Representing a Two-Stage Chance Process
In The Multiplication Principle for Counting Outcomes, you learned to count complete outcomes by identifying the choices made at each stage. A tree diagram gives a different kind of information: it displays the possible results at each stage and assigns probabilities to the branches. Following branches from left to right shows one complete outcome of the process.
A tree is especially useful when the probabilities at a later stage can depend on what happened earlier. Rather than assuming that every complete outcome is equally likely, the tree records the chance of each step. For a two-stage process, each path from the starting point to an ending point represents one complete outcome.
At each branching point, the probabilities of all branches leaving that point must add to 1. Those branches represent all the possible results at that stage, given what has happened so far. A branch probability at the second stage may therefore be different for different first-stage results.
To find the probability of one complete path, multiply the probabilities along that path. This is a probability version of following one particular sequence of results: the first branch must occur, and then the next branch must occur. If an event consists of several different paths, add their path probabilities. Those paths are mutually exclusive: only one complete path can occur in a single run of the process.
The notation \(P(B\mid A)\) means the probability of \(B\) on the second stage, given that \(A\) happened on the first stage. It reminds you to use the branch that starts at the \(A\) result—not a branch from a different point in the tree. If the second-stage chance is the same regardless of the first result, the relevant branches will have the same probabilities. Do not assume they are the same unless the model supports that assumption.
How to Build and Use the Tree
Start by defining what each stage records and listing its possible results. Then draw a branch for each result and label it with the appropriate probability. For the second stage, make a separate set of branches from every first-stage result. Even if two sets have the same labels, check their probabilities separately.
State what happens first and second, and identify the possible result at each stage.
Label each branch with the probability of that first result. Check that the probabilities leaving the start add to 1.
From every first-stage result, draw all possible second-stage results. Label each branch using the probability that applies after that first result, and check that the probabilities at each node add to 1.
Multiply branch probabilities along a complete path. If the event can occur along more than one path, add those path probabilities.
A tree diagram makes the sequence of the process visible, but it does not by itself justify the probabilities written on the branches. Use the chance model in the question: for example, a stated uniform random choice, a known composition of a container, or provided conditional probabilities. Just as in the earlier tutorial on counting probability, do not treat outcomes as equally likely merely because you can count them.
Worked Examples: Reading Paths and Finding Probabilities
The first example has two stages with equal and independent choices. The next examples show why trees are also valuable when later probabilities depend on earlier results.
Worked Example: A Spinner Followed by a Coin Toss
An invented game uses a spinner with three equal-sized sectors labeled A, B, and C. After the spinner stops, a fair coin is tossed. What is the probability of getting sector B and then tails? What is the probability of getting tails, regardless of the spinner result?
Define the stages: The first stage is the spinner result, A, B, or C. The second stage is the coin result, heads (H) or tails (T). The spinner result does not change the coin’s chance of heads or tails, so each first-stage result has the same two second-stage branch probabilities.
Draw and label the tree: The first-stage branches each have probability \(1/3\). From each one, the coin branches have probability \(1/2\) for heads and \(1/2\) for tails.
- Spinner A, \(1/3\)
- Heads, \(1/2\): path A then H
- Tails, \(1/2\): path A then T
- Spinner B, \(1/3\)
- Heads, \(1/2\): path B then H
- Tails, \(1/2\): path B then T
- Spinner C, \(1/3\)
- Heads, \(1/2\): path C then H
- Tails, \(1/2\): path C then T
The first-stage probabilities add to \(1/3+1/3+1/3=1\). At each spinner result, the coin branch probabilities add to \(1/2+1/2=1\).
Find the probability of B then tails: This event is one complete path, so multiply the branch probabilities along that path.
Find the probability of tails: Tails occurs on three different complete paths: A then T, B then T, or C then T. These paths cannot happen together in one play, so add their probabilities.
Conclude: The probability of B followed by tails is \(1/6\). The probability of tails regardless of the spinner result is \(1/2\), because the event includes all three tails paths.
Worked Example: Drawing Two Tokens Without Replacement
A bag contains 3 green tokens and 2 yellow tokens. Two tokens are drawn one after the other without replacement. What is the probability that one token is green and the other is yellow, in either order?
First-stage branches: There are 5 tokens initially, so the probability of green on the first draw is \(3/5\), and the probability of yellow is \(2/5\). These add to 1.
Second-stage branches after green: If the first token is green, 2 green and 2 yellow tokens remain out of 4. Thus, the second-draw probabilities are \(2/4\) for green and \(2/4\) for yellow.
