TR

Binomial and Poisson distributions

When to use which is explained in Chapter 5. In short: binomial when the number of trials is fixed, Poisson when only the average number of events is known.

Binomial

Example:

A coin is tossed 4 times. What is the probability of getting tails 1 or 2 times?

00.1880.375P(X = 0) = 0.06250P(X = 1) = 0.251P(X = 2) = 0.3752P(X = 3) = 0.253P(X = 4) = 0.06254Ortalama 2Number of successes (k)

Dark bars are the values included; the dashed line is the mean (2). Hover over a bar to see its probability.

Result

P(1≤X≤2)P(1 \le X \le 2) = 0.625

  1. Each trial has two outcomes (success / failure) and the success probability is the same in every trial. The probability of exactly k successes in n trials:

    P(X=k)=(nk) pk q n−k,q=1−pP(X = k) = \binom{n}{k}\, p^{k}\, q^{\,n-k}, \qquad q = 1 - p
  2. n=4,p=0.5,q=1−0.5=0.5n = 4, \quad p = 0.5, \quad q = 1 - 0.5 = 0.5
  3. P(1≤X≤2)=(41)0.51 0.53+(42)0.52 0.52P(1 \le X \le 2) = \binom{4}{1} 0.5^{1}\, 0.5^{3} + \binom{4}{2} 0.5^{2}\, 0.5^{2}
  4. =0.25+0.375=0.625= 0.25 + 0.375 = 0.625

  5. Expected (mean) number of successes n·p = 2, variance n·p·q = 1.

Poisson

Example:

On average 3 planes land at an airport every 15 minutes. What is the probability that at most 2 planes land in a given 15 minutes?

00.1120.224P(X = 0) = 0.04980P(X = 1) = 0.14941P(X = 2) = 0.2242P(X = 3) = 0.2243P(X = 4) = 0.1684P(X = 5) = 0.10085P(X = 6) = 0.05046P(X = 7) = 0.02167P(X = 8) = 0.00818P(X = 9) = 0.00279P(X = 10) = 0.000810P(X = 11) = 0.000211P(X = 12) = 0.000112P(X = 13) = 013P(X = 14) = 014P(X = 15) = 015Ortalama 3Number of events (k)

Dark bars are the values included; the dashed line is the mean (3). Hover over a bar to see its probability.

Result

P(X≤2)P(X \le 2) = 0.4232

  1. If λ events happen on average in a given time or area, the probability of exactly k events:

    P(X=k)=e−λ λkk!P(X = k) = \dfrac{e^{-\lambda}\, \lambda^{k}}{k!}
  2. λ=3,e−λ=e−3=0.049787\lambda = 3, \quad e^{-\lambda} = e^{-3} = 0.049787
  3. P(X≤2)=e−3 300!+e−3 311!+e−3 322!P(X \le 2) = \dfrac{e^{-3}\, 3^{0}}{0!} + \dfrac{e^{-3}\, 3^{1}}{1!} + \dfrac{e^{-3}\, 3^{2}}{2!}
  4. =0.0498+0.1494+0.224=0.4232= 0.0498 + 0.1494 + 0.224 = 0.4232

  5. In a Poisson distribution the mean and the variance are equal: both are λ = 3.