Engineering Statistics I

Second Semester 2023/2024

Lecturer: Dr Mohannad Jreissat

Department of Industrial Engineering

School of Engineering

CHAPTER TWO: PROBABILITY

Textbook:
Applied Statistics and Probability for Engineers, by
D. Montgomery and G. Runger, 6th edition, Wiley.

Email: Drjreissat@gmail.com
Office: IE
 What is Probability?
• In Chapters 1, we used graphs and numerical
measures to describe data sets which were
usually samples.
• We measured “how often” using
Relative frequency = f/n

• As n gets larger,
Sample Population
And “How often”
Probability
= Relative frequency
• An experiment is the process by which an
observation (or measurement) is obtained.

• An event is an outcome of an experiment, usually

denoted by a capital letter.

– The basic element to which probability is applied

– When an experiment is performed, a particular event

either happens, or it doesn’t!
Definition:
Outcomes are possible results of a hypothetical experiment

Examples:
✓ Dice roll:
Outcomes = 1, 2, 3 , 4, 5 6 (integers)

✓ Items coming off production line.

Outcomes = “defective” and “ non-defective” ( Categorical data )

✓ Change in ventricular Volume:

Outcomes = A;; possible real numbers
Any recording of a specific outcome whether it is categorical or
numerical is called an observation
Sample Space
Definition:
Sample space: the set of all possible outcomes of a statistical
experiment. It is represented by the symbol S.
** Each outcome in a sample space is called an element or a member
of the sample space, or simply a sample point

Example:
Two ways to define the sample space of a 6 –sided dice roll.
1- Consider the experiment of rolling one die, the sample space is
S1 = {1, 2, 3, 4, 5, 6}

2- If we are interested only in whether the number

is even or odd, the sample space is simply
S2 = {even, odd}
So, This Shows that more than one sample space can be used to
describe the outcomes of an experiment
Example:
The sample space S, of possible outcomes when a coin is tossed may
be written as

where H and T correspond to “heads” and “tails”, respectively.

Example:
The sample space S, of possible outcomes when 2 consecutive coin
are tossed may be written as
Example: Suppose that three items are selected at random from a
manufacturing process. Each item is inspected and classified
defective, D, or nondefective, N. As we proceed along each possible
outcome, we see that the sample space is

S = DDD, DDN , DND, DNN , NDD, NDN , NND, NNN 

❖ Sample spaces with a large or infinite number of sample points

are best described by a statement or rule.

Examples:
1- if the possible outcomes of an experiment are the set of cities in
the world with a population over million, the sample space is written
S =  x x is a city with a population over 1 million

2- If S is the set of all points (x, y) on the boundary or the interior of a

circle of radius 2 with center at the origin, we write

S = ( x, y ) x 2 + y 2  4
Example 2.2: An experiment consists of flipping a coin and then
flipping it a second time if a head occurs. If a tail occurs on the first
flip, then a die is tossed once, the sample space is

S = {HH, HT, T1, T2, T3, T4, T5, T6}

** In some experiments,
it is helpful to list the
elements of the sample
space systematically by
means of a tree diagram
 Figure 2.2 Tree Diagram for
Example 2.3
Three items are selected at random from a manufacturing process.
Each item is inspected and classified defective, D, or non-defective, N.
Events
➢ A set is a collection of unique objects.
➢ A set A is a subset of another set B if every element of A is also an
element of B. We denote this as A  B.
➢ Event: a subset of a sample space. We are interested in
probabilities of events.
The event A that the outcome when a die is tossed is divisible by 3 is
the subset of the sample space S1, and can be expressed as
A = 3, 6

The event B that the number of defectives is greater than 1 in the

example on the previous slide can be written as
B = DDD, DDN , DND, NDD 

Given the sample space S = {t | t ≥ 0}, where t is the life in years of a

certain electronic components, then the event A that the component
fails before the end of the fifth year is the subset
A ={t |0 ≤ t < 5}.
➢ Null set: a subset that contains no elements at all. It is denoted by
the symbol 
B =  x x is an even factor of 7 = 

➢ The complement of an event A with respect to S is the subset of

all elements of S that are not in A. We denote the complement of
A by the symbol A’.
Let R be the event that a red card is selected from an ordinary deck of
52 playing cards, and let S be the entire deck. Then R ’ is the event
that the card selected from the deck is not a red but a black card.

A A’
➢ The intersection of two events A and B, denoted by A ∩ B, is the
event containing all elements that are common to A and B.

