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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.

Start H T ½ ½ H T ½ ½ ½ ½ HH HT TH TT ½×½ = ¼ ½×½ = ¼ ½×½ = ¼ ½×½ = ¼ Sum = 4 × ¼ = 1 ✓

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

Tree diagramA branching diagram showing outcomes at each stage of a compound experiment
BranchA line in a tree diagram representing one possible outcome at a stage; the probability is written on it
StageOne level of the tree corresponding to one part of the experiment (e.g., first flip, second flip)
Multiplication rule$P(A \text{ then } B) = P(A) \times P(B)$ for independent events — multiply along the branches
Independent eventsEvents where the outcome of one does not affect the other (e.g., two coin flips)
With replacementAfter drawing, the item is returned — probabilities stay the same for each draw
Without replacementAfter drawing, the item is NOT returned — the total decreases and probabilities change

⚠ 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.

Write probability on each branch; multiply along a path
All endpoint probabilities must sum to 1 — always verify
For "or" outcomes: add the path probabilities together
With replacement: same probabilities at every stage
Without replacement: denominator decreases, numerator changes
$P(\text{at least one}) = 1 - P(\text{none})$ — complement shortcut

Badges This Lesson

Tree Builder
Branch Multiplier
Product Principle Pioneer
Independent Investigator
Replacement Reasoner
Probability Pro

Daily Challenge

From memory: draw a tree for drawing 2 cards from a bag of 5 (3R, 2B) without replacement. Find $P(\text{both same colour})$.

Next Up

Lesson 19 — Experimental Probability: what happens when we actually run the experiment instead of just calculating?

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