Binomial Distribution — SL 4.6 Binomial distribution - IB Mathematics Analysis and Approaches

SECTION 1

Historical Context & Motivation

Long before statisticians had computers or calculators, mathematicians were fascinated by games of chance. They noticed that when you repeat a simple experiment — such as flipping a coin or rolling a die — the number of "wins" follows a predictable pattern. The quest to describe that pattern led to one of the most widely used probability models in all of statistics: the binomial distribution. Understanding its origins helps us appreciate why this model is so powerful and where it naturally applies.

1654 • Pascal & Fermat's Correspondence

Blaise Pascal and Pierre de Fermat exchanged letters about gambling problems. Their work laid the foundation for combinatorics and the concept of counting outcomes systematically.

1713 • Bernoulli's Ars Conjectandi

Jacob Bernoulli published his landmark treatise posthumously, formally defining repeated independent trials with two outcomes — what we now call Bernoulli trials — and connecting them to the binomial expansion.

1733 • De Moivre's Approximation

Abraham de Moivre showed that as the number of trials grows large, the binomial distribution is well approximated by the normal (bell) curve, foreshadowing the Central Limit Theorem.

1800s–present • Modern Applications

The binomial distribution became indispensable in quality control, medical testing, polling, genetics, and machine learning. It remains a core topic in every statistics curriculum, including IB Mathematics.

The central question the binomial distribution answers is deceptively simple: If I repeat an experiment a fixed number of times, each time with the same probability of success, how likely is it that I get exactly k successes? Answering this question precisely is what Section 4.6 of the IB syllabus is all about.

SECTION 2

Core Principles & Definitions

Before we can use the binomial distribution, we need to confirm that a situation meets a specific set of conditions. These conditions are often remembered by the acronym FIST — Fixed number of trials, Independent trials, only two outcomes (Success or failure), and the same probability (constant Trial probability). If all four conditions hold, the random variable counting the number of successes follows a binomial distribution.

Fixed Number of Trials (n)

The experiment is performed a set number of times, decided in advance. For example, you flip a coin exactly 10 times — not "until you get heads."

Independent Trials

The outcome of one trial does not affect any other trial. Getting heads on flip 3 has no influence on what happens on flip 4.

Two Outcomes per Trial

Each trial results in either a "success" or a "failure." These labels are arbitrary — "success" simply means the outcome you are counting.

Constant Probability (p)

The probability of success, p, remains the same from trial to trial. If you are rolling a fair die and counting sixes, p = 1/6 on every roll.

When these conditions are satisfied, we write X ~ B(n, p), which is read as "X follows a binomial distribution with parameters n and p." Here, n is the number of trials and p is the probability of success on any single trial. The random variable X can take integer values from 0 to n.

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KEY TAKEAWAY

Think of the binomial distribution like a basketball free-throw drill. You shoot a fixed number of free throws (n), each shot is independent, you either make it or miss it, and your shooting percentage (p) stays roughly constant. The binomial model tells you the probability of making exactly k out of n shots.

SECTION 3

Visual Explanation

Probability Distribution for X ~ B(8, 0.4)

Each bar represents P(X = k) for k = 0, 1, 2, …, 8 when n = 8 and p = 0.4. Notice the distribution is slightly right-skewed because p < 0.5. The tallest bar is at k = 3, which is the mode. The mean equals np = 8 × 0.4 = 3.2.

The bar chart above illustrates a key feature of every binomial distribution: the probabilities rise from k = 0, peak near the mean, and then fall again. When p = 0.5 the distribution is perfectly symmetric; when p < 0.5 it skews to the right (as shown), and when p > 0.5 it skews to the left. The sum of all bar heights always equals 1, because they represent every possible outcome.

SECTION 4

Mathematical Framework

The binomial distribution is built on three mathematical ingredients: the probability of a specific sequence of successes and failures, the number of ways to arrange those successes, and the multiplication of these two factors. Let's look at each formula you need for the IB exam.

