acalculator

Poisson distribution: how likely is k?

Enter the average number of events λ and a count k. The probabilities of exactly k, fewer, at most, more, and at least k events update as you type, with a chart of every count.

Your numbers

P(X = k)
0.100819

With an average of 3, the chance of exactly 5 events is 0.100819, and of at most 5 is 0.916082.

P(X < k)
0.815263
P(X ≤ k)
0.916082
P(X > k)
0.083918
P(X ≥ k)
0.184737
Mean (λ)
3
Variance (λ)
3
Standard deviation (√λ)
1.732051

P(X = k): 0.100819. With an average of 3, the chance of exactly 5 events is 0.100819, and of at most 5 is 0.916082.

How likely is each number of events?

How to calculate

Computes Poisson probabilities P(X = k), P(X < k), P(X ≤ k), P(X > k), and P(X ≥ k) for a mean rate λ, with the mean, variance, and standard deviation.

Example with the default inputs (Average rate (λ) 3, Number of events (k) 5): With an average of 3, the chance of exactly 5 events is 0.100819, and of at most 5 is 0.916082.

Method: P(X = k) = e^−λ × λᵏ ÷ k!; P(X ≤ k) = Σ P(X = i) for i = 0 … k; mean = variance = λ.

  • Events happen one at a time, independently, at a constant average rate λ over the interval.
  • P(X ≤ k) and P(X ≥ k) come from the regularized incomplete gamma function, which equals the sum of the terms; each tail is computed directly, not as 1 minus the other, so tiny tails keep their precision.

Machine-readable copies: Markdown, JSON.

Worked examples

Each example is checked against the calculator on every build.

  1. Average rate (λ) 3, Number of events (k) 5 gives P(X = k) 0.100819, P(X ≤ k) 0.916082, P(X < k) 0.815263, P(X > k) 0.083918, P(X ≥ k) 0.184737, Standard deviation (√λ) 1.732051.Source: NIST/SEMATECH e-Handbook of Statistical Methods, §1.3.6.6.19 Poisson Distribution. https://www.itl.nist.gov/div898/handbook/eda/section3/eda366j.htm
  2. Average rate (λ) 2, Number of events (k) 0 gives P(X = k) 0.135335, P(X < k) 0, P(X ≥ k) 1, P(X ≤ k) 0.135335.Source: NIST/SEMATECH e-Handbook of Statistical Methods, §1.3.6.6.19 Poisson Distribution. https://www.itl.nist.gov/div898/handbook/eda/section3/eda366j.htm
  3. Average rate (λ) 1, Number of events (k) 2 gives P(X = k) 0.18394, P(X ≤ k) 0.919699, P(X > k) 0.080301.Source: NIST/SEMATECH e-Handbook of Statistical Methods, §1.3.6.6.19 Poisson Distribution. https://www.itl.nist.gov/div898/handbook/eda/section3/eda366j.htm
  4. Average rate (λ) 0.5, Number of events (k) 10 gives P(X = k) 0, P(X ≥ k) 0.Source: NIST/SEMATECH e-Handbook of Statistical Methods, §1.3.6.6.19 Poisson Distribution. https://www.itl.nist.gov/div898/handbook/eda/section3/eda366j.htm
  5. Average rate (λ) 100, Number of events (k) 100 gives P(X = k) 0.039861, P(X ≤ k) 0.526562.Source: NIST/SEMATECH e-Handbook of Statistical Methods, §1.3.6.6.19 Poisson Distribution. https://www.itl.nist.gov/div898/handbook/eda/section3/eda366j.htm

How it works

For an average rate λ (more than 0) and a whole number k (0 or more), the Poisson distribution gives (NIST/SEMATECH):

  • P(X = k) = e^−λ × λᵏ ÷ k!
  • P(X ≤ k) = P(X = 0) + P(X = 1) + … + P(X = k)
  • P(X < k) = P(X ≤ k − 1), which is 0 when k = 0
  • P(X > k) = 1 − P(X ≤ k)
  • P(X ≥ k) = 1 − P(X < k), which is 1 when k = 0
  • mean = variance = λ, and standard deviation = √λ

