Probability of At Least One in Independent Trials

From One Attempt to Several

In Why Disjoint Events Are Not Independent, you saw that independence describes whether knowing one event occurred changes the probability of another. Now we will use independence in a repeated-trial setting. If a shooter takes several shots, or a machine makes several attempts, what is the probability of at least one success?

The phrase “at least one” includes one success, two successes, and every larger number of successes up to the total number of attempts. Instead of calculating each of those possibilities separately, it is usually simpler to use the complement. The complement of “at least one success” is “no successes”—in other words, every attempt fails.

Definition: A success is the outcome of interest on one attempt. If the probability of success on each of $n$ independent attempts is $p$, then the probability of at least one success is the complement of the probability that all $n$ attempts fail.

The probability of failure on one attempt is $1-p$. If attempts are independent, the multiplication rule from The Multiplication Rule for Independent Events says to multiply the failure probabilities to find the probability that every attempt fails. Since the attempts have the same failure probability, this is $(1-p)^n$. Subtract that probability from 1 to get the probability of at least one success.

P(\text{at least one success}) =1-P(\text{no successes}) =1-(1-p)^n

Here, $p$ is the probability of success on a single attempt, and $n$ is the number of attempts. The formula relies on two important features: the attempts are independent, and the probability of success is the same for every attempt. The probability of failure on each attempt is therefore also the same.

Why the Complement Is Convenient

Suppose a basketball player takes four free throws and you want the probability of making at least one. “At least one” includes making exactly one, exactly two, exactly three, or all four. Finding each probability and adding them would require several calculations. The complement is just one possibility: missing all four shots.

The complement method does not mean that “at least one” and “no successes” are disjoint events whose probabilities you add. They are complementary events: one of them must happen, and they cannot both happen. As in Using Complements with Disjoint Events, their probabilities add to 1. That is why subtracting the probability of no successes gives the probability of at least one.

Formula: For $n$ independent attempts with the same probability $p$ of success on each attempt, $$ P(\text{at least one success})=1-(1-p)^n. $$ The complement, “no successes,” requires every attempt to fail, so its probability is $(1-p)^n$.

A common tempting shortcut is to calculate $np$, as if the probability of success on at least one attempt were the single-attempt probability multiplied by the number of attempts. That is not the right rule. Success on different attempts can occur together, so simply adding the success probabilities can count outcomes more than once. The complement formula accounts for the possible overlap by considering the single, clear event that every attempt fails.

The formula also gives a useful qualitative check. With $p$ fixed, adding independent attempts makes “no success” less likely, so “at least one success” becomes more likely. If $p$ is larger and $n$ stays fixed, each attempt is more likely to succeed, so the chance of at least one success increases as well.

Check the Trial Model Before Calculating

A trial is one attempt in the repeated chance process. Before using the formula, define what counts as success and make sure the model fits the situation. “A shot goes in,” “a machine correctly detects a fault,” or “a test identifies a signal” can each be a success, depending on the question.

Conditions: Use $1-(1-p)^n$ when:

There is a fixed number $n$ of attempts.
Each attempt has the same probability $p$ of success.
The attempts are independent, so one attempt’s result does not change the probability of success on another.

The independence condition needs a reason in context. Separate attempts might reasonably be modeled as independent if the process resets and nothing about one outcome changes the next attempt. But if a shooter becomes tired, learns from a previous shot, or changes technique, the chance of success might change. Likewise, a machine may need adjustment after a failed attempt. In those situations, do not assume the repeated attempts have identical, independent probabilities without justification.

This differs from sampling without replacement, discussed in Independence When Sampling Without Replacement. When an item is not replaced, the contents of the group change after each draw, which can change the probability on the next draw. Repeated attempts use this formula only when the setting supports the stated independence and same-probability assumptions.

Worked Examples

Worked Example: A Shooter Takes Four Shots

In a hypothetical practice drill, a shooter has a $0.25$ probability of hitting the target on each shot. Assume the shots are independent and the shooter’s probability stays the same. Find the probability of hitting the target at least once in four shots.

State. We want the probability of at least one hit in four shots.

Plan. The probability of a hit on one shot is $p=0.25$, so the probability of a miss is $1-0.25=0.75$. Since the four shots are independent and have the same hit probability, find the probability of missing all four and subtract it from 1.