A B

A B

➢ Two events A and B are mutually exclusive, or disjoint if A ∩ B =  ,

that is, if A and B have no elements in common.

A B

A B = 
Disjoint Events

Two events A and B are mutually exclusive, or disjoint,

if A ∩ B = , that is , A and B have no elements in common.

• Two events are mutually exclusive if, when one event

occurs, the other cannot, and vice versa.

Example:
Let V = {a, e, i, o, u } and
C = {l, r, s, t };
then it follows that V ∩ C = Φ.
That is, V and C have no elements in common and,
therefore, cannot both simultaneously occur.
•Example: Experiment: Toss a die

Not Mutually
Exclusive
–A: observe an odd number
–B: observe a number greater than 2
–C: observe a 6
–D: observe a 3
B and C?
Mutually
B and D?
Exclusive
➢ The union of two events A and B, denoted by A ∪ B, is the event
containing all elements that belong to A or B or both.

A B

A B

Let A = {a, b, c} and B = {b, c, d, e}; then

A ∩ B = {b, c}
A ∪ B = {a, b, c, d, e}

If M = {x |3 < x < 9} and N = {y | 5 < y < 12}; then

M ∪ N = {z | 3 < z <12}
M∩N =?
If S = {x | 0 < x < 12}, A = {x | 1 ≤ x < 9}, and B = {x | 0 < x < 5},
determine
(a) A ∪ B
(b) A ∩ B
(c) A’ ∪ B’

(a) A ∪ B = {x | 0 < x < 9}

(b) A ∩ B = {x | 1 ≤ x < 5}
(c) A ’ ∪ B ’ = (A ∩ B)’
= {x | 0 < x < 1, 5 ≤ x <12}
Venn Diagram
Like already seen previously, the relationship between events and
the corresponding sample space can be illustrated graphically by
means of Venn diagrams.
S
A B
2
7 6
1
4 3
5
8
C

A B = {1, 2}
B C = {1, 3}
A C = {1, 2, 3, 4, 5, 7}
B’ A = {4, 7}
AB C = {1}
(A  B )  C ’ = {2, 6, 7}
Experiments and Events

• Experiment: Record an age

– A: person is 30 years old

– B: person is older than 65

• Experiment: Toss a die

– A: observe an odd number

– B: observe a number greater than 2

 Basic Concepts

✓ An event that cannot be decomposed is called a ---------------

✓ Answer: Simple event. Denoted by E with a subscript.
✓ Each simple event will be assigned a ---------------, measuring
“how often” it occurs.
✓ Answer: Probability
✓ The set of all simple events of an experiment is called the -----
----------
✓ Answer: Sample space, S.
Example
• The die toss:

Simple events: Sample space:

1 E1 S ={E1, E2, E3, E4, E5, E6}
2
E2
S
3 E3 •E1 •E3
4
E4 •E5
5
E5 •E2 •E4 •E6
6 E6
Basic Concepts

• An event is a collection of one or more simple events.

S
•E1 •E3
•The die toss:
–A: an odd number A •E5
–B: a number > 2 B
•E2 •E4 •E6

A ={E1, E3, E5}

B ={E3, E4, E5, E6}
The Probability of an Event

• The probability of an event A measures “how often” A will

occur. We write P(A).
• Suppose that an experiment is performed n times. The
relative frequency for an event A is

N u m b e r o f t im e s A o c c u r s f
=
n n
• If we let n get infinitely large,

f
P ( A ) = lim
n→  n
 The Probability of an Event

• P(A) must be between ---- and ----. [ 0 and 1]

- If event A can never occur, P(A) = [0]

- If event A always occurs when the experiment is performed, P(A)= 1

• The sum of the probabilities for all simple events in S equals 1

• The probability of an event A is found by

adding the probabilities of all the simple events contained in A.
Finding Probabilities

• Probabilities can be found using

– Estimates from empirical studies
– Common sense estimates based on equally likely
events.

• Examples:
–Toss a fair coin. P(Head) = 1/2

– Suppose that 10% of the U.S. population has red hair. Then
for a person selected at random,
P(Red hair) = 0.10
Using Simple Events

• The probability of an event A is equal to

the sum of the probabilities of the simple events contained in A
• If the simple events in an experiment are equally likely, you
can calculate

n A number of simple events in A

P( A) = =
N total number of simple events
Example

Toss a fair coin twice. What is the probability of observing at least

one head?

