Binomial Distribution — Edexcel A-level Mathematics (Applied)

SPEC 4.1 • STATISTICAL DISTRIBUTIONS

The Binomial Distribution

Modelling the number of successes in a fixed number of independent trials

What is the Binomial Distribution?

The binomial distribution models situations where we count the number of successes in a fixed number of independent trials, each with the same probability of success.

Notation

If X follows a binomial distribution with n trials and probability p of success:

X \sim B (n, p)

Real-World Examples

Number of heads when flipping a coin 10 times

Number of defective items in a batch of 50

Number of patients responding to treatment out of 20

Number of sales from 100 cold calls

The Four Conditions

For a binomial distribution to be valid, all four conditions must be satisfied:

1. Fixed Number of Trials

There must be a fixed number n of trials decided in advance.

2. Independent Trials

The outcome of each trial must not affect any other trial.

3. Two Outcomes Only

Each trial has exactly two outcomes: success or failure.

4. Constant Probability

The probability p of success must be the same for every trial.

Remember with:

F I T S

Fixed trials, Independent, Two outcomes, Same probability

The Binomial Formula

The probability of getting exactly r successes in n trials is:

Formula

P (X = r) = (r n) p^{r} (1 - p)^{n - r}

SymbolMeaning

(r n)

or

^{n} C_{r}

Number of ways to choose r successes from n trials =

\frac{n !}{r ! ( n - r )!}

p^{r}

Probability of r successes

(1 - p)^{n - r}

Probability of (n-r) failures

Tip

The term (1-p) is often written as q, so the formula becomes

P (X = r) = (r n) p^{r} q^{n - r}

01Worked Example

A factory produces light bulbs with a 3% defect rate. A quality inspector randomly selects 20 bulbs from a large batch. Let X be the number of defective bulbs.

(a) Explain why X can be modelled by a binomial distribution.

(b) Write down the distribution of X.

(a) Checking the four conditions (FITS):

F:

Fixed number of trials: n = 20 bulbs selected

I:

Independent: Selection is random from a large batch, so each selection is independent

T:

Two outcomes: Each bulb is either defective (success) or not defective (failure)

S:

Same probability: p = 0.03 for each bulb (constant defect rate)

(b) X ~ B(20, 0.03)

Worked Example02

A coin is biased so that the probability of heads is 0.6. The coin is flipped 8 times. Find the probability of getting exactly 5 heads.

1

Identify: X ~ B(8, 0.6) where X = number of heads

2

We need P(X = 5)

3

Using formula:

P (X = 5) = (5 8) (0.6)^{5} (0.4)^{3}

4

Calculate

(5 8) = \frac{8 !}{5 ! \times 3 !} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56

5

P (X = 5) = 56 \times (0.6)^{5} \times (0.4)^{3}

6

= 56 \times 0.07776 \times 0.064 = 0.2787...

Answer: P(X = 5) = 0.279 (3 s.f.)

Cumulative Probabilities

Often we need probabilities involving inequalities. The key is P(X ≤ k) which calculators give directly.

Converting Inequalities

You WantConvert To

P (X \leq k)

Direct from calculator (Bcd)

P (X < k)

P (X \leq k - 1)

P (X \geq k)

1 - P (X \leq k - 1)

P (X > k)

1 - P (X \leq k)

Common Error

For P(X ≥ k), do NOT use 1 - P(X ≤ k). You must use 1 - P(X ≤ k-1) to include k itself.

Interpreting Probability Phrases

Exam questions use different phrases for probabilities. Since X takes integer values only, converting correctly is essential.

Phrase Conversion Table

PhraseMeansCalculation"greater than 5"

X > 5

1 - P (X \leq 5)

"at least 7"

X \geq 7

1 - P (X \leq 6)

"fewer than 10"

X < 10

P (X \leq 9)

"no more than 3"

X \leq 3

P (X \leq 3)

"at most 8"

X \leq 8

P (X \leq 8)

"more than 4"

X > 4

1 - P (X \leq 4)

Key Insight

Since X only takes whole numbers: P(X < 6) = P(X ≤ 5) and P(X > 6) = P(X ≥ 7)

Worked Example03

In a medical trial, a drug has a 70% success rate. 12 patients are given the drug. Let X be the number of patients for whom the drug is successful. Find P(X ≤ 8).

1

X ~ B(12, 0.7)

2

P(X ≤ 8) means probability of 8 or fewer successes

3

Using calculator: Bcd(8, 12, 0.7)

4

= 0.5075...

Answer: P(X ≤ 8) = 0.507 (3 s.f.)

Worked Example04

Using the same medical trial: X ~ B(12, 0.7). Find the probability that at least 10 patients respond successfully to the drug.

1

"At least 10" means X ≥ 10

2

P(X ≥ 10) = 1 - P(X ≤ 9)

3

Using calculator: P(X ≤ 9) = Bcd(9, 12, 0.7) = 0.7472...

