SL 4.8 — Binomial Distribution

Term / Concept Definition / Explanation
Binomial Setting A situation in which we repeat an experiment a fixed number of times and each trial has two outcomes:
“success” or “failure.” Examples: flipping a coin 10 times, testing 20 bulbs for defects.
Parameters (n, p) n: number of trials (fixed in advance).
p: probability of success on each trial (constant for all trials).
Conditions for Binomial Model A binomial distribution applies only if:
• The number of trials n is fixed.
• Each trial results in success or failure.
• The probability of success p stays constant.
• Trials are independent.
Binomial Random Variable (X) X counts the number of successes in n independent trials, each with probability p of success.
Its possible values are 0, 1, 2, …, n.
Binomial Probability The probability of exactly k successes is found using technology (GDC).
Students are not required to memorise or derive the formula.
Mean & Variance • Mean of X: n × p
• Variance of X: n × p × (1 − p)
Proof is not required.
Identifying the Binomial Model Used when counting successes out of n repeated, identical, independent trials. Must check all
conditions before applying.

1. What is a Binomial Distribution?

A random variable X follows a binomial distribution when we count the number of “successes” in a fixed number of
repeated trials, each trial having only two possible outcomes (success / failure).
We write this as X ~ B(n, p), where:

  • n = number of trials (fixed in advance)
  • p = probability of success on each trial (constant)
  • Each trial is independent of the others
  • Each trial has only two outcomes: success (with probability p) or failure (with probability 1 − p)
  • X counts how many successes occur in n trials → X takes values 0, 1, 2, …, n

If any of these conditions is clearly broken (e.g. probability changes, trials not independent), the binomial model may not
be appropriate and another distribution should be considered.

🌍 Real-World Connection:
Binomial models appear in:

  • Quality control: number of defective items in a batch
  • Medicine: number of patients responding positively to a treatment
  • Marketing: number of customers who buy after receiving an advert
  • Sports: number of successful penalty kicks out of n attempts

2. Binomial Probability Formula

For X ~ Bin(n, p), the probability that X takes the value k (i.e. exactly k successes) is

P(X = k) = C(n, k) pk (1 − p)n − k    for k = 0, 1, 2, …, n

  • C(n, k) (also written nCk) is the number of different ways to choose which k trials are successes.
  • pk gives the probability of those k successes.
  • (1 − p)n − k gives the probability of the remaining n − k failures.

In IB exams you do not need to derive this formula, but you must know how to:

  • Write down the correct expression for P(X = k)
  • Interpret “at least”, “at most”, “no more than”, “no fewer than” using sums of binomial terms or GDC

Example 1 – Exact probability

The probability that a machine produces a defective item is 0.1. In a batch of n = 8 items, let X be the number of defectives.
Find P(X = 2).

P(X = 2) = C(8, 2) (0.1)2 (0.9)6 = 28 × 0.01 × 0.531441 ≈ 0.148.

📱 GDC Tips — Binomial Distribution

Exam rule: Always define the random variable before using calculator commands, for example:
“Let X be the number of successes in n independent trials.”


TI-Nspire CX II

  • To find P(X = k), open a Calculator page and enter:

    binompdf(n, p, k)

    This gives the probability of exactly k successes.

  • To find P(X ≤ k), enter:

    binomcdf(n, p, k)

    This directly gives the cumulative probability up to and including k.

  • To find P(X ≥ k), use the complement rule:

    1 − binomcdf(n, p, k − 1)

    This avoids calculator errors caused by incorrect bounds.

  • In exams, always write the command used (e.g. binomcdf(10, 0.3, 4)) before quoting the final probability.

Casio fx-CG50 / fx-CG100

  • Press MENU → STAT → DIST → BINOM.
  • For P(X = k), choose Bpd, then enter:

    Number of trials (n), probability (p), value (k).

  • For P(X ≤ k), choose Bcd, and enter:

    Lower bound = 0, Upper bound = k.

  • For P(X ≥ k), calculate:

    1 − Bcd(n, p, k − 1),
    since Bcd always computes cumulative probabilities from below.


⚠️ Common IB Exam Mistakes to Avoid

  • Forgetting to subtract 1 when calculating P(X ≥ k) using cumulative probability.
  • Using cumulative probability when the question explicitly asks for P(X = k).
  • Not defining the random variable X before using calculator output.
  • Copying calculator values without interpretation in context (loses communication marks).
🧠 Examiner Tip:
Marks are often lost by:

  • Not stating X ~ B(n, p) before calculating probabilities
  • Using the wrong n or p (e.g. confusing success with failure)
  • Mistreating “at least / at most” — write the probability sum explicitly or show the GDC command clearly

3. Mean and Variance of X ~ B(n, p)

For a binomial random variable X ~ Bin(n, p):

  • Mean (expected value): E(X) = n p
  • Variance: Var(X) = n p (1 − p)
  • Standard deviation: σ = √[n p (1 − p)]

These results link directly to SL 4.5: if the probability of success is p and there are n trials, the expected number of successes is n p.

