Probability of At Least 1: Formula, Worked Example and Code

The probability of at least 1 success in n trials is found by the complement rule: take the probability that every trial fails and subtract it from 1. For independent trials that each succeed with probability p, the result is 1 − (1 − p)ⁿ. This page gives the formula, a worked example with the arithmetic shown, what to do when the trials are not identical or not independent, and how to compute it in Excel, R and Python.

Definition in one paragraph

“At least one” means one success or more: 1, 2, 3, and so on up to n. Adding all of those outcomes is slow, so the standard shortcut uses the one outcome that is left over, namely no successes at all. In OpenStax Introductory Statistics 2e, the complement A′ of an event A consists of all outcomes that are not in A, and P(A) + P(A′) = 1. If A is “at least one success”, then A′ is “zero successes”, so P(at least one) = 1 − P(zero successes).

Probability of at least one success: the formula and its symbols

For n independent trials with the same success probability p on every trial:

P(at least 1) = 1 − (1 − p)ⁿ

• p: the success probability on a single trial, a number from 0 to 1.
• 1 − p: the probability that a single trial fails (the complement of success).
• n: the number of trials.
• (1 − p)ⁿ: the probability that all n trials fail. This uses the multiplication rule for independent events, P(A and B) = P(A)P(B), which OpenStax states for events where knowing that one occurred does not change the chance of the other.

Success probability and the binomial link

The same result comes from the binomial distribution. The NIST/SEMATECH e-Handbook gives the probability mass function P(x; p, n) = C(n, x) p^x (1 − p)^{n − x} for x = 0, 1, …, n, and notes that the distribution assumes p is fixed for all trials. Setting x = 0 gives C(n, 0) = 1, so P(x = 0) = (1 − p)ⁿ. Then P(at least 1) = 1 − P(x = 0) = 1 − (1 − p)ⁿ. The x = 0 step is our own derivation from the published formula. For the “at least k” and “at most k” variants, see the binomial probability calculator guide.

Worked example with the arithmetic shown

A process succeeds with probability p = 0.2 on each of n = 5 independent attempts. What is the probability of at least one success?

1. Probability that one attempt fails: 1 − 0.2 = 0.8.
2. Probability that all five fail: 0.8⁵ = 0.8 × 0.8 × 0.8 × 0.8 × 0.8 = 0.32768.
3. Probability of at least one success: 1 − 0.32768 = 0.67232.

The answer is about 0.672, or 67.2%. A smaller case can be checked by listing outcomes. OpenStax Example 3.6 flips two fair coins and lists “at least one tail” as {HT, TH, TT}, so the probability is 3/4. The complement route agrees: only HH has no tails, so 1 − (0.5)² = 1 − 0.25 = 0.75 (the complement check is ours, not the textbook’s).

Interpretation: what the result means

The result is the chance that the run of n trials contains one or more successes. It is not the expected number of successes, which for a binomial is np (NIST). The table shows results computed from the formula above.

p per trial	n trials	1 − (1 − p)n	Reading
0.2	5	0.6723	Better than two chances in three
0.02	10	0.1829	Small per-trial chance, still under one in five over ten trials
0.04	3	0.1153	About one in nine
0.1	10	0.6513	Ten 10% chances do not add up to 100%

The probability of at least one success rises with every added trial but never reaches 1 for 0 < p < 1.

At least one in probability when p differs or trials depend

Different probabilities on each trial

If trial i has its own success probability p_i and the trials are independent, multiply the individual failure probabilities: P(at least 1) = 1 − (1 − p₁)(1 − p₂)…(1 − p_n). For p = 0.1, 0.2 and 0.3: 1 − (0.9)(0.8)(0.7) = 1 − 0.504 = 0.496.

Dependent trials, such as sampling without replacement

When one outcome changes the odds of the next, the power (1 − p)ⁿ no longer applies. The all-fail probability becomes a product of conditional probabilities. Suppose 3 of 10 items are defective and 4 are drawn without replacement. The chance that none of the four is defective is (7/10)(6/9)(5/8)(4/7) = 840/5040 = 1/6, so P(at least one defective) = 1 − 1/6 = 5/6 ≈ 0.8333. This is the zero-successes term of the hypergeometric distribution; see the hypergeometric distribution guide.

