IB Maths AA HL Topic 4 — Statistics & Probability Paper 1 & 2 ~7 min read

Independent & Mutually Exclusive Events

Two events can be mutually exclusive (can’t both happen) or independent (one doesn’t affect the other). Spotting which applies tells you exactly which shortcut formula to reach for — and getting that wrong is the single biggest source of lost marks in this sub-section.

📘 What you need to know

Mutually exclusive: P(A ∩ B) = 0 — they share no outcomes.
Mutually exclusive union: P(A ∪ B) = P(A) + P(B).
Independent: P(A | B) = P(A) — knowing B tells you nothing about A.
Independent intersection: P(A ∩ B) = P(A) · P(B).
Test for independence: check whether P(A) · P(B) equals P(A ∩ B).
Mutually exclusive ≠ independent — they are different ideas, not the same thing.
Without replacement ⟹ events are not independent (probabilities change after each draw).
“At least one” + independent ⟹ use the complement: 1 − P(neither happens).

The two relationships at a glance

Mutually exclusive

P(A ∩ B) = 0

can’t both happen — Venn circles don’t overlap

Independent

P(A ∩ B) = P(A) · P(B)

one tells you nothing about the other

Quick gut-check: mutually exclusive is about shape (no overlap on a Venn). Independent is about information (one outcome gives no clue about the other). Different ideas — never blur them.

Mutually exclusive events

Two events are mutually exclusive if they share no outcomes. On a six-sided die, “rolled a 1” and “rolled a 4” are mutually exclusive — but “rolled a 1” and “rolled an odd number” are not, because the outcome 1 sits in both.

Mutually exclusive — intersection & union P(A ∩ B) = 0 ⟹ P(A ∪ B) = P(A) + P(B)

The −P(A ∩ B) term in the general union formula vanishes — there’s no overlap to subtract.

Independent events

Two events are independent if knowing one happened gives you no information about the other. Two coin flips, two die rolls, drawing with replacement — these are the standard setups.

Independence — multiplication rule (also the test) P(A ∩ B) = P(A) · P(B)

This formula does double duty: use it to compute the intersection when you already know events are independent, and use it to test independence by checking whether both sides match.

Property	Mutually exclusive	Independent
Definition	can’t both happen	don’t influence each other
P(A ∩ B)	= 0	= P(A) · P(B)
Useful for	adding probabilities (union)	multiplying probabilities (intersection)
Venn picture	circles don’t overlap	circles may overlap

🧭 Recipe — testing independence

Find P(A) and P(B) from the question.
Find P(A ∩ B) — given directly, or from a Venn / table / tree.
Compute P(A) · P(B).
Compare the two values: equal ⟹ independent; not equal ⟹ not independent.
State your conclusion with both numbers shown — examiners want the comparison written out.

Worked examples

WE 1

Mutually exclusive — union

At a random moment, a traffic light shows red with probability 0.42, amber with probability 0.08, and green otherwise. Find the probability that the light is showing red or amber.

A light shows ONE colour at a time → red and amber mutually exclusive P(red ∩ amber) = 0 Apply the mutually exclusive union formula P(red ∪ amber) = P(red) + P(amber) = 0.42 + 0.08 = 0.50 P(red or amber) = 0.50 single-attribute outcomes (one colour, one face of a die) are nearly always mutually exclusive

WE 2

Independent — multiplication rule

A spinner is divided into 5 equal sectors numbered 1 to 5. The spinner is spun, and a card is drawn from a standard 52-card deck. Find the probability that the spinner lands on 3 AND the card drawn is a heart.

Two physically separate experiments → independent P(spinner = 3) = 1/5 P(heart) = 13/52 = 1/4 Apply the multiplication rule P(3 ∩ heart) = (1/5) × (1/4) = 1/20 P(3 and heart) = 1/20 = 0.05 two unrelated experiments (different objects) are almost always independent

WE 3

Find the intersection from the union

In a school survey, the probability a student plays an instrument is 0.65, the probability they do a sport is 0.50, and the probability they do at least one of the two is 0.90. Find the probability a randomly chosen student does both.

Use the general union formula P(I ∪ S) = P(I) + P(S) − P(I ∩ S) Substitute and rearrange 0.90 = 0.65 + 0.50 − P(I ∩ S) P(I ∩ S) = 1.15 − 0.90 = 0.25 P(does both) = 0.25 because P(I ∩ S) ≠ 0, the events are NOT mutually exclusive

WE 4

Test whether two events are independent

In a city survey, P(uses phone for navigation) = 0.6, P(uses public transit) = 0.25, and P(does both) = 0.15. Determine whether the two events are independent.

Compute P(A) · P(B) 0.6 × 0.25 = 0.15 Compare with P(A ∩ B) P(A ∩ B) = 0.15 (given) P(A) · P(B) = 0.15 = P(A ∩ B) ✓ The events ARE independent always SHOW the comparison — both numbers, then the conclusion

WE 5

Find a probability when events are independent

The probability it rains in city M on a given day is 0.45. The events “rain in M” and “rain in city N” are independent, and the probability it rains in both cities is 0.18. Find the probability it rains in city N.

Independent → use the multiplication rule P(M ∩ N) = P(M) · P(N) Substitute and solve 0.18 = 0.45 × P(N) P(N) = 0.18 / 0.45 = 0.4 P(N) = 0.4 if you’re TOLD events are independent, use the multiplication rule directly

WE 6

“At least one” with independent events

An archer hits the target with probability 0.75 on her first arrow and 0.60 on her second arrow. The two shots are independent. Find the probability she hits the target with at least one arrow.

“At least one” → use the complement P(at least one hit) = 1 − P(misses both) Find P(miss) for each shot P(miss 1) = 1 − 0.75 = 0.25 P(miss 2) = 1 − 0.60 = 0.40 Multiply (independent) for P(misses both) P(miss both) = 0.25 × 0.40 = 0.10 Subtract from 1 P(at least one hit) = 1 − 0.10 = 0.90 P(at least one hit) = 0.90 complement + independence is one of the highest-yield combos in the topic

💡 Top tips

Read for keywords: “with replacement” / “two unrelated experiments” → independent. “Either…or” (one option only) → mutually exclusive.
Test independence numerically — never assume it. Always compute P(A) · P(B) and compare to P(A ∩ B).
“Without replacement” kills independence — use conditional probability instead.
Show your comparison — examiners want both numbers written out before your conclusion.
Combine independence with the complement for “at least one” questions — far cleaner than summing every case.

⚠ Common mistakes

Treating mutually exclusive and independent as the same thing — they’re almost opposites for non-zero probabilities.
Multiplying P(A) · P(B) without checking independence — only valid if events truly are independent.
Adding P(A) + P(B) without subtracting the overlap — only valid if mutually exclusive.
Forgetting “without replacement” makes events dependent — leads to wrong answers in marble/card questions.
Stating “independent” without numerical justification — examiners require the P(A)·P(B) vs P(A ∩ B) comparison.

Next: Conditional Probability. The formal version of P(A | B) — the gateway to tree diagrams, Bayes’ theorem, and most of the harder probability questions on Paper 2.

Need help with Statistics & Probability?

Get 1-on-1 help from an IB examiner who knows exactly what Paper 1 & 2 are looking for.

Book Free Session →