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Relative Frequency & Probability · L2 of 4 ~50 min MST-12-S2-09 ⚡ +90 XP available

Multistage Events, Arrays and Tree Diagrams

When two or more stages happen in sequence, listing outcomes systematically, with arrays or tree diagrams, is the only reliable way to avoid missing combinations. The multiplication rule then turns branch probabilities into exact answers.

Today's hook, A bag has 3 blue and 2 red marbles. You draw one marble, replace it, and draw again. Write your gut answers: How many different (colour, colour) combinations are possible? Are all these combinations equally likely? What do you think P(blue then red) equals?

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Orient and prepare

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Recall, your gut answer first

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A bag has 3 blue and 2 red marbles. You draw one marble, replace it, and draw again. Write your gut answers to these, no calculating yet:

How many different (colour, colour) combinations are possible?
Are all these combinations equally likely?
What do you think P(blue then red) equals?

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The key formulas you need to own

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Counting outcomes for a multistage event uses the multiplication principle: $n_1 \times n_2 \times n_3 \times \ldots$. The probability of any path through a probability tree is the product of its branch probabilities: $P(A \text{ and } B) = P(A) \times P(B)$.

For a two-stage experiment, multiply the number of outcomes at each stage to find $n(S)$. Each path through a probability tree gives one outcome; the probability of that outcome is the product of the two branch probabilities.

$n(S) = n_1 \times n_2$ | $P(\text{path}) = P(\text{branch}_1) \times P(\text{branch}_2)$ | $\sum \text{all paths} = 1$

Replacement matters

With replacement, probabilities stay the same on the second draw. Without replacement, the denominator decreases by 1, and so do the favourable outcomes if you drew one.

Tree vs array

An array (grid) is fastest for two stages. A tree diagram is more flexible for three or more stages, or when "without replacement" changes branch probabilities.

Check your tree

The probabilities along any complete set of branches must sum to 1. If they don't, you've made an error somewhere.

What you'll master

Know

Key facts

Multiplication principle: $n(S) = n_1 \times n_2$
$P(A \text{ and } B) = P(A) \times P(B)$ for independent events
With vs without replacement changes branch probabilities

Understand

Concepts

Why arrays and tree diagrams prevent outcome-counting errors
Why probabilities multiply along tree paths
How "without replacement" changes every branch from stage 2 onward

Can do

Skills

Construct an array for a two-stage experiment
Construct a tree diagram and label branch probabilities
Calculate P(A and B) for multistage events using the multiplication rule

Key terms

Multistage eventA chance experiment that consists of two or more sequential stages, each with its own outcomes.

Array (sample-space grid)A grid with one stage's outcomes across the top and another stage's outcomes down the side. Every cell is one combined outcome.

Tree diagramA branching diagram where each branch represents one outcome at each stage. A complete path from left to right is one combined outcome.

Probability treeA tree diagram with the probability of each branch written on it. The probability of a path = product of its branch probabilities.

Branch probabilityThe probability written on a single branch of a probability tree. All branches leaving one node must sum to 1.

Independent eventsEvents where the outcome of one does not affect the probability of the other, e.g. drawing with replacement. $P(A \text{ and } B) = P(A) \times P(B)$.

With replacementAfter each draw, the item is returned to the group before the next draw. Stage-2 branch probabilities are identical to stage-1.

Without replacementAfter each draw, the item is NOT returned. Stage-2 branch probabilities change: denominator decreases by 1, and numerator decreases by 1 if the same type was drawn.

Arrays for Two-Stage Experiments

Build the first core idea, then test it with a quick check.

Arrays for Two-Stage Experiments

core concept

Before you can calculate a probability, you need to list all possible outcomes. For two-stage experiments, an array (a grid) does this mechanically, rows for one stage, columns for the other, every cell is one outcome.

How to build and use an array:

Set up the grid: Write one stage's outcomes across the top (columns) and the other stage's outcomes down the side (rows). Each cell is an outcome pair.
Example, die and coin: Roll a die (columns 1–6) and flip a coin (rows H, T). The array is 2 × 6 = 12 cells; each cell is e.g. (H,1), (H,2), …, (T,6).
Finding probability from an array: Count cells matching your event and divide by total cells.