BINOMIAL PROBABILITY FORMULA

P(X = k) = C(n, k) × p^k × (1 − p)^(n−k)

where n = number of trials, k = desired number of successes, p = probability of success on one trial, 1 − p (often called q) = probability of failure on one trial, and C(n, k) = the binomial coefficient ("n choose k").

BINOMIAL COEFFICIENT

C(n, k) = n! / (k! × (n − k)!)

This counts the number of distinct ways to choose which k trials out of n will be successes. For example, C(5, 2) = 5! / (2! × 3!) = 10 ways.

EXPECTED VALUE (MEAN)

E(X) = μ = n × p

The mean of a binomial distribution is simply the number of trials multiplied by the probability of success. If you flip a fair coin 100 times, the expected number of heads is 100 × 0.5 = 50.

VARIANCE AND STANDARD DEVIATION

Var(X) = σ² = n × p × (1 − p) ⟹ σ = √(n × p × (1 − p))

The variance measures how spread out the distribution is. Notice it depends on both p and (1 − p), so the spread is largest when p = 0.5 and smallest when p is close to 0 or 1.

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GDC / Calculator Tip

On the IB exam you are expected to use your GDC. Most calculators have built-in functions: binompdf(n, p, k) returns P(X = k) and binomcdf(n, p, k) returns P(X ≤ k). Make sure you know which function to use for "at most," "at least," and "exactly" questions.

SECTION 5

How Parameters Shape the Distribution

The two parameters n and p completely determine every aspect of a binomial distribution — its shape, center, and spread. Changing either parameter transforms the distribution in predictable ways. The diagram below shows three different binomial distributions on the same axes so you can compare them directly.

Three binomial distributions plotted on the same axes. The cyan curve B(10, 0.3) peaks at k = 3. The pink curve B(5, 0.2) is strongly skewed right with its peak at k = 0 (since np = 1). The amber curve B(10, 0.7) is skewed left, peaking near k = 7. Increasing p shifts the peak to the right; increasing n spreads the distribution wider.

Parameter Change	Effect on Mean (np)	Effect on Shape
Increase n, hold p constant	Mean increases	Distribution spreads wider and becomes more symmetric (bell-shaped)
Increase p toward 0.5, hold n constant	Mean increases	Distribution becomes more symmetric
p = 0.5	Mean = n/2	Perfectly symmetric for any n
p close to 0 or 1	Mean near 0 or n	Strongly skewed

How n and p affect the binomial distribution

SECTION 6

Worked Example

A fair six-sided die is rolled 12 times. Let X be the number of times a "6" appears. Find (a) the probability of getting exactly 3 sixes, (b) the expected number of sixes, and (c) P(X ≤ 2).

Rolling a Die 12 Times

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Step 1 — Verify Binomial Conditions

There is a fixed number of trials (n = 12). Each roll is independent. Each roll has two outcomes: "6" (success) or "not 6" (failure). The probability of rolling a 6 is p = 1/6 on every roll. All four conditions are met, so X ~ B(12, 1/6).

X ~ B(12, 1/6)
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Step 2 — Calculate P(X = 3)

Using the binomial formula: P(X = 3) = C(12, 3) × (1/6)³ × (5/6)⁹. First, C(12, 3) = 12! / (3! × 9!) = 220. Then (1/6)³ = 1/216 ≈ 0.004630. And (5/6)⁹ ≈ 0.19381. Multiplying: P(X = 3) = 220 × 0.004630 × 0.19381 ≈ 0.1974.

P(X = 3) ≈ 0.197
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Step 3 — Find E(X)

The expected value formula gives E(X) = n × p = 12 × (1/6) = 2. On average, you would expect to roll two sixes in twelve rolls.

