IB Math AA HL Topic 4.5: Probability basics | Notes & Flashcards

IB Math AA HL — Probability basics

Topic 4.5 of IB Mathematics: Analysis and Approaches covers Probability basics, which is part of Unit 4: Statistics & Probability. Students explore key concepts including Sample space and probability, Tables, experimental probability and expected number. A strong understanding of probability basics is essential for IB Math AA HL exams and builds the foundation for connected topics across the syllabus.

Higher Level students should use this topic hub as a map: start with the shared sub-topics, then follow the HL-only extensions and exam-skill links where this topic asks for deeper analysis.

Key concepts in Probability basics

Key Idea: Probability measures how likely an event is, from 0 (impossible) to 1 (certain). In 4.5 you count outcomes in a sample space, use "not A", read probabilities from tables and data, and find an expected number. Topic 4.6 then combines two or more events.

Core rules

Rule	Formula	When to use
Basic probability	P(A) = n(A) ÷ n(U)	Equally likely outcomes: count A, divide by all.
Complement	P(A′) = 1 − P(A)	"Not A".
From a table	how many you want ÷ the total	A cell, a row or a column of a two-way table.
Experimental probability	frequency ÷ number of trials	Results of an experiment or survey.
Expected number	number of trials × P	How many times you expect it. It need not be whole.

Key Idea: 12 of 30 students walk to school: P(walk) = 12/30 = 0.4, so P(not walk) = 0.6. Over 190 school days, a student who walks with probability 0.4 is expected to walk 0.4 × 190 = 76 days.

Use n(A) ÷ n(U) only when the outcomes are equally likely. Two dice: count the 36 pairs, not the 11 sums.

Show the fraction, e.g. 66/200, before the decimal: it earns the method mark if a later number slips. Give non-exact answers to 3 significant figures.

IB-style question [5 marks]

A café records the drinks bought by 200 customers.

	Coffee	Tea
Under 25	46	24
25 or over	88	42

One of these customers is chosen at random. (a) Find the probability that the customer bought tea.

Solution:

Add the tea column, then divide by the total in the question.
$P(\text{tea})=\frac{24+42}{200}=\frac{66}{200}=0.33$

Final answer:

66/200 = 0.33

Part (b)

(b) Find the probability that the customer did not buy tea.

Solution:

"Not tea" is the complement: subtract from 1.
$P(\text{tea}')=1-0.33=0.67$

Final answer:

0.67

Part (c)

Solution:

Expected number = number of customers × probability.
$500\times0.33=165$

Final answer:

165 customers

IB-style question — single event and its complement

A drawer holds 7 black and 3 grey socks. One sock is taken at random. (a) Find the probability it is grey. (b) Find the probability it is not grey.

Step by step:

Total socks, then favourable ÷ total for grey.
$n(U) = 7 + 3 = 10, \quad P(\text{grey}) = \frac{3}{10}$
'Not grey' is the complement: 1 − P(grey).
$P(\text{not grey}) = 1 - \frac{3}{10} = \frac{7}{10}$

Final answer:

P(grey) = 3/10; P(not grey) = 7/10.

IB-style question — sample space and 'at least one'

Two fair dice are rolled. (a) Find the probability the two scores are equal. (b) Find the probability of getting at least one five.

Step by step:

Grid has 6 × 6 = 36 outcomes. Equal pairs: (1,1)…(6,6) — 6 of them.
$P(\text{equal}) = \frac{6}{36} = \frac{1}{6}$
For 'at least one five', flip to P(no five on either die).
$P(\text{no five}) = \frac{5}{6}\times\frac{5}{6} = \frac{25}{36}$
Take the complement.
$P(\text{at least one five}) = 1 - \frac{25}{36} = \frac{11}{36}$

Final answer:

P(equal) = 1/6; P(at least one five) = 11/36.

IB-style question — drawing without replacement

A jar contains 5 lime and 4 cherry sweets. Two are eaten, one after the other. Find the probability both are lime.

Step by step:

First lime: 5 of the 9 sweets.
$P(\text{1st lime}) = \frac{5}{9}$
Now 4 lime of 8 left. Multiply along the chain.
$\frac{5}{9}\times\frac{4}{8} = \frac{20}{72} = \frac{5}{18}$

Final answer:

P(both lime) = 5/18.

IB-style question — expected number over many trials

A spinner has 10 equal sectors, 4 of them yellow. The spinner is spun 150 times. Find the expected number of yellows.

Step by step:

Probability of yellow on one spin.
$P(\text{yellow}) = \frac{4}{10} = 0.4$
Expected number = n × P.
$150 \times 0.4 = 60$

Final answer:

60 yellows expected.

IB Math AA HL — Probability basics

Higher Level students should use this topic hub as a map: start with the shared sub-topics, then follow the HL-only extensions and exam-skill links where this topic asks for deeper analysis.

