Two special situations simplify the general probability formulas. Mutually exclusive events can’t both happen, so the overlap is zero and union becomes a clean sum. Independent events don’t affect each other, so the intersection is a clean product. Both have tests you can apply to check if events qualify, and both unlock shortcut formulas the exam loves.
Two events are mutually exclusive if they share no outcomes. Examples: rolling a 1 and rolling a 6 on the same die; getting a heart and getting a spade on the same card draw.
Two events are independent if the occurrence of one doesn’t change the probability of the other. Examples: tossing two coins (the first doesn’t influence the second); rolling two separate dice.
Independence test: compute P(
A) × P(
B). If it equals the given P(
A ∩
B), the events are independent. If they differ, the events are NOT independent.
Knock-on effect: if A and B are independent, then so are A‘ and B‘, A and B‘, A‘ and B. Once you’ve established independence, you can multiply probabilities of any combination of the events and their complements.
🧠Recipe — identifying and using the right formula
- Read the question: do the events involve separate trials / independent objects? ⇒ likely independent. Are they two outcomes of the same trial that exclude each other? ⇒ likely mutually exclusive.
- If asked to TEST: for mutual exclusivity check P(A ∩ B) = 0; for independence check P(A)P(B) = P(A ∩ B).
- Choose the right formula: mutually exclusive ⇒ P(A ∪ B) = P(A) + P(B). Independent ⇒ P(A ∩ B) = P(A)P(B).
- For “exactly one” questions: use P(only A) + P(only B), or use 1 − P(both) − P(neither).
- Always sanity-check: the sum of P(both), P(only A), P(only B), P(neither) must equal 1.
Worked examples
WE 1Test for mutual exclusivity
Two events A and B satisfy P(A) = 0.3, P(B) = 0.4, P(A ∪ B) = 0.7. (a) Decide whether A and B are mutually exclusive. (b) Repeat for the alternative case P(A) = 0.3, P(B) = 0.5, P(A ∪ B) = 0.65.
use P(A ∩ B) = P(A) + P(B) − P(A ∪ B)
(a) substitute
P(A ∩ B) = 0.3 + 0.4 − 0.7 = 0
since P(A ∩ B) = 0 ⇒ mutually exclusive ✓
(b) substitute
P(A ∩ B) = 0.3 + 0.5 − 0.65 = 0.15
P(A ∩ B) ≠ 0 ⇒ NOT mutually exclusive
(a) yes, mutually exclusive · (b) no, overlap = 0.15
the test is purely algebraic: rearrange the union formula to isolate P(A ∩ B) and check whether it’s zero. The CONTEXT might suggest the answer, but the numbers decide it.
WE 2Test for independence
Events A and B have P(A) = 0.4, P(B) = 0.5, P(A ∩ B) = 0.2. Determine whether A and B are independent.
test: compare P(A)·P(B) with given P(A ∩ B)
P(A)·P(B) = 0.4 × 0.5 = 0.20
P(A ∩ B) = 0.20 (given)
0.20 = 0.20 ✓
A and B are INDEPENDENT
the test is a single multiplication. If the product P(A)P(B) matches the given intersection, you have independence. If not, you have dependence — one event affects the other.
WE 3Use independence to find a missing probability
Events A and B are independent. P(A) = 0.4 and P(A ∩ B) = 0.12. (a) Find P(B). (b) Find P(A ∪ B).
(a) use P(A ∩ B) = P(A)·P(B)
0.12 = 0.4 × P(B)
P(B) = 0.12 / 0.4 = 0.3
(b) standard union formula
P(A ∪ B) = P(A) + P(B) − P(A ∩ B)
= 0.4 + 0.3 − 0.12
= 0.58
(a) P(B) = 0.3 · (b) P(A ∪ B) = 0.58
notice “independent” is the key word that unlocks P(A ∩ B) = P(A)P(B). Without it, you cannot get P(B) from these numbers alone.
