The Complement of a Conditional Event

Why a Complement Can Simplify a Conditional Question

A question about a group often asks for the proportion of its members with a particular feature. But sometimes it is easier to find the proportion with that feature first, then calculate the proportion without it. For example, if 35% of the members of a specified group completed a task, then 65% did not. The important point is that both percentages refer to the same group.

Conditional probability notation makes this precise. If $A$ is an event of interest and $B$ is the condition that defines the group, then $P(\text{not }A\mid B)$ is the probability that $A$ does not occur, given that $B$ has occurred. This tutorial focuses on finding and interpreting that complement.

Definition: The complement of event $A$, written $A^c$ or “not $A$,” consists of all outcomes in the sample space that are not in $A$. If $P(B)>0$, the conditional complement rule is $$ P(A^c\mid B)=1-P(A\mid B). $$ The condition $B$ is unchanged: both conditional probabilities concern outcomes within $B$.

Why the Conditional Complement Rule Works

Once we restrict attention to outcomes in $B$, each such outcome either belongs to $A$ or does not. These two possibilities cannot happen together, and together they account for all outcomes in $B$. In symbols, $B\cap A$ and $B\cap A^c$ are disjoint, and their union is $B$.

Conditional probability is defined by $P(C\mid B)=P(C\cap B)/P(B)$, provided $P(B)>0$. Applying this definition to the two parts of $B$ gives a short derivation:

\begin{aligned} P(A\mid B)+P(A^c\mid B) &=\frac{P(A\cap B)}{P(B)}+\frac{P(A^c\cap B)}{P(B)}\\ &=\frac{P((A\cap B)\cup(A^c\cap B))}{P(B)}\\ &=\frac{P(B)}{P(B)}=1. \end{aligned}

The second equality uses the fact that the two events in the union are disjoint. Rearranging the result gives $P(A^c\mid B)=1-P(A\mid B)$. The only requirement for these conditional probabilities to be defined is that $P(B)>0$.

This reasoning is also visible in a two-way table. Among the observations in group $B$, the count that is in $A$ plus the count that is not in $A$ equals the total count in $B$. Dividing each count by that total produces two proportions that add to 1.

Worked Example: Students Who Attended a Workshop

In a hypothetical group of 120 students who attended a study workshop, 48 completed a practice set afterward. If one of these 120 students is selected at random, what is the probability that the student did not complete the practice set, given that the student attended the workshop?

Let $B$ be the event that a student attended the workshop and $A$ the event that the student completed the practice set. The condition $B$ restricts the group to the 120 workshop attendees. Of those, $120-48=72$ did not complete the set.

Directly from the counts, $$ P(A^c\mid B)=\frac{72}{120}=0.600. $$

The complement method gives the same answer. First, $$ P(A\mid B)=\frac{48}{120}=0.400, $$ so $$ P(A^c\mid B)=1-P(A\mid B)=1-0.400=0.600. $$

Interpretation: Among the 120 workshop attendees, 60.0% did not complete the practice set. This is a conditional proportion for workshop attendees, not for all students.

Using Counts and Conditional Probabilities

When a question gives counts, identify the conditioned group before calculating. The denominator is the number of observations in that group; the numerator for the complement is the number in that group who do not have the event of interest. If the conditional probability of $A$ is already given, subtraction from 1 is usually the quickest method.

These are two ways of expressing the same calculation: $$ P(A^c\mid B)=\frac{\text{number in }B\text{ and not in }A}{\text{number in }B} =1-\frac{\text{number in }B\text{ and }A}{\text{number in }B}. $$

The count form applies when the data are observed counts and the selection is uniform within the conditioned group. More generally, the probability rule follows from the conditional probability definition, whether probabilities come from counts or from a probability model.

Worked Example: Employees on the Night Shift

In a hypothetical workplace, 80 employees worked the night shift on a particular day. Of those employees, 68 arrived by the scheduled start time. Select one of the night-shift employees at random. Find the probability that this employee did not arrive by the scheduled start time.

Let $B$ be the event that an employee worked the night shift and $A$ the event that the employee arrived by the scheduled start time. There are $80-68=12$ night-shift employees who did not arrive by the start time. The conditioned group has 80 employees, so $$ P(A^c\mid B)=\frac{12}{80}=0.150. $$

Alternatively, the proportion who arrived by the start time is $$ P(A\mid B)=\frac{68}{80}=0.850. $$ Therefore, $$ P(A^c\mid B)=1-0.850=0.150. $$

Interpretation: Among these 80 night-shift employees, 15.0% did not arrive by the scheduled start time. The result does not describe employees on other shifts.

