Year 9 Mathematics | Victorian Curriculum 2.0
Compound probability
Topic 15 | Statistics & Probability | Practice

What you will learn

  • list outcomes of two-step experiments using tree diagrams and tables,
  • distinguish between experiments with replacement and without replacement,
  • calculate probabilities of compound events using the “and” (multiplication) and “or” (addition) rules,
  • connect experimental (relative frequency) results with theoretical probability,
  • use simulations to estimate probabilities when calculation is difficult.
Why compound probability?

Many real situations involve more than one event: drawing two cards, rolling two dice, or taking a medical test that can give false results. Single-event probability is not enough — you need rules for combining probabilities across multiple steps. The tree diagram is the key tool: it organises all possibilities and makes the multiplication and addition rules visible.

Where you'll see this
  • Medicine: two-stage screening tests — the probability of a true positive depends on the disease rate and the test accuracy.
  • Quality control: inspecting two items from a batch without replacement to decide if the batch passes.
  • Games: the chance of drawing two aces from a deck depends on whether you replace the first card.
  • Weather: the probability of rain on both Saturday and Sunday is found by combining daily forecasts.
Worked example 0 Real-world example: quality control

A box contains 888 good batteries and 222 faulty ones. Two batteries are selected at random without replacement. What is the probability that both are good?

  1. P(1st good)=810P(\text{1st good}) = \dfrac{8}{10}P(1st good)=108​.
  2. After removing one good battery, 777 good remain out of 999 total: P(2nd good∣1st good)=79P(\text{2nd good} \mid \text{1st good}) = \dfrac{7}{9}P(2nd good∣1st good)=97​.
  3. P(both good)=810×79=5690=2845≈0.622P(\text{both good}) = \dfrac{8}{10} \times \dfrac{7}{9} = \dfrac{56}{90} = \dfrac{28}{45} \approx 0.622P(both good)=108​×97​=9056​=4528​≈0.622.

Key idea: without replacement, the second probability changes because the total and the count of favourable outcomes both decrease.

1. Two-step experiments and tree diagrams

A compound event involves two or more stages. A tree diagram shows every possible path from start to finish, with probabilities written on the branches.

3/5R2/5B2/42/4RB3/41/4RBRR: 3/5 × 2/4 = 6/20RB: 3/5 × 2/4 = 6/20BR: 2/5 × 3/4 = 6/20BB: 2/5 × 1/4 = 2/201st draw2nd draw (without replacement)
Tree diagram: drawing 2 marbles without replacement from a bag containing 3 red and 2 blue marbles.

To find the probability of any single path, multiply the probabilities along the branches. The four outcomes sum to 111:

620+620+620+220=2020=1.\dfrac{6}{20} + \dfrac{6}{20} + \dfrac{6}{20} + \dfrac{2}{20} = \dfrac{20}{20} = 1.206​+206​+206​+202​=2020​=1.

2. With replacement vs without replacement

FeatureWith replacementWithout replacement
The item is returned before the next drawYesNo
Total stays the sameYesNo — decreases by 1 each draw
Probabilities change between drawsNoYes
Events areIndependentDependent
Worked example 1 With replacement

A bag contains 444 red and 666 blue marbles. Two marbles are drawn with replacement. Find P(both red)P(\text{both red})P(both red).

  1. P(1st red)=410=25P(\text{1st red}) = \dfrac{4}{10} = \dfrac{2}{5}P(1st red)=104​=52​.
  2. The marble is replaced, so P(2nd red)=410=25P(\text{2nd red}) = \dfrac{4}{10} = \dfrac{2}{5}P(2nd red)=104​=52​.
  3. P(both red)=25×25=425P(\text{both red}) = \dfrac{2}{5} \times \dfrac{2}{5} = \dfrac{4}{25}P(both red)=52​×52​=254​.
Worked example 2 Without replacement

Same bag (444 red, 666 blue). Two marbles drawn without replacement. Find P(both red)P(\text{both red})P(both red).

