How to use a tree diagram for probability

What a tree diagram is doing

A tree diagram lays a multi-stage random process out as a set of complete paths, so a probability you cannot see directly becomes a product you can compute. Each split in the tree is one stage. Each complete path from the start to a tip is one outcome of the whole process.

Two rules run everything:

1. Multiply along a path. The probability of one complete outcome is the product of the branch probabilities you pass through.
2. Add across paths. The probability of an event is the sum of the path probabilities for every path that satisfies it.

Those rules work because the branches out of a single split are mutually exclusive and cover every possibility, and because each branch after the first carries a conditional probability. A tree and a two-way table hold the same information. The tree is easier when the process arrives in stages with rates attached, and the table is easier when you already have counts.

Branch probabilities are conditional probabilities

The branches leaving the starting point carry plain unconditional probabilities. Every branch after that carries a probability conditional on everything upstream of it.

That is worth saying with symbols, because it is what makes the multiplication legal. If the first stage lands on $A$ and a second-stage branch is labeled $B$, the number written on that branch is $P(B \mid A)$, read "the probability of B given A", not $P(B)$.

The probabilities leaving any one split add to 1. That holds at the start and at every later split, because at each split you have listed every way the next stage can go for that particular history. When a split does not add to 1, you have either lost a branch or miscounted what remains.

Multiply along a path

Multiplying along a path is the general multiplication rule applied one stage at a time. The AP formula sheet writes it as

where $P(A \cap B)$, read "the probability of A and B", is the joint probability that both happen.

Notice what the rule does not say. It does not say $P(A) \cdot P(B)$. That shorter version is a special case, true only when the events are independent, which is exactly the case where $P(B \mid A) = P(B)$.

Because a tree writes the conditional probability on the branch, you get the general rule for free and never have to decide whether the shortcut applies. When to multiply probabilities works through that decision.

For a three-stage process the pattern continues: $P(A \cap B \cap C) = P(A) \cdot P(B \mid A) \cdot P(C \mid A \cap B)$. Each factor is conditional on everything to its left, which is the same as everything upstream on the path.

Add across the paths that fit the event

Once every path has a probability, most questions reduce to deciding which paths belong.

Adding is legal because distinct paths are mutually exclusive: one run of the process ends at exactly one tip. Disjoint events add with no correction term, so $P(A \cup B) = P(A) + P(B)$, where $\cup$ is read "or", and there is no overlap to subtract. Disjoint vs independent events sorts out that distinction.

This is how you get a total probability, meaning the probability of a later-stage outcome no matter which earlier branch produced it. Collect every path ending in that outcome and add. A motor is defective whether it came from supplier A or supplier B, so $P(\text{defective})$ is the sum of the two defective paths.

The same move handles compound events. "At least one" is the sum of every path with one or more, or faster, 1 minus the single path with none.

Why the second stage changes when you sample without replacement

Sampling without replacement means the item you drew is not available on the next draw. The denominator of every second-stage branch drops by one, and the numerator drops on the branch that matches what you took, so which numerator changes depends on which path you are standing on.

Start with 10 granola bars in a bag, 6 peanut butter and 4 oat, and take two without looking.

The first split is $\frac{6}{10}$ for peanut butter and $\frac{4}{10}$ for oat. After that first bar is gone only 9 remain, so every second-stage branch has 9 on the bottom. What sits on top depends on the path:

• After a peanut butter bar, 5 peanut butter and 4 oat are left, so the branches read $\frac{5}{9}$ and $\frac{4}{9}$.
• After an oat bar, 6 peanut butter and 3 oat are left, so the branches read $\frac{6}{9}$ and $\frac{3}{9}$.

Each pair still adds to 1, which is the check that you counted the remaining bars correctly. With replacement, both second-stage splits would read $\frac{6}{10}$ and $\frac{4}{10}$ again, identical to the first split, and the two draws would be independent. With replacement vs without replacement compares the two setups in more detail.

Working backwards to a first-stage branch

A tree is drawn forwards in time, but questions often run backwards: you are told how the process finished and asked about the stage you did not see. That reversed conditional is still one path divided by a sum of paths.

That is the definition $P(A \mid B) = \frac{P(A \cap B)}{P(B)}$ with the numerator read off one path and the denominator read off a group of paths. No new formula is needed, and you do not have to invoke Bayes' theorem by name.

The answer often surprises people, because the reversed conditional can sit a long way from the forward one. A supplier that ships most of the parts can still account for less than half of the defective parts, which is what the first worked example below shows. P(A|B) vs P(B|A) explains why the two directions come apart.

One check comes free. The reversed conditionals across all the first-stage branches add to 1, because the item came from somewhere.

Three checks that catch tree mistakes

Trees fail in a small number of ways, and three quick checks catch nearly all of them.

1. Every split adds to 1. Check the first split, then check each second-stage split on its own. A split that adds to anything else means a missing branch or a miscount of what is left.
2. All path products add to 1. The tips of the tree are every possible outcome of the process, so their probabilities form a probability distribution. This one catches an arithmetic slip inside a single path.
3. Each set of reversed conditionals adds to 1. If $P(A \mid \text{defective})$ and $P(B \mid \text{defective})$ do not add to 1 when $A$ and $B$ are the only two suppliers, something upstream is wrong.

One habit prevents most arithmetic slips: keep fractions as fractions until the last line. $\frac{6}{10} \times \frac{5}{9} = \frac{30}{90} = \frac{1}{3} \approx 0.3333$ is exact. Rounding $\frac{5}{9}$ to $0.56$ first and computing $0.6 \times 0.56 = 0.336$ is already wrong in the third decimal, and that error compounds across a three-stage tree.

AP note: where tree diagrams show up

Tree diagrams are scratch work rather than a listed topic. The Fall 2026 course framework does not name them in the required content for Unit 2. What it does require is the arithmetic a tree performs.

• Topic 2.6, conditional probability, gives both $P(A \mid B) = \frac{P(A \cap B)}{P(B)}$ and the general multiplication rule $P(A \cap B) = P(A) \cdot P(B \mid A)$. Every branch after the first, and every path product, is one of those two.
• Topic 2.7, independent events and unions of events, is where the shortcut $P(A \cap B) = P(A) \cdot P(B)$ lives, together with $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.

Unit 2 is 15 to 25 percent of the multiple-choice section, and the free-response section rewards organized work, so a labeled tree is worth drawing even when you could do the algebra in your head. Put probabilities on the branches, not just names, and write the path product at each tip. Then practice on the conditional probability set. The official framework is at AP Central.