A two-way table can help keep the condition and the event of interest separate. In the next example, the question asks for the probability of not having a feature within a specified category. The complement must be calculated from the total in that category, not from the entire table.

Worked Example: Orders With Standard Shipping

A hypothetical order summary records shipping type and whether each order arrived within three days. Among 150 orders with standard shipping, 105 arrived within three days. If an order is selected at random from the standard-shipping orders, find the probability that it did not arrive within three days.

Let $B$ be the event that an order used standard shipping and $A$ the event that it arrived within three days. The conditioned group contains 150 orders. The number in that group that did not arrive within three days is $$ 150-105=45. $$

Thus the conditional probability is $$ P(A^c\mid B)=\frac{45}{150}=0.300. $$

Checking by complementation, $P(A\mid B)=105/150=0.700$, and $$ 1-P(A\mid B)=1-0.700=0.300. $$

Interpretation: Of the orders with standard shipping, 30.0% did not arrive within three days. This calculation gives no result for express-shipping orders unless their counts or probabilities are also known.

Keeping the Condition in Place

The complement is taken inside the condition. The expression $P(A^c\mid B)$ asks about outcomes that are in $B$ but not in $A$. It is not the same as the unconditional probability $P(A^c)$, which uses the whole sample space, and it is not the same as $P(B\mid A^c)$, which conditions on not $A$ and asks about $B$.

For instance, in the workshop example, $P(A^c\mid B)=72/120$ uses the 120 attendees as its denominator. The probability $P(A^c)$ would require information about all students, including those who did not attend. The reversed conditional $P(B\mid A^c)$ would require knowing how many students did not complete the practice set in total and how many of those attended.

A useful check is to say the notation aloud in its original order: “not $A$, given $B$.” First identify the group after “given.” Then divide the number in that group who meet the requested condition by the group’s total. This verbal check helps prevent swapping the two events.

Name the events.
Define $A$ as the event whose absence is requested, and $B$ as the condition that specifies the relevant group.

Check the conditioned group.
Confirm that $P(B)>0$, or that the supplied data include at least one observation in group $B$.

Find the probability of $A$ within $B$.
Use the given conditional probability, or divide the count in both $A$ and $B$ by the count in $B$.

Take the complement.
Calculate $1-P(A\mid B)$, keeping $B$ as the condition.

Interpret in context.
Describe the result as the probability or proportion of outcomes not in $A$, among outcomes in $B$.

Common Mistakes and Why the Rule Matters

The complement rule is useful because a “not” question may be answered quickly when the probability of the event itself is known. It is also a strong check on a calculation: the conditional probabilities of $A$ and $A^c$ within $B$ must add to 1. If they do not, either the arithmetic or the choice of denominator needs attention.

Using the wrong denominator. A learner may divide the number in $B$ and not $A$ by the total number of observations. The correction is to divide by the total in $B$, because $B$ is the condition.
Subtracting the condition’s probability. A learner may reason that $P(A^c\mid B)=1-P(B)$. This changes the event being complemented. The correct identity is $P(A^c\mid B)=1-P(A\mid B)$.
Reversing the order of the events. $P(A^c\mid B)$ and $P(B\mid A^c)$ ask different questions and generally have different values. Read “given $B$” as the group being considered.
Assuming the complement is a different category of the same name everywhere. “Not $A$” means every outcome outside $A$ in the full sample space. Once conditioned on $B$, only the part of that complement inside $B$ is counted.

Key takeaway: Within the group specified by $B$, event $A$ and its complement divide the group into two exhaustive, non-overlapping parts. Their conditional probabilities sum to 1, so $P(A^c\mid B)=1-P(A\mid B)$, provided $P(B)>0$.

Check Your Understanding

For each question, state the conditioned group and show how the complement is obtained.

Among 200 randomly selected library visitors, 130 borrowed at least one book. If a visitor is selected from this group, what is the probability that the visitor did not borrow a book?
Suppose $P(A\mid B)=0.72$. Find $P(A^c\mid B)$ and explain what group the answer describes.
Explain why the denominator for $P(A^c\mid B)$, when calculated from counts, is the total number in $B$, not the total number in the sample.
In words, distinguish $P(A^c\mid B)$ from $P(B\mid A^c)$.
A class has 40 students, of whom 28 submitted an assignment. If $B$ is the event that a selected student is in that class and $A$ is the event that the student submitted the assignment, calculate $P(A^c\mid B)$. State the result in context.