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When do the binomial and Poisson distributions apply, and how do we compute their probabilities?

Model situations with the binomial and Poisson distributions, state the conditions for each, and compute probabilities, means and variances

A focused answer to the H2 Mathematics outcome on the binomial and Poisson distributions. The conditions for each model, their probability functions, means and variances, and choosing the right model.

Generated by Claude Opus 4.810 min answer

Reviewed by: AI editorial process; not yet individually human-reviewed

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  1. What this dot point is asking
  2. The answer
  3. Examples in context
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What this dot point is asking

SEAB wants you to recognise when the binomial and Poisson distributions apply, state the conditions for each, and compute probabilities, means and variances, as well as choose the correct model for a given situation.

The answer

The binomial distribution

XB(n,p)X \sim \mathrm{B}(n, p) models the number of successes in nn independent trials, each with constant success probability pp. The conditions: a fixed number of trials, two outcomes per trial, independent trials, and constant pp. The probability function is

P(X=r)=(nr)pr(1p)nr,r=0,1,,n.\mathrm{P}(X = r) = \binom{n}{r}p^r(1 - p)^{n - r}, \qquad r = 0, 1, \ldots, n.

Its mean and variance are E(X)=np\mathrm{E}(X) = np and Var(X)=np(1p)\operatorname{Var}(X) = np(1 - p).

The Poisson distribution

XPo(λ)X \sim \mathrm{Po}(\lambda) models the number of events in a fixed interval when events occur independently at a constant average rate λ\lambda, with no fixed upper limit. The probability function is

P(X=r)=eλλrr!,r=0,1,2,\mathrm{P}(X = r) = \frac{e^{-\lambda}\lambda^r}{r!}, \qquad r = 0, 1, 2, \ldots

A defining feature: the mean and variance are equal, E(X)=Var(X)=λ\mathrm{E}(X) = \operatorname{Var}(X) = \lambda.

Choosing the model

  • Use the binomial when there is a fixed number of trials and you count successes.
  • Use the Poisson when you count occurrences over an interval of time or space with a known average rate and no natural maximum.

The Poisson is also the limit of the binomial when nn is large and pp small with np=λnp = \lambda moderate.

Combining Poisson variables

If XPo(λ1)X \sim \mathrm{Po}(\lambda_1) and YPo(λ2)Y \sim \mathrm{Po}(\lambda_2) are independent, then X+YPo(λ1+λ2)X + Y \sim \mathrm{Po}(\lambda_1 + \lambda_2): rates over combined intervals add.

Examples in context

Example 1. Exam multiple choice. The number correct by pure guessing on 3030 four-option questions is B(30,0.25)\mathrm{B}(30, 0.25), with expected score 30×0.25=7.530 \times 0.25 = 7.5, the baseline against which real performance is judged.

Example 2. Rare faults on a cable. Flaws occurring randomly along a cable at 0.20.2 per metre follow a Poisson model; the probability of a flawless 1010 m length uses λ=2\lambda = 2 and P(X=0)=e2\mathrm{P}(X = 0) = e^{-2}.

Try this

Q1. State the mean and variance of B(50,0.2)\mathrm{B}(50, 0.2). [2 marks]

  • Cue. Mean =10= 10, variance =50(0.2)(0.8)=8= 50(0.2)(0.8) = 8.

Q2. For XPo(4)X \sim \mathrm{Po}(4), find P(X=0)\mathrm{P}(X = 0). [2 marks]

  • Cue. e40.0183e^{-4} \approx 0.0183.

Q3. State one condition distinguishing when to use the Poisson rather than the binomial. [1 mark]

  • Cue. Events occur over an interval with a known rate and no fixed maximum number of trials.

Exam-style practice questions

Practice questions written in the style of SEAB exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Original4 marksA fair coin is tossed 88 times. Find the probability of obtaining exactly 33 heads.
Show worked answer →

Let XX be the number of heads, XB(8,0.5)X \sim \mathrm{B}(8, 0.5).

P(X=3)=(83)(0.5)3(0.5)5=(83)(0.5)8=56×1256=56256=7320.219\mathrm{P}(X = 3) = \binom{8}{3}(0.5)^3 (0.5)^5 = \binom{8}{3}(0.5)^8 = 56 \times \dfrac{1}{256} = \dfrac{56}{256} = \dfrac{7}{32} \approx 0.219.

Markers reward identifying the binomial model, the probability formula with the binomial coefficient, and the numerical value.

Original4 marksCalls arrive at a switchboard at an average rate of 33 per minute, following a Poisson distribution. Find the probability of exactly 22 calls in a given minute.
Show worked answer →

Let XX be the number of calls per minute, XPo(3)X \sim \mathrm{Po}(3).

P(X=2)=e3322!=e3×92=4.5e34.5×0.04980.224\mathrm{P}(X = 2) = \dfrac{e^{-3} 3^2}{2!} = \dfrac{e^{-3} \times 9}{2} = 4.5 e^{-3} \approx 4.5 \times 0.0498 \approx 0.224.

Markers reward identifying the Poisson model with λ=3\lambda = 3, the Poisson formula, and the numerical probability.

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