Tree Diagrams and Multi-Stage Experiments

Tree diagrams represent multi-stage experiments by branching through possible outcomes and attaching probabilities to each conditional step. In brief: two independent fair coin tosses have four paths, each with probability 1/2 × 1/2 = 1/4. The probability of exactly one head is the sum of the HT and TH paths, giving 1/2.

Trees are especially useful when later probabilities depend on earlier results, as in draws without replacement. The two central rules are multiply along a complete path and add mutually exclusive qualifying paths. This guide explains why those rules work, how to label branches and how to check that the complete tree accounts for every possibility.

What does a probability tree diagram show?

A tree begins with the starting situation and splits into branches for the first-stage outcomes. Each endpoint then splits again for the next stage, continuing until a complete experimental outcome is recorded. A root-to-leaf path represents one full sequence of results.

Each branch carries the probability of its outcome given the results already reached on that path. At the first stage, there are no earlier outcomes to condition on. At later stages, the branch probability may change because the state of the experiment has changed.

Branches leaving the same node must represent mutually exclusive and collectively exhaustive alternatives. Their probabilities therefore sum to 1. A node with branches red and blue is complete only if every possible next result is one of those colours; any additional category must be included or explicitly excluded by the model.

Why do you multiply probabilities along a path?

The multiplication rule states P(A and B) = P(A)P(B given A). A tree encodes this directly: the first branch gives P(A), and the second branch after A gives the conditional probability of B in that situation. Their product is the probability of following both branches.

For three stages, multiply the first probability by the second conditional on the first and the third conditional on the first two. The general principle does not require independence. Independence only allows later branch probabilities to remain equal to their unconditional values when earlier outcomes provide no information about them.

For independent coin tosses with head probability 0.5, the HH path has probability 0.5 × 0.5 = 0.25. For a dependent drawing experiment, the second factor may be different after each first result. The tree helps keep those different conditional values attached to the correct history.

Why do you add probabilities of different qualifying paths?

Distinct complete paths describe mutually exclusive sequences: one run cannot finish as both HT and TH. If an event includes several such paths, its probability is the sum of their path probabilities. This addition follows the disjoint-event rule rather than an assumption that all events can be added without overlap.

For two fair coin tosses, exactly one head includes HT and TH. Each path has probability 1/4, so the event probability is 1/4 + 1/4 = 1/2. At least one head includes HH as well, giving 3/4.

Choose complete paths or another disjoint partition before summing. Adding a partial event such as first toss is H to the complete path HH would double-count HH. A well-labelled tree makes it easier to see whether the selected branches overlap or represent separate final outcomes.

How do you build a tree for drawing without replacement?

A bag contains 3 red and 2 blue counters, and two are drawn without replacement. The first-stage probabilities are red 3/5 and blue 2/5. After a red draw, 2 red and 2 blue remain, so the second-stage probabilities are 2/4 and 2/4.

After a blue draw, 3 red and 1 blue remain, so the second-stage probabilities are 3/4 and 1/4. The denominator changes from 5 to 4 because one counter has been removed. The numerators depend on which colour was removed, so the second-stage branches cannot all use the same fractions.

The complete path probabilities are RR = (3/5)(2/4) = 3/10, RB = 3/10, BR = (2/5)(3/4) = 3/10 and BB = (2/5)(1/4) = 1/10. They sum to 1, confirming that the four colour sequences cover the experiment.

How do you calculate an event from the completed tree?

Using the counter tree, the event one counter of each colour includes RB and BR. Add their disjoint probabilities: 3/10 + 3/10 = 3/5. The order is not specified in the event, so both possible orders must be included.

The event at least one red includes RR, RB and BR, giving 9/10. A shorter route uses its complement, no red, which is the BB path: 1 − 1/10 = 9/10. The two methods agree and provide an independent check on the event selection.

The event same colour includes RR and BB, giving 3/10 + 1/10 = 2/5. It is the complement of one of each colour, whose probability is 3/5. Translating the wording before calculating helps distinguish exactly one, at least one and both, which select different sets of paths.

1. Define each stage and its possible outcomes.
2. Label each branch with its conditional probability.
3. Multiply along complete paths.
4. Select the paths that satisfy the event wording.
5. Add their probabilities and check with a complement where convenient.

How does replacement change a probability tree?

If each counter is replaced and the bag is remixed before the next draw, the colour counts return to 3 red and 2 blue. Under the usual independent-draw model, every second-stage red branch has probability 3/5 and every blue branch 2/5, regardless of the first colour.

The path probabilities become RR = 9/25, RB = 6/25, BR = 6/25 and BB = 4/25. The probability of one of each colour is 12/25, different from 3/5 without replacement. The experiment’s physical procedure changes the model and therefore the result.

Replacement alone should be accompanied by the intended randomisation assumption. The standard model assumes the next selection is made independently from the restored composition. State the procedure clearly so the tree reflects the actual experiment rather than carrying over branch probabilities from a superficially similar problem.

How do trees handle three or more stages?

Continue branching from every reachable node, updating probabilities according to its history. For three independent fair coin tosses, there are 2³ = 8 equally likely paths. Exactly two heads occurs on HHT, HTH and THH, giving probability 3/8.

Trees can become large as the number of stages increases. Repeated independent trials with only success and failure can often be handled more efficiently by a binomial formula, which groups paths according to their success count. The tree remains the underlying explanation for why a path probability is multiplied by the number of suitable arrangements.

Some experiments stop at different stages depending on earlier results. In that case, only extend paths that continue under the stated rule. A finished path is a complete outcome even if it has fewer branches than another, and all terminal-path probabilities should still sum to 1 when the model covers every possible ending.

What mistakes should you avoid with probability trees?

The background purpose of a tree is to organise conditional information. It records not only what can happen but what is known at each stage. This makes it useful for sequential selections, reliability models and staged decisions whenever later probabilities depend on earlier outcomes.

Do not add probabilities along a path or multiply separate alternative paths together. Multiply to require all successive branch events, and add disjoint complete paths to allow alternatives. Check each node’s outgoing probabilities sum to 1, then check the final path probabilities sum to 1.

Finally, update counts after draws without replacement and include every order allowed by the event. Missing BR when calculating one red and one blue is a common error. Clear path labels, conditional denominators and a final complement check make multi-stage calculations easier to verify and explain.