Iii if a 1 a n are mutually exclusive events then p a

(iii). If A
1
, . . . , A
n
are mutually exclusive events, then
P ( A
1
∪ . . . ∪ A
n
) = P ( A
1
) + . . . + P ( A
n
) .
A set of events are mutually exclusive if at most one of
the events can occur in a given experiment.
This axiom says that to calculate the probability of the
union of distinct events we can simply add their individual
probabilities.
2.3.2 Complement Law
If A is an event, the set of all outcomes that are not in A is
called the complement of the event A, denoted A
{
.
This is (pronounced ‘A complement’).

Probability I 79
The rule is
A
{
= 1 - P ( A ) (Law 1)
Example 2.4: Complements
Let S (the sample space) be the set of students at
Oxford. We are picking a student at random.
Let A = The event that the randomly selected
student suffers from depression
We are told that 8% of students suffer from depres-
sion, so P ( A ) = 0 . 08 . What is the probability that a
student does not suffer from depression?
The event { student does not suffer from depression }
is A
{
. If P ( A ) = 0 . 08 then P ( A
{
) = 1 - 0 . 08 = 0 . 92 .

Probability I 80
2.3.3 Addition Law (Union)
Suppose,
A = The event that a randomly selected student from the class has b
B = The event that a randomly selected student from the class has b
What is the probability that a student has brown eyes OR blue
eyes?
This is the union of the two events A and B, denoted A ∪ B
(pronounced ‘A or B’)
We want to calculate P(A ∪ B).
In general for two events
P(A ∪ B) = P(A) + P(B) - P(A ∩ B) (Addition Law)
To understand this law consider a Venn diagram of the situa-
tion (below) in which we have two events A and B . The event
A ∪ B is represented in such a diagram by the combined sam-

Probability I 81
ple points enclosed by A or B . If we simply add P ( A ) and
P ( B ) we will count the sample points in the intersection A ∩ B
twice and thus we need to subtract P ( A ∩ B ) from P ( A )+ P ( B )
to calculate P ( A ∪ B ) .
A B
A ∩ B
S
Example 2.5: SNPs
Single nucleotide polymorphisms (SNPs) are nu-
cleotide positions in a genome which exhibit varia-
tion amongst individuals in a species. In some stud-
ies in humans, SNPs are discovered in European
populations. Suppose that of such SNPs, 70% also

Probability I 82
show variation in an African population, 80% show
variation in an Asian population and 60% show vari-
ation in both the African and Asian population.
Suppose one such SNP is chosen at random, what
is the probability that it is variable in either the African
or the Asian population?
Write A for the event that the SNP is variable in
Africa, and B for the event that it is variable in Asia.
We are told
P ( A ) = 0 . 7
P ( B ) = 0 . 8
P ( A ∩ B ) = 0 . 6 .

Probability I 83
We require P ( A ∪ B ) . From the addition rule:
P ( A ∪ B ) = P ( A ) + P ( B ) - P ( A ∩ B )
= 0 . 7 + 0 . 8 - 0 . 6
= 0 . 9 .

3 Probability II
3.1 Independence and the
Multiplication Law
If the probability that one event A occurs doesn’t affect the
probability that the event B also occurs, then we say that A
and B are independent . For example, it seems clear than
one coin doesn’t know what happened to the other one (and if
it did know, it wouldn’t care), so if A
1
is the event that the first
coin comes up heads, and A
2
the event that the second coin
comes up heads, then
84

Probability II 85
Example 3.1: One die, continued