4

P(X ≥ 10) = 1 - 0.7472... = 0.2528...

Answer: P(X ≥ 10) = 0.253 (3 s.f.)

Calculator Methods

Your calculator has built-in functions for binomial probabilities:

Casio ClassWiz

Menu → Distribution → Binomial

Bpd (Binomial probability distribution): P(X = r)

Bcd (Binomial cumulative distribution): P(X ≤ r)

Input Required

ParameterMeaningExamplex (or r)Number of successes5N (or n)Number of trials8pProbability of success0.6

Exam Tip

Always show your distribution notation X ~ B(n, p) and state what probability you're finding, even when using a calculator.

Worked Example05

A multiple choice test has 15 questions, each with 4 options (one correct). A student guesses every answer randomly. Find the probability that the student gets exactly 6 questions correct.

1

Define X = number of correct answers

2

X ~ B(15, 0.25) since p = 1/4 = 0.25 for guessing correctly

3

Need P(X = 6)

4

Using calculator: Bpd(6, 15, 0.25)

5

= 0.09174...

Answer: P(X = 6) = 0.0917 (3 s.f.)

Worked Example06

A fair die is rolled 18 times. Let X be the number of sixes obtained. Find P(3 ≤ X ≤ 6).

1

X ~ B(18, 1/6) since P(six) = 1/6

2

P(3 ≤ X ≤ 6) = P(X ≤ 6) - P(X ≤ 2)

3

P(X ≤ 6) = Bcd(6, 18, 1/6) = 0.9642...

4

P(X ≤ 2) = Bcd(2, 18, 1/6) = 0.3096...

5

P(3 ≤ X ≤ 6) = 0.9642... - 0.3096... = 0.6546...

Answer: P(3 ≤ X ≤ 6) = 0.655 (3 s.f.)

Using Cumulative Binomial Tables

The formula booklet contains cumulative binomial tables giving P(X ≤ x) directly.

Formula Booklet Extract

The tabulated value is P(X ≤ x), where X has a binomial distribution with index n and parameter p.p =0.050.100.150.200.250.300.35n=5, x=00.77380.59050.44370.32770.23730.16810.116010.97740.91850.83520.73730.63280.52820.428420.99880.99140.97340.94210.89650.83690.7648For X ~ B(5, 0.3): P(X ≤ 2) = 0.8369

Finding P(X = x) from Tables

P (X = x) = P (X \leq x) - P (X \leq x - 1)

Table Coverage

Tables cover n = 5, 6, 7, 8, 10, 12, 15, 20, 25, 30, 40, 50 and p = 0.05, 0.10, ... 0.50. For other values, use your calculator.

07Worked Example

X ~ B(20, 0.35). Using the cumulative binomial tables, find:

(a) P(X ≤ 5)

(b) P(X = 7)

(c) P(X ≥ 10)

1

Find n = 20 section, p = 0.35 column in tables:

(a) P(X ≤ 5) = 0.2454 (read directly)

(b) P(X = 7) = P(X ≤ 7) - P(X ≤ 6) = 0.6010 - 0.4166 = 0.1844

(c) P(X ≥ 10) = 1 - P(X ≤ 9) = 1 - 0.8782 = 0.1218

Finding Critical Values

A common exam question type asks you to find a value of k or r that satisfies a probability condition.

Types of Critical Value Questions

Find the largest k such that P(X ≤ k) < 0.05

Find the smallest r such that P(X ≥ r) < 0.01

Find n such that P(X ≥ 1) > 0.95

Method

Using tables: Look down the column until you find the first value that crosses the threshold.

Using calculator: Try successive values of k until the condition is just satisfied.

Problem-Solving Tip

When finding the first probability greater or less than a threshold, tables are often quicker than repeated calculator use.

08Worked Example

A spinner has probability 0.3 of landing on red. Jane spins it 12 times. Let X be the number of reds.

(a) Find P(X ≤ 2)

(b) Find P(X ≥ 5)

(c) Jane wins a prize if she gets r or more reds, where the probability of winning is less than 0.05. Find the smallest value of r.

X ~ B(12, 0.3)

(a) P(X ≤ 2) = 0.2528

(b) P(X ≥ 5) = 1 - P(X ≤ 4) = 1 - 0.7237 = 0.2763

(c) Need P(X ≥ r) < 0.05, so P(X ≤ r-1) > 0.95

From table: P(X ≤ 5) = 0.8822 ✗ P(X ≤ 6) = 0.9614 > 0.95 ✓

So r - 1 = 6, therefore r = 7

Check: P(X ≥ 7) = 1 - 0.9614 = 0.0386 < 0.05 ✓

Mean and Variance

For a binomial distribution X ~ B(n, p), the mean and variance have simple formulas:

Formulas

E (X) = n p

Va r (X) = n p (1 - p) = n pq

Key Properties

The mean np represents the expected number of successes

Variance is always less than the mean (since 1-p < 1)

Standard deviation:

σ = n p (1 - p)

Maximum variance occurs when p = 0.5

Quick Check

For X ~ B(n, p): Var(X) = E(X) × (1-p). If variance ≥ mean, it's NOT binomial!

