OCR A Level Mathematics A: Discrete random variables and the binomial distribution (H240) – Revision Notes

These revision notes condense the full study guide for the OCR A Level Mathematics A (H240) specification, Version 3.1 (August 2026), first assessment 2018. They cover 2.04a to 2.04c (Stage 1, common with AS) and 2.04d (Stage 2, A Level only). Use the practice set afterwards, tick items off on the checklist, see the course hub and the wider Statistics revision notes. Not sure where you stand? Try the free diagnostics.

Key definitions

Term	Meaning
Random variable X	A number produced by a chance process. Capital X is the variable, lower-case x is a value.
Discrete	Takes separate values, usually whole numbers
Probability distribution	Every value x with its P(X = x), as a table or formula
Finite	A fixed list of possible values
Trial	One repetition of the experiment
Success, failure	The two outcomes of a trial. Success is the outcome you are counting.
p, q	p = P(success), q = 1 − p = P(failure)
X ~ B(n, p)	X is binomial: n trials, success probability p

Formulae and results

Result	Statement	Stage
Total probability	ΣP(X = x) = 1, and 0 ≤ P(X = x) ≤ 1	Stage 1
Binomial probability	P(X = x) = nCx pˣ qⁿ⁻ˣ for x = 0, 1, …, n	Stage 1
Cumulative probability	P(X ≤ r) = P(X = 0) + … + P(X = r)	Stage 1
Mean	μ = np	A Level only
Variance	σ² = npq	A Level only
Standard deviation	σ = √(npq)	A Level only

Finding the mean and variance of a general distribution from its table is excluded. Only the binomial formulae above are required.

Method 1: a distribution with an unknown

1. Write out all the probabilities in the table, or substitute each x into the formula.
2. Set the total equal to 1 and solve for the unknown.
3. Substitute back to get every probability and check they are all between 0 and 1.
4. Add the required probabilities. Check whether an inequality includes its end value.

Small reminder: P(X = x) = k(2x − 1) for x = 1, 2, 3 gives k + 3k + 5k = 1, so k = 1/9. Then P(X ≥ 2) = 8/9.

The four conditions

The binomial model needs:

1. a fixed number of trials, n;
2. two outcomes for each trial (success or failure);
3. independent trials;
4. the same probability p of success every time.

Conditions and assumptions

	Condition	Assumption
What it is	A requirement of the model	A statement about the real context that lets a condition be taken to hold
Wording	General	Mentions the situation
Example	The trials are independent	Whether one seed germinates does not affect whether another seed germinates

Quick check of a scenario:

• Sampling without replacement from a small group fails independence and constant p.
• Counting trials until something first happens fails the fixed n.
• A situation where trials affect one another (rain on consecutive days) fails independence.
• A situation with two different groups mixed together (adults and children with different p) fails constant p.

Method 2: binomial probabilities

1. Define X in words and write X ~ B(n, p). Say what a success is.
2. Write the probability you need as P(X = r), P(X ≤ r) or a combination.
3. Convert any other inequality to the form “X ≤ whole number” using the table below.
4. Evaluate with the formula or the calculator’s binomial functions.
5. Round to the accuracy asked (3 significant figures is a sensible default if none is stated) and answer in context.

You need	Use
P(X < r)	P(X ≤ r − 1)
P(X ≥ r)	1 − P(X ≤ r − 1)
P(X > r)	1 − P(X ≤ r)
P(a ≤ X ≤ b)	P(X ≤ b) − P(X ≤ a − 1)
P(X ≥ 1)	1 − P(X = 0) = 1 − qⁿ

Worked reminder: X ~ B(7, 0.2).

P(X = 2)  = 7C2 × 0.2² × 0.8⁵ = 21 × 0.04 × 0.32768 = 0.2753
P(X ≥ 1)  = 1 − 0.8⁷ = 0.7903
P(X ≥ 3)  = 1 − P(X ≤ 2) = 1 − 0.8520 = 0.1480

Worked reminder: building a table

A bag holds 3 mint and 3 toffee sweets. Two are taken without replacement and X is the number of mints.

x	0	1	2
P(X = x)	1/5	3/5	1/5

Working: P(X = 0) = 3/6 × 2/5 = 1/5. P(X = 2) = 3/6 × 2/5 = 1/5. P(X = 1) is the rest, 1 − 1/5 − 1/5 = 3/5. This is not binomial, because taking without replacement changes the probability of a mint.