  1. P(1st red)=410P(\text{1st red}) = \dfrac{4}{10}P(1st red)=104​.
  2. Now 333 red remain out of 999: P(2nd red∣1st red)=39=13P(\text{2nd red} \mid \text{1st red}) = \dfrac{3}{9} = \dfrac{1}{3}P(2nd red∣1st red)=93​=31​.
  3. P(both red)=410×13=430=215P(\text{both red}) = \dfrac{4}{10} \times \dfrac{1}{3} = \dfrac{4}{30} = \dfrac{2}{15}P(both red)=104​×31​=304​=152​.

Note: 215<425\dfrac{2}{15} < \dfrac{4}{25}152​<254​ — removing a red marble on the first draw makes a second red less likely.

3. “And” vs “or” probabilities

Probability rules for compound events

And rule (multiplication)

P(A and B)=P(A)×P(B∣A)P(A \text{ and } B) = P(A) \times P(B \mid A)P(A and B)=P(A)×P(B∣A) If AAA and BBB are independent: P(A and B)=P(A)×P(B)P(A \text{ and } B) = P(A) \times P(B)P(A and B)=P(A)×P(B).

Or rule (addition)

P(A or B)=P(A)+P(B)−P(A and B)P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and } B)P(A or B)=P(A)+P(B)−P(A and B) If AAA and BBB are mutually exclusive: P(A or B)=P(A)+P(B)P(A \text{ or } B) = P(A) + P(B)P(A or B)=P(A)+P(B).

Don't double-count!

When events can overlap (e.g. drawing a card that is both red and a king), the “or” rule subtracts the overlap: P(A or B)=P(A)+P(B)−P(A and B)P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and } B)P(A or B)=P(A)+P(B)−P(A and B). Forgetting to subtract leads to a probability that is too high.

Worked example 3 Using the 'or' rule

A standard deck of 525252 cards. One card is drawn. Find P(heart or king)P(\text{heart or king})P(heart or king).

  1. P(heart)=1352P(\text{heart}) = \dfrac{13}{52}P(heart)=5213​, P(king)=452P(\text{king}) = \dfrac{4}{52}P(king)=524​.
  2. P(heart and king)=152P(\text{heart and king}) = \dfrac{1}{52}P(heart and king)=521​ (the king of hearts).
  3. P(heart or king)=1352+452−152=1652=413P(\text{heart or king}) = \dfrac{13}{52} + \dfrac{4}{52} - \dfrac{1}{52} = \dfrac{16}{52} = \dfrac{4}{13}P(heart or king)=5213​+524​−521​=5216​=134​.

4. Relative frequency and simulations

Theoretical probability is calculated from known, equally likely outcomes. Experimental probability (or relative frequency) is calculated from observed data:

Relative frequency

P(event)≈number of times event occurredtotal number of trialsP(\text{event}) \approx \dfrac{\text{number of times event occurred}}{\text{total number of trials}}P(event)≈total number of trialsnumber of times event occurred​

As the number of trials increases, relative frequency tends to approach theoretical probability — this is the law of large numbers.

A simulation models a random experiment using a tool (coin flips, random number generators, spreadsheets) when theoretical calculation is difficult or impossible.

Worked example 4 Simulation design

Estimate the probability that a family with 333 children has exactly 222 girls.

  1. Model each child as a coin flip: heads === girl, tails === boy.
  2. Flip 333 coins and record whether exactly 222 are heads. This is one trial.
  3. Repeat for 100100100 trials. Suppose 383838 trials gave exactly 222 heads.
  4. Relative frequency =38100=0.38= \dfrac{38}{100} = 0.38=10038​=0.38.
  5. Theoretical probability =(32)×(12)3=3×18=38=0.375= \dbinom{3}{2} \times \left(\dfrac{1}{2}\right)^3 = 3 \times \dfrac{1}{8} = \dfrac{3}{8} = 0.375=(23​)×(21​)3=3×81​=83​=0.375.

The simulation result (0.380.380.38) is close to the theoretical value (0.3750.3750.375).


