IB Maths AA SL
Topic 4 — Probability
Paper 1 & 2
~9 min read
Independent & Mutually Exclusive Events
Two events can either affect each other or be completely separate. Independent events don’t influence each other. Mutually exclusive events can’t both happen at once. Each one simplifies the probability formulas in a really useful way.
📘 What you need to know
- Independent events: one happening doesn’t change the probability of the other.
- For independent events: P(A ∩ B) = P(A) × P(B) (in the formula booklet)
- Mutually exclusive events: they can’t both happen at the same time.
- For mutually exclusive events: P(A ∩ B) = 0, so P(A ∪ B) = P(A) + P(B) (in the formula booklet)
- Test for independence: if P(A ∩ B) = P(A) × P(B), the events are independent. Otherwise, they’re not.
- Independent and mutually exclusive are different things — don’t mix them up!
The two key types of events
Two events have a relationship — and how they relate decides which formula you use. Let’s set them up side-by-side first:
Independent
One event happening doesn’t affect the other. Knowing about A tells you nothing about B.
P(A ∩ B) = P(A) × P(B)
e.g. flipping a coin twice — the second flip doesn’t care about the first.
Mutually exclusive
Both events can’t happen at the same time. If one occurs, the other can’t.
P(A ∩ B) = 0
e.g. on a single dice roll, you can’t get both a 6 AND a 3.
Independent and mutually exclusive sound similar but mean very different things. Independent = “they don’t influence each other”. Mutually exclusive = “they can’t both happen”. Two events can be one, the other, neither — but rarely both.
Mutually exclusive events
Two events are mutually exclusive if they cannot both happen on the same trial.
Examples:
- Rolling a 6 AND rolling a 3 on a single dice roll → can’t both happen.
- A coin landing on heads AND on tails → impossible to happen together.
- “Event A” and its complement “Event A'” → by definition, mutually exclusive.
The formulas for mutually exclusive events
🤔 Why does this simplify the OR formula?
Recall from the last note: P(A ∪ B) = P(A) + P(B) − P(A ∩ B). Since the events are mutually exclusive, the overlap is 0. So you can just add the probabilities — no subtraction needed.
Independent events
Two events are independent if knowing one happened tells you nothing about whether the other did. Their probabilities don’t influence each other.
Examples:
- Flipping a coin twice — the result of flip 2 doesn’t depend on flip 1.
- Picking a card from one deck and rolling a dice — totally separate processes.
- Drawing a ball with replacement — the second draw “starts fresh”.
The formula for independent events
This is much simpler than the general AND formula because there’s no conditional probability to worry about — P(B|A) is just P(B) when events are independent.
📍Two equivalent ways to test independence
Method 1: Check if P(A ∩ B) = P(A) × P(B). Method 2: Check if P(A|B) = P(A) (the conditional probability of A equals the regular probability of A). Either method works — pick whichever has the values you’ve already got.
Independent vs Mutually exclusive — at a glance
| Independent | Mutually exclusive |
|---|
| Can both happen? | Yes | No |
| P(A ∩ B) = | P(A) × P(B) | 0 |
| P(A ∪ B) = | P(A) + P(B) − P(A) × P(B) | P(A) + P(B) |
| Typical example | Two coin flips | Rolling a 3 and a 6 on one dice |
🧠Memory trick: “Independent multiplies. Exclusive adds.”
Independent events for AND → multiply. (P(A) × P(B))
Mutually exclusive events for OR → just add. (P(A) + P(B))
These are the two simplest cases. If neither applies, you have to use the general formulas with the overlap.
One more useful identity
For any two events A and B, you can split A into two mutually exclusive pieces:
In words: “A happens” can be broken into “A happens AND B happens” or “A happens AND B doesn’t happen”. These two pieces are mutually exclusive, so they just add up to P(A).
This identity is super handy when you’re given partial information and need to fill in a gap. It also shows up a lot in Venn diagram and tree diagram problems.
Worked examples
WE 1Find the intersection from the union
A student is chosen at random from a class. The probability they have a dog is 0.8, the probability they have a cat is 0.6, and the probability they have a cat or a dog is 0.9. Find the probability the student has both a dog and a cat.
P(D) = 0.8, P(C) = 0.6, P(D ∪ C) = 0.9
Use the union formula and rearrange to find the intersection.
Use: P(D ∪ C) = P(D) + P(C) − P(D ∩ C)
Substitute: 0.9 = 0.8 + 0.6 − P(D ∩ C)
0.9 = 1.4 − P(D ∩ C)
P(D ∩ C) = 1.4 − 0.9 = 0.5
P(D ∩ C) = 0.5
always rearrange the union formula when you’re given 3 of the 4 values!
