IB Maths AA SLTopic 4 — ProbabilityPaper 1 & 2~10 min read
Tree Diagrams
When events happen one after another — flip a coin, then flip again; pick a ball, then pick another — a tree diagram is the cleanest way to track every possible outcome and its probability. Multiply along branches, add between branches.
📘 What you need to know
A tree diagram shows all possible outcomes of a sequence of events as branching paths.
Multiply along branches to get the probability of a path (e.g. P(A ∩ B)).
Add between branches to combine paths leading to the same overall event.
Each set of branches must add to 1 — they cover every possible next outcome.
For “with replacement”, the second set of branches matches the first.
For “without replacement”, the second set of branches is different — use conditional probabilities.
Tree diagrams are perfect for conditional and sequential probability problems.
What is a tree diagram?
A tree diagram lays out a sequence of events as a branching structure. Each branch is one possible outcome, and you label each branch with its probability. Following a path from start to end (left to right) gives you one combined outcome — like “Heads then Heads” or “Glasses then Right-handed”.
Anatomy of a tree diagram (2 events)
A 2-event tree has 4 paths — exactly the same 4 regions as a Venn diagram. Tree diagrams make sequential events easier to follow; Venn diagrams make overlapping events easier to see.
The two essential rules
✕ Multiply ALONG branches
P(A ∩ B) = P(A) × P(B|A)
To find the probability of one specific path (e.g. A followed by B), multiply each probability along the path.
+ Add BETWEEN branches
P(B) = P(A ∩ B) + P(A‘ ∩ B)
To find the probability of an overall event (e.g. “B happens”), add the probabilities of all paths that lead to it.
🧠
Memory trick: “Across multiplies, down adds”
Move across the tree (along a path) → multiply. Move down the tree (between paths) → add. This single rule handles every tree diagram problem.
📍
Each set of branches must add to 1
If P(A) = 0.2, then P(A‘) = 0.8. The two branches are complementary — together they cover every possibility. Same goes for the second-stage branches: P(B|A) + P(B‘|A) = 1.
How to build a tree diagram
The 5-step method
Draw the first set of branches for the first event (usually 2 branches: the event and its complement).
Label each first branch with its probability.
Draw the second set of branches from each end of the first branches (4 paths total for a 2-event tree).
Label the second branches with conditional probabilities: P(B|A), P(B|A‘), etc.
Multiply along each path to find the probability of each combined outcome. Write the result at the end.
With or without replacement?
How you label the second branches depends on whether the first event affects the second:
With replacement (or independent events): the second set of branches is identical to the first.
Without replacement (or dependent events): the second set of branches is different — use conditional probabilities.
🤔 Why are without-replacement trees so common?
Bag-of-balls problems, card games, picking students from a class — anything where one selection changes what’s left is “without replacement”. The probabilities on the second branches depend on what happened in the first, which is exactly what tree diagrams display so clearly.
Which event goes on the first set of branches?
For two independent events, the order doesn’t matter. But for non-independent events:
If you have P(A|B), put B on the first branches.
If you have P(B|A), put A on the first branches.
The general rule: put the “given” event first, so its branch comes before what you condition on it.
If a question says “30% of glasses-wearers are right-handed”, that’s P(R|G). Put G on the first set of branches because the conditional info is “given G”. Then the second branches show “right-handed or not, given each glasses outcome”.
You don’t always need the full tree
If the question involves many possible outcomes (e.g. drawing colours from a multi-coloured bag), drawing every branch would be a mess. Only draw the parts you care about.
For example, if you’re only interested in red vs not red, just draw branches for “Red” and “Not Red” instead of one branch for every colour.
📍
Speed up by drawing only what’s needed
If a tree has many possible outcomes, group everything you don’t care about into a single “complement” branch. This makes the diagram smaller AND easier to compute. Don’t waste exam time drawing branches you won’t use.
Reverse conditional questions — a tree diagram favourite
A common exam pattern: you build the tree, then the question asks for a “reversed” conditional like P(A|B) when your tree shows P(B|A).
Reverse conditional formula
P(A|B) = P(A ∩ B)P(B)
The reverse-conditional method
Find P(A ∩ B) by multiplying along the right path on the tree.
Find P(B) by adding all paths that end in B (the across-then-down rule).
Divide: P(A|B) = (path) ÷ (sum of all B paths).
This pattern shows up everywhere in IB exams. The classic example (covered in the worked examples) is going from “P(right-handed | wears glasses)” to “P(wears glasses | right-handed)”.
Worked examples
WE 1
Glasses and right-handedness
20% of people in a company wear glasses. 40% of glasses-wearers are right-handed. 50% of non-glasses-wearers are right-handed.
(a) Draw a tree diagram. (b) Find the probability a randomly chosen person is right-handed. (c) Given that the person is right-handed, find the probability they wear glasses.
