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

Tree Diagrams

A tree diagram lays out a multi-stage probability problem branch by branch. Multiply along a path to get the joint probability of that sequence; add up the paths that lead to your event for the final answer. Trees handle “without replacement”, conditional probabilities, and three-stage problems far more cleanly than algebra alone.

📘 What you need to know

Each stage = a set of branches; the probabilities at every fork must sum to 1.
Multiply along a path → joint probability of that exact sequence of outcomes.
Add up paths that produce your event → final probability.
With replacement: second-stage branches mirror the first (independent).
Without replacement: second-stage branches are conditional probabilities (recount what’s left).
Complement trick: P(at least one X) = 1 − P(no X on any path).
Order matters: stage 1 outcomes change stage 2 probabilities when events depend on each other.
Trees feed straight into Bayes’ theorem — joint probabilities live on the paths.

Reading and building a tree

Stage 1 is the leftmost set of branches. Each leaf at the end of stage 1 then sprouts a new set of branches for stage 2 — and so on for any further stages. The probability you write on a stage-2 branch is always conditional on getting to that node from stage 1.

Along a path

multiply

joint probability of the whole sequence

Across paths

add

probability of the event combining all favourable paths

Quick check: at any single fork, the branch probabilities must add to 1. If they don’t, you’ve mislabeled — fix it before computing.

With replacement vs without replacement

Setup	Stage 2 branches	What changes
With replacement	identical to stage 1	nothing — items independent
Without replacement	conditional on stage 1	total drops by 1; the chosen category drops by 1
Independent events	same probabilities every stage	nothing — fully unrelated trials

Path probability — no replacement P(A₁ ∩ A₂) = P(A₁) · P(A₂ | A₁)

This is just the multiplication rule for conditional probability — applied along one branch of the tree. For three or more stages, keep multiplying along the path, with each conditional based on every preceding outcome.

Reverse a tree using conditional probability

If a question gives you the second-stage probabilities and asks for the probability of a stage-1 event given a stage-2 outcome, that’s a conditional going backwards. Compute the joint paths, sum the relevant ones for the denominator, and pick the target path for the numerator.

Reverse-direction conditional from a tree P(A₁ | B₂) = target pathsum of all paths reaching B₂

This is the same engine as Bayes’ theorem — when there are exactly three stage-1 categories, it’s literally the three-event Bayes formula written out from a tree.

🧭 Recipe — solving with a tree diagram

Identify the stages and the outcomes at each stage.
Draw and label every branch — check each fork sums to 1.
Multiply along each path you need to get its joint probability.
Add the path probabilities for every favourable outcome.
Sense-check: result must be in [0, 1]; sum of all leaf paths = 1.

Worked examples

WE 1

Two-stage tree with replacement

A bag contains 4 red balls and 6 blue balls. A ball is drawn at random, then replaced. A second ball is then drawn. Find the probability that both balls drawn are red.

With replacement → stage 2 mirrors stage 1 P(R₁) = 4/10, P(R₂) = 4/10 Multiply along the path (R, R) P(both red) = 4/10 × 4/10 = 16/100 = 4/25 P(both red) = 4/25 = 0.16 “with replacement” = independent stages = same probabilities throughout

WE 2

Without replacement — adding two paths

A box contains 5 white socks and 3 black socks. Two socks are drawn at random without replacement. Find the probability that both socks are the same colour.

Stage 1: 8 socks total. Stage 2: 7 left, recount the chosen colour Path (W, W): both white P(W₁) × P(W₂ | W₁) = 5/8 × 4/7 = 20/56 = 5/14 Path (B, B): both black P(B₁) × P(B₂ | B₁) = 3/8 × 2/7 = 6/56 = 3/28 Add the two favourable paths P(same colour) = 20/56 + 6/56 = 26/56 = 13/28 P(same colour) = 13/28 ≈ 0.464 without replacement → recount each branch from what’s left in the box

WE 3

“At least one” — use the complement

A fair coin is flipped and a fair six-sided die is rolled. The two events are independent. Find the probability of getting heads OR rolling a 6 (or both).

“At least one” → complement (no heads AND no 6) P(T) = 1/2, P(not 6) = 5/6 Multiply along the “neither” path P(neither) = 1/2 × 5/6 = 5/12 Subtract from 1 P(at least one) = 1 − 5/12 = 7/12 P(H or 6) = 7/12 ≈ 0.583 complement is faster than summing 3 favourable paths (H,6 / H,not6 / T,6)

WE 4

Reverse the conditional from a tree

At a school, 60% of students take the bus and 40% walk. The probability of arriving late is 0.05 for bus-takers and 0.15 for walkers. Given that a student arrived late, find the probability they took the bus.

Step 1: Joint probabilities along each path P(bus ∩ late) = 0.6 × 0.05 = 0.03 P(walk ∩ late) = 0.4 × 0.15 = 0.06 Step 2: P(late) = sum of paths leading to “late” P(late) = 0.03 + 0.06 = 0.09 Step 3: Conditional in reverse direction P(bus | late) = P(bus ∩ late) / P(late) = 0.03 / 0.09 = 1/3 P(took bus | late) = 1/3 numerator = the target path; denominator = all paths reaching the same stage-2 outcome

WE 5

Without replacement — different outcomes

A jar contains 7 orange sweets and 3 lemon sweets. Two sweets are drawn at random without replacement. Find the probability that the two sweets are different colours.

Two favourable paths: (O, L) and (L, O) Path (O, L) 7/10 × 3/9 = 21/90 = 7/30 Path (L, O) 3/10 × 7/9 = 21/90 = 7/30 Add P(different) = 7/30 + 7/30 = 14/30 = 7/15 P(different colours) = 7/15 ≈ 0.467 “different colours” can happen two ways — both must be added

WE 6

Three-stage tree

A school admissions process has three rounds. The probability an applicant passes round 1 is 0.6. Given they pass round 1, the probability of passing round 2 is 0.5. Given they pass round 2, the probability of being accepted in round 3 is 0.4. Find the probability that a randomly chosen applicant is accepted.

Acceptance requires passing ALL THREE rounds — multiply along the single success path P(R₁ pass) = 0.6 P(R₂ pass | R₁) = 0.5 P(R₃ pass | R₂) = 0.4 Multiply along the path P(accepted) = 0.6 × 0.5 × 0.4 = 0.12 P(accepted) = 0.12 three-stage chain → just keep multiplying; only branches you need have to be drawn

💡 Top tips

Sum at each fork = 1: a quick label check that catches errors before you compute.
“At least one” → complement: usually one path (the “all bad”) instead of summing several.
Without replacement → recount what’s left: don’t carry stage-1 probabilities into stage 2.
Don’t over-draw: only branches whose outcomes you care about need to appear.
Joint = numerator, marginal = denominator: that’s how trees feed into reverse-direction conditionals (and Bayes).

⚠ Common mistakes

Adding probabilities along a single path — multiplication is the rule along; addition is across.
Multiplying probabilities across different paths — different paths are alternatives, so they’re added.
Using stage-1 probabilities for stage 2 without replacement — second branches must condition on the first.
Branches at a fork not summing to 1 — usually a sign you’ve left a category out.
Using a tree for non-sequential events — for static “and/or” questions on overlapping sets, a Venn is cleaner.

Next: Probability & Types of Events — the chapter opener that sets all the vocabulary (sample space, outcome, event, complement, intersection, union) on a firm footing. Worth circling back to once the sub-topics make sense in practice.

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