Building an array: Die and coin: write columns 1 2 3 4 5 6, rows H T. Fill each cell: (H,1) (H,2) (H,3) (H,4) (H,5) (H,6) / (T,1) (T,2) (T,3) (T,4) (T,5) (T,6). $n(S) = 12$. Event "head and odd": cells (H,1), (H,3), (H,5) → $n(E) = 3$, $P = 3/12 = 1/4$.

Limitation of arrays: Arrays only work well for two stages and when outcomes at each stage are symmetric. For three stages or unequal probabilities, use a tree diagram instead.

An array (two-way table) lists all outcomes of a two-stage experiment in a grid. Rows represent one stage, columns represent the other. Each cell is one equally-likely outcome. Count favourable cells to find probability.

Pause, copy the two-way array structure (rows = one stage outcomes; columns = other stage outcomes; each cell = one combined outcome) and the probability counting rule: P(event) = number of favourable cells ÷ total number of cells into your book.

Quick check: A bag has 4 coloured buttons (red, blue, green, yellow) and a fair coin is flipped. How many outcomes are in the sample space?

Tree Diagrams

Connect the next method to the foundation and check your reasoning.

Tree Diagrams

core concept

A two-way array works well when both stages have the same set of outcomes (e.g., two dice) and the grid structure is compact. When the two stages have different numbers of outcomes, or there are more than two stages, a tree diagram is more flexible: it branches at each stage regardless of how many outcomes there are, and each path through the tree represents one complete outcome.

Arrays work for two stages, but tree diagrams extend to any number of stages and handle changing probabilities (as in "without replacement") naturally.

A tree diagram starts from a single point and branches at each stage. Every possible sequence of outcomes is one complete path from left to right.

How to draw a tree diagram:

Draw a dot (starting point). From it, draw one branch for each outcome at stage 1, labelled with the outcome.
From each stage-1 endpoint, draw branches for each stage-2 outcome, also labelled.
Reading outcomes: Each path (left to right) is one combined outcome. List all paths to get the sample space.
Counting: $n(S) =$ (number of stage-1 branches) × (number of stage-2 branches per branch).

With replacement example: Bag: 3 blue (B), 2 red (R). Draw twice WITH replacement. Stage 1: B or R. Stage 2 from each: same branches (B and R) because the marble went back. Four paths: BB, BR, RB, RR. $n(S) = 2 \times 2 = 4$ distinct outcomes.

Three-stage example: Flip a coin three times: $2 \times 2 \times 2 = 8$ outcomes. A tree has three layers of branches, draw it systematically from left to right: HHH, HHT, HTH, HTT, THH, THT, TTH, TTT.

A tree diagram shows sequential outcomes by branching at each stage. Each branch is labelled with the outcome and its probability. To find the probability of a path, multiply the branch probabilities along that path.

Pause, copy the tree diagram multiplication rule: multiply branch probabilities along a single path to get the probability of that complete sequence, and note each path must be labelled with its outcome and probability into your book.

True or false: When drawing marbles WITH replacement from a bag, the probabilities on the second stage of the tree diagram are identical to the first stage.

Probability Trees and the Multiplication Rule

Apply the third core idea and confirm the key calculation.

Probability Trees and the Multiplication Rule

core concept

A tree diagram labels each branch with its probability, and following one complete path gives one outcome. To find the probability of a specific path, multiply all the branch probabilities along it: P(A and B) = P(A) × P(B|A). When an event can occur via multiple paths, add the probabilities of all valid paths together.

A tree diagram shows outcomes; a probability tree shows both outcomes AND their probabilities. Once you have a probability tree, calculating P(any path) is mechanical.

A probability tree adds the probability of each branch. To find the probability of any complete path (a specific sequence of outcomes), multiply the probabilities along that path.

Using a probability tree:

Labelling branches: Write the probability as a fraction on each branch. From any single node, branch probabilities must sum to 1.
Multiplication rule: $P(A \text{ and } B) = P(A) \times P(B)$. For any path: multiply all branch probabilities along it.
Adding paths: If the event can happen by multiple paths, add the probabilities of all those paths. E.g. P(exactly one red) = P(BR) + P(RB).

