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

This study guide is written against Pearson’s International Advanced Subsidiary/Advanced Level specification for Mathematics, Further Mathematics and Pure Mathematics (YMA01), Issue 3, April 2019. It covers Unit S2 section 1 (topics 1.1 to 1.3). The specification states that Unit S2 is optional for IAS Further Mathematics and optional for IAL Mathematics and Further Mathematics.

Use it with the revision notes and the practice questions. The course hub and the printable checklist show where the unit sits, and the diagnostics help you find gaps first. The specification assumes Unit S1 and binomial coefficients, so the S1 probability guide is useful background.

What this unit covers

Topic	You must be able to
1.1	Use the binomial and Poisson distributions to model real situations and comment critically on whether they suit; find cumulative probabilities by calculation or from tables; use the additive property, so Po(λ) per unit becomes Po(5λ) over five units
1.2	Use the mean and variance of the binomial and Poisson distributions (no derivations are required)
1.3	Use the Poisson distribution as an approximation to the binomial distribution

The specification says calculators may be used in the S2 assessment and that the booklet Mathematical Formulae and Statistical Tables is provided. Questions may ask for a cumulative probability “by calculation or by reference to tables”, so practise both. This page gives table-style values as data in the question and does not rely on any particular table.

The binomial distribution B(n, p)

Use B(n, p) when all of these hold:

• there is a fixed number n of trials
• each trial has two outcomes, success or failure
• the probability of success p is the same every trial
• the trials are independent

Then X, the number of successes, has

P(X = r) = (n choose r) pʳ (1 − p)ⁿ⁻ʳ, for r = 0, 1, …, n

where (n choose r) = n! / (r!(n − r)!). Write X ~ B(n, p).

Worked example 1. A seed supplier claims each seed germinates with probability 0.85, independently. Twelve seeds are sown. The number that germinate is X ~ B(12, 0.85).

Find P(X = 10) and P(X ≥ 10).

P(X = 10) = (12 choose 10) × 0.85¹⁰ × 0.15²
          = 66 × 0.85¹⁰ × 0.0225
          = 0.2924
P(X = 11) = 12 × 0.85¹¹ × 0.15 = 0.3012
P(X = 12) = 0.85¹² = 0.1422
P(X ≥ 10) = 0.2924 + 0.3012 + 0.1422 = 0.7358

So P(X = 10) = 0.292 and P(X ≥ 10) = 0.736 (3 s.f.). Adding three terms was shorter than finding P(X ≤ 9) and subtracting, so choose the shorter route.

Cumulative probabilities

A table or calculator function gives F(r) = P(X ≤ r). Convert every request into this form:

You want	Write it as
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 example 2. X ~ B(20, 0.3). Given F(8) = 0.8867 and F(3) = 0.1071, find P(4 ≤ X ≤ 8).

P(4 ≤ X ≤ 8) = F(8) − F(3)
             = 0.8867 − 0.1071
             = 0.7796

The lower limit 4 gives F(3), not F(4). This is the most common slip with cumulative probabilities.

The Poisson distribution Po(λ)

A variable X ~ Po(λ) counts events in a fixed stretch of time, length or area, provided the events occur

• singly (never two at exactly the same instant)
• independently of each other
• at random, at a constant average rate

with P(X = r) = e⁻λ λʳ / r!, for r = 0, 1, 2, …. There is no upper limit on r.

Worked example 3. Flaws in a roll of fabric occur at an average rate of 0.8 per square metre. Find the probability that in 2.5 m² there are (a) no flaws and (b) at least two flaws.

The mean for 2.5 m² is 0.8 × 2.5 = 2, so X ~ Po(2).

(a) P(X = 0) = e⁻² = 0.1353
(b) P(X ≥ 2) = 1 − P(X = 0) − P(X = 1)
             = 1 − e⁻² − 2e⁻²
             = 1 − 3e⁻²
             = 0.5940

So the answers are 0.135 and 0.594 (3 s.f.).