WE 2Find P(R) given Q and R are independent
Two events Q and R are such that P(Q) = 0.8 and P(Q ∩ R) = 0.1. Given that Q and R are independent, find P(R).
P(Q) = 0.8, P(Q ∩ R) = 0.1, Q and R independent
Independence → use P(A ∩ B) = P(A) × P(B). Rearrange for P(R).
Use: P(Q ∩ R) = P(Q) × P(R)
Substitute: 0.1 = 0.8 × P(R)
Solve: P(R) = 0.10.8 = 0.125
P(R) = 0.125 or 18
independence is the key — without it, you’d need conditional probability
WE 3Find probabilities of mutually exclusive events
Two events S and T are such that P(S) = 2P(T). Given that S and T are mutually exclusive and that P(S ∪ T) = 0.6, find P(S) and P(T).
P(S) = 2P(T), P(S ∪ T) = 0.6, mutually exclusive
Mutually exclusive → P(S ∪ T) = P(S) + P(T). Use the relation P(S) = 2P(T) to solve.
Use: P(S ∪ T) = P(S) + P(T)
0.6 = P(S) + P(T)
Substitute P(S) = 2P(T):
0.6 = 2P(T) + P(T) = 3P(T)
Solve for P(T): P(T) = 0.2
Then: P(S) = 2 × 0.2 = 0.4
P(S) = 0.4 and P(T) = 0.2
two unknowns, two equations — classic exam pattern
WE 4Test whether two events are independent
For two events A and B: P(A) = 0.4, P(B) = 0.3, P(A ∩ B) = 0.12. Are A and B independent?
Test: does P(A ∩ B) = P(A) × P(B)?
Calculate P(A) × P(B): 0.4 × 0.3 = 0.12
Compare with P(A ∩ B): P(A ∩ B) = 0.12 ✓
They match!
Yes — A and B are independent (since P(A ∩ B) = P(A) × P(B))
always justify your conclusion with the matching values
WE 5Use independence to find joint probabilities
A fair coin is flipped twice. Let H1 = “heads on first flip” and H2 = “heads on second flip”.
(a) Are H1 and H2 independent? (b) Find P(H1 ∩ H2). (c) Find P(H1 ∪ H2).
Each flip is its own experiment — they don’t affect each other.part (a)
First flip doesn’t affect second flip → independent.
Yes — independentpart (b) — both heads
Use multiplication for independent events:
P(H₁ ∩ H₂) = P(H₁) × P(H₂) = 0.5 × 0.5 = 0.25
P(H₁ ∩ H₂) = 14part (c) — at least one head
Use union formula (these aren’t mutually exclusive):
P(H₁ ∪ H₂) = 0.5 + 0.5 − 0.25 = 0.75
P(H₁ ∪ H₂) = 34
independent ≠ mutually exclusive — the overlap (both heads) is 0.25, not 0!
💡 Top tips
- Independent & mutually exclusive are NOT the same thing. Independent = no influence on each other. Mutually exclusive = can’t both happen.
- For independent events: P(A ∩ B) = P(A) × P(B) — multiply.
- For mutually exclusive events: P(A ∪ B) = P(A) + P(B) — just add.
- To test independence, check if P(A ∩ B) = P(A) × P(B). Calculate both sides and compare.
- If the question states “independent”, use the multiplication formula straight away — don’t try to derive it.
- “With replacement” usually means independent. “Without replacement” usually means NOT independent.
- If two events both have non-zero probability and are mutually exclusive, they CANNOT also be independent. Knowing one happened tells you the other didn’t!
- For “show that A and B are independent” questions, you must state the criterion (P(A ∩ B) = P(A) × P(B)) and verify with numbers.
⚠ Common mistakes
- Confusing independent with mutually exclusive. They sound similar but mean opposite things — independent allows both events; mutually exclusive forbids both.
- Adding probabilities for AND. AND always involves multiplying (and possibly using P(B|A) if not independent).
- Multiplying probabilities for OR. OR involves adding (with subtraction of overlap when not mutually exclusive).
- Using P(A) × P(B) for non-independent events. Only valid when independence is stated or proven.
- Assuming mutually exclusive events are independent. They’re almost never both — knowing one happened tells you the other didn’t.
- Forgetting to subtract the overlap when events aren’t mutually exclusive in OR questions.
- Not justifying the conclusion when testing independence. You must state the criterion and verify with values.
- Treating “without replacement” as independent. The probability changes after each draw, so events are NOT independent.
Independent and mutually exclusive cover the two simplest cases. But what if events do affect each other and one already happened? That’s where the next note comes in: conditional probability — the maths of “given that…”.
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