A tree diagram maps out a sequence of events — first this, then that. Each set of branches shows the options at one stage, with their probabilities. Two rules do everything: multiply along the branches to get the probability of a path (an “and”), and add the paths that make up an event (an “or”). They’re the natural tool for multi-stage and “with/without replacement” problems — and the second set of branches can carry conditional probabilities.
📘 What you need to know
A tree diagram shows the outcomes of combined events stage by stage.
Each set of branches covers mutually exclusive options (usually an event and its complement) and sums to 1.
Multiply along branches for the probability of a path: P(A ∩ B) = P(A) × P(B|A).
Add the relevant paths for an event (the “or” combinations).
With replacement: the second branches are the same as the first. Without replacement: they change (conditional).
Useful shortcut: P(at least one) = 1 − P(none).
You don’t always need the full tree — draw only the branches you care about (e.g. “red” vs “not red”).
Conditional questions: combine the tree with P(A|B) = P(A ∩ B)P(B).
The two rules: multiply along, add across
Along a branch
multiply (AND)
Follow one path through the tree → multiply the probabilities. Gives P of that combined outcome.
Across paths
add (OR)
An event made of several end-points → add those paths’ probabilities together.
tree diagram structure
Each end-point is found by multiplying along its path. The four end-points cover every combined outcome and add up to 1.
🧠 Memory aid — “along = ×, across = +”
Travel along a single branch path → multiply (it’s an “and”). Combine across different end-points → add (it’s an “or”). And for “at least one”, flip to the complement: 1 − P(none).
With vs without replacement
The only thing that changes between the two is the second set of branches.
With replacement
same branches
The item goes back, so totals don’t change. 2nd branches identical to the 1st (independent).
Without replacement
changed branches
The item is kept out, so totals shrink. 2nd branches are conditional on the 1st pick.
Which event on the first branch? If events are independent, order doesn’t matter. If not, put whichever you have the unconditional probability for first — so the second branches carry the conditional ones.
Worked examples
WE 1
Draw the tree
20% of a company wear glasses. Of those who wear glasses, 40% are right-handed; of those who don’t, 50% are right-handed. Set out the tree.
1st event: glasses G / not G′P(G) = 0.2, P(G′) = 0.82nd event: right-handed R / not R′ (conditional)from G: P(R) = 0.4, P(R′) = 0.6from G′: P(R) = 0.5, P(R′) = 0.5each pair of branches sums to 1
The four end-points (0.08 + 0.12 + 0.40 + 0.40) sum to 1 ✓
WE 2
Find P(right-handed)
Using the tree, find the probability a random person is right-handed.
add the paths that end in RP(G ∩ R) = 0.2 × 0.4 = 0.08P(G′ ∩ R) = 0.8 × 0.5 = 0.40P(R) = sum= 0.08 + 0.40P(R) = 0.48multiply along each R-path, then add them.
WE 3
Conditional from a tree
Given a person is right-handed, find the probability they wear glasses.
P(G|R) = P(G ∩ R) / P(R)= 0.08 / 0.48P(G|R) = 1/6numerator = the G-and-R path; denominator = total P(R) from WE 2.
WE 4
Without replacement
A bag has 6 red and 4 blue (10 total). Two are drawn without replacement. Find P(both red).
multiply along the red–red path1st red: 6/102nd red (conditional): 5/9P(both red)= 6/10 × 5/9 = 30/90P(both red) = 1/3totals shrink: 10 → 9 and 6 → 5.
WE 5
“At least one” via the complement
For the same bag (6 red, 4 blue, two drawn without replacement), find P(at least one red).
P(at least one red) = 1 − P(no red)P(no red) = both blue= 4/10 × 3/9 = 12/90 = 2/15subtract from 11 − 2/15P(at least one red) = 13/15“at least one” is far quicker via the complement than adding paths.
💡 Top tips
Multiply along branches (AND), add across paths (OR).
Each set of branches sums to 1 — a quick error check.
Without replacement → conditional second branches; totals shrink after the first pick.
“At least one” → 1 − P(none) is almost always faster.
You don’t need the whole tree — draw only the branches you care about.
Conditional questions: find P(A ∩ B) and P(B) from the tree, then divide.
⚠ Common mistakes
Adding along a branch instead of multiplying. Along a path = multiply.
Multiplying across paths instead of adding. Combining outcomes = add.
Keeping the same branches for “without replacement”. The second set must change.
Branches not summing to 1 — recheck the complement probabilities.
Forgetting the complement shortcut for “at least one”.
Wrong denominator in a conditional — it’s the total P of the given event, summed over all its paths.
That completes the Probability unit! You can now work with events and their combinations, handle independent, mutually exclusive and conditional events, and use Venn and tree diagrams to organise multi-stage problems. These foundations lead straight into probability distributions — the binomial, Poisson and normal — where you’ll attach probabilities to whole patterns of outcomes.
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