IB Maths AI SL Probability Paper 1 & 2 Tree diagrams ~7 min read

Tree Diagrams

A tree diagram lays out a sequence of events — draw a counter, then another; flip, then flip again. Each set of branches splits into an event and its complement. The two golden rules: multiply along a path to get the probability of a combined outcome, and add between paths to get the probability of an event. Conditional probabilities sit naturally on the second set of branches.

📘 What you need to know

Branches: each split shows an event and its complement (e.g. A and A‘). The probabilities on one split add to 1.
Multiply ALONG branches: P(A ∩ B) = P(A) × P(B|A) — trace one path, multiply the probabilities.
Add BETWEEN paths: to get the probability of an event, add the combined outcomes that include it.
Second branches can be conditional: their probabilities may depend on the first outcome (especially without replacement).
With replacement: the second branches match the first. Without replacement: the second branches change (totals shrink).
All end-outcomes sum to 1: the four (or more) path probabilities at the tips always total 1 — a built-in check.

How a tree diagram works

Read a tree left to right. The first split is the first event; each branch then splits again for the second event. Follow any path from start to tip and you’ve described one combined outcome.

Each path from start to tip is one combined outcome; its probability is the product of the branch probabilities along the way. To find the probability of an event (say “B happens”), add the path probabilities that end in B: P(B) = P(A∩B) + P(A‘∩B).

Multiplying and adding — the two operations

Multiply along branches to get a joint outcome. The second branch is a conditional, so a path multiplies P(first) by P(second given first):

The two tree operations P(A ∩ B) = P(A) × P(B|A) P(B) = P(A ∩ B) + P(A‘ ∩ B)

Add between paths to get an event. “B happened” could occur via the A route or the A‘ route — so add both. For “at least one”, it’s often faster to use the complement: P(at least one) = 1 − P(none).

Two ways to find a union: P(A ∪ B) = P(A ∩ B) + P(A ∩ B‘) + P(A‘ ∩ B) — add the three paths that include something. Or P(A ∪ B) = 1 − P(A‘ ∩ B‘) — subtract the one “neither” path.

With replacement, without replacement, and reversing

With replacement: the item goes back, so the second set of branches has the SAME probabilities as the first. Without replacement: the item is kept, so the second branches change — the total shrinks and counts adjust.

Reversing a tree: if a question asks P(first event | second event) — e.g. “given the item is faulty, which machine made it?” — you can’t read it straight off. Use the conditional formula: P(first | second) = P(first ∩ second) ÷ P(second), where P(second) is found by adding paths.

🧭 Recipe — any tree diagram question

Draw the first split: the first event and its complement, with their probabilities (which add to 1).
Draw the second split from each branch. Check replacement: same probabilities if replaced, adjusted if not.
Multiply along each path to get the joint probability of each end-outcome. Check they all sum to 1.
For an event, add the path probabilities that contain it. For “at least one”, use 1 − P(none).
For a reversed conditional, use P(first | second) = P(first ∩ second) / P(second).

Worked examples

WE 1

Independent events on a tree

A driver passes two traffic lights. Each light is green with probability 0.6, independently of the other. Find (a) P(both green), (b) P(both red), (c) P(exactly one green).

independent ⇒ second branches = first branches P(green) = 0.6, P(red) = 0.4 (a) both green — multiply along the path P(GG) = 0.6 × 0.6 = 0.36 (b) both red P(RR) = 0.4 × 0.4 = 0.16 (c) exactly one green — two paths (GR and RG) P(GR) = 0.6 × 0.4 = 0.24 P(RG) = 0.4 × 0.6 = 0.24 P(exactly one) = 0.24 + 0.24 = 0.48 (a) 0.36 · (b) 0.16 · (c) 0.48 check: 0.36 + 0.16 + 0.48 = 1 ✓. “Exactly one” always has TWO paths — green-then-red AND red-then-green.

WE 2

Two draws WITH replacement

A bag has 7 red and 3 blue counters. A counter is drawn, its colour noted, and replaced; then a second is drawn. Find (a) P(both red), (b) P(both blue), (c) P(one of each colour).

WITH replacement ⇒ probabilities stay the same P(red) = 7/10, P(blue) = 3/10 on BOTH draws (a) both red P(RR) = 7/10 × 7/10 = 49/100 (b) both blue P(BB) = 3/10 × 3/10 = 9/100 (c) one of each — two paths P(RB) + P(BR) = 21/100 + 21/100 = 42/100 = 21/50 (a) 49/100 · (b) 9/100 · (c) 21/50 check: 49/100 + 9/100 + 42/100 = 100/100 = 1 ✓. With replacement, the bag is restored, so the second draw is identical to the first.

