Probability: complementary & compound events

What you will learn

use the complement rule $P(A) + P(\text{not } A) = 1$ ,
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.

1. Complementary events

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

Complement rule

P(A) + P(A') = 1, \qquad P(A') = 1 - P(A).

Worked example 1 Complement

The probability of rain tomorrow is $0.35$ . The probability it does not rain is $1 - 0.35 = 0.65$ .

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:

Tree diagram for flipping two coins. Each branch has probability 1/2. The four leaves give the sample space.

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

$P(\text{at least one head}) = \tfrac{3}{4}$ . (Three of the four outcomes include an $H$ .)

Worked example 3 Two dice (table)

Roll two dice and record the sum. A $6 \times 6$ table shows $36$ equally-likely outcomes.

$P(\text{sum} = 7) = \tfrac{6}{36} = \tfrac{1}{6}$ (the six outcomes $(1,6), (2,5), \ldots, (6,1)$ ).

3. Venn diagrams and two-way tables

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

Venn-diagram formulas

Union

P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and } B).

We subtract the overlap because it was counted in both.

Mutually exclusive

If $A$ and $B$ cannot happen together, $P(A \text{ and } B) = 0$ , and $P(A \text{ or } B) = P(A) + P(B)$ .

Worked example 4 Two-way table

In a class of $30$ students, $18$ play netball, $12$ play football, and $6$ play both.

	Plays football	Does not
Plays netball	$6$	$12$
Does not	$6$	$6$

From the table:

$P(\text{netball}) = \dfrac{18}{30} = 0.60$ .
$P(\text{netball and football}) = \dfrac{6}{30} = 0.20$ .
$P(\text{netball or football}) = \dfrac{18 + 12 - 6}{30} = \dfrac{24}{30} = 0.80$ .

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 $1000$ rolls of two dice on a spreadsheet. Count how often the sum equals $7$ . Expected: about $\tfrac{1000}{6} \approx 167$ . Actual result might be $171$ - close, as predicted by the law of large numbers.

Practice: Year 8 core

Fluency

Complementary events

$P(A) = 0.3$ . Find $P(A')$ .
$P(\text{rain}) = 65\%$ . Find $P(\text{no rain})$ .
A bag has $3$ red and $7$ blue balls. $P(\text{not red})$ ?
A spinner has $8$ equal sectors, one labelled “WIN”. $P(\text{not WIN})$ ?
Two dice rolled. $P(\text{sum} \ne 7)$ ?

Fluency

Two-stage experiments

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

Fluency

Venn and two-way tables

In a survey of $40$ people: $24$ like coffee, $20$ like tea, $10$ like both.

How many like only coffee?
How many like only tea?
How many like neither?
$P(\text{likes coffee or tea})$ ?
$P(\text{likes exactly one})$ ?

Reasoning

Explain and spot the mistake

Sam says ” $P(A) + P(B) = 1$ , so $A$ and $B$ are complementary.” Is Sam correct? Give a counter-example.
A student writes “mutually exclusive” events are the same as “independent” events. Are they? Explain with examples.
Using the two-way table for coffee/tea, explain why $P(\text{coffee}) + P(\text{tea}) \ne P(\text{coffee or tea})$ .
After flipping a coin and getting $6$ heads in a row, Ben says the next flip is “more likely tails to balance it out”. Is Ben correct?

Problem solving

Real contexts

A committee of two is chosen from Anna, Ben, Chloe, Dan. List the sample space. What is $P(\text{Anna is chosen})$ ?
A bag has $5$ red and $3$ blue marbles. One is drawn, not replaced, then another is drawn. Using a tree diagram, find $P(\text{both red})$ .
$60\%$ of students play a sport; $40\%$ play an instrument; $25\%$ do both. $P(\text{neither})$ ?
Two fair coins and a die are tossed together. $P(\text{2 heads and a 6})$ ?