WITHOUT replacement: Bag: 3 B, 2 R. Draw twice WITHOUT replacement. Stage 1: $P(B) = 3/5$, $P(R) = 2/5$. Stage 2 IF first was B: 4 marbles left (2B, 2R) → $P(B) = 2/4$, $P(R) = 2/4$. Stage 2 IF first was R: 4 marbles left (3B, 1R) → $P(B) = 3/4$, $P(R) = 1/4$. Note: stage-2 probabilities depend on stage-1 outcome.

Sanity check: Add all four path probabilities (BB + BR + RB + RR). They must sum to 1. If they don't, recheck your fractions.

Probability tree and multiplication rule: P(A and B) = P(A) × P(B|A). For independent events P(B|A) = P(B), so P(A and B) = P(A) × P(B). Add probabilities of all branches that satisfy the required event.

Pause, copy P(A and B) = P(A) × P(B|A), the independence simplification P(A and B) = P(A) × P(B), and the addition rule for multiple paths: add path probabilities when the event can occur in more than one way into your book.

Fill the blanks: A probability tree has branches $P(A) = 1/3$ and $P(B|A) = 1/4$. The probability of path A then B is $P(A) \times P(B|A) =$ . The remaining path probabilities (from A) must sum to since branches from one node sum to 1.

Work through the examples

Follow the worked solutions, then transfer the method to the final quick tasks.

Worked examples · 3 problems

PROBLEM 1 · ARRAY, DIE AND COIN

A fair six-sided die is rolled and a fair coin is flipped. (a) Draw an array showing all outcomes. State $n(S)$. (b) Find $P(\text{head and even number})$. (c) Find $P(\text{tail or number less than 3})$ using the array.

Part (a), construct the array
Columns: 1, 2, 3, 4, 5, 6 Rows: H, T
Cells: (H,1) (H,2) (H,3) (H,4) (H,5) (H,6)
(T,1) (T,2) (T,3) (T,4) (T,5) (T,6)
$n(S) = 6 \times 2 = 12$

One row per coin outcome, one column per die outcome. Every combination is listed exactly once.

PROBLEM 2 · TREE DIAGRAM WITH REPLACEMENT

A bag contains 3 blue and 2 red marbles. A marble is drawn, its colour noted, then REPLACED, and a second marble drawn. (a) Construct a probability tree. (b) Find $P(\text{both same colour})$. (c) Find $P(\text{at least one blue})$.

Part (a), probability tree (with replacement)
Stage 1: B $\frac{3}{5}$, R $\frac{2}{5}$
Stage 2 from B: B $\frac{3}{5}$, R $\frac{2}{5}$ (same, replaced)
Stage 2 from R: B $\frac{3}{5}$, R $\frac{2}{5}$ (same, replaced)
Paths: $BB = \frac{3}{5} \times \frac{3}{5} = \frac{9}{25}$
$BR = \frac{3}{5} \times \frac{2}{5} = \frac{6}{25}$
$RB = \frac{2}{5} \times \frac{3}{5} = \frac{6}{25}$
$RR = \frac{2}{5} \times \frac{2}{5} = \frac{4}{25}$
Check: $9+6+6+4 = 25$ ✓

With replacement the bag is reset each time, so stage-2 branches are identical to stage-1. All four path probabilities must sum to 1.

PROBLEM 3 · WITHOUT REPLACEMENT

A bag contains 2 red and 3 blue marbles. Two marbles are drawn WITHOUT replacement. (a) Construct a probability tree. (b) Find $P(\text{both blue})$. (c) Find $P(\text{one of each colour})$.

