IB Maths AA HL
Topic 4 — Statistics & Probability
Paper 1 & 2
SL & HL
~8 min read
Probability & Types of Events
Probability is just careful counting dressed in formal language. Once you’re fluent in the vocabulary — experiment, outcome, event, sample space — the formulas almost write themselves. This note pins down that vocabulary and the four most-used formulas: complement, union, intersection, and expected number of occurrences.
📘 What you need to know
- Sample space U = the set of all possible outcomes. n(U) = how many there are.
- Event A = a chosen collection of outcomes. n(A) = how many outcomes are in it.
- Theoretical probability (equally likely outcomes): P(A) = n(A)n(U)
- Experimental probability (a.k.a. relative frequency) = frequency of the outcome ÷ number of trials.
- Expected number of occurrences in n trials with probability p = np. (An average, not a guarantee.)
- Complement: P(A) + P(A′) = 1, so P(A′) = 1 − P(A).
- Union (A or B or both): P(A ∪ B) = P(A) + P(B) − P(A ∩ B).
- Intersection (A and B): P(A ∩ B) = P(A) · P(B | A).
- Conditional P(A | B) = “the probability of A given B has happened” — covered fully in the next note.
The vocabulary of probability
Before any formulas, get the words right — examiners use them precisely and so should you.
| Word | What it means | Example (rolling a fair die) |
|---|
| Experiment | A repeatable activity with an observable result. | Rolling the die once. |
| Trial | One single repeat of the experiment. | One roll. |
| Outcome | A possible result of one trial. | 1, 2, 3, 4, 5, or 6. |
| Event | One outcome — or a group of outcomes — you care about. | A = “rolled an even number” = {2, 4, 6}. |
| Sample space U | The set of all possible outcomes. | U = {1, 2, 3, 4, 5, 6}, so n(U) = 6. |
Why this matters: almost every probability question reduces to “list the sample space, count the outcomes in the event, divide”. If you can describe the sample space cleanly, you’ve already done the hard part.
Calculating a basic probability
When every outcome in the sample space is equally likely, the probability of an event is simply the share of outcomes it contains:
Theoretical probability (equally likely outcomes)
P(A) = n(A)n(U)
If outcomes are not equally likely, you fall back to data — counting how often the outcome actually appeared in repeated trials:
Experimental probability (relative frequency)
P(A) ≈ frequency of Atotal number of trials
And if you want to predict how often an event will occur in n trials, multiply:
Expected number of occurrences
Expected count = n · p
“Expected” doesn’t mean it will happen exactly that many times. Toss a fair coin 10 times — the expected number of heads is 5, but you won’t be shocked to see 4 or 6. It’s the long-run average.
Combining events: AND, OR, NOT, GIVEN
Intersection (AND)
A ∩ B
both A and B happen
Union (OR)
A ∪ B
A, B, or both happen
Complement (NOT)
A′
A does not happen
Conditional (GIVEN)
A | B
A happens given B already has
The complement formula
P(A) + P(A′) = 1 ⟹ P(A′) = 1 − P(A)
The complement trick is one of the most useful shortcuts in the whole topic. Whenever a question says “at least one“, flip it to “none” and use 1 − P(none).
The union formula
P(A ∪ B) = P(A) + P(B) − P(A ∩ B)
Why subtract the intersection? Because when you add P(A) and P(B) the overlap gets counted twice — once in each — so you take it off once to get the true total.
The intersection formula
P(A ∩ B) = P(A) · P(B | A)
Read it left-to-right: “the chance that A happens, and then B happens given A already did”. This is the engine behind every tree diagram.
Both formulas are in the booklet — but knowing why each works (overlap counted twice; sequential dependence) is what saves you in tricky questions where you have to rearrange them.
🧭 Recipe — solving a basic probability question
- Identify the experiment and write down what one trial looks like.
- List the sample space U — as a list, table, or two-way grid for combined experiments.
- Count n(U), the total number of equally likely outcomes.
- Highlight the event A in your sample space and count n(A).
- Apply P(A) = n(A) / n(U) — or use complement / union / intersection if events are combined.
- Sense-check: probability must be between 0 and 1.
Worked examples
WE 1Build a sample space and find a basic probability
Two fair spinners are spun. Spinner A has sectors 1, 2, 5 and Spinner B has sectors 3, 4, 6, 8. A two-digit number is formed by writing Spinner A‘s result first and Spinner B‘s result second. Let E = “the two-digit number is even”. Find P(E).
