Open Math›Year 11›Mathematics Advanced›Probability and data›Probability rules

Probability rules

Find probabilities using equally likely outcomes, apply the complement and addition rules, and verify whether two events are independent.

Worked examples

Finding a probability and its complement

Straightforward

Problem

A bag contains 5 red counters, 3 blue counters and 2 yellow counters. A counter is drawn at random.

(a) Find

P(\text{red})

.
(b) Find

P(\text{not red})

Count the total number of counters in the bag.

|S| = 5 + 3 + 2 = 10

Apply

P(A) = \dfrac{|A|}{|S|}

to find

P(\text{red})

P(\text{red}) = \frac{5}{10} = \frac{1}{2}

Apply the complement rule

P(\bar{A}) = 1 - P(A)

to find

P(\text{not red})

P(\text{not red}) = 1 - \frac{1}{2} = \frac{1}{2}

Answer

(a)

P(\text{red}) = \dfrac{1}{2}

. (b)

P(\text{not red}) = \dfrac{1}{2}

Applying the addition rule

Moderate

Problem

In a class, the probability that a student owns a dog is

0.45

, the probability they own a cat is

0.30

, and the probability they own both is

0.15

. Find the probability that a student owns a dog or a cat.

State the addition rule for two events.

P(A \cup B) = P(A) + P(B) - P(A \cap B)

Identify the values:

P(\text{dog}) = 0.45

P(\text{cat}) = 0.30

P(\text{both}) = 0.15

Let

D

= owns a dog and

C

= owns a cat.

P(D) = 0.45, \quad P(C) = 0.30, \quad P(D \cap C) = 0.15

Substitute into the addition rule.

P(D \cup C) = 0.45 + 0.30 - 0.15 = 0.60

Answer

P(\text{dog or cat}) = 0.60

Using a two-way table and checking independence

Challenging

Problem

The two-way table below classifies 80 students by whether they play sport and whether they play a musical instrument.

| | Plays sport | Does not play sport | Total |
|---|---|---|---|
| Plays an instrument | 24 | 16 | 40 |
| Does not play an instrument | 20 | 20 | 40 |
| Total | 44 | 36 | 80 |

(a) Find

P(\text{plays sport AND plays an instrument})

.
(b) Find

P(\text{plays sport OR plays an instrument})

.
(c) Determine whether playing sport and playing an instrument are independent events.

Read the joint frequency from the table to find part (a).

P(\text{sport} \cap \text{instrument}) = \frac{24}{80} = \frac{3}{10}

Find the marginal probabilities, then apply the addition rule for part (b).

P(\text{sport}) = \frac{44}{80} = \frac{11}{20}, \quad P(\text{instrument}) = \frac{40}{80} = \frac{1}{2}

P(\text{sport} \cup \text{instrument}) = \frac{11}{20} + \frac{1}{2} - \frac{3}{10} = \frac{11}{20} + \frac{10}{20} - \frac{6}{20} = \frac{15}{20} = \frac{3}{4}

Test independence by checking whether

P(A \cap B) = P(A) \times P(B)

for part (c).

P(\text{sport}) \times P(\text{instrument}) = \frac{11}{20} \times \frac{1}{2} = \frac{11}{40}

P(\text{sport} \cap \text{instrument}) = \frac{3}{10} = \frac{12}{40}

Since

\dfrac{11}{40} \ne \dfrac{12}{40}

, the events are not independent.

Answer

(a)

\dfrac{3}{10}

. (b)

\dfrac{3}{4}

. (c) Not independent, since

P(\text{sport}) \times P(\text{instrument}) = \dfrac{11}{40} \ne \dfrac{3}{10} = P(\text{sport} \cap \text{instrument})

Practise

Q1·Straightforward

A fair six-sided die is rolled. The sample space is

S = \{1, 2, 3, 4, 5, 6\}

. What is the probability of rolling a number less than 3?

$\dfrac{1}{6}$$\dfrac{1}{3}$$\dfrac{1}{2}$$\dfrac{2}{3}$

Explanation

The outcomes less than 3 are

\{1, 2\}

, giving 2 favourable outcomes out of 6 equally likely outcomes. So

P(\text{less than 3}) = \dfrac{2}{6} = \dfrac{1}{3}

Q2·Straightforward

A bag contains 4 red marbles, 3 green marbles and 3 blue marbles. A marble is drawn at random. Find the probability of drawing a red marble. Give your answer as a fraction in simplest form.

Explanation

There are

4 + 3 + 3 = 10

marbles in total. There are 4 red marbles, so

P(\text{red}) = \dfrac{4}{10} = \dfrac{2}{5}

Q3·Straightforward

The probability that a student passes a test is

0.72

. Find the probability that the student does not pass.

