The complement rule: how to calculate “at least one”

The quick answer

If an event has probability p on each of n independent trials, the chance that it happens at least once is:

P(at least one) = 1 − (1 − p)ⁿ

The key word is independent: what happens on one trial must not change the probability on the next. This formula is a shortcut, not a universal rule for drawing objects without replacement.

Why subtract from one?

Every experiment falls into exactly one of two categories: either the event happens at least once, or it never happens. These outcomes cannot overlap and together cover every possibility. Their probabilities must therefore add up to one.

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

Usually “none” is easier to calculate than “one, or two, or three, and so on.” For repeated independent trials, the probability of missing every time is the product of the individual miss probabilities.

Worked example 1: two coin flips

Flip two fair coins. What is the chance of seeing at least one head?

1. One flip misses heads with probability 1/2.
2. Both flips miss heads with probability (1/2)² = 1/4.
3. Subtract from one: 1 − 1/4 = 3/4 = 75%.

You could enumerate the four equally likely sequences, but the complement method reaches the same answer in three lines.

Worked example 2: three dice rolls

A six-sided die is rolled three times. Find the probability that a six appears at least once.

On one roll, the probability of not seeing a six is 5/6. Three independent misses have probability (5/6)³ = 125/216. The desired probability is therefore:

1 − (5/6)³ = 91/216 ≈ 42.13%

Notice how the answer is substantially above the single-roll 1/6 probability while remaining below one half. More independent chances make at least one success increasingly likely.

What changes without replacement?

Suppose a bag contains two red counters and three blue counters. Draw two counters without replacement. The probability of getting at least one red counter is still one minus the probability of getting none. But the probability of missing changes after the first draw.

P(no red) = (3/5) × (2/4) = 3/10

P(at least one red) = 1 − 3/10 = 7/10

It would be wrong to use (3/5)², because the first blue draw changes the number of blue counters left. The complement idea survives; the independent-trials shortcut does not.

Three common mistakes

• Adding the same success probability repeatedly: For four trials, 4p is not generally the chance of at least one success. It double-counts outcomes with multiple successes.
• Confusing “at least one” with “exactly one”: At least one includes two, three and more successes. Exactly one requires a different calculation.
• Assuming independence: If objects are sampled without replacement, multiply the appropriate changing conditional probabilities instead.

Try it yourself

Five independent fair coins are tossed. What is the probability of getting at least one tail?

Reveal the worked answer

The only way to avoid tails is five heads, with probability (1/2)⁵ = 1/32. Thus the answer is 31/32 = 96.875%.

For formal definitions of complementary and conditional events, see OpenStax’s introductory probability reference.