How to find P(at least one) with the complement rule

Q: What is the formula for the probability of at least one?

In general, P(at least one) = 1 - P(none). When n trials are independent with the same probability p each, P(none) = (1-p)^n, so P(at least one) = 1 - (1-p)^n. If the trials have different probabilities, multiply the individual (1-p_i) terms instead.

Q: How many trials until at least one success is likely?

Solve 1 - (1-p)^n ≥ your target. With p = 0.1 and a target of 0.95, you need (0.9)^n ≤ 0.05, which gives n ≥ (ln 0.05)/(ln 0.9) ≈ 28.4, so n = 29 trials. Check it: 1 - 0.9²⁹ ≈ 0.9529, while n = 28 gives only 0.9477. The AP exam will not ask you to solve for n, but the pattern explains why rare events become near certainties over long runs.

By Jude Wallis · Published July 28, 2026

Find P(at least one) as 1 minus P(none). The complement of "at least one" is "none", not "exactly one". If n trials are independent and each produces the event with probability p, then P(at least one) = 1 - (1-p)^n.

AP Statistics: Unit 2 (topics 2.4 Introduction to Probability, 2.7 Independent Events and Unions of Events, 2.10 The Binomial Distribution). The complement rule behind P(at least one) is Unit 2 topic 2.4 of the Fall 2026 AP Statistics course, the independent-trials product and the union rule are topic 2.7, and the same subtraction returns for binomial counts in topic 2.10.

P(at least one) = 1 minus P(none)

"At least one" means one or more. There is exactly one way for that to fail: the thing never happens. So the fast route to the answer runs through the case you do not want.

P(\text{at least one}) = 1 - P(\text{none})

That is the complement rule from topic 2.4, $P(A^c) = 1 - P(A)$ , aimed at the complement that saves the most work. Adding up "exactly one, exactly two, exactly three" and onward can run to a dozen terms. "None" is always one term, no matter how many trials there are.

Students start adding cases because the word "one" is in the question. Do not follow that word. Count the ways the event can fail to happen at all, find that probability, subtract from 1, done.

The complement is "none", not "exactly one"

This is the mistake that costs the most points. Let $X$ count how many times the event happens. Then "at least one" is $X \ge 1$ , which covers $X = 1$ , $X = 2$ , $X = 3$ , and so on up to $n$ . The only outcome left outside that set is $X = 0$ . So the complement of $X \ge 1$ is $X = 0$ , and

P(X \ge 1) = 1 - P(X = 0)

Writing $1 - P(X = 1)$ instead does two wrong things at once. It throws away an outcome you wanted ( $X = 1$ ), and it keeps the one outcome you did not want ( $X = 0$ ). The numbers below in the first worked example show how far off it lands.

Translate the phrase into a boundary before you compute anything:

Phrase	In symbols	What to compute
At least one	$X \ge 1$	$1 - P(X = 0)$
At least two	$X \ge 2$	$1 - P(X \le 1)$
At least $k$	$X \ge k$	$1 - P(X \le k-1)$
None	$X = 0$	direct
Exactly one	$X = 1$	direct

Only the top row collapses to a single subtraction. "At least two" already needs two terms inside the bracket, which is why "at least one" is the phrase worth memorizing.

The shortcut 1 minus (1-p) to the n

When you repeat the same trial $n$ times and the event happens on each trial independently with the same probability $p$ , the complement has a closed form. One trial misses with probability $1-p$ . Independence lets you multiply, so $n$ misses in a row is $(1-p)$ multiplied by itself $n$ times:

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

Three conditions have to hold, and if any one fails the formula stops being exact. How far off it lands depends on how badly the condition breaks:

The number of trials $n$ is fixed before you start.
The probability $p$ is the same on every trial.
The trials are independent, so one result does not change the odds for another.

One consequence surprises people. Run a 1-in-100 event for 100 trials and you do not get a sure thing: $1 - 0.99^{100} = 1 - 0.3660 = 0.6340$ , about a 63.4% chance. As $n$ grows, $(1-p)^n$ shrinks toward $0$ but stays above it for any finite $n$ , so the probability of at least one occurrence climbs toward 1 without ever reaching it.