Second-stage branches after yellow: If the first token is yellow, 3 green and 1 yellow token remain out of 4. Thus, the second-draw probabilities are \(3/4\) for green and \(1/4\) for yellow.
- First draw green, \(3/5\)
- Second draw green, \(2/4\): path GG
- Second draw yellow, \(2/4\): path GY
- First draw yellow, \(2/5\)
- Second draw green, \(3/4\): path YG
- Second draw yellow, \(1/4\): path YY
Check each branching point: \(2/4+2/4=1\) after a green draw, and \(3/4+1/4=1\) after a yellow draw. The second-draw probabilities differ because the first token is not replaced.
Find each favorable path: One green and one yellow can occur as GY or YG. Multiply along each path.
Add the favorable paths:
Conclude: The probability of drawing one green and one yellow token, in either order, is \(3/5\), or \(0.60\). The two orders are separate paths, and both must be included.
Worked Example: Different Groups, Different Second-Stage Chances
A community center uses two shuttle routes. In an invented model, a randomly selected shuttle trip uses Route North with probability \(0.60\) and Route South with probability \(0.40\). A trip on Route North is late with probability \(0.10\); a trip on Route South is late with probability \(0.25\). What is the probability that a randomly selected trip is late?
Worked Example: Finding the Overall Probability of a Late Trip
State: The two stages are selecting the route for a trip and then recording whether the trip is late or on time. The event of interest is that the trip is late, on either route.
Plan: Use a tree with one route branch for each route. From each route, add late and on-time branches using the probabilities specified for that route. Check that the branch probabilities at each node add to 1. Multiply to find the two late-path probabilities, then add them because they are separate ways for a trip to be late.
Do—label the tree: The first-stage probabilities \(0.60\) and \(0.40\) add to 1. On Route North, the probability of being on time is \(1-0.10=0.90\), so its two second-stage branches add to \(0.10+0.90=1\). On Route South, the probability of being on time is \(1-0.25=0.75\), and its branches add to \(0.25+0.75=1\).
- Route North, \(0.60\)
- Late, \(0.10\): path North then late
- On time, \(0.90\): path North then on time
- Route South, \(0.40\)
- Late, \(0.25\): path South then late
- On time, \(0.75\): path South then on time
The probability of a late trip on Route North is:
The probability of a late trip on Route South is:
Do—combine the paths: A trip uses only one route, so the two late paths are mutually exclusive. Add their probabilities.
Conclude: Under this model, the probability that a randomly selected shuttle trip is late is \(0.16\), or 16%. The route-specific late probabilities are different, so it would not be appropriate to use just one of them as the overall probability without accounting for how often each route is used.
Common Mistakes and AP Exam Tips
- Using the wrong second-stage branch. On a tree, follow the branch that begins at the result that actually occurred first. In the token example, after a first-draw green, the probability of a second-draw green is \(2/4\), not \(3/4\).
- Adding along a path. A complete outcome requires the first result and then the second result, so multiply the probabilities on that path.
- Multiplying separate ways an event can occur. If an event includes different paths, such as GY or YG, add their path probabilities after finding each one.
- Assuming later probabilities are unchanged. Without replacement, or whenever the model gives route-specific chances, the first-stage result can affect the second-stage probabilities. Label the branches separately.
- Forgetting to check the branches at each node. The probabilities leaving each branching point should add to 1. Check the first stage and every distinct second-stage branching point.
- Confusing counts with probabilities. The multiplication principle from the earlier tutorial counts possible outcomes. A probability tree instead labels branches with chances; complete paths need not be equally likely.
For a full-credit response, state the two stages, label each branch with the probability that applies at that point, and show the multiplication for every path used. If the event includes several paths, identify them and add their probabilities. Interpret the result in the situation rather than giving only a decimal.
Check Your Understanding
For each question, identify the stages and show which branches or paths are needed.
- A fair six-sided die is rolled, then a fair coin is tossed. What is the probability of rolling a 4 and getting heads?
- A jar contains 4 red and 1 blue bead. Two beads are drawn without replacement. What probabilities label the second-draw branches after a first-draw red?
- Using the jar in Question 2, what is the probability of drawing one red and one blue in either order?
- An invented bus model uses the East route with probability \(0.70\) and the West route with probability \(0.30\). The probabilities of being delayed are \(0.20\) on East and \(0.10\) on West. What is the overall probability of a delay?
- Why should the probabilities on the branches leaving each node add to 1?