens and 2 gold tokens. A token is drawn, returned to the bag, and then a second token is drawn. Every ordered pair of colors for the two draws has probability determined by the five equally likely tokens on each draw. Let $A$ be “the first draw is green” and $B$ be “the second draw is green.” Determine whether the events are independent.

There are 25 equally likely ordered pairs of individual tokens. The first draw is green in 15 pairs, so $P(A)=15/25=3/5$. The second draw is green in 15 pairs, so $P(B)=3/5$. Both draws are green in 9 pairs, giving $P(A\cap B)=9/25$. Hence $$ P(A\mid B)=\frac{P(A\cap B)}{P(B)} =\frac{9/25}{3/5} =\frac{9}{25}\times\frac{5}{3} =\frac{3}{5}. $$

Since $P(A\mid B)=3/5=P(A)$, the events are independent. In this setup, learning that the second draw was green leaves the probability that the first draw was green unchanged. Returning the token is essential to the specified probability model: it keeps the same five tokens available for the second draw.

Interpreting Differences Carefully

The independence decision applies to events under a stated probability model or to a fully specified finite population such as the 80-student roster above. In observed data, a difference between a conditional proportion and an overall proportion can be described directly, but it does not automatically establish a general relationship in a larger population. A separate inferential analysis would be needed to assess how much a difference might reflect sampling variability.

Independence is about whether knowing one event occurred changes the probability of the other; it is not a statement that the events have no connection in every possible sense. Keep the interpretation tied to the events, the population or model, and the probabilities actually compared.

Key takeaway: To check independence using conditional probabilities, compare $P(A\mid B)$ with $P(A)$, provided $P(B)>0$. Use the full outcome set for $P(A)$, the $B$ outcomes for $P(A\mid B)$, and interpret any difference in the context of the model or data.

Common Mistakes

Using the wrong denominator. For $P(A\mid B)$, the denominator is the probability or count for $B$, not the total for $A$ or the whole table. In the student table, it is $18/40$, not $18/30$.
Comparing the wrong quantities. Independence is checked by comparing $P(A\mid B)$ with $P(A)$, not by comparing $P(A\mid B)$ with $P(B)$.
Ignoring a zero-probability condition. If $P(B)=0$, the expression $P(A\mid B)$ is undefined. Use the product definition of independence instead.
Treating a finite-data difference as a causal result. A difference in observed rates describes the data, but it does not by itself show that one event causes the other or establish a population-level effect.

Check Your Understanding

Use the conditional-probability comparison where it is defined.

If $P(A)=0.30$ and $P(A\mid B)=0.30$, with $P(B)>0$, what conclusion follows about $A$ and $B$?
A two-way table has 24 outcomes in event $A$, 40 in event $B$, 12 in both, and 100 in total. Calculate $P(A)$ and $P(A\mid B)$. Are the events independent?
Why must $P(B)>0$ when checking independence by comparing $P(A\mid B)$ and $P(A)$?
In the die example, what did learning that the result was greater than 3 do to the probability of an even result?
A data set shows different rates of an outcome between two groups. Give one reason this difference alone does not establish that group membership caused the outcome.