09Worked Example

A call centre knows that 15% of calls result in a sale. On a typical day, an agent makes 40 calls.

(a) Find the expected number of sales.

(b) Find the standard deviation of the number of sales.

(c) Interpret these values in context.

1

Let X = number of sales. X ~ B(40, 0.15)

(a)

2

E(X) = np = 40 × 0.15 = 6

(b)

3

Var(X) = np(1-p) = 40 × 0.15 × 0.85 = 5.1

4

σ = √5.1 = 2.258...

(c)

5

On average, the agent makes 6 sales per day. The number typically varies by about 2.26 sales either side of this average.

Answers: (a) 6 sales (b) σ = 2.26 (3 s.f.)

10

Worked Example

A factory produces electronic components with a 0.15 probability of being faulty. Each box contains 10 components. A box passes inspection if it contains at most 1 faulty component.

(a) Find the probability that a randomly selected box passes inspection.

A shipment contains 8 boxes.

(b) Find the probability that exactly 6 boxes in the shipment pass inspection.

(c) Find the probability that at least 7 boxes pass inspection.

Part (a):

Let X = number of faulty components in a box. X ~ B(10, 0.15)

•

Box passes if X ≤ 1

•

P(X ≤ 1) = Bcd(1, 10, 0.15) = 0.5443 (4 s.f.)

Part (b):

Let Y = number of boxes that pass inspection

•

From (a), p = 0.5443. So Y ~ B(8, 0.5443)

•

P(Y = 6) = Bpd(6, 8, 0.5443) = 0.147 (3 s.f.)

Part (c):

P(Y ≥ 7) = 1 - P(Y ≤ 6)

•

= 1 - Bcd(6, 8, 0.5443) = 1 - 0.9602... = 0.0398 (3 s.f.)

The probability from the first binomial (0.5443) becomes the parameter p for the second binomial distribution.

Practice

Question 1:

(a) State the distribution of X, the number of defective bolts.

(b) Find the probability that exactly 1 bolt is defective.

(c) Find the probability that at most 2 bolts are defective.

(a) X ~ B(25, 0.02)

(b) P(X = 1) = Bpd(1, 25, 0.02) = 0.3079...

Answer: 0.308 (3 s.f.)

(c) P(X ≤ 2) = Bcd(2, 25, 0.02) = 0.9868...

Answer: 0.987 (3 s.f.)

Practice

Question 2:

(a) Find the probability that more than 20 households have solar panels.

(b) Find the expected number and standard deviation.

X ~ B(50, 0.35)

(a) P(X > 20) = 1 - P(X ≤ 20)

= 1 - Bcd(20, 50, 0.35) = 1 - 0.8139... = 0.1861...

Answer: 0.186 (3 s.f.)

(b) E(X) = np = 50 × 0.35 = 17.5

Var(X) = 50 × 0.35 × 0.65 = 11.375

σ = √11.375 = 3.373...

Answer: E(X) = 17.5, σ = 3.37 (3 s.f.)

Practice

Question 3:

(a) Find the probability that she makes at least 6 free throws in a game.

She plays 10 games in a season.

(b) Find the probability that she makes at least 6 free throws in exactly 8 of the 10 games.

(c) Find the probability that she makes at least 6 free throws in more than 8 games.

(a) Let X = free throws made per game. X ~ B(8, 0.75)

P(X ≥ 6) = 1 - P(X ≤ 5) = 1 - 0.3215... = 0.679 (3 s.f.)

(b) Let Y = games with ≥6 free throws. Y ~ B(10, 0.6785)

P(Y = 8) =

(8 10)

(0.6785)⁸(0.3215)² = 0.199 (3 s.f.)

(c) P(Y > 8) = P(Y = 9) + P(Y = 10)

= 1 - P(Y ≤ 8) = 1 - 0.9092... = 0.0908 (3 s.f.)

Summary

Key Concepts

ConceptFormula/Detail

NotationX ~ B(n, p)

Conditions (FITS)Fixed n, Independent, Two outcomes, Same p

P(X = r)

(r n) p^{r} (1 - p)^{n - r}

or Bpd

P(X ≤ r)Bcd (cumulative)

P(X ≥ r)1 - P(X ≤ r-1)

MeanE(X) = np

VarianceVar(X) = np(1-p)

Exam Skills

Always state X ~ B(n, p) before calculating

Check all four conditions when asked to justify binomial model

Convert inequalities carefully before using calculator

Interpret mean and variance in context