Do. The probability of four misses is the product of four miss probabilities:

P(\text{four misses})=(0.75)^4=0.31640625

As a check, $0.75^2=0.5625$, and $0.5625^2=0.31640625$. Therefore,

P(\text{at least one hit}) =1-(0.75)^4 =1-0.31640625 =0.68359375 \approx 0.6836

Conclude. Under the stated assumptions, the probability that the shooter hits the target at least once in four shots is about $0.6836$, or $68.36\%$.

Worked Example: A Machine Checks for Misalignment

In a hypothetical production process, a machine checks an item for misalignment at each pass. Suppose the probability that one pass correctly detects a misaligned item is $0.12$. Assume eight passes are independent and have the same detection probability. What is the probability that the machine detects the misalignment at least once?

State. The event of interest is at least one successful detection in eight passes.

Plan. Let success mean detecting the misalignment. On one pass, the probability of no detection is $1-0.12=0.88$. Under the independence and same-probability assumptions, calculate the probability of no detection on all eight passes, then use its complement.

Do. The probability of no detection on all eight passes is:

P(\text{no detection})=(0.88)^8

To verify the power, $0.88^2=0.7744$, $0.88^4=0.7744^2=0.59969536$, and $0.88^8=0.59969536^2=0.3596345248$, rounded. Thus,

P(\text{at least one detection}) =1-(0.88)^8 =1-0.3596345248 =0.6403654752 \approx 0.6404

Conclude. Under the model assumptions, the probability that the machine detects the misalignment at least once in eight passes is about $0.6404$, or $64.04\%$.

Worked Example: A Sensor Tests for a Signal

A hypothetical sensor has a $0.40$ probability of detecting a weak signal on each test. The sensor runs five independent tests, with the same detection probability each time. Find the probability that it detects the signal at least once.

State. We want the probability of one or more detections in five tests.

Plan. The probability of no detection on one test is $1-0.40=0.60$. Because the five tests are independent and have the same probability of detection, use the complement of five failures.

Do. The probability of no detection in all five tests is:

P(\text{no detection})=(0.60)^5=0.07776

For a check, $0.60^2=0.36$, $0.60^4=0.1296$, and $0.60^5=0.1296(0.60)=0.07776$. Therefore,

P(\text{at least one detection}) =1-(0.60)^5 =1-0.07776 =0.92224 \approx 0.9222

Conclude. Under the stated assumptions, the probability that the sensor detects the signal at least once in five tests is about $0.9222$, or $92.22\%$.

Common Mistakes and AP Exam Tips

Using $np$ as the probability of at least one success. The success events can overlap when successes happen on multiple attempts. Use $1-(1-p)^n$, which correctly includes all ways to have one or more successes.
Forgetting to use the failure probability. In the complement, each attempt must fail. If the success probability is $p$, the failure probability is $1-p$, not $p$.
Calculating “no successes” and stopping. The question asks for at least one success, so subtract the probability of no successes from 1.
Using the formula without checking independence. Say why the attempt outcomes can be treated as independent in context. Repeated attempts alone do not guarantee independence.
Assuming $p$ changes meaning partway through. In this formula, $p$ is the probability of success on one attempt, not the probability of at least one success across all attempts.
Rounding too early. Keep the power and subtraction unrounded until the end, then report a suitably rounded probability.

AP Exam Tip: A complete explanation names the success, identifies $p$ and $n$, justifies the independence and same-probability assumptions, shows $1-(1-p)^n$, and interprets the result in context. Do not report only a calculator value.

Key Takeaway

For repeated independent attempts with the same success probability, the complement of “at least one success” is “no successes.” Find the chance that every attempt fails, then subtract that probability from 1.

Key takeaway: If each of $n$ independent attempts has probability $p$ of success, then $P(\text{at least one success})=1-(1-p)^n$. Check that the attempts are independent and share the same success probability before using the formula.

Check Your Understanding

For each question, identify the probability of failure on one attempt and show how the complement is used.

A hypothetical archer has a $0.30$ probability of hitting a target on each of three independent shots. Find the probability of at least one hit.
A machine has a $0.10$ probability of detecting a fault on each of six independent checks. Find the probability of no detections, then the probability of at least one detection.
A student says that if the probability of success is $0.20$ on each of five attempts, the probability of at least one success is $5(0.20)=1$. Explain the error and find the correct probability.
Give one reason repeated attempts by the same shooter might not be independent.
In $1-(1-p)^n$, explain what $p$, $1-p$, and $n$ represent in context.