The Binomial Distribution

This lesson covers the binomial distribution as required by the Edexcel A-Level Mathematics specification (9MA0), Paper 3 Section A -- Statistics. You need to understand the conditions for a binomial model, calculate probabilities using the formula, use cumulative probabilities, and know the mean and variance.

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Conditions for a Binomial Distribution

A random variable X follows a binomial distribution B(n, p) if all four of the following conditions are met:

1. There is a fixed number of trials (n).
2. Each trial has exactly two outcomes: success or failure.
3. The probability of success (p) is constant for each trial.
4. The trials are independent of each other.

Notation

X ~ B(n, p) where n = number of trials and p = probability of success on each trial.

Example

A fair coin is flipped 10 times. Let X = number of heads.

• n = 10 (fixed). Two outcomes per flip. P(heads) = 0.5 (constant). Flips are independent.

So X ~ B(10, 0.5).

Counter-example (not binomial)

Drawing 4 balls without replacement from a bag of 5 red and 3 blue -- the probability changes with each draw, so it is not binomial.

Exam Tip: Always check all four conditions and state clearly which are or are not satisfied.

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The Binomial Probability Formula

P(X = r) = C(n, r) x p^r x (1 - p)^(n - r)

where C(n, r) = n! / (r!(n - r)!) is the binomial coefficient ("n choose r").

Breaking down the formula

• C(n, r) = number of ways to choose which r of the n trials are successes.
• p^r = probability that those r trials are all successes.
• (1 - p)^(n - r) = probability that the remaining trials are all failures.

Example

X ~ B(6, 0.3). Find P(X = 2).

C(6, 2) = 6! / (2! x 4!) = 15

P(X = 2) = 15 x 0.09 x 0.2401 = 0.3241 (to 4 d.p.)

Special cases

• P(X = 0) = (1 - p)^n
• P(X = n) = p^n
• P(X ≥ 1) = 1 - P(X = 0) = 1 - (1 - p)^n

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Cumulative Probabilities

P(X ≤ r) = P(X = 0) + P(X = 1) + ... + P(X = r)

Finding other probabilities from cumulative tables

• P(X < r) = P(X ≤ r - 1)
• P(X > r) = 1 - P(X ≤ r)
• P(X ≥ r) = 1 - P(X ≤ r - 1)
• P(a ≤ X ≤ b) = P(X ≤ b) - P(X ≤ a - 1)

Example

X ~ B(8, 0.4). Given P(X ≤ 3) = 0.5941 and P(X ≤ 4) = 0.8263:

• P(X = 4) = 0.8263 - 0.5941 = 0.2322
• P(X > 4) = 1 - 0.8263 = 0.1737
• P(X ≥ 4) = 1 - 0.5941 = 0.4059

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Mean and Variance

For X ~ B(n, p):

• Mean: E(X) = np
• Variance: Var(X) = np(1 - p) = npq, where q = 1 - p
• Standard deviation: SD(X) = sqrt(np(1 - p))

Example

X ~ B(20, 0.35): E(X) = 7, Var(X) = 4.55, SD(X) = 2.133

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Using the Calculator

• Binomial PD: calculates P(X = r).
• Binomial CD: calculates P(X ≤ r).

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Solving Problems Involving Inequalities

A common question gives a condition like P(X ≥ 1) ≥ 0.9 and asks for the minimum n.

Method

P(X ≥ 1) = 1 - (1 - p)^n ≥ 0.9

(1 - p)^n ≤ 0.1

n ≥ log(0.1) / log(1 - p)

Example

X ~ B(n, 0.15). Find smallest n such that P(X ≥ 1) ≥ 0.95.

(0.85)^n ≤ 0.05 --> n ≥ log(0.05)/log(0.85) = 18.43 --> n = 19.

Exam Tip: Always round up when finding the minimum number of trials.

