Tree Diagrams
Draw a tree diagram for a two-stage experiment, multiply along the branches, and see how the second stage’s probabilities change when you do not replace the first item.
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Think First
You flip a coin twice. Draw a branching diagram to show all possible outcomes. How many branches do you have in total at the end?
Tree Diagrams
A branching diagram where each stage splits into outcomes. Write the probability on each branch, then multiply along branches to find compound probabilities.
What You'll Master
- Draw a tree diagram for a two-stage or three-stage compound experiment
- Write probabilities on branches and multiply along branches to find compound probabilities
- Verify that all endpoint probabilities sum to 1
- Distinguish between experiments with and without replacement, and adjust branch probabilities accordingly
Words You Need
⚠ Spot the Trap
Do not add probabilities along a branch, you multiply them. Adding is for combining separate paths to the same outcome (e.g., "at least one head" can happen via HT, TH, or HH, you multiply along each path, then add the path results together).
1. Tree Diagram Structure and the Multiplication Rule
Each branch is labelled with its probability. To find the probability of a complete path (sequence of outcomes), multiply the probabilities of each branch along that path.
Two fair coin flips:
- $P(HH) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$
- $P(HT) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$
- $P(TH) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$
- $P(TT) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$
Verification: all endpoint probabilities must sum to 1: $\frac{1}{4} + \frac{1}{4} + \frac{1}{4} + \frac{1}{4} = 1$ ✓
$P(\text{at least one head}) = P(HH) + P(HT) + P(TH) = \frac{3}{4}$
2. Unequal Branch Probabilities, With Replacement
A bag has 3 red (R) and 2 blue (B) marbles. Two marbles are drawn with replacement (marble is returned before the second draw).
Branch probabilities: $P(R) = \frac{3}{5}$, $P(B) = \frac{2}{5}$ at every stage.
- $P(RR) = \dfrac{3}{5} \times \dfrac{3}{5} = \dfrac{9}{25}$
- $P(RB) = \dfrac{3}{5} \times \dfrac{2}{5} = \dfrac{6}{25}$
- $P(BR) = \dfrac{2}{5} \times \dfrac{3}{5} = \dfrac{6}{25}$
- $P(BB) = \dfrac{2}{5} \times \dfrac{2}{5} = \dfrac{4}{25}$
Check: $\dfrac{9+6+6+4}{25} = \dfrac{25}{25} = 1$ ✓
$P(\text{one of each colour}) = P(RB) + P(BR) = \dfrac{6}{25} + \dfrac{6}{25} = \dfrac{12}{25}$
3. Without Replacement, Probabilities Change
The same bag: 3 red, 2 blue. Two marbles drawn without replacement. After the first draw, there are only 4 marbles left.
From the first draw: $P(R_1) = \frac{3}{5}$, $P(B_1) = \frac{2}{5}$
Second draw branches, depend on first result:
- If first was R: 2R and 2B remain (4 total) → $P(R_2 | R_1) = \frac{2}{4}$, $P(B_2 | R_1) = \frac{2}{4}$
- If first was B: 3R and 1B remain (4 total) → $P(R_2 | B_1) = \frac{3}{4}$, $P(B_2 | B_1) = \frac{1}{4}$
Outcomes:
- $P(RR) = \dfrac{3}{5} \times \dfrac{2}{4} = \dfrac{6}{20} = \dfrac{3}{10}$
- $P(RB) = \dfrac{3}{5} \times \dfrac{2}{4} = \dfrac{6}{20} = \dfrac{3}{10}$
- $P(BR) = \dfrac{2}{5} \times \dfrac{3}{4} = \dfrac{6}{20} = \dfrac{3}{10}$
- $P(BB) = \dfrac{2}{5} \times \dfrac{1}{4} = \dfrac{2}{20} = \dfrac{1}{10}$
Check: $\dfrac{3+3+3+1}{10} = \dfrac{10}{10} = 1$ ✓
Common Pitfalls
- Adding instead of multiplying along a path, always multiply branch probabilities
- Using the same denominator in without-replacement problems, the total decreases after each draw
- Forgetting to check that all endpoint probabilities sum to 1, this catches errors
- Missing a branch, use the counting principle to know how many endpoints to expect
Copy This Into Your Book
Tree diagram rules:
- Write the probability on each branch
- Multiply along a path to find $P(\text{that outcome sequence})$
- Add path probabilities to combine routes to the same result
- All endpoint probabilities must sum to 1
Without replacement: the denominator decreases by 1 after each draw, and the numerator changes based on what was drawn.
Two fair coins are flipped. Using a tree diagram, what is $P(HH)$?
What probability is written on each branch of a tree diagram?
Two fair coins are flipped. What is $P(\text{at least one head})$?
In a without-replacement experiment, how do the second-stage branch probabilities compare to the first?
The probabilities at all endpoints of a tree diagram must sum to:
Q6. A coin is flipped three times. Draw a tree diagram (describe or sketch it). Find $P(\text{exactly 2 heads})$.
Q7. A bag has 4 red and 1 blue marble. Two marbles are drawn with replacement. Draw a tree diagram and find: (a) $P(\text{both red})$, (b) $P(\text{one of each colour})$.
Q8. A bag has 3 yellow and 2 green marbles. Two marbles are drawn without replacement. Find: (a) $P(\text{both yellow})$, (b) $P(\text{at least one green})$.
Show Answers
Q6
8 endpoints: HHH, HHT, HTH, HTT, THH, THT, TTH, TTT each with probability $\frac{1}{8}$.
Exactly 2 heads: HHT, HTH, THH, 3 paths.
$P(\text{exactly 2 heads}) = 3 \times \frac{1}{8} = \dfrac{3}{8}$
Q7
Branch probabilities: $P(R) = \frac{4}{5}$, $P(B) = \frac{1}{5}$ at every stage (with replacement).
(a) $P(RR) = \dfrac{4}{5} \times \dfrac{4}{5} = \dfrac{16}{25}$
(b) $P(\text{one of each}) = P(RB) + P(BR) = \dfrac{4}{25} + \dfrac{4}{25} = \dfrac{8}{25}$
Q8
Without replacement: first draw from 5, second from 4.
(a) $P(YY) = \dfrac{3}{5} \times \dfrac{2}{4} = \dfrac{6}{20} = \dfrac{3}{10}$
(b) $P(\text{at least one green}) = 1 - P(YY) = 1 - \dfrac{3}{10} = \dfrac{7}{10}$
Stretch Challenge
A biased coin has $P(H) = 0.6$ and $P(T) = 0.4$. The coin is flipped three times. (a) Draw the tree diagram showing all 8 endpoints and their probabilities. (b) Find $P(\text{exactly 2 heads})$. (c) Compare to a fair coin: which gives a higher $P(\text{exactly 2 heads})$? Show your calculation for both.