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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.

अद्यतनित तपासलेली उदाहरणे: १२

As a decimal between 0 and 1; 30% also works.
वापरून पाहा
संभाव्यता
संभाव्यता: 0.416371
कमाल दशांश स्थाने: ६; जवळचे मूल्य; समान अंतरावर शून्यापासून दूर
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
गणना कशी केली जाते 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. निकाल

    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

Binomial & Poisson distribution calculator बद्दल

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.

सोडवलेली उदाहरणे

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

पडताळणीचा स्रोत: 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
संभाव्यता
0.416371
Mean E[X]
6
Variance
4.2

पडताळणीचा स्रोत: 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
संभाव्यता
0.224042
Variance
3

पडताळणीचा स्रोत: 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
संभाव्यता
0.184737

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

प्रश्न

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” किती अचूक आहे?

अचूकता तुमच्या इनपुटवर आणि पद्धतीच्या गृहीतकांवर अवलंबून असते. दशांश गणना 50 सार्थ अंक वापरते, पण अंदाज, संख्यात्मक पद्धती आणि मूळ डेटा कमी अचूक असू शकतात; दाखवलेल्या मूल्यांचे पूर्णांकन केल्याने या मर्यादा दूर होत नाहीत. स्वतंत्र स्रोतांतील सोडवलेल्या उदाहरणांशी पडताळणी: १२. उदाहरणार्थ, “Binomial n = 10, p = 0.5, P(X = 5)” ची पडताळणी C(10,5)/2¹⁰ = 252/1024 = 63/256 (Python fractions) याच्याशी केली आहे.

या पद्धतीचा स्रोत कोणता?

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).

या कॅल्क्युलेटरबद्दल

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}

स्रोत

  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)

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या कॅल्क्युलेटरमध्ये स्वतंत्र स्रोतांमधील उत्तरांसह १२ सोडवलेली उदाहरणे आहेत. ती चाचणी संचात चालतात आणि तुम्ही ती येथेही चालवू शकता.

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