Step 1: List sample space (two-way table)
Rows = Spinner A; Columns = Spinner B
{13, 14, 16, 18, 23, 24, 26, 28, 53, 54, 56, 58}
n(U) = 3 × 4 = 12
Step 2: Identify event E (even = ends in 4, 6, or 8)
{14, 16, 18, 24, 26, 28, 54, 56, 58} → n(E) = 9
Step 3: Apply formula
P(E) = 9/12 = 3/4
P(E) = 3/4
a two-way table is the safest way to list every outcome — never miss one
WE 2Use the complement to save time
A fair six-sided die is rolled three times. Find the probability of getting at least one six.
“At least one” → use the complement
P(at least one 6) = 1 − P(no sixes)
P(no six on one roll) = 5/6
P(no six in three rolls) = (5/6)³ = 125/216
Apply complement
P(at least one 6) = 1 − 125/216 = 91/216
P(at least one 6) = 91/216 ≈ 0.421
listing all “one six, two sixes, three sixes” cases would take five times longer
WE 3Expected number of occurrences
A factory produces light bulbs and 3% are faulty. In a batch of 500 bulbs, how many faulty bulbs would you expect?
Identify n and p
n = 500, p = 0.03
Apply expected count = np
Expected = 500 × 0.03 = 15
Expect ≈ 15 faulty bulbs
this is an average — the actual batch may have more or fewer
WE 4Use the union formula
In a class, P(student plays football) = 0.55, P(student plays basketball) = 0.40, and P(student plays both) = 0.20. Find the probability that a randomly chosen student plays at least one of the two sports.
Identify the formula
P(F ∪ B) = P(F) + P(B) − P(F ∩ B)
Substitute
= 0.55 + 0.40 − 0.20
= 0.75
P(plays at least one) = 0.75
“at least one of the two” is exactly P(F ∪ B) — the union
WE 5Intersection with conditional probability
A bag has 5 red and 3 blue marbles. Two are drawn one after the other, without replacement. Find the probability that both are red.
Identify the events
R₁ = “1st marble is red”, R₂ = “2nd marble is red”
P(R₁) — straight from the bag
P(R₁) = 5/8
P(R₂ | R₁) — one red is gone, so 4 red of 7 left
P(R₂ | R₁) = 4/7
Apply intersection formula
P(R₁ ∩ R₂) = P(R₁) · P(R₂ | R₁) = (5/8)(4/7) = 20/56 = 5/14
P(both red) = 5/14
“without replacement” is your signal that the second probability is conditional
WE 6Rearranging the union formula
40 students were surveyed: 22 own a cat, 18 own a dog, 6 own neither. Find the probability that a randomly chosen student owns both a cat and a dog.
Step 1: Find P(C ∪ D) — owns at least one
P(neither) = 6/40 = 0.15
P(C ∪ D) = 1 − 0.15 = 0.85
Step 2: Apply the union formula and rearrange
P(C ∪ D) = P(C) + P(D) − P(C ∩ D)
0.85 = 22/40 + 18/40 − P(C ∩ D)
0.85 = 0.55 + 0.45 − P(C ∩ D)
P(C ∩ D) = 1.00 − 0.85 = 0.15
P(owns both) = 0.15 (i.e. 6 students)
“neither” + “at least one” always gives the full sample — that’s how you crack it
💡 Top tips
- Always sketch the sample space first — list, two-way table, or Venn — before reaching for a formula.
- Translate words to symbols early: “and” → ∩, “or” → ∪, “not” → ′, “given” → |.
- “At least one” almost always means use the complement — it’s the single biggest time-saver.
- Probabilities live in [0, 1]. If your answer is negative or above 1, you’ve miscounted somewhere.
- Without replacement = conditional probability on the second draw.
⚠ Common mistakes
- Forgetting to subtract the intersection in the union formula — gives a probability greater than 1.
- Treating “expected number” as exact — it’s a long-run average, not a guarantee.
- Confusing P(A ∩ B) with P(A) · P(B) — that shortcut only works when events are independent (next note).
- Missing outcomes when listing the sample space — use a systematic table, not a freeform list.
- Reading “or” as exclusive — in maths, “A or B” includes “both” unless the question says otherwise.
That’s the foundation: vocabulary, the four core formulas, and the complement shortcut. The next sub-section sharpens two special cases — independent events (one doesn’t affect the other) and mutually exclusive events (they can’t both happen). Both unlock simpler versions of the formulas above.
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