# Binomial & Poisson distribution calculator

> Binomial distribution calculator: exact P(X = k), P(X ≤ k), tails and ranges for binomial, Poisson, geometric, negative binomial and hypergeometric.

Etkileşimli sürüm: https://www.calcopenly.com/tr/statistics/binomial-poisson-distribution-calculator
Konu: İstatistik ve olasılık hesaplayıcıları

A discrete distribution gives the probability of each whole-number count. The binomial counts successes in n independent trials with the same success probability p: P(X = k) = C(n, k)·pᵏ·(1 − p)ⁿ⁻ᵏ. The Poisson counts events at an average rate λ, the geometric and negative binomial count trials until the first or r-th success, and the hypergeometric counts successes drawn without replacement. Cumulative and range probabilities add the individual terms with exact fraction arithmetic, switching to the incomplete beta or gamma function when the fractions grow too large.

The default asks how likely 5 or fewer successes are in 20 trials with p = 0.3: 41.64%, against a mean of 6. With λ = 3 calls an hour, the Poisson gives a 22.4% chance of exactly 2; the hypergeometric gives a 34.1% chance of at least one ace in a 5-card hand.

Each model assumes independent events with a constant probability or rate. Events that cluster, such as defects arriving in batches, vary more than the model predicts.

## Girdiler

- **Distribution** (seçenekler: Binomial, Poisson, Geometric, Negative binomial, Hypergeometric)
- **Number of trials n**
- **Probability of success p**: As a decimal between 0 and 1; 30% also works.
- **Mean number of events λ**
- **Successes needed r**
- **X counts** (seçenekler: Trials, including the success, Failures before the success)
- **Population size N**
- **Successes in the population K**
- **Number drawn n**
- **Find** (seçenekler: P(X = k), P(X ≤ k), P(X < k), P(X ≥ k), P(X > k), P(a ≤ X ≤ b))
- **Value k**
- **From a**
- **To b**

## Sonuçlar

- Olasılık — ana sonuç
- Exact fraction
- As a percentage
- Mean E[X]
- Variance
- Standard deviation

## Formül

$$
\begin{gathered}\text{Binomial: } \binom{n}{k}p^k(1-p)^{n-k}\qquad \text{Poisson: } \frac{e^{-\lambda}\lambda^k}{k!}\\ \text{Hypergeometric: } \frac{\binom{K}{k}\binom{N-K}{n-k}}{\binom{N}{n}}\end{gathered}
$$

## Çözümlü örnekler

### Binomial n = 10, p = 0.5, P(X = 5)

- Distribution: Binomial
- Number of trials n: 10
- Probability of success p: 0.5
- Find: P(X = k)
- Value k: 5
- **Olasılık: 0.246094**
- **Exact fraction: 63/256**
- **Mean E[X]: 5**
- **Variance: 2.5**
- Doğrulama kaynağı: C(10,5)/2¹⁰ = 252/1024 = 63/256 (Python fractions)

### Binomial n = 20, p = 0.3, P(X ≤ 5) (defaults)

- Distribution: Binomial
- Number of trials n: 20
- Probability of success p: 0.3
- Find: P(X ≤ k)
- Value k: 5
- **Olasılık: 0.416371**
- **Mean E[X]: 6**
- **Variance: 4.2**
- Doğrulama kaynağı: Python fractions: Σ_{k≤5} C(20,k)(3/10)^k(7/10)^(20−k) = 0.4163708291

### Poisson λ = 3, P(X = 2)

- Distribution: Poisson
- Mean number of events λ: 3
- Find: P(X = k)
- Value k: 2
- **Olasılık: 0.224042**
- **Variance: 3**
- Doğrulama kaynağı: 4.5·e^(−3) = 0.2240418077 (Python math.exp)

### Poisson λ = 3, P(X ≥ 5)

- Distribution: Poisson
- Mean number of events λ: 3
- Find: P(X ≥ k)
- Value k: 5
- **Olasılık: 0.184737**
- Doğrulama kaynağı: 1 − e^(−3)(1 + 3 + 4.5 + 4.5 + 3.375), Python math.exp

### Hypergeometric: at least one ace in 5 cards

- Distribution: Hypergeometric
- Population size N: 52
- Successes in the population K: 4
- Number drawn n: 5
- Find: P(X ≥ k)
- Value k: 1
- **Olasılık: 0.341158**
- **Exact fraction: 18472/54145**
- Doğrulama kaynağı: 1 − C(48,5)/C(52,5) = 1 − 1712304/2598960 (Python math.comb, fractions)