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Summary

• Conditions: fixed n, two outcomes per trial, constant p, independent trials.
• Formula: P(X = r) = C(n, r) x p^r x (1 - p)^(n - r).
• Cumulative: P(X ≤ r). Use this to find P(X > r), P(X ≥ r), etc.
• Mean: E(X) = np. Variance: Var(X) = np(1 - p).
• For minimum n problems, solve (1 - p)^n ≤ target using logarithms and round up.

A-Level Deep Dive: The Binomial Distribution

Spec mapping

Edexcel 9MA0-03 specification, Paper 3 — Statistics and Mechanics, Section 6: The binomial distribution. The specification requires students to "understand and use the binomial distribution as a model; calculate probabilities using the binomial distribution"; recognise the conditions under which a binomial model is appropriate; use $X \sim B(n, p)$X∼B(n,p) notation; compute $P(X = r)$P(X=r), $P(X \leq r)$P(X≤r) and $P(X \geq r)$P(X≥r) using a calculator's statistical functions; and quote the mean $E(X) = np$E(X)=np and variance $\mathrm{Var}(X) = np(1-p)$Var(X)=np(1−p) as standard results. The binomial distribution is examined in 9MA0-03 Section A (Statistics) and is the gateway to Section 7 — Hypothesis testing for a binomial proportion. The Edexcel formula booklet lists $P(X = r) = \binom{n}{r}p^r(1-p)^{n-r}$P(X=r)=(rn​)pr(1−p)n−r but does not list $E(X)$E(X) or $\mathrm{Var}(X)$Var(X) — these must be memorised.

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Worked example with full mark scheme

Question (8 marks):

A factory machine produces components. Past data suggests that 15% of components are defective. A quality inspector takes a random sample of 20 components from a single production batch. Let $X$X denote the number of defective components in the sample.

(a) State two conditions, beyond those given, required for $X$X to be modelled by a binomial distribution. (2)

(b) Assuming $X \sim B(20, 0.15)$X∼B(20,0.15), find $P(X \leq 3)$P(X≤3). (2)

(c) Find $P(X \geq 5)$P(X≥5). (2)

(d) State the mean and variance of $X$X. (2)

Solution with mark scheme:

(a) B1 — each defect is independent of the others (whether one component is defective does not affect another).

B1 — the probability of a defect is constant at $p = 0.15$p=0.15 for every component in the sample (e.g. the production process does not drift mid-batch).

Common error: stating "there are two outcomes" or "fixed $n$n" — these are given in the question stem ("defective or not", "sample of 20") and earn nothing. The mark is for the conditions the candidate must add: independence and constancy of $p$p.

(b) Using a calculator's binomial cumulative function with $n = 20$n=20, $p = 0.15$p=0.15:

$P(X \leq 3) = \sum_{r=0}^{3} \binom{20}{r}(0.15)^r(0.85)^{20-r}$P(X≤3)=∑r=03​(r20​)(0.15)r(0.85)20−r

M1 — correct cumulative set-up (sum from $r = 0$r=0 to $r = 3$r=3, or calculator notation $\mathrm{B.cd}(20, 0.15, 3)$B.cd(20,0.15,3)).

A1 — $P(X \leq 3) = 0.6477$P(X≤3)=0.6477 (to 4 d.p.).

(c) $P(X \geq 5) = 1 - P(X \leq 4)$P(X≥5)=1−P(X≤4).

M1 — using the complement $1 - P(X \leq 4)$1−P(X≤4) (a single subtraction; not $1 - P(X \leq 5)$1−P(X≤5), which is the most common slip and earns zero).

$P(X \leq 4) = 0.8298$P(X≤4)=0.8298, so $P(X \geq 5) = 1 - 0.8298 = 0.1702$P(X≥5)=1−0.8298=0.1702.

A1 — $P(X \geq 5) = 0.1702$P(X≥5)=0.1702 (to 4 d.p.).

(d) $E(X) = np = 20 \times 0.15 = 3$E(X)=np=20×0.15=3. B1

$\mathrm{Var}(X) = np(1-p) = 20 \times 0.15 \times 0.85 = 2.55$Var(X)=np(1−p)=20×0.15×0.85=2.55. B1

Total: 8 marks (B2 + M1 A1 + M1 A1 + B2).