Explanation

Using the complement rule:

P(\text{not pass}) = 1 - P(\text{pass}) = 1 - 0.72 = 0.28

Q4·Straightforward

In a school of 200 students, 125 catch the bus to school. A student is chosen at random. Find the probability that the chosen student does not catch the bus. Give your answer as a fraction in simplest form.

Explanation

The number of students who do not catch the bus is

200 - 125 = 75

. So

P(\text{not bus}) = \dfrac{75}{200} = \dfrac{3}{8}

Q5·Moderate

For events

A

and

B

P(A) = 0.5

P(B) = 0.4

and

P(A \cap B) = 0.2

. Find

P(A \cup B)

Explanation

Using the addition rule:

P(A \cup B) = P(A) + P(B) - P(A \cap B) = 0.5 + 0.4 - 0.2 = 0.7

Q6·Moderate

Events

C

and

D

are mutually exclusive.

P(C) = 0.3

and

P(D) = 0.45

. Find

P(C \cup D)

Explanation

Since

C

and

D

are mutually exclusive,

P(C \cap D) = 0

. Therefore:

P(C \cup D) = P(C) + P(D) = 0.3 + 0.45 = 0.75

Q7·Moderate

For events

A

and

B

P(A) = 0.6

P(B) = 0.35

and

P(A \cup B) = 0.8

. Find

P(A \cap B)

Explanation

Rearranging the addition rule:

P(A \cap B) = P(A) + P(B) - P(A \cup B) = 0.6 + 0.35 - 0.8 = 0.15

Q8·Moderate

Events

E

and

F

have

P(E) = 0.3

P(F) = 0.5

and

P(E \cup F) = 0.8

. Which statement is correct?

$E$ and $F$ are mutually exclusive since $P(E \cap F) = 0$.$E$ and $F$ are not mutually exclusive since $P(E \cup F) = 0.8$.$E$ and $F$ are independent since $P(E) + P(F) = 0.8$.$E$ and $F$ overlap since $P(E \cup F) < 1$.

Explanation

From the addition rule:

P(E \cap F) = P(E) + P(F) - P(E \cup F) = 0.3 + 0.5 - 0.8 = 0

. Since

P(E \cap F) = 0

, the events cannot occur together, so

E

and

F

are mutually exclusive.

Q9·Challenging

A fair coin is tossed and a fair four-sided spinner with faces numbered 1 to 4 is spun. The sample space has 8 equally likely outcomes. Find the probability of getting a tail and an even number. Give your answer as a fraction in simplest form.

Explanation

The 8 equally likely outcomes are:

(H,1), (H,2), (H,3), (H,4), (T,1), (T,2), (T,3), (T,4)

. The outcomes with a tail and an even number are

(T,2)

and

(T,4)

— that is, 2 favourable outcomes. So

P(\text{tail and even}) = \dfrac{2}{8} = \dfrac{1}{4}

Q10·Challenging

In a survey of 200 adults, 120 regularly read a newspaper, 80 regularly listen to a podcast, and 50 do both. Find the probability that a randomly selected adult reads a newspaper or listens to a podcast. Give your answer as a fraction in simplest form.

Explanation

Using the addition rule:

|\text{newspaper} \cup \text{podcast}| = 120 + 80 - 50 = 150

P(\text{newspaper or podcast}) = \frac{150}{200} = \frac{3}{4}

Q11·Challenging

Events

A

and

B

have

P(A) = 0.4

P(B) = 0.5

and

P(A \cap B) = 0.2

. Which statement correctly describes

A

and

B

$A$ and $B$ are independent because $P(A \cap B) = P(A) \times P(B)$.$A$ and $B$ are not independent because $P(A \cap B) \ne P(A) + P(B)$.$A$ and $B$ are mutually exclusive because $P(A) + P(B) = 0.9$.$A$ and $B$ are independent because $P(A \cup B) = 1$.

Explanation

Checking the independence condition:

P(A) \times P(B) = 0.4 \times 0.5 = 0.20

. Since

P(A \cap B) = 0.20 = P(A) \times P(B)

, events

A

and

B

are independent.

Q12·Challenging

The two-way table below shows 100 students classified by exercise habit and breakfast habit.

| | Eats breakfast | Skips breakfast | Total |
|---|---|---|---|
| Exercises regularly | 45 | 15 | 60 |
| Does not exercise | 20 | 20 | 40 |
| Total | 65 | 35 | 100 |

Find the probability that a randomly selected student both exercises regularly and eats breakfast. Give your answer as a fraction in simplest form.

Explanation

From the table, 45 out of 100 students both exercise regularly and eat breakfast. So

P(\text{exercises and eats breakfast}) = \dfrac{45}{100} = \dfrac{9}{20}