E(X) = 2
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Step 4 — Calculate P(X ≤ 2)

P(X ≤ 2) = P(X = 0) + P(X = 1) + P(X = 2). Calculating each term: P(X = 0) = C(12,0) × (1/6)⁰ × (5/6)¹² = 1 × 1 × 0.1122 = 0.1122. P(X = 1) = C(12,1) × (1/6)¹ × (5/6)¹¹ = 12 × 0.1667 × 0.1346 = 0.2692. P(X = 2) = C(12,2) × (1/6)² × (5/6)¹⁰ = 66 × 0.02778 × 0.1615 = 0.2961. Therefore P(X ≤ 2) = 0.1122 + 0.2692 + 0.2961 = 0.6775. On a GDC you could simply enter binomcdf(12, 1/6, 2).

P(X ≤ 2) ≈ 0.678

SECTION 7

Strengths & Limitations

The binomial distribution is one of the most useful discrete probability models, but it does not fit every situation. Understanding when it works well and when it breaks down will help you choose the right tool on the IB exam and in real-world analysis.

Strengths	Limitations
Simple to set up: only two parameters (n, p) are needed.	Trials must be independent — fails for sampling without replacement from small populations.
Widely applicable: quality control, medicine, sports, genetics.	p must be constant. If success probability changes from trial to trial, binomial is not appropriate.
Built into every scientific calculator and GDC, making computation fast.	Only counts successes — if you need the number of trials until first success, use the geometric distribution.
Clear formulas for mean (np) and variance (npq) give quick summaries.	For very large n and small p, the Poisson distribution is often more practical.

When the binomial model shines and when to look for alternatives

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REMEMBER

Before applying the binomial model, always check the four conditions (FIST). If any condition fails — for instance, you're drawing cards from a deck without replacing them — the binomial distribution will give inaccurate results. In the IB exam, identifying whether a situation is binomial is itself a commonly tested skill.

SECTION 8

Connection to Other Distributions

The binomial distribution does not exist in isolation. It connects naturally to several other probability models that you may encounter later in your IB studies or at university. Understanding these connections deepens your statistical thinking and helps you recognise when a different model is more suitable.

Distribution	Relationship to Binomial	When to Use Instead
Bernoulli	A Bernoulli trial is a single binomial trial (n = 1). X ~ B(1, p) is the same as a Bernoulli distribution.	When there is exactly one trial.
Normal (Gaussian)	For large n (np ≥ 5 and nq ≥ 5), B(n, p) is well approximated by N(np, npq). This is the basis of the normal approximation.	When n is very large and exact binomial calculations become impractical.
Poisson	When n is large and p is very small, B(n, p) ≈ Poisson(np). The Poisson acts as a limit of the binomial.	Modelling rare events over time/space (e.g., defects per batch).
Geometric	Counts the number of trials until the first success. It uses the same Bernoulli trials but asks a different question.	"How many attempts until the first success?"

The binomial distribution and its relatives

In the IB HL course and at university, you will study the normal approximation to the binomial in greater depth. For now, the key point is that the binomial distribution is a fundamental building block: master it, and the transition to more advanced distributions will feel natural.

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

A teacher draws 5 names from a hat of 30 students (without replacement) and counts how many are left-handed. Explain why this situation does not exactly satisfy the conditions for a binomial distribution.

Show answer

The binomial distribution requires independent trials with a constant probability of success. Here the sampling is done without replacement, which means each draw changes the composition of the remaining students. The probability of selecting a left-handed student shifts after each draw, violating both the independence and constant-probability conditions. If the class were very large (say 3000), the probability change would be negligible and a binomial model could serve as a reasonable approximation, but with only 30 students the effect is significant. The correct model here would be the hypergeometric distribution.

Question 2: BASIC CALCULATION

A multiple-choice quiz has 6 questions, each with 4 options (only one correct). A student guesses randomly on every question. Let X be the number of correct answers. Find P(X = 2) and E(X).

Show answer

Since the student guesses randomly, each question is an independent trial with p = 1/4 = 0.25 and n = 6, so X ~ B(6, 0.25). Using the formula: P(X = 2) = C(6,2) × (0.25)² × (0.75)⁴ = 15 × 0.0625 × 0.3164 ≈ 0.2966. The expected value is E(X) = np = 6 × 0.25 = 1.5. So the student can expect, on average, to get 1.5 questions correct by pure guessing, and there is roughly a 30% chance of getting exactly 2 right.

