Binomekundige Probity Sakrekenaar
Kry die waarskynlikheid van ksukses in geen onafhanklike beproewinge nie.
Resultate werk op soos jy tik.
Aangaande hierdie sakrekenaar
The binomial probability calculator uses the binomial distribution to model the number of successes in a fixed number of independent trials, each with the same success probability — the classic "how many heads in ten coin flips" problem. It returns the probability of exactly k successes, P(X = k) = C(n, k)·pᵏ·(1 − p)ⁿ⁻ᵏ, along with the cumulative probabilities of at most k and at least k successes.
The term C(n, k) is the number of ways to choose which k of the n trials succeed, pᵏ is the chance those k succeed and (1 − p)ⁿ⁻ᵏ the chance the rest fail. It also reports the distribution’s mean (n·p), variance (n·p·(1 − p)) and standard deviation, and shows the full probability distribution as a table and a chart.
Worked example: for n = 10 flips of a fair coin (p = 50%) the chance of exactly k = 3 heads is C(10, 3) × 0.5³ × 0.5⁷ = 120 × 0.5¹⁰ ≈ 11.7%. The expected number of heads is 10 × 0.5 = 5, with a standard deviation of √(10 × 0.5 × 0.5) ≈ 1.58.
Dikwels gevra vrae
Wanneer kan ek die houeromale verspreiding gebruik?
Gebruik dit wanneer daar'n vasgestelde aantal onafhanklike beproewinge is, elke verhoor het net twee resultate (sukses of mislukking) en die waarskynlikheid van sukses is dieselfde elke keer as ium Ã2m soos om'n muntstuk 10 keer om te skiet of foutiewe items in'n klomp te tel.
Wat is die verskil tussen Pex = k) en Penix 0°) k)?
P(X = k) is the chance of exactly k successes, while P(X ≤ k) adds up the chances of 0, 1, … up to k successes. The cumulative version answers questions like "what is the probability of at most 3 successes?"
Wat is "n keuse k"?
Cern, k), lees "n kies k," is die aantal verskillende maniere om te kies watter k van die n beproewinge die suksesse is. Vir 10 beproewinge en 3 suksesse is daar Ceb10, 3) = 120 sulke kombinasies, en elkeen dra by tot die totale waarskynlikheid.
Wat is die beteken en standaard afwyking van'n binomale verspreiding?
The mean (expected number of successes) is n·p and the variance is n·p·(1 − p), so the standard deviation is √(n·p·(1 − p)). For 10 fair coin flips the mean is 5 and the standard deviation about 1.58.
Wanneer kan ek die binomale met 'n normale verspreiding benader?
Wanneer n groot en beide n·p en n·ec1 -1 -1 p) ten minste ongeveer 10 is, word die binomial goed gegrens deur 'n normale verspreiding met dieselfde gemeen en standaard afwyking. Dit is die basis van talle groot- posle verhouding toetse.
Hoe vind ek die moontlikheid dat ek ten minste ksukses sal behaal?
P(X ≥ k) sums the probabilities of k, k+1, … up to n successes. It equals 1 − P(X ≤ k − 1), and this binomial probability calculator reports it directly alongside the "at most" value.
API ← gebruik hierdie sakrekenaar van kode
Bel hierdie sakrekenaar as 'n vry JSON-punt takies wat nodig is. Stuur die veldwaardes hieronder as navraag parameters of JSON. Enigiets wat jy weglaat gebruik dieselfde verstek hierdie bladsy is vooraf gevul met; 'n onbekende parameter is 'n 400, nooit' n stil nul. Lees die volle API docs →
Einde punt:
GET https://calculator.free/api/v1/binomial-probability/curl
curl "https://calculator.free/api/v1/binomial-probability/?n=10&k=3&p=50"JavaScript fetch()
const r = await fetch(
"https://calculator.free/api/v1/binomial-probability/?" + new URLSearchParams({
"n": "10",
"k": "3",
"p": "50"
}));
const data = await r.json();
console.log(data.results);Die resultate is skattings van algemene leiding, nie net finansiële, mediese of belastingvoorligting nie.