1st Coin 2nd Coin Ei P(Ei)

H HH
1/4 P(at least 1 head)
H
T HT 1/4 = P(E1) + P(E2) + P(E3)

H = 1/4 + 1/4 + 1/4 = 3/4

TH 1/4
T
T TT 1/4
Example
A bowl contains three M&Ms®, one red, one blue and one green.
A child selects two M&Ms at random. What is the probability
that at least one is red?
P(at least 1 red) =
P(RB)
1st M&M 2nd M&M Ei P(Ei)
+
m RB 1/6 P(BR)
m
m RG 1/6 +
m BR P(RG)
m 1/6
m +
BG 1/6
m P(GR)
m GB 1/6
m = 4/6 = 2/3
GR
1/6
 Example

The sample space of throwing a pair of dice is

Example: solution

Event Simple events Probability

Dice add to 3 (1,2),(2,1) 2/36

Dice add to 6 (1,5),(2,4),(3,3), 5/36

(4,2),(5,1)
Red die show 1 (1,1),(1,2),(1,3), 6/36
(1,4),(1,5),(1,6)

Green die show 1 (1,1),(2,1),(3,1), 6/36

(4,1),(5,1),(6,1)
Counting Sample Points
Counting Sample Points
➢ Goal: to count the number of points in the sample space without
actually enumerating each element.
➢ Multiplication Rule| If an operation can be performed in n1 ways,
and if for each of these a second operation can be performed in
n2 ways, then the two operations can be performed together in
n1·n2 ways.

Example:
How may sample points are in the sample space when a pair of dice is
thrown once?

n1 = 6, n2 = 6
→ n1  n2 = 36 possible ways
Counting Sample Points
Sam is going to assemble a computer by himself. He has the choice of
ordering chips from two brands, a hard drive from four, memory from
three and an accessory bundle from five local stores. How many
different ways can Sam order the parts?

Since n1 = 2, n2 = 4, n3 = 3, and n4 = 5,

there are n1·n2·n3·n4 = 2·4·3·5 = 120 different ways to order the parts
Counting Sample Points

Example: How many even four-digit numbers can be formed from the
digits 0, 1, 2, 5, 6, and 9 if each number can be used only once?`
For even numbers, there are n1 = 3 choices for units position.
However, the thousands position cannot be 0.
If units position is 0, n1 = 1, then we have n2 = 5 choices for thousands
position, n3 = 4 for hundreds position, and n4 = 3 for tens position. In
this case, totally n1·n2·n3·n4 = 1·5·4·3 = 60 numbers.
If units position is not 0, n1 = 2, then we have n2 = 4, n3 = 4, and n4 = 3.
In this case, totally n1·n2·n3·n4 = 2·4·4·3 = 96 numbers.
The total number of even four-digit numbers can be calculated by
60 + 96 = 156.
?
How if each number can
be used more than once?
Permutation

➢ A permutation is an arrangement of all or part of a set of objects.

Example: Consider the three letters a, b, and c. There are 6 distinct

arrangements of them: abc, acb, bac, bca, cab, and cba.

There are n1 = 3 choices for the first position,

then n2 = 2 for the second, and n3 = 1 choice for the last position,
giving a total n1·n2·n3 = 3·2·1 = 6 permutations.
Permutation

➢ In general, n distinct objects can be arranged in

n (n–1)(n–2) · · · (3)(2)(1) ways.

➢ This product is represented by the symbol n!, which is read “n

factorial.”

➢ The number of permutations of n distinct objects is n!

➢ The number of permutations of n distinct objects taken r at a time
is :
n!
n Pr =
(n − r )!
Permutation
Consider the four letters a, b, c, and d. Now consider the number of
permutations that are possible by taking 2 letters out of 4 at a time.
The possible permutations are ab, ac, ad, ba, bc, bd, ca, cb, cd, da,
db, and dc.

There are n1 = 4 choices for the first position, and n2 = 3 for the second,
giving a total n1·n2 = 4·3 = 12 permutations.

Another way, by using formula,

4!
4 P2 =
= 12
(4 − 2)!
Permutation

Example:
Three awards (research, teaching and service) will be given one year
for a class of 25 graduate students in a statistics department.
If each student can receive at most one award, how many possible
selections are there?
Since the awards are distinguishable, it is a permutation problem.
The total number of sample points is

25! 25! 25  24  23  22!

