Complements and At-Least-One Events

Why “At Least One” Often Calls for a Complement

An event such as “at least one item is defective” includes several possibilities: exactly one item is defective, exactly two are defective, and so on. Adding the probabilities of those cases can be tedious, especially when there are many trials. Often it is simpler to calculate the probability of none being defective and subtract that probability from one.

This method is not an assumption about how trials behave. It follows from the way events and probabilities are defined. The calculation of “none,” however, depends on the probability model. In particular, multiplying individual failure probabilities is justified only when the model supports that multiplication.

Definition: Let $\Omega$ be the sample space, and let $A_1,\ldots,A_n$ be events in $\Omega$. The event that at least one of them occurs is $\bigcup_{i=1}^{n} A_i$. The event that none occurs is $\bigcap_{i=1}^{n} A_i^c$, where $A_i^c=\Omega\setminus A_i$ is the complement of $A_i$ relative to $\Omega$.

The sample space matters because a complement is always relative to a specified set of possible outcomes. If a die is rolled twice, for example, the outcomes are ordered pairs, not just the two face values considered separately. The event “at least one roll is a six” is a subset of that ordered-pair sample space; its complement consists precisely of pairs with no six.

The Complement Rule for At-Least-One Events

Theorem: For any finite collection of events $A_1,\ldots,A_n$ in a probability space, $$ P\left(\bigcup_{i=1}^{n} A_i\right) =1-P\left(\bigcap_{i=1}^{n} A_i^c\right). $$ No independence assumption is required.

Proof. An outcome is not in $\bigcup_{i=1}^{n} A_i$ exactly when it is in none of $A_1,\ldots,A_n$. Thus the complement of the union is $\bigcap_{i=1}^{n}A_i^c$. The union and its complement are disjoint and together equal $\Omega$. By additivity for disjoint events and $P(\Omega)=1$, $$ P\left(\bigcup_{i=1}^{n} A_i\right) +P\left(\bigcap_{i=1}^{n} A_i^c\right)=1. $$ Rearranging gives the stated identity. $\square$

This theorem turns an “at least one” question into a “none” calculation. It does not, by itself, tell us how to calculate the probability of none. For that, we must use the model: perhaps outcomes are equally likely and can be counted, or perhaps a sequence of trials has specified joint probabilities.

Worked Example: At Least One Six in Two Rolls

Roll a fair six-sided die twice. Find the probability of at least one six. Use the sample space of ordered pairs $\Omega=\{(i,j):i,j\in\{1,2,3,4,5,6\}\}$. Under the model, all $6^2=36$ ordered pairs are equally likely.

Let $A_1$ be the event that the first roll is a six, and $A_2$ the event that the second roll is a six. The event of interest is $A_1\cup A_2$. Its complement is the event that neither roll is a six. There are five choices, $1,\ldots,5$, for each roll in that complement, giving $5\cdot5=25$ ordered pairs.

Therefore, $$ P(\text{at least one six}) =1-P(\text{no six}) =1-\frac{25}{36} =\frac{11}{36}. $$

As a check, the 11 favorable pairs are the six pairs whose first entry is 6 plus the six whose second entry is 6, with $(6,6)$ counted in both groups. Thus $6+6-1=11$, agreeing with $11/36$.

When the Trials Are Independent

In some models, a sequence of trials is specified to be independent, with a success probability $p_i$ on trial $i$. For such a model, the probability of any specified sequence of successes and failures is the product of the corresponding trial probabilities. In particular, the probability that every trial fails is $\prod_{i=1}^{n}(1-p_i)$.

Theorem: Suppose $n$ independent trials have success probabilities $p_1,\ldots,p_n$, and $A_i$ is success on trial $i$. Then $$ P\left(\bigcup_{i=1}^{n}A_i\right) =1-\prod_{i=1}^{n}(1-p_i). $$

Proof. By the complement rule, the probability of at least one success is one minus the probability of failure on every trial. In the independent-trial model, the joint probability of the specified sequence “failure, failure, ..., failure” is the product of its trial probabilities. The probability of failure on trial $i$ is $1-p_i$, so $$ P\left(\bigcap_{i=1}^{n}A_i^c\right) =\prod_{i=1}^{n}P(A_i^c) =\prod_{i=1}^{n}(1-p_i). $$ Substitution into the complement rule proves the formula. $\square$

If all trials have the same success probability $p$, the formula becomes $1-(1-p)^n$. Both the independence assumption and the probability $p$ must belong to the model; the formula is not a general rule for any collection of events.