The sums are computed with the regularized incomplete gamma functions, which equal them exactly: P(X ≤ k) = Q(k + 1, λ) and P(X ≥ k) = P(k, λ) for k ≥ 1. Each tail comes from the side of the function that is accurate for it, so a tail of 10⁻¹⁰ is not lost in rounding. P(X = k) is computed in logarithms, as exp(k × ln λ − λ − ln k!), so large λ and k do not overflow.

The chart shows P(X = j) for each count j from about λ − 6√λ to λ + 6√λ (at least 0 to 10, and always including k), with k marked.

Rules

  • λ must be more than 0 and at most 10,000.
  • k must be a whole number from 0 to 100,000.
  • Events are independent and happen at a constant average rate over the interval.
  • Results are shown to at most 6 decimal places; a result below 0.0001 shows 6 significant digits as a plain decimal, so 1.6323 × 10⁻¹⁰ shows as 0.000000000163226.

Worked examples by hand

λ = 3, k = 5. P(X = 5) = e^−3 × 3⁵ ÷ 5! = 0.0497871 × 243 ÷ 120 = 0.100819. Adding P(X = 0) to P(X = 5): e^−3 × (1 + 3 + 4.5 + 4.5 + 3.375 + 2.025) = e^−3 × 18.4 = 0.916082 for P(X ≤ 5). P(X < 5) = e^−3 × 16.375 = 0.815263, and P(X > 5) = 1 − 0.916082 = 0.083918.

λ = 2, k = 0. P(X = 0) = e^−2 = 0.135335. P(X < 0) = 0 and P(X ≥ 0) = 1.

λ = 1, k = 2. P(X = 2) = e^−1 ÷ 2 = 0.183940. P(X ≤ 2) = e^−1 × (1 + 1 + 0.5) = 2.5 e^−1 = 0.919699, so P(X > 2) = 0.080301.

λ = 0.5, k = 10. P(X = 10) = e^−0.5 × 0.5¹⁰ ÷ 10! = 1.6323 × 10⁻¹⁰. P(X ≥ 10), the sum of the terms from k = 10 up, is 1.7097 × 10⁻¹⁰.

λ = 100, k = 100. P(X = 100) = e^−100 × 100¹⁰⁰ ÷ 100! = 0.039861, and P(X ≤ 100) = 0.526562.

Other questions people ask

What is a Poisson distribution?

It gives the chance of each number of events in a fixed interval when events happen one at a time, independently, at a steady average rate λ. Examples are calls to a help desk per hour, typos per page, or cars through a toll gate per minute.

What is the Poisson formula?

P(X = k) = e^−λ × λᵏ ÷ k!. With λ = 3 and k = 5, that is e^−3 × 243 ÷ 120 = 0.1008.

What is the difference between P(X ≤ k) and P(X < k)?

P(X ≤ k) includes k itself; P(X < k) stops at k − 1. They differ by P(X = k). With λ = 3, P(X ≤ 5) = 0.9161 and P(X < 5) = 0.8153.

What are the mean and variance of a Poisson distribution?

Both equal λ, so the standard deviation is √λ. If a shop gets 3 customers a minute on average, the count per minute has mean 3 and standard deviation 1.732.

When can I use the Poisson distribution instead of the binomial?

When there are many trials, each with a small chance, the binomial with n trials and chance p is close to a Poisson with λ = n × p. A common rule of thumb is n of 20 or more and p of 0.05 or less.

Can λ be a decimal?

Yes. λ is an average, so 2.5 events per hour is fine. k must be a whole number, because you count events.

How precise are very small probabilities?

Each tail is computed directly from the incomplete gamma function, not as 1 minus a number close to 1, so a tail like P(X ≥ 10) with λ = 0.5 keeps its digits (1.7097 × 10⁻¹⁰).