The recurring error worth flagging is the boundary handling in (c). For a discrete distribution, $P(X \geq 5) = 1 - P(X \leq 4)$P(X≥5)=1−P(X≤4) — the inequality flips and the boundary moves down by one. Candidates who write $P(X \geq 5) = 1 - P(X \leq 5)$P(X≥5)=1−P(X≤5) are computing $P(X \geq 6)$P(X≥6) and lose both marks.

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Specimen question modelled on the Edexcel 9MA0 Paper 3 format

Question (6 marks): A telesales operator estimates that 30% of cold calls result in a successful sale. On a particular morning the operator makes 12 calls. Let $Y$Y be the number of successful sales.

(a) State an assumption required to model $Y \sim B(12, 0.3)$Y∼B(12,0.3). (1)

(b) Find $P(Y = 4)$P(Y=4). (2)

(c) Find the probability that the operator makes at least 3 sales but fewer than 7. (3)

Mark scheme decomposition by AO:

(a)

• B1 (AO3.3) — a context-appropriate assumption: e.g. "the success of one call is independent of the others" or "the probability of a sale is constant at 0.3 across all 12 calls". Generic statements ("trials are random") earn nothing.

(b)

• M1 (AO1.1a) — correct binomial point formula $P(Y = 4) = \binom{12}{4}(0.3)^4(0.7)^8$P(Y=4)=(412​)(0.3)4(0.7)8 or calculator notation $\mathrm{B.pd}(12, 0.3, 4)$B.pd(12,0.3,4).
• A1 (AO1.1b) — $P(Y = 4) = 0.2311$P(Y=4)=0.2311 (to 4 d.p.).

(c)

• M1 (AO2.5) — translating "at least 3 but fewer than 7" into $P(3 \leq Y \leq 6) = P(Y \leq 6) - P(Y \leq 2)$P(3≤Y≤6)=P(Y≤6)−P(Y≤2). The common slip is $P(Y \leq 7) - P(Y \leq 3)$P(Y≤7)−P(Y≤3), which mishandles both boundaries.
• M1 (AO1.1b) — correct calculator values: $P(Y \leq 6) = 0.9614$P(Y≤6)=0.9614, $P(Y \leq 2) = 0.2528$P(Y≤2)=0.2528.
• A1 (AO1.1b) — $P(3 \leq Y \leq 6) = 0.9614 - 0.2528 = 0.7086$P(3≤Y≤6)=0.9614−0.2528=0.7086 (to 4 d.p.).

Total: 6 marks split AO1 = 4, AO2 = 1, AO3 = 1. Edexcel uses Paper 3 binomial questions to test (i) procedural fluency with the cumulative tables / calculator, and (ii) careful translation of English phrasing into discrete inequalities — that translation is where AO2 marks live.

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Synoptic links

Connects to:

1. Section 5 — Probability: the binomial sits on top of independent-trials probability. $P(X = r)$P(X=r) counts the $\binom{n}{r}$(rn​) orderings of $r$r successes among $n$n trials, each ordering having probability $p^r(1-p)^{n-r}$pr(1−p)n−r. Without confidence in independent compound probability, the multiplication of $p^r(1-p)^{n-r}$pr(1−p)n−r has no justification.

2. Section 4 (Pure) — Binomial expansion: the binomial coefficient $\binom{n}{r}$(rn​) is the same object that appears in $(a + b)^n = \sum_{r=0}^{n}\binom{n}{r}a^{n-r}b^r$(a+b)n=∑r=0n​(rn​)an−rbr. Substituting $a = (1-p)$a=(1−p), $b = p$b=p gives $\sum_{r=0}^{n} P(X = r) = ((1-p) + p)^n = 1^n = 1$∑r=0n​P(X=r)=((1−p)+p)n=1n=1 — a one-line proof that the binomial probabilities sum to one. This is a beautiful synoptic identity Edexcel rewards if cited.

