Practice Maths

Multi-Step Probability and Tree Diagrams

Key Ideas

Key Terms

multi-stage experiment
Involves two or more events performed in sequence (e.g. spin then flip).
tree diagram
Shows all possible outcomes of a multi-stage experiment. Each branch represents one outcome at that stage, labelled with its probability.
sample space
The complete list of all possible outcomes. For a tree, list all end-of-branch paths.
independent events
A and B: P(A and B) = P(A) × P(B). Multiply probabilities along branches.
replacement
Probabilities stay the same on every stage. Without replacement: the pool shrinks, so probabilities change.
Hot Tip Always check that all branch probabilities at each stage sum to 1. If they don’t, you have made an error in the tree.

Worked Example

Question: A spinner has three equal sectors: Red (R), Blue (B), and Green (G). After spinning, a fair coin is flipped. Draw a tree diagram and find P(Red AND Heads).

Stage 1 — Spinner: P(R) = 1/3, P(B) = 1/3, P(G) = 1/3

Stage 2 — Coin: P(H) = 1/2, P(T) = 1/2  (from each spinner branch)

Sample space (6 outcomes): RH, RT, BH, BT, GH, GT — each has probability 1/3 × 1/2 = 1/6

P(Red AND Heads) = P(R) × P(H) = 1/3 × 1/2 = 1/6

Check: all 6 outcomes each have probability 1/6, and 6 × 1/6 = 1. ✓

Multi-Step Experiments

A multi-step experiment involves two or more stages, such as flipping a coin twice, drawing two cards from a deck, or choosing two students from a class. For each stage, the possible outcomes branch out, and we need to track all combinations systematically.

The total number of outcomes in a two-step experiment is found by multiplying: if Step 1 has a outcomes and Step 2 has b outcomes, there are a × b outcomes in total (this is the counting principle).

Drawing Tree Diagrams

A tree diagram is the most reliable way to list all outcomes of a multi-step experiment. At each branch, write the outcome and its probability on the branch itself.

Example: A bag has 3 red (R) and 2 blue (B) marbles. One is drawn, then replaced, then another is drawn. The first-level branches are R (prob 3/5) and B (prob 2/5). From each, second-level branches are again R (3/5) and B (2/5).

To find the probability of any complete path (e.g. RR), multiply along the branches: P(RR) = 3/5 × 3/5 = 9/25.

To find the probability of an event that can happen multiple ways (e.g. "one of each colour"), add the relevant branch products: P(one of each) = P(RB) + P(BR) = (3/5 × 2/5) + (2/5 × 3/5) = 6/25 + 6/25 = 12/25.

With vs Without Replacement

With replacement: After drawing an item, it is put back before the next draw. This means the probabilities on the second set of branches are the same as the first. The draws are independent.

Without replacement: The item is not returned. This changes the probabilities on the second set of branches because there is now one fewer item in the total. The draws are dependent.

Example (without replacement): Same bag of 3R and 2B. Draw two without replacing. If first draw is R (prob 3/5), now there are 2R and 2B left (4 total). P(second is R | first was R) = 2/4 = 1/2. So P(RR) = 3/5 × 1/2 = 3/10.

Systematic Listing as an Alternative

For small experiments, you can list all equally-likely outcomes in a table or list rather than a tree diagram. For example, flipping two coins: {HH, HT, TH, TT} gives 4 outcomes, each equally likely (prob 1/4).

Tree diagrams become more efficient when probabilities differ between branches (i.e. when outcomes are not equally likely), because they let you calculate combined probabilities directly without listing every outcome.

The Addition Rule for "Or"

When an event can occur via multiple paths in a tree diagram, add the probabilities of those paths. For mutually exclusive events A and B (they can't both happen at the same time): P(A or B) = P(A) + P(B). Each separate branch that leads to the desired outcome contributes to the total probability.

Key tip: Always check whether the problem is "with replacement" or "without replacement" — this changes the second-branch probabilities. With replacement: same fractions each time. Without replacement: subtract 1 from the numerator (if drawing the same type) and always subtract 1 from the denominator.

