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
- 확률
- 0.246094
- Exact fraction
- 63/256
- Mean E[X]
- 5
- Variance
- 2.5
검증 출처: C(10,5)/2¹⁰ = 252/1024 = 63/256 (Python fractions)
Binomial distribution calculator: exact P(X = k), P(X ≤ k), tails and ranges for binomial, Poisson, geometric, negative binomial and hypergeometric.
업데이트 검증한 예제: 12
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).
Binomial(n = 20, p = 0.3)
Exact rational arithmetic (BigInt numerator and denominator), converted to 50 significant digits at the end.
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.
검증 출처: C(10,5)/2¹⁰ = 252/1024 = 63/256 (Python fractions)
검증 출처: Python fractions: Σ_{k≤5} C(20,k)(3/10)^k(7/10)^(20−k) = 0.4163708291
검증 출처: 4.5·e^(−3) = 0.2240418077 (Python math.exp)
검증 출처: 1 − e^(−3)(1 + 3 + 4.5 + 4.5 + 3.375), Python math.exp
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.
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 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.
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.
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.
정확도는 입력값과 계산 방법의 가정에 따라 달라집니다. 십진 연산은 유효숫자 50자리를 사용하지만, 추정값·수치해석 방법·원본 데이터의 정밀도는 더 낮을 수 있습니다. 표시값을 반올림해도 이러한 한계는 사라지지 않습니다. 독립적인 출처의 풀이와 대조한 계산 예시: 12. 예를 들어 “Binomial n = 10, p = 0.5, P(X = 5)”은 C(10,5)/2¹⁰ = 252/1024 = 63/256 (Python fractions)와 대조해 확인합니다.
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).
이 계산기에는 독립적인 출처에서 답을 얻은 계산 예제가 12개 있습니다. 테스트 모음에서 실행되며 여기에서도 실행할 수 있습니다.
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