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IB Diploma Mathematics: Analysis & Approaches HL · Statistics & Probability
Mini-Lesson

Statistics & Probability

This HL mini-lesson extends probability: conditional probability, independence, the law of total probability, Bayes' theorem, expectation and variance, and continuous random variables via probability density functions.

Answer as you go and collect ⭐ stars. Press Start when ready.

Conditional

Conditional probability

The probability of A given B is:

P(A | B) = P(A ∩ B) / P(B)rearranges to P(A ∩ B) = P(A|B)·P(B)
Quick check

Bayes

?Bayes’ theorem is used to:
Independence

Independence & mutual exclusivity

Independent: P(A ∩ B) = P(A)P(B), equivalently P(A|B) = P(A). Mutually exclusive: P(A ∩ B) = 0.

Don't confuse them: mutually exclusive events (except trivial ones) are not independent — if one happens the other cannot.

Calculate

Bayes

A disease has P(D)=0.01. A test gives P(+|D)=0.99 and P(+|D′)=0.05. Find P(D|+) to 3 d.p.
Hint: P(+) = 0.99(0.01) + 0.05(0.99); divide 0.0099 by it.
Total probability

The law of total probability

If B₁, B₂, … partition the sample space, then for any event A:

P(A) = Σ P(A | Bᵢ)·P(Bᵢ)the denominator in Bayes' theorem
Quick check

Continuous variable

?For a continuous random variable X, P(X = a) for a single value a equals:
Calculate

Conditional

Given P(A ∩ B) = 0.12 and P(B) = 0.3, find P(A | B).
Hint: P(A|B) = P(A∩B)/P(B) = 0.12/0.3.
Bayes

Bayes’ theorem

Bayes' theorem reverses the conditioning:

P(D | +) = P(+ | D)·P(D) / P(+)
Worked example

P(D) = 0.01, P(+|D) = 0.99, P(+|D′) = 0.05. Then P(+) = 0.99(0.01) + 0.05(0.99) = 0.0594, so P(D|+) = 0.0099/0.0594 = 1/6 ≈ 0.167.

Sort it

Which idea does it belong to?

Tap a formula, then its family.

🔗 Conditional / Bayes

📊 Expectation

🔔 Distribution

Expectation

Expectation & variance

For a discrete random variable:

E(X) = Σ x·P(X = x) · Var(X) = E(X²) − [E(X)]²
Calculate

Density constant

f(x) = kx is a pdf on 0 ≤ x ≤ 2. Find k so the total area is 1.
Hint: ∫₀² kx dx = k·2 = 1.
Quick check

Total probability

?If B₁,…,Bₙ partition the sample space, P(A) equals:
Continuous

Continuous random variables

A continuous variable has a probability density function f(x) with total area 1. Probabilities and expectation are integrals:

∫ f(x) dx = 1 · E(X) = ∫ x·f(x) dxand P(X = a) = 0 for any single value a
Calculate

Continuous mean

For f(x) = 0.5x on [0, 2], find E(X) = ∫₀² x·f(x) dx to 3 d.p.
Hint: 0.5 ∫₀² x² dx = 0.5 × 8/3.
Match it

Match each quantity to its formula

Tap a statement on the left, then its matching partner on the right.

Quantity
Formula
Distributions

Binomial & normal at HL

Binomial X ~ B(n, p): P(X = r) = ₙCᵣ pʳ(1−p)ⁿ⁻ʳ, E(X) = np, Var(X) = np(1−p). The normal N(μ, σ²) is standardised with z = (x − μ)/σ.

Calculate

Variance

A variable has E(X) = 2 and E(X²) = 6. Find Var(X).
Hint: Var(X) = E(X²) − [E(X)]² = 6 − 4.
Quick check

Variance

?Variance can be computed as:
Strategy

Reasoning with probability

Draw a tree or table for two-stage experiments; use total probability for the denominator and Bayes to invert a condition. For continuous variables, integrate the density.

Exam habit: check probabilities sum to 1 and lie in [0, 1]; a Bayes answer should respect the base rate.

Recap

The big ideas to know

Conditional: P(A|B)=P(A∩B)/P(B); independent if =P(A)

Total probability: P(A)=Σ P(A|Bᵢ)P(Bᵢ)

Bayes: P(D|+)=P(+|D)P(D)/P(+)

Expectation: E(X)=Σ xP(x); Var=E(X²)−E(X)²

Continuous: ∫f=1, E(X)=∫xf dx, P(X=a)=0

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