Further Statistics: Poisson and Geometric Distributions
The Poisson distribution models the number of independent random events occurring in a fixed interval of time or space, when events happen at a constant average rate lambda with no fixed upper limit on the count; if X ~ Po(lambda) then P(X=x) = e^(-lambda) * lambda^x / x! for x = 0, 1, 2, ..., and E(X) = Var(X) = lambda. The geometric distribution models the number of independent trials, each with the same success probability p, needed up to and including the first success; if X ~ Geo(p) then P(X=x) = (1-p)^(x-1) * p for x = 1, 2, 3, ..., with E(X) = 1/p and Var(X) = (1-p)/p^2. A key further result is that if X ~ Po(lambda1) and Y ~ Po(lambda2) are independent, then X + Y ~ Po(lambda1 + lambda2).
Before you start
Make sure you're comfortable with these topics first:
Method
- Decide which model fits the context: Poisson for a COUNT of rare, independent events over a fixed interval of time, length or area; geometric for the NUMBER OF TRIALS up to and including the first success in repeated independent trials with constant success probability p.
- For X ~ Po(lambda), find a single probability with P(X=x) = e^(-lambda) * lambda^x / x!, and remember the mean scales with the size of the interval - doubling the length doubles lambda.
- For X ~ Geo(p), use P(X=x) = (1-p)^(x-1) * p for 'exactly on trial x', and use P(X>x) = (1-p)^x for 'more than x trials needed' (no success in the first x trials).
- Quote E(X) = lambda and Var(X) = lambda for a Poisson variable, and E(X) = 1/p and Var(X) = (1-p)/p^2 for a geometric variable, rather than summing the defining series from scratch unless a derivation is asked for.
- When two or more independent Poisson variables are added, add their means: if X ~ Po(lambda1) and Y ~ Po(lambda2) are independent, then X + Y ~ Po(lambda1 + lambda2).
- Check every probability found lies between 0 and 1, and check which inequality a cumulative probability table or complement calculation actually refers to before combining results.
Worked example
The number of flaws in a metre of fabric follows a Poisson distribution with mean 0.6 flaws per metre. (a) Find the probability that a randomly chosen metre of fabric has exactly 2 flaws. (b) Find the probability that a 3 metre length of the same fabric has at least 1 flaw.
- Let X be the number of flaws in one metre, so X ~ Po(0.6).
- Use P(X=x) = e^(-0.6) * 0.6^x / x! with x = 2: P(X=2) = e^(-0.6) * 0.6^2 / 2!.
- Simplify the constants: 0.6^2 = 0.36 and 2! = 2, so P(X=2) = 0.18 * e^(-0.6) = 0.0988 (3 s.f.).
- For a 3 metre length the mean scales up: 3 x 0.6 = 1.8, so the number of flaws Y ~ Po(1.8).
- Use the complement to find 'at least 1': P(Y>=1) = 1 - P(Y=0) = 1 - e^(-1.8).
- Final answer: P(X=2) = 0.0988 (3 s.f.); P(Y>=1) = 1 - e^(-1.8) = 0.835 (3 s.f.).
Practice questions
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Q1State the two conditions needed for the number of events in a fixed interval to be well modelled by a Poisson distribution.Show answer
Answer: The events occur independently of each other, and at a constant average rate, so the mean number of events in an interval is proportional to the length of that interval.
Q2X ~ Po(3). Find P(X=0), giving your answer to 3 significant figures.Show answer
Answer: P(X=0) = e^(-3) = 0.0498 (3 s.f.)
Q3X ~ Geo(0.25). Find P(X=3), the probability that the first success occurs on the third trial.Show answer
Answer: P(X=3) = (0.75)^2(0.25) = 9/64 = 0.140625
Q4Given X ~ Po(4), write down E(X) and Var(X).Show answer
Answer: E(X) = 4 and Var(X) = 4 (both equal to lambda for a Poisson variable)
Q5Given Y ~ Geo(0.2), find E(Y) and Var(Y).Show answer
Answer: E(Y) = 1/0.2 = 5; Var(Y) = (1-0.2)/0.2^2 = 0.8/0.04 = 20
Q6Independent random variables X ~ Po(2.5) and Y ~ Po(1.5). State the distribution of X + Y.Show answer
Answer: X + Y ~ Po(4), since 2.5 + 1.5 = 4
Q7A spinner lands on red with probability 1/6 on each independent spin. Find the probability that red first occurs on the 4th spin.Show answer
Answer: P(X=4) = (5/6)^3(1/6) = 125/1296 (approximately 0.0965)
Exam-style questions
Written in the style of a A Level Further Maths exam paper, with a full mark scheme.
Text messages arrive on Amy's phone at a constant average rate of 2.5 per hour, independently of each other. (a) Find the probability that Amy receives exactly 3 text messages in a given hour. (b) Find the probability that Amy receives at most 1 text message in a given hour.
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A biased coin has probability 0.3 of landing heads on each toss, independently of other tosses. Rav tosses the coin repeatedly until it lands heads for the first time, and X is the number of tosses needed. (a) Find P(X>4), the probability that more than 4 tosses are needed. (b) Find E(X).
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Faults on a length of cable occur independently and at a constant average rate of 0.4 faults per metre. (a) Find the probability that a randomly selected 5 metre length of cable has at least 2 faults. (b) Two independent 5 metre lengths of the cable are selected. Find the probability that the total number of faults on the two lengths combined is exactly 3.
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