Some probability problems involve more than one stage β draw a card, then draw another; test a batch, then test again; a customer browses, then decides whether to buy. Once a problem has two or more sequential steps, keeping every branch straight in your head gets error-prone fast. A probability tree (or tree diagram) solves this by laying every possible sequence of outcomes out visually, branch by branch, turning a problem that's easy to get wrong mentally into one that's mechanical to work through on paper.
- How to construct a probability tree, stage by stage
- The two core rules for reading one: multiply along branches, add across paths
- A full worked example with sampling without replacement
- A second worked example showing how a tree organizes a conditional-probability problem
How a Probability Tree Is Built
Each stage of the problem becomes a new set of branches. From a single starting point, draw one branch for every possible outcome at the first stage, and label each branch with its probability. From the end of each of those branches, draw a new set of branches for every possible outcome at the second stage β and label those with the probability of that outcome given everything that happened on the branches leading to it. Repeat for as many stages as the problem has. The result is a tree that widens at each stage, with every path from the start to a final branch tip representing one complete, distinct sequence of outcomes.
The Two Rules for Reading a Tree
Once the tree is built, two simple rules extract any probability you need from it:
Along a single path (following one route from start to a tip): multiply the probabilities on each branch you pass through. This is just the multiplication rule from basic probability, applied one stage at a time.
Across multiple paths (when more than one path leads to the outcome you care about): add the probabilities of every qualifying path together. This is the addition rule, applied to a set of mutually exclusive paths β no two distinct paths through a tree can happen at once, so adding them never double-counts anything.
Every complete path's probability, once multiplied out, should sum to 1 across all paths in the tree β a quick sanity check worth running on any tree before trusting its answer.
Worked Example: Drawing Without Replacement
Problem: A bag contains 5 red balls and 3 blue balls (8 total). Two balls are drawn one at a time, without replacement. What's the probability of drawing exactly one red and one blue ball, in either order? What's the probability of drawing at least one blue ball?
First stage: P(Red) = 5/8, P(Blue) = 3/8.
Second stage (probabilities change because a ball has been removed, and removal without replacement affects what remains): if the first ball was red, 4 red and 3 blue remain out of 7 total, so P(Red | Red first) = 4/7 and P(Blue | Red first) = 3/7. If the first ball was blue, 5 red and 2 blue remain out of 7, so P(Red | Blue first) = 5/7 and P(Blue | Blue first) = 2/7.
Multiplying along each of the four paths:
P(Red, Red) = 5/8 Γ 4/7 = 20/56 = 5/14 β 0.357
P(Red, Blue) = 5/8 Γ 3/7 = 15/56 β 0.268
P(Blue, Red) = 3/8 Γ 5/7 = 15/56 β 0.268
P(Blue, Blue) = 3/8 Γ 2/7 = 6/56 = 3/28 β 0.107
Check: 5/14 + 15/56 + 15/56 + 3/28 = 20/56 + 15/56 + 15/56 + 6/56 = 56/56 = 1 β exactly what a valid, complete tree should give.
Exactly one red and one blue, in either order: two paths qualify β (Red, Blue) and (Blue, Red) β so add them: 15/56 + 15/56 = 30/56 = 15/28 β 0.536.
At least one blue: three of the four paths include at least one blue ball. Rather than adding all three, it's faster to use the complement: P(at least one blue) = 1 β P(no blue) = 1 β P(Red, Red) = 1 β 5/14 = 9/14 β 0.643.
Worked Example: A Two-Stage Manufacturing Problem
Problem: A factory has two machines. Machine A produces 60% of all items and has a 5% defect rate; Machine B produces the remaining 40% and has an 8% defect rate. An item is picked at random from the full output. What's the probability it's defective? Given that it's defective, what's the probability it came from Machine A?
The tree's first stage splits by machine: P(A) = 0.60, P(B) = 0.40. The second stage splits by outcome, conditional on which machine: P(Defective | A) = 0.05, P(Not defective | A) = 0.95, P(Defective | B) = 0.08, P(Not defective | B) = 0.92.
Two paths lead to "defective," so the overall defect probability adds them together (this is exactly the law of total probability β a defective item must have come from either A or B, with no overlap):
P(A and Defective) = 0.60 Γ 0.05 = 0.03
P(B and Defective) = 0.40 Γ 0.08 = 0.032
P(Defective) = 0.03 + 0.032 = 0.062
So 6.2% of all items are defective. Now for the second question β given a defective item, how likely is it from Machine A β the tree makes the logic visual even though the calculation is Bayes' theorem underneath: divide the one path that matches both conditions (A and defective) by the total probability of defective (both qualifying paths added together):
P(A | Defective) = P(A and Defective) / P(Defective) = 0.03 / 0.062 = 15/31 β 0.484
Even though Machine A produces the majority of items (60%) and has the lower defect rate, a randomly selected defective item is only very slightly more likely to have come from A than B (48.4% vs. 51.6%) β Machine B's higher defect rate pulls its share of the defective pile up disproportionately to its smaller output share. This is exactly the kind of result that's easy to get backwards without a tree laid out explicitly.
When a Tree Is the Right Tool
Trees are most valuable when a problem has a small number of sequential stages (two or three is typical β beyond that, the tree becomes unwieldy to draw by hand) and when at least one stage's probabilities depend on what happened at an earlier stage, as in both worked examples above. For problems with many independent trials of the identical type β ten coin flips, a hundred survey respondents β a tree with every branch drawn out becomes impractical, and the binomial distribution formula (built on exactly the same multiplication and addition logic, just generalized) is the more practical tool.
Common Mistakes
Reusing first-stage probabilities at the second stage after sampling without replacement. The whole point of the without-replacement worked example above is that the second stage's probabilities genuinely change β using 5/8 again instead of the correct 4/7 or 5/7 is one of the most common tree-diagram errors.
Adding along a single path instead of multiplying. A single path represents one event and the next and the next β that's multiplication, not addition. Addition is reserved for combining separate, complete paths, not steps within one path.
Forgetting to check that all path probabilities sum to 1. This check catches most arithmetic slips immediately and takes only a few seconds β skipping it means an error can go unnoticed all the way to a final wrong answer.
Do the branches at each stage always have to sum to 1?
How is a probability tree different from a Venn diagram?
Can a probability tree have more than two branches at a stage?
Does a probability tree work for independent events too?
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