Pearson Edexcel International Advanced Level Mathematics: The Binomial and Poisson distributions and the Poisson approximation to the binomial (YMA01) – Revision Notes

These revision notes condense Unit S2 section 1 (topics 1.1 to 1.3) of Pearson’s International Advanced Subsidiary/Advanced Level Mathematics, Further Mathematics and Pure Mathematics specification (YMA01), Issue 3, April 2019. The specification states that Unit S2 is optional for IAS Further Mathematics and for IAL Mathematics and Further Mathematics. For full explanations and worked examples, read the study guide.

Test yourself with the practice questions. The course hub, the printable checklist and the diagnostics help you plan the final weeks.

Key results

Result	B(n, p)	Po(λ)
Probability	P(X = r) = (n choose r) pʳ (1 − p)ⁿ⁻ʳ	P(X = r) = e⁻λ λʳ / r!
Values of r	0, 1, …, n	0, 1, 2, … with no upper limit
Mean	np	λ
Variance	np(1 − p)	λ
Sum	no result needed	independent Po(a) + Po(b) = Po(a + b)

No derivations of the means or variances are required.

Model checklists

Binomial. All four must hold, and you must say them in context:

• fixed number of trials
• two outcomes (success or failure)
• constant probability of success
• independent trials

Poisson. Events occur

• singly
• independently
• at random
• at a constant average rate

Quick contrast. A count with a known maximum (items in a box, patients in a ward of 20) points to binomial. A count of events with no natural maximum (calls, flaws, particles) points to Poisson. A Poisson model has variance equal to its mean, while a binomial variance is below its mean.

Method: cumulative probabilities

Let F(r) = P(X ≤ r).

Request	Rewrite
P(X = r)	F(r) − F(r − 1)
P(X < r)	F(r − 1)
P(X ≥ r)	1 − F(r − 1)
P(X > r)	1 − F(r)
P(a ≤ X ≤ b)	F(b) − F(a − 1)

Worked reminder. X ~ B(10, 0.3). Find P(X ≥ 4).

P(X ≥ 4) = 1 − P(X ≤ 3) = 1 − 0.6496 = 0.3504

Keep four decimal places in given cumulative values until the last line.

Method: scaling a Poisson mean

1. Write the rate with its unit, for example 1.2 per 4 minutes.
2. Convert to the interval in the question: 1.2 ÷ 4 × 10 = 3 per 10 minutes.
3. State the distribution: X ~ Po(3).
4. Calculate, using P(X ≥ r) = 1 − P(X ≤ r − 1) when needed.

Worked reminder. X ~ Po(6). Then P(3 ≤ X ≤ 5) = F(5) − F(2) = 0.4457 − 0.0620 = 0.3837, and P(X ≥ 7) = 1 − F(6) = 0.3937.

If the interval is five times as long, the mean is five times as large: Po(λ) becomes Po(5λ). The probability is not multiplied by five.

Method: finding n, p or λ from information

• From a mean and variance of a binomial: divide the variance by the mean to get 1 − p, then n = mean ÷ p.
• From a Poisson statement such as P(X = 0) = 0.2: e⁻λ = 0.2, so λ = ln 5.
• From P(X = r) = P(X = r + 1): cancel e⁻λ and solve. For example λ²/2 = λ³/6 gives λ = 3.

Method: the Poisson approximation to B(n, p)

1. Check n is large and p is small, as your teacher’s conditions require.
2. Use Po(np).
3. Calculate with the Poisson distribution and state that the answer is an approximation.

Worked reminder. X ~ B(300, 0.01). Then np = 3, and P(X = 2) ≈ e⁻³ × 3²/2 = 0.2240. The exact value is 0.2244.

If p is near 0.5, the Poisson approximation is poor, and Po(np) has the wrong variance.

Worked reminders: mixed

Poisson then binomial. Faults occur on a cable at 0.5 per 100 m. For 400 m the mean is 2, so X ~ Po(2) and P(X = 0) = e⁻² = 0.1353. If five separate 400 m cables are tested, the number with no faults is B(5, 0.1353). The probability that exactly one of the five has no faults is 5 × 0.1353 × 0.8647⁴ = 0.378. Work out the Poisson probability first, keep it unrounded for the next step, and name the new distribution before you use it. That is the usual route when a question links the two distributions.

