IB Applications and Interpretation SL · Probability · Section 1 of 4

Sample space and events: union, intersection and Venn diagrams

3 worked examples · By Álvaro López

Key idea

Before calculating a probability you need to know what can happen and what you are counting. This section gives you the vocabulary: random experiment, sample space and events, and the operations on them (union, intersection, complement). In the IB exam you will see it mostly as Venn diagrams and outcome tables, and it is the foundation of the whole topic.

Random experiment and sample space

An event $A$ occurs when the result of the experiment is one of the elements of $A$ . With a six-sided die, $U = {1, 2, 3, 4, 5, 6}$ and $n (U) = 6$ . The event $A$ : "a prime number is rolled" is $A = {2, 3, 5}$ , with $n (A) = 3$ . If a 5 comes up, $A$ has occurred; if a 4 comes up, it has not.

Two events are special:

The certain event is $U$ : it always occurs.
The impossible event is the empty set, $\emptyset$ : it never occurs. It has no elements, so $n (\emptyset) = 0$ .

An event is called compound if it has more than one outcome (such as $A = {2, 3, 5}$ ) and elementary if it has only one.

Writing down the sample space

You must list the outcomes without forgetting any and without repeating them. For an experiment with one stage a list is enough. For one with two stages (two throws, two draws…) each outcome is an ordered pair, and the safest option is an outcome table (a two-way table): the rows are the outcomes of the first stage, the columns those of the second, and each cell is an outcome of $U$ .

For example, two 4-sided dice (numbered 1 to 4) are thrown and the sum is recorded:

First die \ Second die	1	2	3	4
1	2	3	4	5
2	3	4	5	6
3	4	5	6	7
4	5	6	7	8

The table has $4 \times 4 = 16$ cells, so $n (U) = 16$ (each cell is a different pair, such as $(1, 4)$ or $(4, 1)$ ). The event "the sum is 5" has 4 cells: ${(1, 4), (2, 3), (3, 2), (4, 1)}$ .

Operations with events

Events are sets, so they combine like sets. Let $A$ and $B$ be two events of the same experiment.

In everyday language: "or" is union, "and" is intersection and "not" is complement. The word "or" in mathematics does not exclude the case in which both occur.

Four Venn diagrams of A and B: A ∪ B (both circles shaded), A ∩ B (only the common region), A′ (everything except circle A) and A ∩ B′ (the part of A that is not in B)

With a die, "even" and "a 5" are mutually exclusive: there is no outcome that satisfies both. By contrast, "even" and "greater than 3" are not mutually exclusive, because 4 and 6 satisfy both.

De Morgan's laws

Two identities let you pass from the complement of a union or of an intersection to the complements of each event.

De Morgan's laws with Venn diagrams: (A ∪ B)′ is what lies outside both circles and (A ∩ B)′ is everything except the common region

Venn diagrams with quantities

When the question gives how many elements there are in each set, fill in the Venn diagram with numbers in each region. A diagram with two events has four regions: only $A$ , only $B$ , both, and neither.

Write in each region the number of elements, not the total of the circle: a very common mistake is to put the whole $n (A)$ in the "only $A$ " region.

Worked examples

Example 1 Basic

Tokens numbered 1 to 12

A bag contains 12 tokens numbered from 1 to 12. One token is drawn at random, and the events $A$ : "the number is a multiple of 3" and $B$ : "the number is even" are considered.

a) Write $A \cap B$ .

b) Write $A \cup B$ .

c) Write $A^{'}$ .

d) Write $A \cap B^{'}$ .

Solution

The sample space is $U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}$ , and the events are $A = {3, 6, 9, 12} B = {2, 4, 6, 8, 10, 12}$

a) $A \cap B$ are the numbers that are in both: $A \cap B = {6, 12}$ .

b) $A \cup B$ gathers those in one or the other (without repetition): $A \cup B = {2, 3, 4, 6, 8, 9, 10, 12}$ .

c) $A^{'}$ are the elements of $U$ that are not in $A$ : $A^{'} = {1, 2, 4, 5, 7, 8, 10, 11}$ .

d) $A \cap B^{'}$ are the multiples of 3 that are not even: $A \cap B^{'} = {3, 9}$ .