WE 3

Two draws WITHOUT replacement

A bag has 5 green and 3 yellow marbles. Two are drawn at random without replacement. Find (a) P(both green), (b) P(both yellow), (c) P(one of each colour).

WITHOUT replacement ⇒ second branches change (a) both green P(GG) = 5/8 × 4/7 = 20/56 = 5/14 (b) both yellow P(YY) = 3/8 × 2/7 = 6/56 = 3/28 (c) one of each — two paths P(GY) = 5/8 × 3/7 = 15/56 P(YG) = 3/8 × 5/7 = 15/56 P(one of each) = 15/56 + 15/56 = 30/56 = 15/28 (a) 5/14 · (b) 3/28 · (c) 15/28 notice the second-draw denominators are 7, not 8 — one marble has been removed. The numerator drops too if you drew that colour first (5 → 4 green).

WE 4

Find an event’s probability by adding paths

A factory has two machines. Machine A makes 70% of items, Machine B makes 30%. Of Machine A’s items, 3% are faulty; of Machine B’s items, 8% are faulty. Find the probability that a randomly chosen item is faulty.

first branches: which machine P(A) = 0.7, P(B) = 0.3 second branches: faulty given the machine P(faulty|A) = 0.03, P(faulty|B) = 0.08 multiply along the two “faulty” paths P(A ∩ faulty) = 0.7 × 0.03 = 0.021 P(B ∩ faulty) = 0.3 × 0.08 = 0.024 add the paths that end in “faulty” P(faulty) = 0.021 + 0.024 = 0.045 P(faulty) = 0.045 “faulty” can be reached two ways — via Machine A or via Machine B. Add BOTH paths. This adding-of-paths is how you collapse a tree down to a single event probability.

WE 5

“At least one” — use the complement

A student takes 3 buses to get to school. Each bus is late with probability 0.2, independently of the others. Find the probability that at least one bus is late.

“at least one late” is awkward directly — flip it opposite event = “no bus is late” (all 3 on time) P(a bus is on time) = 1 − 0.2 = 0.8 multiply along the all-on-time path P(none late) = 0.8 × 0.8 × 0.8 = 0.512 use the complement P(at least one late) = 1 − 0.512 = 0.488 P(at least one late) = 0.488 “at least one” via direct counting would need 7 separate paths (1, 2 or 3 late). The complement — ONE path for “none late” — is far quicker. Always reach for it.

WE 6

Reversing a tree — a “given” question

Using the factory from WE 4 (Machine A 70%, Machine B 30%; faulty rates 3% and 8%): given that a randomly chosen item is faulty, find the probability it was made by Machine B.

want P(B | faulty) — this REVERSES the tree use the conditional formula P(B | faulty) = P(B ∩ faulty) / P(faulty) from WE 4: P(B ∩ faulty) = 0.024 P(faulty) = 0.045 substitute P(B | faulty) = 0.024 / 0.045 = 8/15 ≈ 0.533 P(B | faulty) = 8/15 ≈ 0.533 the tree gives P(faulty | machine) directly; the REVERSE, P(machine | faulty), needs the conditional formula. Numerator = the one matching path; denominator = the total probability of the condition (all paths to “faulty”).

💡 Top tips

Multiply along, add between: one path ⇒ multiply; combining paths for an event ⇒ add.
Check the tips sum to 1: every end-outcome probability added together must equal 1 — a free error-check.
“At least one” ⇒ complement: P(at least one) = 1 − P(none). One path beats counting many.
Without replacement: adjust the second branches — the denominator drops by 1 and the relevant count drops too.
Don’t draw the whole tree if you don’t need it: if only one colour matters, draw just “that colour” and “not that colour” branches.

⚠ Common mistakes

Adding along a path instead of multiplying: a single combined outcome needs the branch probabilities MULTIPLIED, not added.
Forgetting the second “exactly one” path: “one of each” / “exactly one” has TWO routes — include both.
Keeping the same probabilities without replacement: the second draw’s totals must shrink when an item isn’t replaced.
Reading a reversed conditional straight off the tree: P(machine|faulty) is NOT a branch — use P(first ∩ second) / P(second).
Branch probabilities not summing to 1: each split (an event and its complement) must total 1 — if not, recheck.

That completes the Probability chapter. You’ve built the full toolkit: the language of events and sample spaces, the special cases of independent and mutually exclusive events, conditional probability, and the two great organising pictures — Venn diagrams for “AND/OR/NOT” and tree diagrams for sequences. Exam questions mix and match these freely, so the instinct to ask “Venn or tree? with or without replacement? is a condition involved?” is what carries you through. Next chapter, the maths turns geometric — but the habit of drawing the situation first stays just as valuable.

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 →

IB Demystified is a trusted online learning platform led by certified IB examiners and educators.