Practice

Fluency

Tier 1: basic skills

    1. A coin is tossed twice. List all outcomes using a tree diagram.
    2. Two dice are rolled. How many outcomes are in the sample space?
    3. A bag has 555 red and 333 green marbles. One marble is drawn and replaced, then another is drawn. Find P(both red)P(\text{both red})P(both red).
    4. Repeat Q3 but without replacement.
    5. A spinner has sections labelled 1,2,3,41, 2, 3, 41,2,3,4 (equally likely). It is spun twice. Find P(both even)P(\text{both even})P(both even).
    6. From a standard deck of 525252 cards, one card is drawn. Find P(spade or ace)P(\text{spade or ace})P(spade or ace).
    7. A coin is tossed 200200200 times and lands heads 114114114 times. Calculate the relative frequency of heads.
    8. Are the events “rolling a 666” and “rolling an odd number” on a single die mutually exclusive? Explain.
    9. A tree diagram has two branches at the first stage (AAA and BBB) and two at the second stage (CCC and DDD from each). How many outcomes are there in total?
    10. In 505050 trials of a simulation, an event occurred 181818 times. Estimate the probability of the event.
Reasoning

Tier 2: mixed practice

    1. A bag contains 666 red, 444 blue, and 222 green marbles. Two marbles are drawn without replacement. Draw a full tree diagram and find P(both blue)P(\text{both blue})P(both blue).
    2. Two cards are drawn without replacement from a standard deck. Find P(both aces)P(\text{both aces})P(both aces).
    3. A box has 333 defective items out of 202020. Two items are selected at random without replacement. Find the probability that (a) both are defective, (b) neither is defective, (c) exactly one is defective.
    4. Events AAA and BBB are independent with P(A)=0.3P(A) = 0.3P(A)=0.3 and P(B)=0.5P(B) = 0.5P(B)=0.5. Find P(A and B)P(A \text{ and } B)P(A and B) and P(A or B)P(A \text{ or } B)P(A or B).
    5. A student rolls a die and flips a coin. Find the probability of getting an even number and heads.
    6. In a class of 303030, there are 121212 students who play sport and 888 who play music. Of these, 333 play both. A student is chosen at random. Find P(sport or music)P(\text{sport or music})P(sport or music).
    7. Explain the difference between independent events and mutually exclusive events, using an example of each.
    8. Design a simulation to estimate the probability that when 444 coins are tossed, at least 333 are tails. Describe the steps clearly.
Reasoning

Tier 3: explain and apply

    1. A medical test has a 95%95\%95% chance of correctly detecting a disease (sensitivity) and a 90%90\%90% chance of correctly identifying a healthy person (specificity). If 2%2\%2% of the population has the disease, draw a tree diagram and find the probability that a person who tests positive actually has the disease.
    2. Two events satisfy P(A)=0.4P(A) = 0.4P(A)=0.4, P(B)=0.5P(B) = 0.5P(B)=0.5, and P(A or B)=0.7P(A \text{ or } B) = 0.7P(A or B)=0.7. Determine whether AAA and BBB are independent. Justify.
    3. Three marbles are drawn without replacement from a bag of 555 red and 444 white. Find the probability of drawing at least one red marble.
    4. A game costs $2 to play. You roll two dice: if the sum is 777 you win $10, otherwise you win nothing. Find the expected profit per game and decide whether the game is fair.
    5. Explain why P(A or B)≤P(A)+P(B)P(A \text{ or } B) \leq P(A) + P(B)P(A or B)≤P(A)+P(B), and state when equality holds.

Challenge

Reasoning

Harder reasoning

    1. Five cards numbered 111 to 555 are placed face down. Two cards are selected at random without replacement. Find the probability that the sum of the two cards is even.
    2. A bag contains nnn red and 333 blue marbles. Two marbles are drawn without replacement. If P(both red)=12P(\text{both red}) = \dfrac{1}{2}P(both red)=21​, find nnn.
    3. In a best-of-three game series, Team A has a 0.60.60.6 probability of winning each game (games are independent). Draw a tree diagram and find the probability that Team A wins the series.
    4. Three students independently attempt a problem. Their probabilities of solving it are 0.70.70.7, 0.80.80.8, and 0.90.90.9 respectively. Find the probability that at least one student solves the problem.
Year 9 Mathematics study companion | Practice