WE 4Mutually exclusive with a probability ratio
Events X and Y are mutually exclusive, with P(X) = 3 P(Y) and P(X ∪ Y) = 0.8. Find P(X) and P(Y).
mutually exclusive ⇒ P(X ∪ Y) = P(X) + P(Y)
P(X) + P(Y) = 0.8
use the ratio P(X) = 3 P(Y)
3 P(Y) + P(Y) = 0.8
4 P(Y) = 0.8
P(Y) = 0.2
find P(X)
P(X) = 3 × 0.2 = 0.6
P(Y) = 0.2, P(X) = 0.6
“mutually exclusive” eliminates the − P(A ∩ B) term in the union formula, turning two unknowns into a clean linear equation.
WE 5Independent components — multi-stage problem
A factory tests components; each is defective with probability 0.05, independently of the others. Two components are tested. Find (a) P(both defective), (b) P(neither defective), (c) P(exactly one defective).
independence ⇒ multiply
(a) both defective
P(both) = 0.05 × 0.05 = 0.0025
(b) neither defective — use complement
P(not def) = 1 − 0.05 = 0.95
P(neither) = 0.95 × 0.95 = 0.9025
(c) exactly one defective
P(D, ND) = 0.05 × 0.95 = 0.0475
P(ND, D) = 0.95 × 0.05 = 0.0475
P(exactly one) = 0.0475 + 0.0475 = 0.095
(a) 0.0025 · (b) 0.9025 · (c) 0.095
sanity check: 0.0025 + 0.9025 + 0.095 = 1 ✓. The three outcomes (both, neither, exactly one) are mutually exclusive and exhaust the sample space.
WE 6Splitting an event using P(A) = P(A ∩ B) + P(A ∩ B‘)
In a sixth-form class, 60% of students take French and 35% take BOTH French and German. Find the probability that a randomly chosen student takes French but NOT German.
let F = French, G = German
split F into “French AND German” + “French only”
P(F) = P(F ∩ G) + P(F ∩ G’)
0.60 = 0.35 + P(F ∩ G’)
P(F ∩ G’) = 0.25
P(French only) = 0.25 (i.e. 25% of students)
this splitting works for ANY two events — no need for independence or mutual exclusivity. Every outcome in F is either ALSO in G or NOT in G; nothing else is possible.
💡 Top tips
- Spot the keyword: “mutually exclusive” / “cannot both” ⇒ subtraction-free union. “Independent” / “doesn’t affect” / “separate trials” ⇒ multiplication for intersection.
- Two tests, two formulas: P(A ∩ B) = 0 tests mutual exclusivity; P(A)P(B) = P(A ∩ B) tests independence.
- Use the complement: “at least one” = 1 − “neither”. For independent events P(neither) = P(A‘)P(B‘).
- Build the four-way split: for any pair of events, P(both) + P(only A) + P(only B) + P(neither) = 1. Use this as a self-check.
- Mutually exclusive events with positive probability are NOT independent — one happening forces the other to NOT happen, which is the opposite of “doesn’t affect”.
âš Common mistakes
- Treating “mutually exclusive” and “independent” as the same: they’re opposite extremes. Mutually exclusive ⇒ total overlap = 0; independent ⇒ specific positive overlap = P(A)P(B).
- Multiplying P(A) and P(B) without checking independence: this only works for independent events. Otherwise use P(A ∩ B) = P(A)P(B|A).
- Adding P(A) + P(B) without subtracting overlap: this only works for mutually exclusive events. Otherwise the union formula’s −P(A ∩ B) is essential.
- Forgetting “exactly one” has TWO arrangements: P(A, then B‘) AND P(A‘, then B). Add both.
- Using the wrong P(A) value after a conditional event: independence preserves P(A); without independence, P(A|B) generally differs from P(A).
Next up: Conditional Probability — the formula P(A|B) = P(A ∩ B) / P(B). When events aren’t independent, “A given that B has happened” is genuinely different from P(A). The next note treats this rigorously and shows how a reduced sample space makes the maths intuitive.
Need help with AI SL Probability?
Get 1-on-1 help from an IB examiner who knows exactly what Paper 1 & 2 are looking for.
Book Free Session → 