3. Section 7 — Statistical hypothesis testing: the binomial is the underlying distribution for tests of a population proportion $p$p. A test of $H_0: p = p_0$H0​:p=p0​ versus $H_1: p \neq p_0$H1​:p=p0​ uses the binomial sampling distribution under $H_0$H0​ to compute the $p$p-value. Without binomial fluency the hypothesis-testing topic collapses.

4. Year 2 Section 4 — Normal approximation: when $n$n is large and $p$p is not too close to $0$0 or $1$1, $X \sim B(n, p)$X∼B(n,p) is approximately $N(np, np(1-p))$N(np,np(1−p)). The continuity correction $P(X \leq k) \approx P(Y \leq k + 0.5)$P(X≤k)≈P(Y≤k+0.5) is the bridge between discrete and continuous, and it is exactly the link Edexcel exploits in Year 2 Statistics.

5. Modelling and contextual judgement (AO3): real situations rarely meet binomial conditions exactly — sampling without replacement violates independence, drift in a production process violates constant $p$p. Recognising when the binomial model breaks is an AO3 skill that distinguishes A* from A.

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Mark-scheme literacy

Binomial questions on 9MA0-03 split AO marks roughly as follows:

AO	Typical share	Earned by
AO1 (knowledge / procedure)	55–65%	Stating $X \sim B(n, p)$X∼B(n,p), computing $P(X = r)$P(X=r), $P(X \leq r)$P(X≤r) and $P(X \geq r)$P(X≥r), quoting $E(X) = np$E(X)=np and $\mathrm{Var}(X) = np(1-p)$Var(X)=np(1−p)
AO2 (reasoning / interpretation)	20–30%	Translating "at least", "at most", "more than", "fewer than" into discrete inequalities; using complements correctly; combining $P(a \leq X \leq b)$P(a≤X≤b) as $P(X \leq b) - P(X \leq a-1)$P(X≤b)−P(X≤a−1)
AO3 (modelling)	10–20%	Stating context-appropriate independence and constancy assumptions; identifying when binomial conditions fail; commenting on suitability of the model

Examiner-rewarded phrasing: "let $X$X be the number of [successes] in $n$n trials, so $X \sim B(n, p)$X∼B(n,p)"; "since the trials are independent and $p$p is constant, the binomial model is appropriate"; "the probability of a success is constant in this context because …". Phrases that lose marks: "binomial because two outcomes" (insufficient — independence and constancy missing); answers given to 1 d.p. when 4 d.p. is conventional for binomial probabilities; using $P(X \leq r)$P(X≤r) when $P(X < r) = P(X \leq r-1)$P(X<r)=P(X≤r−1) is required.

A specific Edexcel pattern to watch: questions that say "find the probability that more than 5 succeed" mean $P(X > 5) = P(X \geq 6) = 1 - P(X \leq 5)$P(X>5)=P(X≥6)=1−P(X≤5). Questions that say "at least 5 succeed" mean $P(X \geq 5) = 1 - P(X \leq 4)$P(X≥5)=1−P(X≤4). The boundary shifts by one — read the wording precisely.

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Grade-band model answers

3-mark question

Question: $X \sim B(15, 0.4)$X∼B(15,0.4). Find $P(X = 6)$P(X=6).

Grade C response (~180 words):

Using the binomial formula with $n = 15$n=15, $p = 0.4$p=0.4, $r = 6$r=6:

$P(X = 6) = \binom{15}{6}(0.4)^6(0.6)^9$P(X=6)=(615​)(0.4)6(0.6)9.

$\binom{15}{6} = 5005$(615​)=5005. $(0.4)^6 = 0.004096$(0.4)6=0.004096. $(0.6)^9 = 0.010077696$(0.6)9=0.010077696.

So $P(X = 6) = 5005 \times 0.004096 \times 0.010077696 = 0.2066$P(X=6)=5005×0.004096×0.010077696=0.2066 (to 4 d.p.).