"At least one of A or B" is a union

The same phrase gets used for a different setup: two named events rather than repeated trials. "At least one of $A$ or $B$ happens" is the union $A \cup B$ , and you have two ways to get it.

The union rule from topic 2.7 handles it directly:

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

The complement handles it too, since "at least one of them" fails only when neither happens:

P(A \cup B) = 1 - P(\text{neither } A \text{ nor } B)

They agree. Take independent events with $P(A) = 0.4$ and $P(B) = 0.3$ . The union rule gives $0.4 + 0.3 - (0.4)(0.3) = 0.7 - 0.12 = 0.58$ . The complement gives $1 - (0.6)(0.7) = 1 - 0.42 = 0.58$ .

Prefer the complement once you have three or more events. The union rule for three events subtracts the three pairwise intersections and then adds the triple intersection back:

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

That is seven terms, four of them intersections you have to find, while $1 - P(\text{none of them})$ is still one product. See union vs intersection if the notation is the part slowing you down.

When the trials are not independent

Drawing objects without replacement breaks independence, because each draw changes what is left. The target does not change: you still want $1 - P(\text{none})$ . What changes is how you build $P(\text{none})$ .

Chain the conditional probabilities with the general multiplication rule, shrinking the pool as you go:

P(\text{none}) = P(\text{1st misses}) \times P(\text{2nd misses} \mid \text{1st missed}) \times \cdots

Both the numerator and the denominator drop by one on each draw. The second worked example runs this end to end.

There is an escape hatch the AP course uses constantly. If your sample is no more than 10% of the population, the draws are close enough to independent that $1 - (1-p)^n$ is a fine approximation. That is the 10% condition. Above 10%, the shortcut drifts, and the size of the drift grows with the fraction of the population you take. With replacement vs without replacement sets out the two multiplication rules side by side.

Mistakes that show up on every homework set

Subtracting P(exactly one). The complement of "at least one" is "none". Nothing else.
**Multiplying $n$ by $p$ , which is the same move as adding $p$ across trials.** For four rolls of a die, $4 \times \frac{1}{6}$ is the same sum as $\frac{1}{6} + \frac{1}{6} + \frac{1}{6} + \frac{1}{6} \approx 0.667$ , and it is the expected number of sixes, not a probability. Adding would only be valid if the trials were disjoint, and they are not: two sixes in four rolls is possible, so the sum double counts. With ten rolls it gives $1.667$ , and no probability exceeds 1, which is the tell.
**Using $(1-p)^n$ when $p$ changes.** If trial $i$ has its own probability $p_i$ , then $P(\text{none}) = (1-p_1)(1-p_2)\cdots(1-p_n)$ . The exponent form is just the special case where every $p_i$ matches.
Stopping at P(none). You computed the thing you did not want. Subtract from 1 before you write the answer.
Rounding the intermediate power. Round $(1-p)^n$ to two decimals and the final answer inherits that error. Carry four decimals until the last step.

AP note: where this sits and what graders want

The complement rule $P(A^c) = 1 - P(A)$ is Unit 2 topic 2.4, the independent-trials product is topic 2.7, and the same trick returns for counts in topic 2.10, where $P(X \ge 1) = 1 - P(X = 0)$ for a binomial. On a graphing calculator that is 1 - binomcdf(n, p, 0), and the binomial probability calculator takes the same three inputs.

Free response earns communication points, so name the move in words before you compute: "at least one" fails only when none occur, so I will find $P(\text{none})$ and subtract from 1. State the independence assumption too, because a grader reading $(1-p)^n$ with no justification cannot tell whether you checked it or guessed.

For more repetitions of this pattern, work through the binomial probability guide or the conditional probability practice set.

At least one six in four rolls of a die

You roll a fair six-sided die 4 times. Find the probability of getting at least one six.