Let G = “wears glasses”, R = “right-handed”. The conditional info is given G, so put G on the first branches.part (a) — tree diagrampart (b) — p(r)Add ALL paths ending in R:P(R) = P(G ∩ R) + P(G’ ∩ R)P(R) = 0.2 × 0.4 + 0.8 × 0.5P(R) = 0.08 + 0.40 = 0.48P(R) = 0.48part (c) — p(g | r)Use formula:P(G | R) = P(G ∩ R)P(R)P(G | R) = 0.080.48 = 16P(G | R) = 16multiply ALONG paths, then ADD between paths — that’s the whole game!
WE 2
Without replacement (changing branches)
A bag contains 4 red and 6 blue balls. Two are drawn without replacement.
(a) Draw a tree diagram. (b) Find the probability that both balls are red. (c) Find the probability of one red and one blue (in either order).
10 balls total. After drawing 1, only 9 left → 2nd branches change.part (a) — treepart (b) — both redMultiply along the R-then-R path:410 × 39 = 1290 = 215P(both red) = 215part (c) — one of eachTwo paths: R then B, OR B then R. Add them:410 × 69 + 610 × 49= 2490 + 2490 = 4890 = 815P(one of each) = 815“in either order” → add both paths (R-B and B-R)
WE 3
“At least one” using a tree
A fair coin is flipped 3 times. Find the probability of getting at least one head.
“At least one head” = 1 − P(no heads) = 1 − P(all tails). Tree gives the all-tails path.Each flip independent: P(T) = 0.5Multiply along the all-tails path:P(TTT) = 0.5 × 0.5 × 0.5 = 0.125Use complement:P(at least 1 head) = 1 − 0.125 = 0.875P(at least 1 head) = 0.875 = 78“at least one” almost always uses the complement trick — saves drawing all 8 paths!
WE 4
Tree for independent events
The probability that Alex passes a maths test is 0.7, and the probability he passes a physics test is 0.6. The two tests are independent. Find the probability he passes both tests, and the probability he passes exactly one.
Independent → second branches stay the same regardless of the first outcome.P(M) = 0.7, P(M’) = 0.3P(P) = 0.6, P(P’) = 0.4 (same after either outcome)both passP(M ∩ P) = 0.7 × 0.6 = 0.42P(both) = 0.42exactly one passesTwo paths: M then P’, OR M’ then P:0.7 × 0.4 + 0.3 × 0.6= 0.28 + 0.18 = 0.46P(exactly one) = 0.46“exactly one” = pass-fail OR fail-pass — add both paths!
WE 5
Reverse conditional probability
A factory has two machines. Machine A produces 60% of items with a 5% defect rate. Machine B produces 40% with a 2% defect rate. An item is found to be defective. What’s the probability it came from Machine A?
We have P(D|A) and P(D|B), but we want P(A|D). Use the reverse-conditional formula!Path A → D:0.6 × 0.05 = 0.03Path B → D:0.4 × 0.02 = 0.008Total P(D):0.03 + 0.008 = 0.038Apply formula:P(A | D) = P(A ∩ D)P(D) = 0.030.038≈ 0.789P(A | D) ≈ 0.789classic exam pattern: tree gives P(D|A), then question flips it to P(A|D)
💡 Top tips
Multiply ALONG branches, ADD between paths. The single golden rule of tree diagrams.
Each set of branches must add to 1. If they don’t, you’ve misread something.
Without replacement = different second branches. Recount what’s left after the first event.
For independent events, the second set of branches is identical to the first.
Write the path probability at the end of each path — makes adding/comparing easier.
For “at least one” questions, use 1 − P(none) — it’s much faster than tracing every “at least one” path.
Don’t draw every branch if there are too many. Bundle uninteresting outcomes into a “complement” branch.
For reverse conditional P(A|B): the bottom is “all paths to B“, the top is “the specific path through A“.
⚠ Common mistakes
Adding along a path instead of multiplying. Each branch is “AND” — multiply. Adding gives nonsense above 1.
Multiplying between paths instead of adding. Different paths are mutually exclusive — they add.
Using the same probabilities on both branches when the events aren’t independent. Without replacement → recount!
Forgetting that branches sum to 1. If P(A) = 0.3, then P(A‘) = 0.7, not 0.3.
Tracing only one path when several paths satisfy the event. “Exactly one of two events” usually has 2 paths.
Confusing “at least one” with “exactly one”. At least one = 1, 2, or more. Exactly one = exactly that count.
Putting the wrong event first when conditional info is given. The “given” event goes on the FIRST set of branches.
Drawing a fully-detailed tree for a multi-coloured bag. Group everything you don’t care about into a “not red” branch.
🎉 You’ve finished the entire Probability subtopic! You can list outcomes, combine events, test independence, find conditional probabilities, and visualise everything with Venn or tree diagrams. The next chunk of Topic 4 dives into discrete random variables — when probability meets numerical outcomes.
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