Example 2 Medium

Two spinners and an outcome table

Two spinners are spun. The first gives a number from 1 to 3 and the second a number from 1 to 4. Each outcome is a pair $(a, b)$ , where $a$ is the number from the first spinner and $b$ the number from the second. Let $A$ be the event "the sum $a + b$ is even" and $B$ the event "the product $a \cdot b$ is greater than or equal to 6".

a) Calculate $n (U)$ .

b) Calculate $n (A)$ and $n (B)$ .

c) Write $A \cap B$ .

d) Calculate $n (A \cup B)$ and the number of outcomes that are in neither $A$ nor $B$ .

Solution

We organise the 12 outcomes in a table, with the pair $(a, b)$ in each cell:

$a$ \ $b$	1	2	3	4
1	(1,1)	(1,2)	(1,3)	(1,4)
2	(2,1)	(2,2)	(2,3)	(2,4)
3	(3,1)	(3,2)	(3,3)	(3,4)

a) The table has $3 \cdot 4 = 12$ cells: $n (U) = 12$ .

b) The sum is even when $a$ and $b$ have the same parity: $A = {(1, 1), (1, 3), (2, 2), (2, 4), (3, 1), (3, 3)} n (A) = 6$ The product is at least 6 for these pairs: $B = {(2, 3), (2, 4), (3, 2), (3, 3), (3, 4)} n (B) = 5$

c) The pairs that are in both lists are $(2, 4)$ and $(3, 3)$ : $A \cap B = {(2, 4), (3, 3)}$

d) Counting the common ones only once: $n (A \cup B) = 6 + 5 - 2 = 9$ . Those in neither are $12 - 9 = 3$ (namely $(1, 2)$ , $(1, 4)$ and $(2, 1)$ ).

Example 3 Hard

A survey with an unknown in the Venn diagram

In a group of 60 people, 34 read digital news and 28 listen to podcasts. There are 10 people who do neither of the two. One person from the group is chosen at random. Let $A$ be the event "reads digital news" and $B$ the event "listens to podcasts".

a) Find the number of people who do both.

b) Calculate $n (A \cap B^{'})$ .

c) Calculate how many people do exactly one of the two.

d) Calculate $n (A^{'} \cup B^{'})$ and interpret the result in words.

Solution

a) Let $x = n (A \cap B)$ and fill in the Venn diagram from the inside out:

Only $A$ : $34 - x$ .
Only $B$ : $28 - x$ .
Neither: $10$ .

The sum of the four regions is the total of the group: $(34 - x) + x + (28 - x) + 10 = 60 ⟹ 72 - x = 60 ⟹ x = 12$ 12 people do both.

b) $A \cap B^{'}$ is "only $A$ ": $n (A \cap B^{'}) = 34 - 12 = 22$ .

c) Only $A$ is 22 and only $B$ is $28 - 12 = 16$ . Since they are mutually exclusive, they are added: $22 + 16 = 38$ .

d) By De Morgan, $A^{'} \cup B^{'} = (A \cap B)^{'}$ , so $n (A^{'} \cup B^{'}) = 60 - n (A \cap B) = 60 - 12 = 48$ These are the people who do not do both things at once (they do one or neither): $22 + 16 + 10 = 48$ , which agrees.

In the exam: you earn marks for drawing the diagram with the four regions and for setting up the equation in which the sum equals the total. If you write $n (A) = 34$ in full in the "only $A$ " region, the rest of the problem goes wrong. Also write which event each quantity you calculate represents.

Common mistakes

Summary

A random experiment has known outcomes, but it is not known which one will occur. The sample space $U$ gathers all the outcomes, and an event is a subset of $U$ ; $n (A)$ counts its elements.
The certain event is $U$ and the impossible event is $\emptyset$ , with $n (\emptyset) = 0$ .
$A \cup B$ is " $A$ or $B$ (or both)", $A \cap B$ is " $A$ and $B$ ", $A^{'}$ is "not $A$ " and $A \cap B^{'}$ is " $A$ but not $B$ ".
$A$ and $B$ are mutually exclusive if $A \cap B = \emptyset$ .
De Morgan's laws: $(A \cup B)^{'} = A^{'} \cap B^{'}$ and $(A \cap B)^{'} = A^{'} \cup B^{'}$ .
With two stages, use an outcome table: $n (U)$ is the product of the possibilities of each stage, and each outcome is an ordered pair.
In a Venn diagram with numbers, fill it in from the inside out, writing in each region the number of elements that belong only to that region; the sum of the four regions is $n (U)$ .

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