Year 8 Mathematics | Victorian Curriculum 2.0
Probability: complementary & compound events
Topic 16 | Statistics & Probability | Practice

What you will learn

  • use the complement rule P(A)+P(not A)=1P(A) + P(\text{not } A) = 1P(A)+P(not A)=1,
  • list outcomes of two-stage experiments using tables and tree diagrams,
  • use Venn diagrams and two-way tables to organise events,
  • recognise mutually exclusive events and calculate “A or B” probabilities,
  • compare predicted and experimental probabilities in simulations.
Where you'll see this
  • Weather forecasts: ”70%70\%70% chance of rain” means P(rain)=0.7P(\text{rain}) = 0.7P(rain)=0.7, P(no rain)=0.3P(\text{no rain}) = 0.3P(no rain)=0.3.
  • Genetics: a Punnett square is a two-way table of parent genotypes.
  • Quality control: factories use P(defect)P(\text{defect})P(defect) from sampled items.
  • Games: “PPP(at least one six in two rolls)” uses the complement rule.

1. Complementary events

Two events are complementary if exactly one of them must happen. The event “not A”, written A′A'A′, is the complement of AAA.

Complement rule
P(A)+P(A′)=1,P(A′)=1−P(A).P(A) + P(A') = 1, \qquad P(A') = 1 - P(A).P(A)+P(A′)=1,P(A′)=1−P(A).
Worked example 1 Complement

The probability of rain tomorrow is 0.350.350.35. The probability it does not rain is 1−0.35=0.651 - 0.35 = 0.651−0.35=0.65.

2. Two-stage experiments

Two coins, a coin and a die, or two draws from a bag - list outcomes systematically.

Worked example 2 Two coins (tree diagram)

Flip two coins. The tree diagram shows all four equally likely outcomes:

1/2H1/2THTHTHH1/4HT1/4TH1/4TT1/4
Tree diagram for flipping two coins. Each branch has probability 1/2. The four leaves give the sample space.

Sample space: {HH,HT,TH,TT}\{HH, HT, TH, TT\}{HH,HT,TH,TT}. Each outcome has probability 14\tfrac{1}{4}41​.

P(at least one head)=34P(\text{at least one head}) = \tfrac{3}{4}P(at least one head)=43​. (Three of the four outcomes include an HHH.)

Worked example 3 Two dice (table)

Roll two dice and record the sum. A 6×66 \times 66×6 table shows 363636 equally-likely outcomes.

P(sum=7)=636=16P(\text{sum} = 7) = \tfrac{6}{36} = \tfrac{1}{6}P(sum=7)=366​=61​ (the six outcomes (1,6),(2,5),…,(6,1)(1,6), (2,5), \ldots, (6,1)(1,6),(2,5),…,(6,1)).

3. Venn diagrams and two-way tables

For two events AAA and BBB defined on the same population, a Venn diagram (two overlapping circles) or a two-way table both show the four regions: in AAA only, in BBB only, in both, in neither.

Venn-diagram formulas

Union
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).

We subtract the overlap because it was counted in both.

Mutually exclusive

If AAA and BBB cannot happen together, P(A and B)=0P(A \text{ and } B) = 0P(A and B)=0, and 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).

Worked example 4 Two-way table

In a class of 303030 students, 181818 play netball, 121212 play football, and 666 play both.

Plays footballDoes not
Plays netball666121212
Does not666666

From the table:

  • P(netball)=1830=0.60P(\text{netball}) = \dfrac{18}{30} = 0.60P(netball)=3018​=0.60.
  • P(netball and football)=630=0.20P(\text{netball and football}) = \dfrac{6}{30} = 0.20P(netball and football)=306​=0.20.
  • P(netball or football)=18+12−630=2430=0.80P(\text{netball or football}) = \dfrac{18 + 12 - 6}{30} = \dfrac{24}{30} = 0.80P(netball or football)=3018+12−6​=3024​=0.80.
30 studentsNetballFootball12666neither
Venn diagram for the same data. The overlap (6) is counted in both circles, which is why we subtract it in the union formula.

4. Simulations

For complex experiments or when theoretical probability is hard to compute, run many repeated trials (by hand, spreadsheet, or random-number generator) and estimate by relative frequency.