Reading data. Counts of text messages in 60 five-minute spells have mean 3.1 and variance 3.0. Both are close to 3, so Po(3.1) is plausible. Say what you tested and give the reason: the mean and variance are nearly equal. State any condition you cannot check from the data, such as independence.

Choosing the model. Four questions settle it:

1. Is there a fixed number of trials n? If yes, think binomial.
2. Is the count a number of events in an interval with no maximum? If yes, think Poisson.
3. Is the probability or rate constant throughout?
4. Are the trials or events independent?

Model comments: templates

• Binomial fails: “The trials are not independent because [one trial changes the probability of the next], so B(n, p) is not suitable.”
• Binomial fails: “The probability is not constant because [it differs between trials], so B(n, p) is not suitable.”
• Poisson fails: “The rate is not constant because [it changes with the time of day or location], so Po(λ) is not suitable.”
• Poisson fails: “Events occur in clusters, not singly or independently, so Po(λ) is not suitable.”
• Model holds: name each condition and give a reason it is reasonable in this context.

Fill the square brackets with the context in the question. A comment that could be pasted into any question earns little.

Must-know distinctions

• Binomial: fixed n, p constant. Poisson: no fixed n, rate constant.
• Binomial variance np(1 − p) is less than the mean. Poisson variance equals the mean.
• Scaling λ changes the mean. Scaling a probability is never valid.
• 1 − F(r − 1) includes r. 1 − F(r) does not.
• An approximation is stated as “≈” and uses Po(np), not B(n, p).
• “Comment on the model” needs a condition and a reason from the context.

Quick self-test

1. X ~ B(10, 0.3). Write down E(X) and Var(X).
2. X ~ Po(4.2). Write down E(X) and Var(X).
3. X ~ B(6, 0.5). Find P(X = 2).
4. X ~ Po(2.5). Find P(X = 0).
5. Calls arrive at 0.9 per minute on average. State the distribution of the number in 5 minutes.
6. X ~ B(15, 0.4) and F(r) = P(X ≤ r). Write P(X ≥ 6) in terms of F.
7. X ~ B(n, p) has mean 6 and variance 4.2. Find n and p.
8. X ~ B(300, 0.01). Use a Poisson approximation to estimate P(X = 2).
9. A and B are independent, A ~ Po(2) and B ~ Po(3). Find P(A + B = 1).
10. A sample of counts has mean 5.1 and variance 5.0. State, with a reason, what this suggests about a Poisson model.
11. X ~ B(12, 0.25). Find P(X ≥ 1).
12. State why the number of faulty items in a box of 10 is better modelled by a binomial than a Poisson distribution.

Answers

1. E(X) = np = 3, Var(X) = np(1 − p) = 2.1.
2. E(X) = 4.2 and Var(X) = 4.2.
3. 15 × 0.5² × 0.5⁴ = 15/64 = 0.2344.
4. e⁻²·⁵ = 0.0821.
5. Mean 0.9 × 5 = 4.5, so Po(4.5).
6. P(X ≥ 6) = 1 − F(5).
7. 1 − p = 4.2 ÷ 6 = 0.7, so p = 0.3 and n = 6 ÷ 0.3 = 20.
8. np = 3, so P(X = 2) ≈ e⁻³ × 3²/2 = 0.224.
9. A + B ~ Po(5), so P = 5e⁻⁵ = 0.0337.
10. Mean and variance are almost equal, which is consistent with a Poisson model. It does not prove the model fits.
11. 1 − 0.75¹² = 0.968.
12. The number of items is fixed at 10 and cannot exceed 10, but a Poisson variable has no upper limit.

Where marks are usually lost

• Using F(r) when F(r − 1) is needed, especially for “at least” and for ranges.
• Not rescaling the Poisson mean, or rescaling by the wrong factor.
• Leaving a model comment generic (“it is random”) without naming the condition and the context.
• Applying the Poisson approximation without checking that n is large and p is small.
• Using the Poisson variance rule for a binomial variable, or the reverse.
• Rounding table values early and losing accuracy in a difference.
• Forgetting that a Poisson variable starts at 0, so P(X ≥ 1) = 1 − P(X = 0).
• Not stating the distribution (for example Po(4.5)) before calculating.

Official syllabus

Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and Pure Mathematics, Specification, Issue 3, April 2019 (Pearson Education Limited). This page covers Unit S2, section 1.