Set it up. Let $X$ be the number of sixes in $n = 4$ rolls. Each roll gives a six with probability $p = \frac{1}{6}$ , the rolls do not affect each other, and $n$ is fixed, so the shortcut applies. You want $P(X \ge 1)$ .
Take the complement. The only way to miss "at least one" is $X = 0$ , so $P(X \ge 1) = 1 - P(X = 0)$ .
One roll misses with probability $1 - \frac{1}{6} = \frac{5}{6}$ .
Four independent misses: $P(X = 0) = \left(\frac{5}{6}\right)^4 = \frac{625}{1296} \approx 0.4823$ .
Subtract from 1: $P(X \ge 1) = 1 - \frac{625}{1296} = \frac{671}{1296} \approx 0.5177$ .
Check the two wrong routes. Multiplying gives $4 \times \frac{1}{6} \approx 0.6667$ , the expected count of sixes rather than a probability. Subtracting exactly one gives $P(X = 1) = \binom{4}{1}\left(\frac{1}{6}\right)\left(\frac{5}{6}\right)^3 = \frac{500}{1296} \approx 0.3858$ , so $1 - 0.3858 = 0.6142$ . Neither equals $0.5177$ .

$P(\text{at least one six}) = \frac{671}{1296} \approx 0.5177$ , about a 51.8% chance. Slightly better than a coin flip, not the 66.7% that multiplying $n$ by $p$ suggests.

At least one dead battery, drawn without replacement

A drawer holds 12 batteries, 3 of which are dead. You grab 2 at random without replacement. Find the probability that at least one of the two is dead, then check what the independent-trials shortcut would have given.

Take the complement. "At least one dead" fails only when both batteries are good, so $P(\text{at least one dead}) = 1 - P(\text{both good})$ .
First draw: 9 of the 12 batteries are good, so $P(\text{1st good}) = \frac{9}{12} = \frac{3}{4}$ .
Second draw, given the first was good: 8 good batteries remain out of 11 total, so $P(\text{2nd good} \mid \text{1st good}) = \frac{8}{11}$ .
Multiply the chain: $P(\text{both good}) = \frac{3}{4} \times \frac{8}{11} = \frac{24}{44} = \frac{6}{11} \approx 0.5455$ .
Subtract from 1: $P(\text{at least one dead}) = 1 - \frac{6}{11} = \frac{5}{11} \approx 0.4545$ .
Now the shortcut, for comparison. With $p = \frac{3}{12} = 0.25$ treated as fixed, $1 - (1 - 0.25)^2 = 1 - 0.5625 = 0.4375$ .
Judge the gap. The shortcut is off by about $0.017$ , a relative error near 3.8%, because the sample of 2 is $\frac{2}{12} \approx 16.7\%$ of the population, past the 10% condition. The draws are dependent enough that the approximation has drifted, so report the exact chained value.

$P(\text{at least one dead}) = \frac{5}{11} \approx 0.4545$ , about 45.5%. The independent shortcut returns $0.4375$ , off by about $0.017$ because 2 of 12 is 16.7% of the population, past the 10% condition, so it is an approximation rather than the exact answer here.

Frequently asked questions

Why is P(at least one) not 1 minus P(exactly one)?

Because "at least one" includes exactly one, exactly two, exactly three, and every larger count. The outcomes outside that set are just $X = 0$ . Subtracting $P(X = 1)$ removes an outcome you wanted and leaves in the one you did not ( $X = 0$ ), so the answer is wrong in both directions.

What is the formula for the probability of at least one?

In general, $P(\text{at least one}) = 1 - P(\text{none})$ . When $n$ trials are independent with the same probability $p$ each, $P(\text{none}) = (1-p)^n$ , so $P(\text{at least one}) = 1 - (1-p)^n$ . If the trials have different probabilities, multiply the individual $(1-p_i)$ terms instead.

Can I use 1 minus (1-p) to the n for sampling without replacement?

Only as an approximation, and only when the sample is at most 10% of the population. Below that threshold the probabilities barely move between draws. Above it, chain conditional probabilities instead: drawing 2 batteries from 12 gives $\frac{5}{11} \approx 0.4545$ exactly, while the shortcut returns $0.4375$ .

How many trials until at least one success is likely?

Solve $1 - (1-p)^n \ge$ your target. With $p = 0.1$ and a target of 0.95, you need $(0.9)^n \le 0.05$ , which gives $n \ge \frac{\ln 0.05}{\ln 0.9} \approx 28.4$ , so $n = 29$ trials. Check it: $1 - 0.9^{29} \approx 0.9529$ , while $n = 28$ gives only $0.9477$ . The AP exam will not ask you to solve for $n$ , but the pattern explains why rare events become near certainties over long runs.