Challenge

Reasoning

Harder reasoning

A fair coin is tossed $3$ times. Find the probability of getting exactly $2$ heads using a tree diagram.
A spinner has sectors $1$ - $5$ (equally likely). It is spun twice. Find $P(\text{sum} \geq 8)$ .
In a class, $P(\text{boy}) = 0.48$ , $P(\text{plays cricket}) = 0.30$ , $P(\text{boy and plays cricket}) = 0.20$ . Find $P(\text{boy or plays cricket})$ and $P(\text{neither})$ .
Two dice are rolled. Using a $6 \times 6$ table, find the probability that the difference of the two faces is $2$ .

Answers

Answer key

Attempt the practice first. When you're ready to check, expand the answers below.

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Year 8 core - answers

Fluency

Complementary events

$0.7$
$35\%$
$\dfrac{7}{10}$
$\dfrac{7}{8}$
$\dfrac{5}{6}$ . Method: $1 - \dfrac{6}{36}$ .

Fluency

Two-stage experiments

$4$
$12$
$\dfrac{1}{4}$ . Method: $\dfrac{3}{6} \times \dfrac{3}{6}$ .
$\dfrac{3}{36} = \dfrac{1}{12}$ . Method: $(4,6), (5,5), (6,4)$ .
$\dfrac{11}{36}$ . Method: $1 - P(\text{no 6}) = 1 - \left(\dfrac{5}{6}\right)^2 = 1 - \dfrac{25}{36}$ .

Fluency

Venn and two-way tables

$14$ .
$10$ .
$6$ . Method: $40 - 24 - 10 - (\text{neither}) = 40 - 24 - 10 + 10 = ?$ . Actually: coffee only $+$ tea only $+$ both $+$ neither $= 40$ ; $14 + 10 + 10 + ? = 40$ ; neither $= 6$ .
$\dfrac{34}{40} = \dfrac{17}{20} = 0.85$ .
$\dfrac{24}{40} = \dfrac{3}{5} = 0.60$ .

Reasoning

Explain and spot the mistake

Not correct. Complementary events must also cover all possibilities (be exhaustive) and not overlap. Counter-example: $A =$ “roll a 1”, $B =$ “roll a 6” on a fair die. $P(A) + P(B) = \tfrac{2}{6} \ne 1$ , and even if both halved the probability to sum to $1$ , they’d need to overlap zero and exhaust the sample space.
Not the same. Mutually exclusive events cannot both happen. Independent events have no influence on each other. Example: “rolling a 6” and “rolling a 1” are mutually exclusive but not independent in a single roll. “Coin lands heads” and “die rolls 6” are independent but not mutually exclusive.
Because the overlap (both coffee and tea) is counted twice when you add $P(\text{coffee}) + P(\text{tea})$ . You subtract $P(\text{both})$ to correct.
No - the gambler’s fallacy. Each flip is independent; the coin has no memory. $P(\text{tails next}) = 0.5$ .

Problem solving

Real contexts

Pairs: AB, AC, AD, BC, BD, CD - six total. $P(\text{Anna}) = \dfrac{3}{6} = \dfrac{1}{2}$ .
$P(\text{both red}) = \dfrac{5}{8} \times \dfrac{4}{7} = \dfrac{20}{56} = \dfrac{5}{14}$ .
$P(\text{neither}) = 1 - P(\text{sport or instrument}) = 1 - (0.60 + 0.40 - 0.25) = 1 - 0.75 = 0.25$ .
$P = \dfrac{1}{2} \times \dfrac{1}{2} \times \dfrac{1}{6} = \dfrac{1}{24}$ .

Challenge - answers

Reasoning

Harder reasoning

$\dfrac{3}{8}$ . Method: $3$ outcomes with exactly $2$ heads out of $8$ total (HHT, HTH, THH).
$\dfrac{6}{25}$ . Method: $P(\text{sum} \geq 8)$ : count pairs (3,5),(4,4),(4,5),(5,3),(5,4),(5,5) - six pairs.
$P(\text{boy or plays cricket}) = 0.48 + 0.30 - 0.20 = 0.58$ ; $P(\text{neither}) = 1 - 0.58 = 0.42$ .
$\dfrac{8}{36} = \dfrac{2}{9}$ . Method: pairs with difference $2$ : $(1,3), (3,1), (2,4), (4,2), (3,5), (5,3), (4,6), (6,4)$ - eight.

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