Worked Example: At Least One Positive Result in Independent Tests

Suppose a model specifies three independent tests, each with probability $0.10$ of a positive result. Find the probability of at least one positive result.

Let $A_i$ be the event of a positive result on test $i$. The probability of a negative result on each test is $1-0.10=0.90$. By the independent-trial model, the probability that all three results are negative is $$ 0.90\cdot0.90\cdot0.90=0.729. $$

Consequently, $$ P(A_1\cup A_2\cup A_3) =1-0.729 =0.271. $$

The result is a probability under the stated model. It does not say that a real set of tests is independent; that assumption would need to be justified separately.

When the Events Are Dependent

The complement rule still works when events are dependent. What changes is the calculation of the probability that none occurs. Drawing without replacement is a common situation in which successive outcomes are dependent: the first draw changes what remains for the second.

Worked Example: Defective Items Drawn Without Replacement

A lot contains four items: two defective and two good. Two items are selected uniformly at random without replacement. Find the probability of selecting at least one defective item. Use ordered pairs of distinct items as the sample space. There are $4\cdot3=12$ equally likely ordered pairs.

Let $A_1$ and $A_2$ denote the events that the first and second selected items, respectively, are defective. “None are defective” means both selected items are good. There are $2\cdot1=2$ ordered pairs of distinct good items, so $$ P(A_1^c\cap A_2^c)=\frac{2}{12}=\frac{1}{6}. $$

The complement rule gives $$ P(A_1\cup A_2)=1-\frac{1}{6}=\frac{5}{6}. $$

The draws are dependent. In fact, each draw position has probability $1/2$ of containing a defective item, but multiplying the two marginal probabilities of a good item would give $(1/2)^2=1/4$, not the correct probability $1/6$ of two good items. The complement method is valid; that unsupported multiplication is not.

A Reliable Calculation Process

Before calculating, translate the verbal event into a precise statement about outcomes. This keeps “at least one,” “exactly one,” and “none” from being confused. Then identify how the model assigns probabilities to the outcomes in the complement event.

Specify the sample space.
State what one outcome records, including order or replacement when these affect the model.

Name the events.
Define $A_i$ as the occurrence of the target event on trial or position $i$. The “at least one” event is $\bigcup_i A_i$.

Describe the complement.
“None” means $\bigcap_i A_i^c$. Count or calculate its probability using the actual model.

Subtract from one.
Use $P(\text{at least one})=1-P(\text{none})$, and check that the result lies between zero and one.

Common Mistakes and Why This Method Matters

Multiplying without independence. The complement rule needs no independence, but a product of failure probabilities generally does. Check the model before using a product.
Using the wrong sample space. If order matters, count ordered outcomes; if items are selected without replacement, do not treat selections as independent.
Adding event probabilities as though they were disjoint. Events such as “success on trial 1” and “success on trial 2” can both occur. Their probabilities cannot simply be added without accounting for overlap.
Confusing “at least one” with “exactly one.” The complement of “at least one” is “none,” not “more than one.”

Complements are useful because they replace a collection of possible success patterns with one clearly described failure pattern. Their main benefit is often computational, but their conceptual benefit is just as important: they separate a valid event identity from the assumptions needed to calculate a particular probability.

Key takeaway: Define “at least one” as a union, identify “none” as its complement in the stated sample space, and calculate the probability of none under the model. Use a product of failure probabilities only when independence justifies it.

Check Your Understanding

For each question, identify the complement event before calculating or explaining.

A fair die is rolled three times. Describe the sample space and the complement of “at least one roll is a 1.”
Two cards are drawn without replacement from a standard deck. Why does the complement method apply to “at least one ace” even though the draws are dependent?
Three independent trials have success probabilities $0.2$, $0.4$, and $0.5$. Write, but do not evaluate, the probability of at least one success.
Two events each have probability $0.3$. Is their probability of both failing necessarily $0.7^2$? Explain what additional condition would justify that calculation.
Explain why the complement of “at least one defective item” is “no defective items,” rather than “exactly one good item.”