Mastery Practice

  1. A bag contains 2 red (R) and 1 blue (B) marble. A marble is drawn, its colour noted, then it is returned to the bag and a second marble is drawn. Fluency

    Draw 1 Draw 2 Outcome R (2/3) B (1/3) R (2/3) RR B (1/3) RB R (2/3) BR B (1/3) BB
    1. Draw a fully labelled tree diagram showing all outcomes for the two draws.
    2. List the sample space.
    3. Find P(Red, Red).
    4. Find P(at least one Blue).
    5. Find P(both the same colour).

  2. Two fair coins are flipped one after the other. Fluency

    1. Draw a tree diagram and list the sample space.
    2. Find P(two Heads).
    3. Find P(exactly one Tail).
    4. Find P(at least one Head).

  3. A spinner has equal sectors labelled 1, 2, 3. A fair die is then rolled. Fluency

    1. How many outcomes are in the sample space? Explain without drawing the full tree.
    2. Find P(spinner shows 2 AND die shows 5).
    3. Find P(spinner and die show the same number).
    4. Find P(the sum of spinner and die is greater than 7).

  4. A student randomly selects one of four elective subjects (Art, Drama, Music, Sport) in Year 9, then one of three language options (French, Japanese, Spanish). Fluency

    1. How many possible combinations are there?
    2. Find P(Art and French).
    3. Find P(Drama or Music, with any language).
    4. Find P(not Sport, with Spanish).

  5. A box holds 3 green (G) and 2 yellow (Y) balls. Two balls are drawn one at a time without replacement. Understanding

    Draw 1 Draw 2 Outcome G (3/5) Y (2/5) G (2/4) GG Y (2/4) GY G (3/4) YG Y (1/4) YY
    1. Draw a tree diagram with probabilities on each branch. (Note how second-stage probabilities change.)
    2. Find P(Green, Green).
    3. Find P(one of each colour).
    4. Find P(at least one Yellow).
    5. Compare your answer to (b) with what you would get if the balls were replaced. Explain why the probabilities differ.

  6. A fair coin is flipped three times. Understanding

    1. How many outcomes are in the full sample space?
    2. Find P(three Heads).
    3. Find P(exactly two Heads).
    4. Find P(at least one Tail).
    5. Find P(Heads on the first flip given the other two are Tails).

  7. A deck has only 5 cards: Ace (A), 2, 3, 4, 5. Two cards are drawn without replacement. Understanding

    1. Find P(Ace drawn first).
    2. Find P(Ace drawn second, given Ace was drawn first).
    3. Find P(Ace on both draws). Is this possible? Explain.
    4. Find P(sum of the two cards is greater than 7).
    5. Find P(at least one card is odd-numbered — where Ace = 1).

  8. In a class of 15 students, 9 are girls and 6 are boys. Two students are selected at random without replacement to represent the class at a maths competition. Understanding

    1. Find P(both representatives are girls).
    2. Find P(both representatives are boys).
    3. Find P(one girl and one boy are selected).
    4. Verify your answers to (a), (b) and (c) sum to 1.

  9. A factory produces light bulbs. 5% of bulbs are defective. A quality inspector randomly selects 2 bulbs from a large batch (treat draws as independent). Problem Solving

    Bulb 1 Bulb 2 Outcome D (0.05) N (0.95) D (0.05) DD N (0.95) DN D (0.05) ND N (0.95) NN
    1. Draw a tree diagram for the two draws (Defective / Not Defective), labelling all branches with probabilities.
    2. Find P(both bulbs are defective).
    3. Find P(exactly one bulb is defective).
    4. Find P(at least one bulb is not defective).
    5. The inspector decides to pass a batch if both tested bulbs are not defective. What is the probability a batch is passed?

  10. The probability it rains on any given day in Brisbane in October is 0.3. Xavier plays cricket if it does not rain, and stays inside if it rains. He also has a 40% chance of having extra practice on any day, independent of weather. Problem Solving

    Weather Practice? Outcome Rain (0.3) No Rain (0.7) Practice (0.4) Rain, P No P (0.6) Rain, NP Practice (0.4) NR, P No P (0.6) NR, NP
    1. Draw a tree diagram showing all outcomes for one day (Rain/No Rain, then Practice/No Practice).
    2. Find P(it rains and Xavier has extra practice).
    3. Find P(Xavier plays cricket and has no extra practice).
    4. Xavier earns 2 points if he plays cricket with extra practice, 1 point if he plays without extra practice, and 0 otherwise. Find his expected number of points on a random October day.
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