Writing it up in context

Answers that name the context are easier to follow and to credit than general statements, so write sentences such as these:

• “Let X be the number of seeds, out of 15, that germinate. Then X ~ B(15, 0.8).”
• “Assume each seed germinates independently of the others.”
• “Assume the probability of germination is 0.8 for every seed.”
• “The model may not be suitable if the seeds come from different packets with different germination rates.”

When a question says “explain why a binomial model may not be appropriate”, pick one condition, say which one, and link it to a feature of the scenario.

Method 3: mean and variance (A Level only)

1. Identify n and p, and work out q = 1 − p.
2. μ = np, σ² = npq.
3. If asked for the standard deviation, take √(npq).
4. To go backwards from a given mean and variance: divide to get q = σ²/μ, then p = 1 − q, then n = μ/p.

Worked reminder: X ~ B(50, 0.1) has μ = 5, σ² = 50 × 0.1 × 0.9 = 4.5. A normal model with the same mean and variance is written N(5, 4.5). The next unit covers how that model is chosen and used.

Must-know distinctions

• Condition and assumption. The condition is the rule of the model. The assumption is the statement in context.
• P(X < r) and P(X ≤ r). They differ by P(X = r). On a discrete variable they are not the same.
• p and q. A question may give q, or the probability of failure, as its headline number.
• Variance and standard deviation. npq is the variance. Square-root it when σ is wanted.
• Distribution table and formula. Both describe the same thing. A table has to sum to 1 to be valid.

Quick self-test

1. A table has P(X = x) values 0.2, 0.3, k, 0.1 for x = 0, 1, 2, 3. Find k.
2. X ~ B(5, 0.4). Find P(X = 0) and P(X = 5).
3. X ~ B(5, 0.4). Find P(X = 2).
4. X ~ B(10, 0.5). Find P(X ≤ 3).
5. X ~ B(10, 0.5). Find P(X ≥ 8).
6. Write P(X ≥ 6) for X ~ B(15, 0.3) in terms of a cumulative probability.
7. State why “the number of cards dealt before the first ace” is not binomial.
8. State the mean and variance of X ~ B(40, 0.25). (A Level only)
9. Y ~ B(n, p) has mean 24 and variance 6. Find n and p. (A Level only)
10. Twelve people share a room during a flu outbreak and X is the number who catch it. Name the condition that could fail.

Answers

1. 1 − 0.6 = 0.4
2. P(X = 0) = 0.6⁵ = 0.0778, and P(X = 5) = 0.4⁵ = 0.0102
3. 10 × 0.4² × 0.6³ = 0.3456
4. 0.1719 (this is 176/1024)
5. 1 − P(X ≤ 7) = 0.0547
6. 1 − P(X ≤ 5)
7. The number of trials is not fixed. The cards are also drawn without replacement, so p changes.
8. Mean 10, variance 7.5
9. q = 6/24 = 0.25, so p = 0.75 and n = 24 ÷ 0.75 = 32. So n = 32, p = 0.75
10. Independence. One person being infected makes it more likely that others are.

Where marks are usually lost

• Not stating the distribution X ~ B(n, p) with the parameters before using it.
• Treating P(X ≥ r) as 1 − P(X ≤ r) instead of 1 − P(X ≤ r − 1).
• Quoting the conditions as a bare list when the question asks for an assumption in context.
• Rounding a probability too early, so that the last digit of the answer is wrong.
• Using the binomial for sampling without replacement from a small group without comment.
• Forgetting that the table of a distribution should add to 1, and not using this to find an unknown.
• Giving npq when the question asks for the standard deviation.
• Swapping p and q in the formula when the failure probability is the one given.

Last-minute checklist

• Can you write the sum-to-one equation for a table or a formula in under a minute?
• Can you list the four conditions from memory and give an assumption in context for each?
• Can you convert P(X < r), P(X ≥ r), P(X > r) and P(a ≤ X ≤ b) into cumulative form without looking?
• Do you know that npq is a variance and that np and npq are A Level only?
• Do you say what a success is, and write X ~ B(n, p), at the start of every binomial answer?

Official syllabus

OCR A Level Mathematics A (H240) specification, Version 3.1 (August 2026), for first assessment in 2018, published by OCR: sections 2.04a to 2.04d.