So, to find the
P
(
A
∪
B
) we can either
• Break up the set into 3 mutually exclusive sets
1.
A
∩
B
c
=
{
1
,
3
,
5
}
2.
A
∩
B
=
{
7
,
9
}
3.
A
c
∩
B
=
{
6
,
8
,
10
}
and by using the axioms
P
(
A
∪
B
) =
P
(
A
∩
B
c
) +
P
(
A
∩
B
) +
P
(
A
c
∩
B
)
• Add
P
(
A
) +
P
(
B
) but we need to subtract the probability of the intersection as that
probability was double counted since it is included within
A
and within
B
, leading to
P
(
A
∪
B
) =
P
(
A
) +
P
(
B
)

P
(
A
∩
B
)
.
Example 2.8
A sample of 1000 aluminum rods is taken and a quality check was per
formed on each rod’s diameter and length.
Diameter
Length
Too Thin
OK
Too Thick
Sum
Too Short
10
3
5
18
OK
38
900
4
942
Too Long
2
25
13
40
Sum
50
928
22
1000
•
P
(Diameter OK) = 928
/
1000 = 0
.
928
•
P
(Length OK) = 942
/
1000 = 0
.
942
•
P
(Diameter OK
∩
Length OK) = 900
/
1000 = 0
.
9
•
P
(Diameter OK
∪
Length OK) = 0
.
928 + 0
.
942

0
.
9 = 0
.
97
2.3
Counting Methods
Lemma
2.1
(Product Rule)
If the first task of an experiment can result in
n
1
possi
ble outcomes and for each such outcome, the second task can result in
n
2
possible
outcomes, and so forth up to
k
tasks, then the total number of ways to perform the
sequence of
k
tasks is
Q
k
i
=1
n
i
22
2.3. Counting Methods
Example 2.9
When buying a certain type and brand of car a buyer has the following
number of choices:
(2) Engine
(5) Color
(4) Interior
Then, the total number of car choices, i.e. number of unique cars, is
2
×
5
×
4 = 40
.
Notation:
In mathematics, the factorial of a nonnegative integer
n
, denoted by
n
!, is the
product of all positive integers less than or equal to
n
, with 0! = 1
.
For example,
5! = 5
×
4
×
3
×
2
×
1 = 120
In
R
:
factorial(5)
2.3.1
Permutations
Definition
2.4
(Permutation)
Permutation
is the number of
ordered
arrangements, or
permutations, of
r
objects selected from
n
distinct objects (
r
≤
n
)
without replacement
.
It is given by,
P
n
r
=
n
!
(
n

r
)!
.
The number of permutations of
n
objects is
n
!
.
Example
2.10
Take the letters
A,B,C
. Then there are 3! = 6 possible permutations.
These are:
ABC,ACB,BAC,BCA,CAB,CBA
In e
ff
ect we have three “slots” in which to place the letters. There are 3 options for the
first slot, 2 for the second and 1 for the third. Hence we have 3
×
2
×
1 = 6 =
P
3
3
.
Example 2.11
Let
n
= 10 with
A,B,C,D,E,F,G,H,I,J
and we wish to select 5 letters at
random, where order is important. Then there are 5 slots with 10 choices for the first
slot, 9 for the second, 8 for the third, 7 for the fourth and 6 for the fifth. That is,
10
×
9
×
8
×
7
×
6 =
10!
5!
=
P
10
5
= 30240
.
Example 2.12
An election is held for a board consisting of 3 individuals out of a pool
of 40 candidates. There will be 3 positions of:
1. President
2. VicePresident
3. Secretary
Chapter 2. Probability and Random Variables