Example 2 – Mean and variance

A basketball player scores a free throw with probability 0.75.
She takes 20 shots in a practice session. Let X be the number of successful shots (assume independence).

  • Model: X ~ B(20, 0.75)
  • E(X) = n p = 20 × 0.75 = 15 → on average she scores 15 shots.
  • Var(X) = n p (1 − p) = 20 × 0.75 × 0.25 = 3.75
  • σ ≈ √3.75 ≈ 1.94 → typical deviation from the mean is about 2 shots.

4. When is a Binomial Model Appropriate?

To Identify if a probability follows a Binomial Model, check for these 4 simple parameters:

  • Is there a fixed number of trials n?
  • Does each trial have only two outcomes (success / failure)?
  • Is the probability of success constant between trials?
  • Are outcomes of trials independent of each other?

If all answers are “yes”, then a binomial model is usually reasonable.

📝 Paper Strategy:
In explanation questions (“justify the use of a binomial model”), list the four key conditions
in short sentences. Examiners look for explicit reference to fixed n, independence, constant p, and two outcomes.
🔍 TOK Perspective:

  • How do we choose between different probability models (binomial vs. normal vs. Poisson)?
  • To what extent is a model “true”, and to what extent is it only a convenient approximation?
  • Does assigning a probability to rare events (e.g. system failures) change how society responds to risk?
🌐 Enrichment / EE Ideas:

  • Hypothesis testing using binomial models (e.g. testing if a coin or die is biased)
  • Comparing theoretical binomial predictions with experimental data
  • Investigating real-world data sets where binomial or related models appear

📌 Binomial Distribution — Practice Questions

Multiple Choice Questions (MCQs)

MCQ 1
Which of the following situations can be appropriately modelled using a binomial distribution?

  • A. The time taken for students to finish an exam
  • B. The number of heads obtained when a fair coin is tossed 10 times
  • C. The heights of students in a class
  • D. The daily temperature in a city
Answer & Explanation

Correct answer: B

A binomial distribution applies when there is a fixed number of trials, each trial has two outcomes
(success or failure), the probability of success is constant, and trials are independent.
Tossing a fair coin 10 times satisfies all these conditions.
The other options involve continuous data or outcomes that are not binary.


MCQ 2
Which situation does NOT satisfy the assumptions of a binomial distribution?

  • A. Inspecting 15 light bulbs and counting how many are defective
  • B. Rolling a fair die 12 times and counting the number of sixes
  • C. Selecting 5 cards without replacement and counting the number of red cards
  • D. Surveying 20 people and recording whether they prefer tea or coffee
Answer & Explanation

Correct answer: C

A binomial model requires the probability of success to remain constant.
When cards are drawn without replacement, the probability of drawing a red card changes after each draw,
so the trials are not independent.
The other scenarios maintain independence and a fixed probability of success.


MCQ 3
For a binomial random variable X ~ B(n, p), which expression gives P(X = k)?

  • A. nCk · p · (1 − p)
  • B. nCk · pk
  • C. nCk · pk · (1 − p)n − k
  • D. pk · (1 − p)k
Answer & Explanation

Correct answer: C

The binomial probability formula is:

P(X = k) = nCk · pk · (1 − p)n − k

This accounts for the number of ways k successes can occur, the probability of k successes,
and the probability of the remaining failures.


Long Answer Questions

Long Question 1

A factory produces electronic components. Each component has a probability of 0.08 of being defective.
A quality inspector randomly selects 12 components.

(a) Define a suitable random variable X and state its distribution.
(b) Find the probability that exactly 2 components are defective.
(c) Find the probability that at least 1 component is defective.

Full Solution

(a) Definition of random variable

Let X be the number of defective components among the 12 selected.
Since there are a fixed number of trials, two outcomes per trial, constant probability, and independence,
X follows a binomial distribution:

X ~ B(12, 0.08)

(b) Probability that exactly 2 are defective

P(X = 2) = 12C2 · (0.08)2 · (0.92)10

12C2 = 66
P(X = 2) ≈ 66 × 0.0064 × 0.434 ≈ 0.183

(c) Probability that at least 1 is defective

P(X ≥ 1) = 1 − P(X = 0) = 1 − (0.92)12 ≈ 0.632


Long Question 2

A basketball player has a probability of 0.75 of scoring a free throw.
The player attempts 8 free throws in a game.

(a) State two assumptions required for a binomial model and explain why they are satisfied.
(b) Calculate the probability that the player scores exactly 6 free throws.
(c) Calculate the probability that the player scores fewer than 6 free throws.

Full Solution
  • Each free throw results in either a score or a miss.
  • The probability of scoring remains constant at 0.75.
  • The attempts are independent.

Let X ~ B(8, 0.75).

P(X = 6) = 8C6 · (0.75)6 · (0.25)2 ≈ 0.311

P(X < 6) = 1 − [P(X = 6) + P(X = 7) + P(X = 8)] ≈ 0.322