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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.

آخر تحديث أمثلة تم التحقق منها: 12

As a decimal between 0 and 1; 30% also works.
جرّب
الاحتمالات
الاحتمالات: 0.416371
الحد الأقصى للمنازل العشرية: 6؛ إلى الأقرب، وعند التعادل بعيدًا عن الصفر
As a percentage
41.6371%
Mean E[X]
6
Variance
4.2
Standard deviation
2.04939

For Binomial(n = 20, p = 0.3), P(X ≤ 5) = 0.416371: a 41.6371% chance, about 1 in 2.4. On average X is 6, typically within 2.0494 of that (one standard deviation).

P(X = k), with P(X ≤ 5) highlighted

0: 01: 12: 23: 34: 45: 56: 0.19267: 78: 89: 910: 1011: 1112: 1213: 1314: 1415: 1516: 16
طريقة الحساب S

  1. Distribution

    P(X=k)=(20k) 0.3k 0.720−kP(X = k) = \binom{20}{k}\,0.3^{k}\,0.7^{20 - k}

    Binomial(n = 20, p = 0.3)

  2. Outcomes counted

    P(X≤5)=∑k=05P(X=k)P(X \le 5) = \sum_{k=0}^{5} P(X = k)

  3. النتيجة

    P(X≤5)=0.416370829447P(X \le 5) = 0.416370829447

    Exact rational arithmetic (BigInt numerator and denominator), converted to 50 significant digits at the end.

  4. Mean and variance

    E[X]=np=6,Var⁡(X)=np(1−p)=4.2E[X] = np = 6,\quad \operatorname{Var}(X) = np(1-p) = 4.2

حول Binomial & Poisson distribution calculator

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.

أمثلة محلولة

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
الاحتمالات
0.246094
Exact fraction
63/256
Mean E[X]
5
Variance
2.5

مصدر التحقق: ⁨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
الاحتمالات
0.416371
Mean E[X]
6
Variance
4.2

مصدر التحقق: ⁨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
الاحتمالات
0.224042
Variance
3

مصدر التحقق: ⁨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
الاحتمالات
0.184737

مصدر التحقق: ⁨1 − e^(−3)(1 + 3 + 4.5 + 4.5 + 3.375), Python math.exp⁩

الأسئلة

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.

ما مدى دقة «⁨Binomial & Poisson distribution calculator⁩»؟

تعتمد الدقة على مدخلاتك وافتراضات الطريقة. يستخدم الحساب العشري 50 رقمًا معنويًا، لكن التقديرات والأساليب العددية وبيانات المصدر قد تكون أقل دقة؛ تقريب القيم المعروضة لا يزيل هذه الحدود. أمثلة محلولة جرى التحقق منها بمصادر مستقلة: 12. مثلًا، يجري التحقق من «⁨Binomial n = 10, p = 0.5, P(X = 5)⁩» بالرجوع إلى ⁨C(10,5)/2¹⁰ = 252/1024 = 63/256 (Python fractions)⁩.

ما مصدر هذه الطريقة؟

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).

حول هذه الحاسبة

Binomial: (nk)pk(1−p)n−kPoisson: e−λλkk!Hypergeometric: (Kk)(N−Kn−k)(Nn)\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}

المصادر

  1. NIST/SEMATECH e-Handbook of Statistical Methods, §1.3.6.6.18 Binomial distribution and §1.3.6.6.19 Poisson distribution
  2. Johnson, Kemp & Kotz, Univariate Discrete Distributions, 3rd ed. (Wiley, 2005)
  3. 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)

تم التحقق بالرجوع إلى المصادر

تتضمن هذه الحاسبة أمثلة محلولة بإجابات من مصادر مستقلة، وعددها 12. تُشغّل ضمن مجموعة الاختبارات، ويمكنك تشغيلها هنا أيضًا.

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