Complements and the Probability of at Least One

Why Look at the Event That Does Not Happen?

Some probability questions are easier to answer by considering the opposite event. If a question asks for the chance that at least one of several trials has a particular result, listing every possible way that could happen can be cumbersome. The alternative is to find the probability that it happens zero times, then subtract that probability from 1.

This approach is based on a basic feature of probability: for any event, either it occurs or it does not. Those two possibilities together cover the entire sample space. The event that it does not occur is called its complement.

Definition: Let $S$ be a sample space and $A$ an event, so $A$ is a subset of $S$. The complement of $A$, written $A^c$ (or “not $A$”), is the event consisting of all outcomes in $S$ that are not in $A$. The complement rule is $$ P(A^c)=1-P(A), $$ or equivalently $P(A)=1-P(A^c)$.

Why the Complement Rule Works

An event and its complement cannot occur together: an outcome cannot both belong and not belong to $A$. They also account for every outcome in $S$. Thus, together, they form a partition of the sample space. Since the probability of the whole sample space is 1, the probabilities of these two parts must add to 1.

$$ P(A)+P(A^c)=P(S)=1 $$

Rearranging this equality gives $P(A^c)=1-P(A)$. This argument uses the basic probability property that probabilities of disjoint parts making up a whole sum to the probability of that whole. The result applies whether probabilities come from equally likely outcomes or from a probability model with unequal outcome probabilities.

For a finite sample space in which all outcomes are equally likely, the same rule can be seen by counting. If $S$ has $N$ outcomes and $A$ has $k$, then $A^c$ has $N-k$. Therefore,

P(A^c)=\frac{N-k}{N}=1-\frac{k}{N}=1-P(A).

The complement is always defined relative to the sample space under consideration. For example, if $A$ is “a die roll is 5 or 6,” its complement is “the die roll is 1, 2, 3, or 4,” not an event about outcomes outside the die-rolling experiment.

Worked Example: Complementing an Event on One Die

A fair six-sided die is rolled once. Let $A$ be the event that the result is 5 or 6. Find $P(A)$ and the probability that $A$ does not occur.

The sample space is $S=\{1,2,3,4,5,6\}$, with six equally likely outcomes. The event $A$ contains the two outcomes $\{5,6\}$, so

P(A)=\frac{2}{6}=\frac{1}{3}\approx 0.3333.

The complement consists of the other four outcomes, $\{1,2,3,4\}$. Direct counting gives

P(A^c)=\frac{4}{6}=\frac{2}{3}\approx 0.6667.

The complement rule gives the same result:

P(A^c)=1-P(A)=1-\frac{1}{3}=\frac{2}{3}\approx 0.6667.

Thus, the probability that the roll is neither 5 nor 6 is $2/3$. The two probabilities sum to 1, as they must.

“At Least One” Means “Not Zero”

An event of the form “at least one success” includes one success, two successes, and every larger possible number of successes. Its complement is the single case in which there are no successes. This is why the complement method is often simpler: it avoids separating the event into many possible counts.

Key result: For any set of trials, let $A$ be the event that at least one trial has the outcome of interest. Then $A^c$ is the event that none of the trials has that outcome, and $$ P(\text{at least one})=1-P(\text{none}). $$

For a finite sample space, identify the “none” outcomes and count them, if outcomes are equally likely. In repeated trials, a familiar formula is available when every trial has the same probability $p$ of the outcome of interest and the trials are independent. Independence here means that the outcomes of different trials do not change one another’s probabilities; in particular, the probability of no successes across all trials is the product of the probability of failure on each trial.

P(\text{at least one in }n\text{ independent trials}) =1-(1-p)^n.

This formula has specific conditions: there must be $n$ trials, each trial must have probability $p$ of the outcome of interest, and the trials must be independent. On each trial, the probability of no occurrence is $1-p$. Independence makes the probability of no occurrence on all $n$ trials equal to $(1-p)^n$; taking its complement gives the displayed result. If the trials are dependent or have different probabilities, this formula cannot be used without further information.

Worked Example: At Least One Six in Two Rolls

A fair six-sided die is rolled twice. Assume the rolls are independent. Find the probability of getting at least one 6.

The event of interest is “a 6 appears on the first roll, the second roll, or both.” Its complement is simpler: neither roll is a 6. On one roll, the probability of not getting a 6 is $5/6$. By independence, the probability that both rolls are not 6 is

P(\text{no six in two rolls})=\frac{5}{6}\times\frac{5}{6} =\frac{25}{36}\approx 0.6944.

