Binomial & Poisson distribution calculator

How to calculate binomial probability

For n independent trials that each succeed with probability p, the chance of exactly k successes is P(X = k) = C(n, k) × pᵏ × (1 − p)ⁿ⁻ᵏ (NIST/SEMATECH e-Handbook §1.3.6.6.18). For 5 heads in 10 fair tosses, C(10, 5) = 252 sequences each have probability 0.5¹⁰ = 1/1024, so P = 252/1024 = 0.2461.

A cumulative probability adds the terms. For 20 trials with p = 0.3, P(X ≤ 5) is the sum for k = 0 through 5:

• P(X = 0) = 0.000798
• P(X = 1) = 0.006839
• P(X = 2) = 0.027846
• P(X = 3) = 0.071604
• P(X = 4) = 0.130421
• P(X = 5) = 0.178863

The total is 0.416371, about a 41.6% chance. This distribution has mean np = 6, variance np(1 − p) = 4.2 and standard deviation √4.2 = 2.05.

Which distribution fits your count

• Binomial: a fixed number of independent trials with the same success probability, such as successes in 20 free throws.
• Poisson: events in a fixed stretch of time or space at a steady average rate, with no set maximum, such as calls to a help desk in an hour.
• Geometric: trials until the first success, such as rolls until the first six.
• Negative binomial: trials until the r-th success, such as tosses until the third head.
• Hypergeometric: successes when drawing without replacement from a finite group, such as aces in a five-card hand.

At most, fewer than, at least, more than

For counts, the wording decides whether k itself is included. With 20 trials and p = 0.3:

• At most 5: P(X ≤ 5) = 0.4164
• Fewer than 5: P(X < 5) = P(X ≤ 4) = 0.2375
• At least 5: P(X ≥ 5) = 1 − P(X ≤ 4) = 0.7625
• More than 5: P(X > 5) = 1 − P(X ≤ 5) = 0.5836

Complements pair ≤ with >, and < with ≥. Adding P(X ≤ 5) and P(X ≥ 5) gives 1.1789, because P(X = 5) = 0.1789 is counted twice. A range P(a ≤ X ≤ b) includes both endpoints.

How to calculate Poisson probability

For events arriving at an average of λ per interval, P(X = k) = e^(−λ) × λᵏ ÷ k!. A help desk averaging 3 calls an hour gets exactly 2 calls with probability e^(−3) × 3² ÷ 2 = 0.2240, and at least 5 with probability 1 − P(X ≤ 4) = 0.1847.

λ must match the interval in the question. Three calls an hour is λ = 1.5 for half an hour, so a call-free half hour has probability e^(−1.5) = 0.2231. Scaling λ this way assumes the rate holds steady across the hour.

A Poisson count has mean and variance both equal to λ. When counts vary far more than their mean, the events cluster or the rate shifts, and a negative binomial often fits better.

Poisson approximation to the binomial

With many trials and a small p, a binomial is close to a Poisson with λ = np. For 100 trials at p = 0.02, P(X = 0) is 0.98¹⁰⁰ = 0.1326 against e^(−2) = 0.1353, and P(X ≤ 2) is 0.6767 under both.

A commonly cited rule allows the swap for n ≥ 20 with p ≤ 0.05, or n ≥ 100 with np ≤ 10. Outside it the answers drift: for 20 trials at p = 0.3, P(X ≤ 5) is 0.4164 exactly but 0.4457 from a Poisson with λ = 6.

Geometric and negative binomial: waiting for a success

The geometric distribution gives the chance that the first success comes on trial k, (1 − p)ᵏ⁻¹ × p, or by trial k, 1 − (1 − p)ᵏ. Rolling a six within 6 rolls has probability 1 − (5/6)⁶ = 31031/46656 = 0.6651, and the average wait is 1/p = 6 rolls.

The negative binomial extends this to the r-th success: P(X = k) = C(k − 1, r − 1) × pʳ × (1 − p)ᵏ⁻ʳ. The third head arriving on the fifth toss has probability C(4, 2) × 0.5⁵ = 6/32 = 0.1875.

Some textbooks count failures before the success instead of trials. The same event is then X = 2 failures rather than X = 5 trials. The probability stays 0.1875, but every threshold moves by r, or by 1 for the geometric.

Hypergeometric: drawing without replacement

Drawing n items from a population of N that holds K successes gives P(X = k) = C(K, k) × C(N − K, n − k) ÷ C(N, n). The chance of at least one ace in five cards is 1 − C(48, 5)/C(52, 5) = 1 − 1,712,304/2,598,960 = 0.3412.

Treating the five cards as independent draws with p = 4/52 gives 0.3298 instead. The binomial shortcut is close only when the sample is a small fraction of the population.

Worked examples

Each row is worked in a section above. "Trial X = 5" counts trials, including the third head.

Entering probabilities and counts

• p can be written as 0.3, 30% or 3/10.
• λ is the mean count for the whole interval you are asking about, and must be above 0.
• Geometric and negative binomial counts can be trials including the success, or failures before it.

Mistakes that change the answer

• Trials that share a cause. Parts from one faulty batch, or students who studied together, tend to pass or fail together. That dependence widens the spread of the count, so the binomial understates the chance of extreme results.
• Expecting the mean to happen. Twenty trials at p = 0.3 average 6 successes, and 6 is the most likely count, yet P(X = 6) is only 0.1916.
• Setting K to the wrong card group. At least one ace needs K = 4; K = 13 is one suit, such as hearts.
• Using a per-hour λ for an interval of another length. Rescale λ first: 3 an hour is 24 in an 8-hour day.

For dice totals, coin runs and poker hands, use the dice, coin and card probability calculator. C(n, k) on its own comes from the permutations and combinations calculator.

Continuous measurements and test statistics belong in the t, chi-square and F distribution calculator.