Key concepts in Probability basics

Key Idea: Probability measures how likely an event is, from 0 (impossible) to 1 (certain). In 4.5 you count outcomes in a sample space, use "not A", read probabilities from tables and data, and find an expected number. Topic 4.6 then combines two or more events.

Core rules

Rule	Formula	When to use
Basic probability	P(A) = n(A) ÷ n(U)	Equally likely outcomes: count A, divide by all.
Complement	P(A′) = 1 − P(A)	"Not A".
From a table	how many you want ÷ the total	A cell, a row or a column of a two-way table.
Experimental probability	frequency ÷ number of trials	Results of an experiment or survey.
Expected number	number of trials × P	How many times you expect it. It need not be whole.

Key Idea: 12 of 30 students walk to school: P(walk) = 12/30 = 0.4, so P(not walk) = 0.6. Over 190 school days, a student who walks with probability 0.4 is expected to walk 0.4 × 190 = 76 days.

Use n(A) ÷ n(U) only when the outcomes are equally likely. Two dice: count the 36 pairs, not the 11 sums.

Show the fraction, e.g. 66/200, before the decimal: it earns the method mark if a later number slips. Give non-exact answers to 3 significant figures.

IB-style question [5 marks]

A café records the drinks bought by 200 customers.

	Coffee	Tea
Under 25	46	24
25 or over	88	42

One of these customers is chosen at random. (a) Find the probability that the customer bought tea.

Solution:

Add the tea column, then divide by the total in the question.
$P(\text{tea})=\frac{24+42}{200}=\frac{66}{200}=0.33$

Final answer:

66/200 = 0.33

Part (b)

(b) Find the probability that the customer did not buy tea.

Solution:

"Not tea" is the complement: subtract from 1.
$P(\text{tea}')=1-0.33=0.67$

Final answer:

0.67

Part (c)

Solution:

Expected number = number of customers × probability.
$500\times0.33=165$

Final answer:

165 customers

IB-style question — single event and its complement

A drawer holds 7 black and 3 grey socks. One sock is taken at random. (a) Find the probability it is grey. (b) Find the probability it is not grey.

Step by step:

Total socks, then favourable ÷ total for grey.
$n(U) = 7 + 3 = 10, \quad P(\text{grey}) = \frac{3}{10}$
'Not grey' is the complement: 1 − P(grey).
$P(\text{not grey}) = 1 - \frac{3}{10} = \frac{7}{10}$

Final answer:

P(grey) = 3/10; P(not grey) = 7/10.

IB-style question — sample space and 'at least one'

Two fair dice are rolled. (a) Find the probability the two scores are equal. (b) Find the probability of getting at least one five.

Step by step:

Grid has 6 × 6 = 36 outcomes. Equal pairs: (1,1)…(6,6) — 6 of them.
$P(\text{equal}) = \frac{6}{36} = \frac{1}{6}$
For 'at least one five', flip to P(no five on either die).
$P(\text{no five}) = \frac{5}{6}\times\frac{5}{6} = \frac{25}{36}$
Take the complement.
$P(\text{at least one five}) = 1 - \frac{25}{36} = \frac{11}{36}$

Final answer:

P(equal) = 1/6; P(at least one five) = 11/36.

IB-style question — drawing without replacement

A jar contains 5 lime and 4 cherry sweets. Two are eaten, one after the other. Find the probability both are lime.

Step by step:

First lime: 5 of the 9 sweets.
$P(\text{1st lime}) = \frac{5}{9}$
Now 4 lime of 8 left. Multiply along the chain.
$\frac{5}{9}\times\frac{4}{8} = \frac{20}{72} = \frac{5}{18}$

Final answer:

P(both lime) = 5/18.

IB-style question — expected number over many trials

A spinner has 10 equal sectors, 4 of them yellow. The spinner is spun 150 times. Find the expected number of yellows.

Step by step:

Probability of yellow on one spin.
$P(\text{yellow}) = \frac{4}{10} = 0.4$
Expected number = n × P.
$150 \times 0.4 = 60$

Final answer:

60 yellows expected.

IB Math AA HL — Probability basics

Key concepts in Probability basics

Core rules

IB-style question [5 marks]

Part (b)

Part (c)

IB-style question — single event and its complement

IB-style question — sample space and 'at least one'

IB-style question — drawing without replacement

IB-style question — expected number over many trials

What you'll learn in Topic 4.5

Study resources — 4.5 Probability basics

Sample space and probability

Tables, experimental probability and expected number

Ready to study Probability basics?

Ready to practice?

IB Math AA HL — Probability basics

Key concepts in Probability basics

Core rules

IB-style question [5 marks]

Part (b)

Part (c)

IB-style question — single event and its complement

IB-style question — sample space and 'at least one'

IB-style question — drawing without replacement

IB-style question — expected number over many trials

What you'll learn in Topic 4.5

Study resources — 4.5 Probability basics

Sample space and probability

Tables, experimental probability and expected number

Ready to study Probability basics?

Ready to practice?