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Poisson and geometric models

Decide when Poisson and geometric models suit a situation, and use their probabilities, means, variances and special properties.

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  • GCE A-Level H2 Further Mathematics 2027
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Enlarged diagram

Two discrete models cover many questions about random events. The Poisson distribution counts how many events occur in a fixed interval when they happen singly, independently and at a constant rate, such as calls to a help line or flaws in a cable. The geometric distribution counts how many independent trials it takes to get the first success. This lesson states the conditions for each in context, scales a Poisson mean to the interval asked about, adds independent Poisson counts, and uses the geometric tail (1-p)^x and its memoryless property.

Counting random events in an interval

A random variable X has a Poisson distribution with mean μ, written X∼Po(μ), when

P(X = x) = (e^(-μ)μ^x)/(x!), x = 0,1,2,…

It models the number of events in a fixed interval of time or space when the events occur

  • singly (two cannot happen at exactly the same instant),
  • independently of one another, and
  • at a constant average rate, so the mean is proportional to the length of the interval.

Its mean and variance are both μ. A model whose data have variance far from the mean is a warning that a condition fails.

For mean 3.00, P(X = 0) is 0.0498. The stems show the probabilities for counts 0 to 16; larger counts remain possible. Variance also equals 3.00.

P(X = 0)
0.0498
Variance = mean
3.00
Probability beyond the plot: X > 16
2.16 × 10⁻⁸
Predict how the most likely count and spread change as μ increases, then compare. A larger μ means a higher rate or a longer interval. The chart stops at 16; the model does not.

Additive property. If X∼Po(λ) and Y∼Po(μ) are independent, then X + Y∼Po(λ + μ).

Worked example 1

Calls to two help lines

Calls to help line A arrive independently at an average of 3 per hour, and calls to line B, independently of A, at 2 per hour.

(a) Find the probability that line A receives at least 2 calls in a 30-minute period.

(b) Find the probability that the two lines together receive exactly 4 calls in a 40-minute period.

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  1. Scale the mean to the interval

    Method

    In 30 minutes, A has mean 3 × 1/2 = 1.5.

    Reason

    A constant rate makes the mean proportional to the length of time.

    Working

    Answer

    With X∼Po(1.5),

    P(X ≥ 2) = 1-P(X = 0)-P(X = 1) = 1-e^(-1.5)(1 + 1.5) ≈ 0.442.

  2. Add independent Poisson variables

    Method

    In 40 minutes the means are 2 for A and 4/3 for B.

    Reason

    The lines are independent, so the total is Poisson with the summed mean.

    Working

    Answer

    T∼Po(10/3) and

    P(T = 4) = (e^(-10/3)(10/3)⁴)/(4!) ≈ 0.184.

Spot the mistake 1

When a Poisson model is not suitable

About 5 min

A student models the number of customers entering a café each minute at lunchtime as Poisson, with mean 1.2. Customers often arrive in groups of friends. Over many minutes, the counts have mean 1.2 and variance 3.1. Explain why the model is unsuitable.

Write your answer, then compare it with the solution

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  1. Check the conditions

    Method

    Customers in a group arrive together, and one arrival makes others in the same group likely.

    Reason

    This breaks both “singly” and “independently”.

    Working

    Answer

    The data agree: a Poisson variable has variance equal to its mean, but 3.1 is much larger than 1.2. Grouped arrivals spread the counts out. Counting groups arriving, rather than people, might suit a Poisson model better.

Waiting for the first success

Repeat independent trials, each a success with the same probability p. The number of trials up to and including the first success, X, has a geometric distribution, X∼Geo(p):

P(X = x) = (1-p)^(x-1)p, x = 1,2,3,…
  • The first x-1 trials fail and trial x succeeds, so P(X > x) = (1-p)^x: the first x trials all fail.
  • E(X) = 1/p and Var(X) = (1-p)/p².
  • Memoryless: P(X > m + n ∣ X > m) = ((1-p)^(m + n))/((1-p)^m) = P(X > n). Earlier failures do not make a success “due”.

It needs independent trials with a constant success probability, and counts trials up to the first success only; a fixed number of trials needs the binomial model instead.

Check your understanding 1

Rolling until a six

A fair die is rolled until a six appears. Let X be the number of rolls.

  1. Find P(X = 4) and P(X > 6).
  2. Find E(X) and Var(X).
  3. After five rolls with no six, what is the probability that at least three more rolls are needed?
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X∼Geo(1/6).

  1. P(X = 4) = (5/6)³1/6 = 125/1296 ≈ 0.0965, and P(X > 6) = (5/6)⁶ ≈ 0.335.
  2. E(X) = 6 and Var(X) = (5/6)/(1/36) = 30.
  3. “At least three more” means the first two of them fail: P(X > 7 ∣ X > 5) = P(X > 2) = (5/6)² = 25/36 ≈ 0.694.

Common mistakes

  • Using the mean for the wrong interval: a rate of 3 per hour gives a mean of 1.5 in 30 minutes.
  • Adding Poisson means for variables that are not independent.
  • Translating “more than 2” as 1-P(X ≤ 1) instead of 1-P(X ≤ 2).
  • Counting failures before the first success with the formula for trials up to and including it, which shifts every value by one.
  • Believing that after many failures a success becomes more likely; geometric waiting is memoryless.

Before you move on

Check your understanding 2

How many rolls to be fairly sure?

A fair die is rolled repeatedly. Find the least number of rolls n for which the probability of getting at least one six in the first n rolls exceeds 0.9.

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Let X∼Geo(1/6) be the roll on which the first six appears. Then P(X ≤ n) = 1-(5/6)ⁿ, so

(5/6)ⁿ < 0.1,; n > (ln 0.1)/(ln 5/6) ≈ 12.63.

Dividing by the negative ln 5/6 reverses the inequality. Check the neighbours: 1-(5/6)¹² ≈ 0.888 and 1-(5/6)¹³ ≈ 0.907. The least n is 13.

Continuous random variables continue in Density functions and exponential models.

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