Probability Trees

A tree diagram is a systematic way to list and calculate the probabilities of combined outcomes when two or more events happen in sequence.

Use a tree diagram when:

• There are two or more events happening one after another.
• Each event has a small number of distinct outcomes.
• You need to find the probability of a particular combination.

Structure of a tree diagram:

Each branch represents one outcome of one event. The probability of that outcome is written along the branch. A complete path from left to right represents one combined outcome.

Key rules	Explanation
Probabilities on branches from the same point sum to 1	All outcomes at each stage must be listed
Multiply along a path to get the combined probability	The AND rule: multiply the probabilities along the path
Add across paths to combine outcomes	The OR rule: add the probabilities of all paths that satisfy the condition

A tree diagram is always drawn from left (first event) to right (later events). Never draw it backwards.

Start from a single point on the left. Draw one branch for each outcome of the first event, labelling each branch with its probability. From the end of each branch, draw another set of branches for the second event.

Worked example — A spinner has outcomes Red (probability 0.3) and Blue (probability 0.7). It is spun twice. Draw the tree diagram.

              ─── Red (0.3)  →  Red, Red
Red (0.3) ─┤
              ─── Blue (0.7) →  Red, Blue

              ─── Red (0.3)  →  Blue, Red
Blue (0.7) ─┤
              ─── Blue (0.7) →  Blue, Blue

Each stage has probabilities that sum to 1: $0.3+0.7=1$ ✓ at every branch point.

Always write the probability on the branch, not at the end of the branch. Placing probabilities at the tips is a common layout error that confuses calculations.

The probability of any combined outcome is found by multiplying the probabilities along that path.

Using the spinner example above:

Combined outcome	Calculation	Probability
Red, Red	$0.3\times 0.3$	$0.09$
Red, Blue	$0.3\times 0.7$	$0.21$
Blue, Red	$0.7\times 0.3$	$0.21$
Blue, Blue	$0.7\times 0.7$	$0.49$

Check: All four outcomes must sum to 1: $0.09+0.21+0.21+0.49=1.00$ ✓

Question — What is the probability of getting at least one Red?

Add the probabilities of all paths containing Red:

$P(\text{at least one Red})=0.09+0.21+0.21=0.51$

Or use the complement: $P(\text{at least one Red})=1-P(\text{Blue, Blue})=1-0.49=0.51$ ✓

When an outcome can happen in more than one way, add the probabilities of all the paths that satisfy the condition.

Worked example — Using the spinner above, find the probability of getting exactly one Red.

The paths with exactly one Red are: (Red, Blue) and (Blue, Red).

$P(\text{exactly one Red})=P(\text{Red, Blue})+P(\text{Blue, Red})=0.21+0.21=0.42$

The rule: Multiply along branches, add across paths.

This mirrors the probability rules:

• AND (events in sequence on the same path) → multiply
• OR (different paths that both satisfy the condition) → add

When items are selected without replacement, the probability on the second branch depends on what happened on the first branch. The branches at the second stage change based on the first outcome.

Worked example — A bag contains 3 red balls and 2 blue balls. Two balls are drawn without replacement. Find the probability that both are red, and the probability of getting one of each colour.

After drawing 1 red: 2 red and 2 blue remain (4 total). After drawing 1 blue: 3 red and 1 blue remain (4 total).

Path	Calculation	Probability
Red then Red	$\frac{3}{5}\times \frac{2}{4}$	$\frac{6}{20}=\frac{3}{10}$
Red then Blue	$\frac{3}{5}\times \frac{2}{4}$	$\frac{6}{20}=\frac{3}{10}$
Blue then Red	$\frac{2}{5}\times \frac{3}{4}$	$\frac{6}{20}=\frac{3}{10}$
Blue then Blue	$\frac{2}{5}\times \frac{1}{4}$	$\frac{2}{20}=\frac{1}{10}$

Check: $\frac{3}{10}+\frac{3}{10}+\frac{3}{10}+\frac{1}{10}=\frac{10}{10}=1$ ✓

$P(\text{both red})=\frac{3}{10}$

$P(\text{one of each})=P(\text{Red, Blue})+P(\text{Blue, Red})=\frac{3}{10}+\frac{3}{10}=\frac{3}{5}$

1. Using the same probabilities on the second set of branches for without-replacement problems

If the question says "without replacement", the probabilities change after the first pick. After drawing 3 red from 5 (red=3, blue=2), the second pick comes from 4 balls — not 5. Using a denominator of 5 throughout is the most common error on dependent events questions.

2. Adding instead of multiplying along a path

"Red AND Blue" requires $P(\text{Red})\times P(\text{Blue})$, not $P(\text{Red})+P(\text{Blue})$. Adding probabilities along a path produces a value greater than either individual probability, which is a reliable sign something has gone wrong.

3. Not listing all paths when using the complement

For "at least one", it is usually faster to use $1-P(\text{none})$. Listing all paths with at least one of the outcome works too, but missing one path produces a wrong answer. The complement method is safer.

4. Branch probabilities not summing to 1

At every branch point, all the branch probabilities must sum to exactly 1. If they do not, a probability has been misread or miscalculated. Always check this before multiplying.

5. Confusing "with replacement" and "without replacement"

Read the question carefully. "Without replacement" means the total number of items decreases after each pick. "With replacement" means the ball (or card, etc.) is returned, so the probabilities stay the same at every stage.