How many trials to reach 95%

To make the probability of at least one success reach a target level L, solve 1 − (1 − p)ⁿ ≥ L for n, giving n ≥ ln(1 − L) / ln(1 − p). For L = 0.95 and p = 0.05, ln(0.05) / ln(0.95) = 58.4, so n = 59 (1 − 0.95⁵⁹ = 0.9515, while n = 58 gives 0.9490). For p = 0.01, the bound is 298.07, so n = 299 (0.9505).

Assumptions and common mistakes

• Independence. OpenStax defines independent events as events where knowing that one occurred does not change the chance of the other. The formula 1 − (1 − p)ⁿ relies on this.
• Constant p. NIST notes the binomial assumes p is fixed for all trials. If p changes, use the product of 1 − p_i above.
• Adding instead of complementing. Five trials at 0.2 each do not give 5 × 0.2 = 1.0 (certainty); the correct value is 0.67232. Adding probabilities is valid only for mutually exclusive events, and successes on separate trials can occur together (see disjoint events).
• Confusing “at least one” with “exactly one”. Exactly one success is C(n, 1) p (1 − p)^{n − 1}; for p = 0.2, n = 5 that is 5 × 0.2 × 0.8⁴ = 0.4096, which is less than 0.67232 because it leaves out two or more successes.
• Using it for dependent events. Drawing without replacement from a small group needs the conditional product shown above.

For the single-trial building block of this formula, see the Bernoulli distribution guide. For how two events combine when they are not independent, see the joint probability formula guide.

How to report it (APA example)

The APA Style numbers and statistics handout (7th edition) says to use a leading zero when a statistic can exceed 1; a probability cannot exceed 1 (the handout lists proportions among such statistics), so write .67 rather than 0.67. A reporting sentence for the worked example: “With a success probability of .20 per attempt, the probability of at least one success in five independent attempts was .67.” State the assumptions (independence, constant p) in the method section, and give the formula or the software function used.

How to calculate it in Excel, R or Python

Each snippet computes 1 − P(x = 0) for p = 0.2 and n = 5 and returns 0.67232.

Excel

=1-BINOM.DIST(0,5,0.2,TRUE)

R

1 - pbinom(0, size = 5, prob = 0.2)

Python

from scipy.stats import binom
1 - binom.pmf(0, n=5, p=0.2)
# or without a library: 1 - (1 - 0.2) ** 5

For unequal probabilities, multiply the failure terms: 1 - prod(1 - p) in R, or 1 - math.prod(1 - p for p in ps) in Python.

FAQ

What is the probability of at least 1 success?

It is 1 minus the probability of zero successes. For independent trials with constant success probability p, that is 1 − (1 − p)ⁿ.

What does “at least one” mean in probability?

It means one or more occurrences of the event, so it covers 1, 2, up to n successes. Its complement is “none”.

Why not add the probabilities of each trial?

Adding works only for mutually exclusive events. Successes on different trials can happen together, so the sum double-counts and can exceed 1.

How do I find the success probability needed for a target?

Rearrange the formula: p = 1 − (1 − L)^1/n. For L = 0.95 and n = 10, p = 1 − 0.05^0.1 = 0.2589 (our calculation from the formula above).

Sources

• OpenStax, Introductory Statistics 2e, 3.1 Terminology (complement, P(A) + P(A′) = 1): openstax.org/books/introductory-statistics-2e/pages/3-1-terminology, checked 2026-10-29.
• OpenStax, Introductory Statistics 2e, 3.2 Independent and Mutually Exclusive Events (independence, multiplication rule, Example 3.6): openstax.org/books/introductory-statistics-2e/pages/3-2-independent-and-mutually-exclusive-events, checked 2026-10-29.
• NIST/SEMATECH e-Handbook of Statistical Methods, binomial distribution: itl.nist.gov/div898/handbook/eda/section3/eda366i.htm, checked 2026-10-29.
• APA Style, numbers and statistics guide (7th edition handout): apastyle.apa.org/instructional-aids/numbers-statistics-guide.pdf, checked 2026-10-29.