Union Probability for Independent Events

When Either of Two Independent Alarms Can Activate

In Probability of a Run of Successes, you used independence to multiply the probabilities of several required successes. Here, we combine that multiplication rule with the addition rule to answer a different question: what is the probability that at least one of two independent alarms activates?

Let $A$ be the event that the first alarm activates, and let $B$ be the event that the second alarm activates. The event “at least one alarm activates” is $A\cup B$. It includes three possibilities: only the first alarm activates, only the second activates, or both activate. The word “or” is inclusive, as in Interpreting a Union Probability in Context.

To find this union, add $P(A)$ and $P(B)$. That counts the overlap, $A\cap B$, twice, because the outcome in which both alarms activate is included in each individual event. Subtract the overlap once. The general addition rule from Common Errors with the Addition Rule is

P(A\cup B)=P(A)+P(B)-P(A\cap B).

When $A$ and $B$ are independent, the multiplication rule from The Multiplication Rule for Independent Events gives $P(A\cap B)=P(A)P(B)$. Substitute that product for the overlap in the addition rule:

Formula: If $A$ and $B$ are independent, then $$ P(A\cup B)=P(A)+P(B)-P(A)P(B). $$ This is the probability that $A$ occurs, $B$ occurs, or both occur.

The independence condition matters: the product $P(A)P(B)$ gives the overlap probability only when the events are independent. If they are not independent, the general addition rule still works, but you need the actual value of $P(A\cap B)$. Do not substitute the product just because the events are described as two alarms.

Why Subtract the Product?

For independent alarms, $P(A)P(B)$ is the chance that both activate. Adding the two individual activation probabilities counts that “both” outcome twice. Subtracting the product once leaves it counted once, which is what the union requires.

There is another useful way to see the formula. “At least one activates” is the complement of “neither activates.” If $A$ and $B$ are independent, the probability that the first does not activate is $1-P(A)$, and the probability that the second does not activate is $1-P(B)$. The probability that neither activates is the product of those two probabilities.

P(A\cup B) =1-P(\text{neither}) =1-[1-P(A)][1-P(B)].

Expanding the product gives the same union formula:

$$ 1-[1-P(A)][1-P(B)] =1-[1-P(A)-P(B)+P(A)P(B)] =P(A)+P(B)-P(A)P(B). $$

These are two ways to calculate the same probability. The addition-rule version makes the overlap explicit. The complement version can be especially convenient when the probabilities that neither alarm activates are easy to find. You can use the complement as a check on your answer.

Conditions: To use $P(A\cup B)=P(A)+P(B)-P(A)P(B)$:

Define $A$ and $B$ as the two events of interest.
Use the product $P(A)P(B)$ for the overlap only if $A$ and $B$ are independent.
Interpret $A\cup B$ as at least one event occurring, including the possibility that both occur.

Worked Examples

Worked Example: Two Smoke Alarms

In a hypothetical building-safety model, the first smoke alarm activates when smoke is present with probability $0.92$, and a second alarm activates with probability $0.85$. Assume their activation events are independent. What is the probability that at least one alarm activates?

State. Let $A$ be the event that the first alarm activates and $B$ the event that the second alarm activates. The target is $P(A\cup B)$.

Plan. The question asks for an inclusive “or.” The model says the events are independent, so find the overlap using $P(A\cap B)=P(A)P(B)$, then use the general addition rule.

Do. The probability that both activate is $0.92(0.85)=0.782$. Therefore,

P(A\cup B) =0.92+0.85-(0.92)(0.85) =1.77-0.782 =0.988.

Check using the complement. The probabilities that the first and second alarms do not activate are $1-0.92=0.08$ and $1-0.85=0.15$. Independence gives $P(\text{neither})=(0.08)(0.15)=0.012$, so $1-0.012=0.988$, matching the addition-rule calculation.

Conclude. Under the model’s independence assumption, the probability that at least one of the two alarms activates when smoke is present is $0.988$, or $98.8\%$.

Worked Example: Checking the Four Outcomes

In another hypothetical system, two independent water-level alarms activate during a rising-water event with probabilities $0.70$ and $0.60$. Find the probability that at least one activates, and check it by accounting for all four possible outcomes.

State. Let $A$ mean the first alarm activates and $B$ mean the second activates. We want $P(A\cup B)$.

Plan. Since the events are independent, first calculate the probability that both alarms activate. Then use the addition rule. As a check, calculate the probabilities of the four non-overlapping outcomes: both, first only, second only, and neither.

Do. The probability that both activate is $(0.70)(0.60)=0.42$. Thus,

P(A\cup B) =0.70+0.60-0.42 =0.88.