Therefore,

P(\text{at least one six}) =1-\frac{25}{36} =\frac{11}{36} \approx 0.3056.

The probability of at least one 6 in the two rolls is about 0.3056, or 30.56%. The assumption of independence is appropriate for separate rolls of a fair die: the result of the first roll does not change the probabilities on the second.

Repeated Independent Trials

The repeated-trial formula is especially useful when the number of ways to get one or more occurrences would be inconvenient to enumerate. For example, a question about at least one occurrence in ten trials could involve many different counts of occurrences. The complement event remains simple: it requires the outcome of interest to be absent on every trial.

A useful way to organize a calculation is to define the event of interest precisely, name its complement in words, calculate the probability of that complement under the stated model, and subtract from 1. The “none” event must match the original event exactly. If the original event concerns at least one success, “none” means zero successes across all the trials in question.

Worked Example: At Least One Flag in Ten Inspections

For a hypothetical quality-control exercise, suppose each inspected package has probability $0.02$ of being flagged for a specified issue. Ten packages are inspected. Assume the outcomes are independent and that the probability is $0.02$ for each package. Find the probability that at least one package is flagged.

Let $A$ be the event that one or more of the ten packages are flagged. Then $A^c$ is the event that none is flagged. For any one package, the probability of not being flagged is

$$ 1-0.02=0.98. $$

Because the ten inspection outcomes are assumed independent and have the same probabilities, the probability that none of the ten packages is flagged is

P(A^c)=(0.98)^{10}=0.8171 \quad \text{(approximately)}.

So the probability that at least one package is flagged is

P(A)=1-P(A^c) =1-(0.98)^{10} \approx 1-0.8171 =0.1829.

Under this hypothetical model, there is about a 0.1829 probability, or 18.29%, that at least one of the ten packages is flagged. This result depends on the model assumptions; it is not justified merely by knowing the per-package probability if the inspection outcomes are dependent.

A Short Justification for the Repeated-Trial Formula

The formula follows by applying the complement rule and the meaning of independence. Suppose there are $n$ independent trials, each with probability $p$ of the outcome of interest. The probability of no occurrence on any one trial is $1-p$. Independence gives the probability of no occurrence on every trial as the product of these $n$ probabilities:

P(\text{none})= \underbrace{(1-p)(1-p)\cdots(1-p)}_{n\text{ factors}} =(1-p)^n.

The event “at least one” is the complement of “none.” Substitution into the complement rule therefore gives

P(\text{at least one})=1-P(\text{none})=1-(1-p)^n.

The exponent $n$ counts the number of independent trials. It does not mean that the probability of at least one occurrence is $np$; the complement calculation accounts for the fact that more than one occurrence can happen and that probabilities cannot exceed 1.

Common Mistakes and Checks

Choosing the wrong complement. The complement of “at least one” is “none,” not “exactly one.” “Exactly one” leaves out outcomes with two or more occurrences.
Subtracting the wrong probability. The complement rule is $1-P(A)$ for the event $A$ whose complement you want. State both events in words before substituting a number.
Assuming independence without a reason. The expression $(1-p)^n$ requires independent trials with the same probability $p$. If outcomes influence one another, the probability of no occurrences may not be that power.
Using the formula when trial probabilities differ. If independent trials have different probabilities $p_1,\ldots,p_n$, the probability of none is instead $(1-p_1)\cdots(1-p_n)$. If independence is not given or justified, more information is needed.
Forgetting the sample space. A complement contains all outcomes in the specified experiment that are outside the event, so define the experiment and event clearly before counting.

A final check is to confirm that the answer is between 0 and 1 and has the intended direction. If the chance of no occurrences is large, the chance of at least one should be small; if the chance of no occurrences is small, the chance of at least one should be large. For $p=0$, the repeated-trial formula gives 0, and for $p=1$ with at least one trial it gives 1, as expected.

Check Your Understanding

For each question, identify the event’s complement before calculating.

A fair die is rolled once. Let $A$ be the event that the result is 5 or 6. What is $A^c$, and what are $P(A)$ and $P(A^c)$?
A fair coin is tossed three times. What is the complement of getting at least one head? Use it to find the probability of at least one head.
In five independent trials, each with probability $0.10$ of success, write the expression for the probability of at least one success. What assumption makes the expression valid?
A machine produces items with a 0.03 probability of a specified flaw per item. Explain why knowing this probability alone may not be enough to use $1-(1-0.03)^8$ for eight items.
Why is “exactly one success” not the complement of “at least one success”?