Part (a), probability tree (without replacement)
Stage 1: R $\frac{2}{5}$, B $\frac{3}{5}$
Stage 2 IF R first: 4 marbles left (1R, 3B)
→ R $\frac{1}{4}$, B $\frac{3}{4}$
Stage 2 IF B first: 4 marbles left (2R, 2B)
→ R $\frac{2}{4} = \frac{1}{2}$, B $\frac{2}{4} = \frac{1}{2}$
Paths: $RR = \frac{2}{5} \times \frac{1}{4} = \frac{2}{20} = \frac{1}{10}$
$RB = \frac{2}{5} \times \frac{3}{4} = \frac{6}{20} = \frac{3}{10}$
$BR = \frac{3}{5} \times \frac{2}{4} = \frac{6}{20} = \frac{3}{10}$
$BB = \frac{3}{5} \times \frac{2}{4} = \frac{6}{20} = \frac{3}{10}$
Check: $2+6+6+6 = 20$ ✓

Without replacement the denominator drops from 5 to 4. The numerator also drops by 1 if the same colour was drawn. Critically, stage-2 branches differ depending on what was drawn first.

Match each setup with its sample space size:

Roll one die and flip one coin
Draw two cards from a 5-card set WITHOUT replacement (ordered)
Spin a 3-section spinner twice
Flip a coin three times

Top 3 list: What THREE things must you check when you have drawn a probability tree?

Show what you can do

Answer the exam-style questions, then compare your working with the model answers.

Short answer, exam-style questions

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ApplyBand 33 marks

SA 1. A fair die is rolled and a fair coin is flipped simultaneously. (a) State the total number of outcomes $n(S)$. (b) Find $P(\text{a number greater than 4 and heads})$. (c) Find $P(\text{tails})$, does it match the expected value of $1/2$? Explain using the array. (3 marks)

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ApplyBand 44 marks

SA 2. A bag contains 4 green and 1 yellow marble. A marble is drawn, noted and REPLACED, then drawn again. (a) Construct a probability tree showing all four paths with their probabilities. (b) Find $P(\text{green both times})$. (c) Find $P(\text{at least one yellow})$ using the complement. (4 marks)

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AnalyseBand 4–54 marks

SA 3. A bag has 3 red and 2 white balls. Two balls are drawn WITHOUT replacement. (a) Draw a probability tree for the two draws. (b) Calculate the probability of drawing one red and one white ball (in any order). (c) Is it more likely to get two reds or one of each colour? Show your working. (4 marks)

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📖 Comprehensive answers (click to reveal)

SA 1 (3 marks): (a) $n(S) = 6 \times 2 = 12$ [1]. (b) Numbers >4 are 5 and 6; $P(\text{>4 and H}) = 2/12 = 1/6$ [1]. (c) Tails: 6 cells with T in the array, $P = 6/12 = 1/2$, matches the expected $1/2$, confirming the symmetry of the sample space [1].

SA 2 (4 marks): (a) $GG = 4/5 \times 4/5 = 16/25$; $GY = 4/5 \times 1/5 = 4/25$; $YG = 1/5 \times 4/5 = 4/25$; $YY = 1/5 \times 1/5 = 1/25$. Check: $16+4+4+1 = 25/25$ ✓ [2]. (b) $P(GG) = 16/25$ [1]. (c) $P(\text{at least one Y}) = 1 - P(GG) = 1 - 16/25 = 9/25$ [1]. (Or add $GY + YG + YY = 4/25 + 4/25 + 1/25 = 9/25$.)

SA 3 (4 marks): (a) Stage 1: R($3/5$), W($2/5$). Stage 2 if R: R($2/4$)=$1/2$, W($2/4$)=$1/2$; Stage 2 if W: R($3/4$), W($1/4$) [2]. (b) $P(RW) = 3/5 \times 2/4 = 6/20 = 3/10$; $P(WR) = 2/5 \times 3/4 = 6/20 = 3/10$; $P(\text{one each}) = 3/10 + 3/10 = 3/5$ [1]. (c) $P(\text{two reds}) = 3/5 \times 2/4 = 6/20 = 3/10 < 3/5$. More likely to get one of each colour [1].

Retrieve, reflect and finish

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Revisit your thinking

Bag: 3 blue, 2 red, with replacement. Stage 1 branches: B(3/5), R(2/5). Stage 2: same probabilities since marble was replaced. $P(\text{blue then red}) = 3/5 \times 2/5 = 6/25$. All four combinations are NOT equally likely, BB has probability 9/25 while RR has 4/25. The key insight: equal probabilities at each stage does not mean all multi-stage paths are equally likely unless all single-stage probabilities are equal.

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