For the four-outcome check, the probability that only the first activates is $0.70(1-0.60)=0.70(0.40)=0.28$. The probability that only the second activates is $(1-0.70)(0.60)=0.30(0.60)=0.18$. The probability that neither activates is $(0.30)(0.40)=0.12$. Adding the three outcomes in the union gives $0.42+0.28+0.18=0.88$. All four outcomes sum to $0.42+0.28+0.18+0.12=1.00$, as they should.

The complement gives the same check: $1-P(\text{neither})=1-0.12=0.88$.

Conclude. In this model, the probability that at least one water-level alarm activates during a rising-water event is $0.88$, or $88\%$.

Worked Example: At Least One Nuisance Alert

A hypothetical monitoring setup has two independent alarms. During a period with no hazard, the first alarm gives a nuisance alert with probability $0.12$, and the second gives one with probability $0.08$. What is the probability of at least one nuisance alert?

State. Let $A$ be a nuisance alert from the first alarm and $B$ a nuisance alert from the second. The target is $P(A\cup B)$.

Plan. The question asks whether either alarm or both give an alert. Use the independent-events union formula. A complement calculation will verify the result.

Do. The chance both alarms give nuisance alerts is $(0.12)(0.08)=0.0096$. So,

P(A\cup B) =0.12+0.08-0.0096 =0.20-0.0096 =0.1904.

For a second calculation, the probability that neither gives a nuisance alert is $(1-0.12)(1-0.08)=(0.88)(0.92)=0.8096$. Therefore, $P(A\cup B)=1-0.8096=0.1904$. The result rounded to four decimal places is $0.1904$.

Conclude. Under the stated independence model, the probability of at least one nuisance alert during the period is $0.1904$, or $19.04\%$.

Independence Is Not Mutual Exclusivity

An important point from Mutually Exclusive Versus Independent Events is that independent events can occur together. In the alarm examples, both alarms can activate; independence lets us calculate the probability of that overlap by multiplying their probabilities.

Mutually exclusive events, by contrast, cannot happen together. If two events with positive probabilities are mutually exclusive, their intersection probability is zero, not the product of their probabilities. In that situation, using the independent-events formula would generally be wrong. Do not infer independence from the fact that the events are different or come from separate devices; the model or information in the question must support it.

The independent union probability should be at least as large as either individual probability, because $A$ and $B$ are both included in the union. It also cannot exceed $1$. These are useful reasonableness checks. For instance, a calculated union of $0.50$ would be impossible if one alarm alone activates with probability $0.70$.

Keep full precision in the overlap calculation and round the final probability. As with other probability calculations, a clear solution names the events, identifies the independence assumption, shows the calculation, and interprets the result in context.

Common Mistakes and AP Exam Tips

Adding without subtracting the overlap. $P(A)+P(B)$ counts the outcome where both occur twice. Subtract $P(A\cap B)$ once.
Using the product as the union. $P(A)P(B)$ is the probability that both independent events occur, not the probability that at least one occurs.
Forgetting that “or” includes both. The union includes the overlap. It is not the event that exactly one alarm activates.
Assuming independence automatically. State that the alarms’ events are modeled as independent. Separate alarms are not necessarily independent in every real setting; for example, a shared power failure could affect both.
Calling independent events disjoint. Independent events may occur together. For positive event probabilities, disjoint events are not independent.
Giving a probability without context. Say what event the probability describes, under what assumption, and in the situation being studied.

AP Exam Tip: Define the two events, state why independence is given or reasonable in the model, show $P(A)+P(B)-P(A)P(B)$, and interpret the result as the chance that at least one event occurs. A complement calculation, $1-[1-P(A)][1-P(B)]$, is a useful check.

Key Takeaway

For two independent events, multiply their probabilities to find the chance that both occur, then subtract that overlap from the sum of their individual probabilities. This gives the chance that at least one occurs. Equivalently, subtract the probability that neither occurs from $1$.

Key takeaway: If $A$ and $B$ are independent, then $P(A\text{ or }B)=P(A)+P(B)-P(A)P(B)$. The union includes both events occurring, so account for the overlap exactly once.

Check Your Understanding

For each question, define the events, show the calculation, and interpret the probability in context.

Two independent backup alarms activate with probabilities $0.75$ and $0.50$. Find the probability that at least one activates.
Two independent sensors each give a warning with probability $0.10$. Find the probability that both give a warning, then find the probability that at least one does.
For two independent alarms with activation probabilities $0.40$ and $0.30$, find the probability that neither activates. Use it to find the probability that at least one activates.
Explain why adding $P(A)$ and $P(B)$ without subtracting $P(A\cap B)$ overcounts the union.
Explain why knowing that two alarms are separate devices is not, by itself, enough to establish that their activation events are independent.