Worked example 5 Simulation of two dice

Simulate 100010001000 rolls of two dice on a spreadsheet. Count how often the sum equals 777. Expected: about 10006≈167\tfrac{1000}{6} \approx 16761000​≈167. Actual result might be 171171171 - close, as predicted by the law of large numbers.


Practice: Year 8 core

Fluency

Complementary events

    1. P(A)=0.3P(A) = 0.3P(A)=0.3. Find P(A′)P(A')P(A′).
    2. P(rain)=65%P(\text{rain}) = 65\%P(rain)=65%. Find P(no rain)P(\text{no rain})P(no rain).
    3. A bag has 333 red and 777 blue balls. P(not red)P(\text{not red})P(not red)?
    4. A spinner has 888 equal sectors, one labelled “WIN”. P(not WIN)P(\text{not WIN})P(not WIN)?
    5. Two dice rolled. P(sum≠7)P(\text{sum} \ne 7)P(sum=7)?
Fluency

Two-stage experiments

    1. Flip two coins. Sample space size?
    2. Flip a coin and roll a die. Sample space size?
    3. Two dice rolled. P(both show even)P(\text{both show even})P(both show even)?
    4. Two dice rolled. P(sum=10)P(\text{sum} = 10)P(sum=10)?
    5. Two dice rolled. P(at least one 6)P(\text{at least one 6})P(at least one 6)? (Hint: consider the complement.)
Fluency

Venn and two-way tables

    In a survey of 404040 people: 242424 like coffee, 202020 like tea, 101010 like both.

    1. How many like only coffee?
    2. How many like only tea?
    3. How many like neither?
    4. P(likes coffee or tea)P(\text{likes coffee or tea})P(likes coffee or tea)?
    5. P(likes exactly one)P(\text{likes exactly one})P(likes exactly one)?
Reasoning

Explain and spot the mistake

    1. Sam says ”P(A)+P(B)=1P(A) + P(B) = 1P(A)+P(B)=1, so AAA and BBB are complementary.” Is Sam correct? Give a counter-example.
    2. A student writes “mutually exclusive” events are the same as “independent” events. Are they? Explain with examples.
    3. Using the two-way table for coffee/tea, explain why P(coffee)+P(tea)≠P(coffee or tea)P(\text{coffee}) + P(\text{tea}) \ne P(\text{coffee or tea})P(coffee)+P(tea)=P(coffee or tea).
    4. After flipping a coin and getting 666 heads in a row, Ben says the next flip is “more likely tails to balance it out”. Is Ben correct?
Problem solving

Real contexts

    1. A committee of two is chosen from Anna, Ben, Chloe, Dan. List the sample space. What is P(Anna is chosen)P(\text{Anna is chosen})P(Anna is chosen)?
    2. A bag has 555 red and 333 blue marbles. One is drawn, not replaced, then another is drawn. Using a tree diagram, find P(both red)P(\text{both red})P(both red).
    3. 60%60\%60% of students play a sport; 40%40\%40% play an instrument; 25%25\%25% do both. P(neither)P(\text{neither})P(neither)?
    4. Two fair coins and a die are tossed together. P(2 heads and a 6)P(\text{2 heads and a 6})P(2 heads and a 6)?

Challenge

Reasoning

Harder reasoning

    1. A fair coin is tossed 333 times. Find the probability of getting exactly 222 heads using a tree diagram.
    2. A spinner has sectors 111-555 (equally likely). It is spun twice. Find P(sum≥8)P(\text{sum} \geq 8)P(sum≥8).
    3. In a class, P(boy)=0.48P(\text{boy}) = 0.48P(boy)=0.48, P(plays cricket)=0.30P(\text{plays cricket}) = 0.30P(plays cricket)=0.30, P(boy and plays cricket)=0.20P(\text{boy and plays cricket}) = 0.20P(boy and plays cricket)=0.20. Find P(boy or plays cricket)P(\text{boy or plays cricket})P(boy or plays cricket) and P(neither)P(\text{neither})P(neither).
    4. Two dice are rolled. Using a 6×66 \times 66×6 table, find the probability that the difference of the two faces is 222.
Year 8 Mathematics study companion | Practice