Scaling the mean

If the number of events per unit has mean λ and the counts in separate units are independent, the count over t units has mean tλ, so Po(tλ). This is the additive property in the specification: Po(λ) per minute gives Po(5λ) per five minutes. It works down as well as up.

Worked example 4. Emails arrive at an average of 1.5 per 10 minutes. Find the probability that (a) at least 10 arrive in one hour and (b) at most 3 arrive in 20 minutes.

(a) One hour is six lots of 10 minutes: mean 6 × 1.5 = 9.
    Y ~ Po(9).  P(Y ≥ 10) = 1 − P(Y ≤ 9) = 1 − 0.5874 = 0.4126
(b) 20 minutes: mean 2 × 1.5 = 3.
    W ~ Po(3).  P(W ≤ 3) = e⁻³(1 + 3 + 4.5 + 4.5) = 13e⁻³ = 0.6472

The answers are 0.413 and 0.647 (3 s.f.). Always state the new distribution before you calculate. The sum of independent Poisson variables is also Poisson: if A ~ Po(a) and B ~ Po(b) are independent, A + B ~ Po(a + b).

Mean and variance (topic 1.2)

Distribution	Mean	Variance
B(n, p)	np	np(1 − p)
Po(λ)	λ	λ

No derivations are required. Know them, and use them in both directions.

Worked example 5. A binomial random variable has mean 15 and variance 10.5. Find n and p.

np = 15 and np(1 − p) = 10.5
Divide: 1 − p = 10.5 / 15 = 0.7, so p = 0.3
Then n = 15 / 0.3 = 50

So n = 50 and p = 0.3.

A binomial variance is always smaller than its mean, because 1 − p < 1. A Poisson variance equals its mean. This gives a quick test of a Poisson model: if a data set of counts has mean 2.3 and variance 2.2, a Poisson model looks plausible. If the mean is 2.6 and the variance is 5.9, the counts are too spread out, which suggests events cluster or the rate varies.

Modelling and commenting critically (topic 1.1)

A model answer names the condition and ties it to the context. Compare:

Situation	Model	Comment
Number of 15 plants in one greenhouse that get a disease	B(15, p) is doubtful	Disease spreads, so trials are not independent
Number of customers entering a shop each minute all day	Po(λ) is doubtful	The rate is higher at lunchtime, so it is not constant
Number of heads in 40 fair coin tosses	B(40, 0.5)	Fixed number of independent trials, constant p
Number of typing errors in a 500-word passage	Po(λ) is reasonable	Errors are rare, independent and at a steady rate, with no fixed maximum

A phrase such as “it is not independent” earns little. Say which events depend on which, and why.

The Poisson approximation to the binomial (topic 1.3)

When n is large and p is small, B(n, p) behaves like Po(np). The two have mean np, and the variance np(1 − p) is close to np when p is small. The specification lists the topic without numerical limits, so follow the conditions your teacher gives, which are typically large n and small p.

Worked example 6. X ~ B(150, 0.02). Estimate P(X ≤ 2) with a Poisson approximation, then compare it with the exact value.

np = 150 × 0.02 = 3, so use Y ~ Po(3)
P(Y ≤ 2) = e⁻³(1 + 3 + 4.5) = 8.5e⁻³ = 0.4232
Exact: 0.98¹⁵⁰ + 150 × 0.02 × 0.98¹⁴⁹ + 11175 × 0.02² × 0.98¹⁴⁸ = 0.4209

The approximation 0.423 is close to the exact 0.421. When p is not small the approximation fails. For X ~ B(20, 0.4), P(X ≤ 5) = 0.1256, but Po(8) gives 0.1912.

Common errors

• Writing P(X ≥ r) as 1 − F(r) instead of 1 − F(r − 1).
• Forgetting to rescale λ when the interval changes, or rescaling a probability instead of the mean.
• Using Po(λ) for a count with an upper limit (for example, defective items in a box of 10), when binomial fits.
• Using the Poisson approximation when p is not small.
• Saying “the events are not random” without explaining which condition fails.
• Giving mean and variance for B(n, p) as np and np(1 + p).

Next steps

Try the practice questions, then condense the unit with the revision notes.

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.