### Geometric: a six within 6 rolls

- Distribution: Geometric
- Probability of success p: 1/6
- X counts: Trials, including the success
- Find: P(X ≤ k)
- Value k: 6
- **Olasılık: 0.665102**
- **Exact fraction: 31031/46656**
- **Mean E[X]: 6**
- Doğrulama kaynağı: 1 − (5/6)⁶ = 31031/46656 (Python fractions)

## Sorular

### How do you calculate binomial probability?

Multiply the number of ways to place k successes among n trials by the chance of any one such sequence: P(X = k) = C(n, k)·pᵏ·(1 − p)ⁿ⁻ᵏ. For 5 heads in 10 fair tosses, C(10, 5) = 252 and 0.5¹⁰ = 1/1024, so P = 252/1024 = 0.2461. A cumulative probability such as P(X ≤ 5) adds the terms for k = 0 to 5.

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

For whole-number counts they differ by the single term P(X = k), so P(X < 5) = P(X ≤ 4). With n = 20 and p = 0.3, P(X ≤ 5) = 0.4164 but P(X < 5) = 0.2375. In words, 'at most 5' is ≤ 5, 'fewer than 5' is < 5, 'at least 5' is ≥ 5 and 'more than 5' is > 5.

### When can the Poisson distribution replace the binomial?

When n is large and p is small, using λ = np. A commonly cited rule is n ≥ 20 with p ≤ 0.05, or n ≥ 100 with np ≤ 10. Outside those limits the approximation drifts: for n = 20 and p = 0.3 the binomial gives P(X ≤ 5) = 0.4164, while a Poisson with λ = 6 gives 0.4457. This calculator computes the binomial exactly, so the shortcut is only needed by hand.

### When should I use the hypergeometric distribution instead of the binomial?

Use it when you draw without replacement from a finite population, so each draw changes the odds for the next. The chance of at least one ace in a 5-card hand is 34.12% by the hypergeometric (N = 52, K = 4, n = 5). Treating the draws as independent binomial trials with p = 4/52 gives 32.98%. The two agree closely only when the sample is a small fraction of the population.

### What are the mean and variance of a binomial distribution?

The mean is np and the variance np(1 − p). For 20 trials with p = 0.3 the mean is 6 successes, the variance 4.2 and the standard deviation √4.2 = 2.05. For a Poisson distribution the mean and the variance both equal λ, so counts whose variance is well above their mean are overdispersed and often fit a negative binomial better.

### “Binomial & Poisson distribution calculator” ne kadar doğru sonuç verir?

Doğruluk, girdilerinize ve yöntemin varsayımlarına bağlıdır. Ondalık aritmetik 50 anlamlı basamak kullanır; ancak tahminler, sayısal yöntemler ve kaynak veriler daha az hassas olabilir. Gösterilen değerin yuvarlanması bu sınırları ortadan kaldırmaz. Bağımsız kaynaklarla doğrulanan çözümlü örnek sayısı: 12. Örneğin “Binomial n = 10, p = 0.5, P(X = 5)”, C(10,5)/2¹⁰ = 252/1024 = 63/256 (Python fractions) ile karşılaştırılarak doğrulanır.

### Yöntemin kaynağı nedir?

NIST/SEMATECH e-Handbook of Statistical Methods, §1.3.6.6.18 Binomial distribution and §1.3.6.6.19 Poisson distribution; Johnson, Kemp & Kotz, Univariate Discrete Distributions, 3rd ed. (Wiley, 2005); Abramowitz & Stegun, Handbook of Mathematical Functions, 26.5.24 (binomial CDF as an incomplete beta) and 6.5.13 (Poisson CDF as an incomplete gamma).

## Kaynaklar

- [NIST/SEMATECH e-Handbook of Statistical Methods, §1.3.6.6.18 Binomial distribution and §1.3.6.6.19 Poisson distribution](https://www.itl.nist.gov/div898/handbook/eda/section3/eda366i.htm)
- Johnson, Kemp & Kotz, Univariate Discrete Distributions, 3rd ed. (Wiley, 2005)
- Abramowitz & Stegun, Handbook of Mathematical Functions, 26.5.24 (binomial CDF as an incomplete beta) and 6.5.13 (Poisson CDF as an incomplete gamma)