Question 3: INTERMEDIATE

A basketball player makes 70% of her free throws. During a game she takes 10 free throws. Find the probability that she makes at least 8 of them.

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Let X ~ B(10, 0.7). We need P(X ≥ 8) = P(X = 8) + P(X = 9) + P(X = 10). Calculating each: P(X = 8) = C(10,8) × (0.7)⁸ × (0.3)² = 45 × 0.05765 × 0.09 = 0.2335. P(X = 9) = C(10,9) × (0.7)⁹ × (0.3)¹ = 10 × 0.04035 × 0.3 = 0.1211. P(X = 10) = C(10,10) × (0.7)¹⁰ × (0.3)⁰ = 1 × 0.02825 × 1 = 0.0282. Therefore P(X ≥ 8) = 0.2335 + 0.1211 + 0.0282 ≈ 0.383. On a GDC you could compute 1 − binomcdf(10, 0.7, 7) to get the same result.

Question 4: APPLIED

A factory produces light bulbs with a 5% defect rate. An inspector randomly selects 20 bulbs from a large production run. Find the probability that (a) none are defective, and (b) more than 2 are defective. Also calculate the standard deviation of the number of defective bulbs.

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Let X ~ B(20, 0.05). (a) P(X = 0) = C(20,0) × (0.05)⁰ × (0.95)²⁰ = (0.95)²⁰ ≈ 0.3585. (b) P(X > 2) = 1 − P(X ≤ 2) = 1 − [P(X=0) + P(X=1) + P(X=2)]. P(X=1) = 20 × 0.05 × (0.95)¹⁹ ≈ 0.3774. P(X=2) = C(20,2) × (0.05)² × (0.95)¹⁸ = 190 × 0.0025 × 0.3972 ≈ 0.1887. So P(X ≤ 2) ≈ 0.3585 + 0.3774 + 0.1887 = 0.9246, giving P(X > 2) ≈ 1 − 0.9246 = 0.0754. The standard deviation is σ = √(npq) = √(20 × 0.05 × 0.95) = √0.95 ≈ 0.975. This means the number of defective bulbs typically varies by about 1 from the mean of 1.

Question 5: CRITICAL THINKING

A pharmaceutical company claims its new drug is effective for 80% of patients. In a trial of 15 patients, only 9 experienced a positive effect. Using the binomial distribution, calculate P(X ≤ 9) and discuss whether this result provides evidence against the company's claim.

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If the company's claim is correct, then X ~ B(15, 0.8). The expected number of successes is E(X) = 15 × 0.8 = 12. We need P(X ≤ 9). Using a GDC: binomcdf(15, 0.8, 9) ≈ 0.0611. This means that if the drug truly works 80% of the time, there is only about a 6.1% chance of seeing 9 or fewer successes in 15 patients. In statistical terms, this is a moderately low probability. At a 10% significance level, we would reject the company's claim; at the conventional 5% level, we would just barely fail to reject it (since 0.061 > 0.05). So the data casts doubt on the 80% claim but does not provide overwhelming evidence against it. This type of reasoning — comparing observed results to what a model predicts — is the foundation of hypothesis testing, which you will study further in the IB course.

SUMMARY

Lesson Summary

Summary

The binomial distribution models the number of successes in a fixed number of independent trials, each with the same probability of success p. We write X ~ B(n, p) and compute individual probabilities using the formula P(X = k) = C(n, k) × $p^k$ × $(1 − p)^{n−k}$ . The expected value is E(X) = np, the variance is Var(X) = np(1 − p), and the distribution's shape is governed by the values of n and p.

Before applying the model, always verify the four conditions: Fixed n, Independence, two outcomes (Success/failure), and constant Trial probability. On your GDC, use binompdf for P(X = k) and binomcdf for P(X ≤ k). Mastering the binomial distribution lays the groundwork for the normal approximation and hypothesis testing topics you will encounter later in the IB course.