25 P3 = = = = 13,800
(25 − 3)! 22! 22!
Permutation

A president and a treasurer are to be chosen from a student club

consisting of 50 people. How many different choices of officers are
possible if
(a) There are no restrictions
(b) A will serve only if he is president
(c) B and C will serve together or not at all
(d) D and E will not serve together

(a) 50P2
? For detailed explanation
(b) 49P1 + 49P2
read the e-book.
(c) 2P2 + 48P2
(d) 50P2 – 2 or {2·2P1·48P1 + 48P2}
a) n = 50 , r = 2 …… then nPr >>>> 50 P 2 = 2450
b) (i) A is made President >>> 49 – 1 position
n = 49 , r = 1 >>>>> 49 P1= 49! / 48! = 49
(ii) A is not made President
n = 49 , r = 2 >>>>>> 49P2= 2352
(c) (i) B : President C : Treasurer …. 1
C : President B : Treasurer …. 1
___
2
(ii) n = 48 , r = 2 >>>>>>>

48P2 = 2256

THEN THE ANSWER IS 2 + 2256 = 2258

(d) D serves (president) 48 = 48 choices
E serves (president) 48 = 48 choices
_______
96
………………………………………….
n = 48 , r = 2 >>>>>>>> 48P2 = 2256
THE ANSWER IS ( 2*96) + 2256 = 2448
Permutation

➢ The number of distinct permutations of n things of which n1 are of

one kind, n2 of a second kind, …, nk of a kth kind is

n!
n1 !n2 ! nk !

Example: How many distinct permutations can be made from the

letters a, a, b, b, c, and c?
6!
= 90
2!2!2!
Permutation

Example: In a college football training session, the defensive

coordinator needs to have 10 players standing in a row. Among
these 10 players, there are 1 freshman, 2 sophomore, 4 juniors,
and 3 seniors, respectively. How many different ways can they be
arranged in a row if only their class level will be distinguished?

10!
1!2!4!3! = 12, 600
Theorem

Example: In how many ways can 7 graduate students be assigned

to 1 triple and 2 double hotel rooms during a conference?

The total number of possible partitions would be

 7  7!
  = = 210
 3 2 2  (3!)(2!)(2!)
Combination
➢ In many problems we are interested in the number of ways of
selecting r objects from n without regard to order. These selections
are called combinations.
➢ The number of combinations of n distinct objects taken r at a time
is
n!
n Cr =
r !(n − r )!
Example: A young boy asks his mother to get five game-boy
cartridges from his collection of 10 arcade and 5 sport games. How
many ways are there that his mother will get 3 arcade and 2 sports
games, respectively?
The number of ways of selecting 3 arcade games is 10C3.
The number of ways of selecting 2 sports games is 5C2.
Using the multiplication rule,
10! 5!
10 C3  5 C2 =  = 1200 ways
3!(10 − 3)! 2!(5 − 2)!
Probability of an Event
➢ The likelihood of the occurrence of an event resulting from such a
statistical experiment is evaluated by means of a set of real
numbers called weights or probabilities ranging from 0 to 1.

◼ The probability of an event A is the sum of the weights of all sample

points in A. Therefore,
0  P ( A)  1, P () = 0, P( S ) = 1
`
◼ Furthermore, if A1, A2, A3, ... is a sequence of mutually exclusive
events, then
P( A1  A2  A3  ) = P( A1 ) + P( A2 ) + P( A3 ) +

◼ If an experiment can result in any one of N different equally likely

outcomes, and if exactly n of these outcomes correspond to event A,
then the probability of event A is
n
P( A) =
N
Probability of an Event

Example :
A coin is tossed twice. What is the probability that at least one head
occurs?

S = {HH , HT , TH , TT }  Sample space of the experiment, 4 events

A = {HH , HT , TH }  Events of interest, at least one head occurs
3 ** If the coin is balanced, each of these outcomes is
P( A) =
4 equally likely to occur. Therefore, we assign a
probability of ω to each sample point.
Then 4ω = 1, or ω = 1/4. If A represents the event of
at least 1 head occurring, then
A = {HH ,HT, TH } and
P(A) =(1/4) + (1/4) + (1/4) = 3/4
Probability of an Event

Example: A dice is loaded in such a way that an even number is twice

as likely to occur as an odd number. If E is the event that a number
less than 4 occurs on a single toss of the dice, find P (E).
S = {1, 2,3, 4,5, 6} P (even) = 2 9
E = {1, 2,3} P (odd) = 1 9
P( E ) = 1 9 + 2 9 + 1 9 = 4 9

Example: As the last example, let A be the event that an even

number turns up and let B be the event that a number divisible by 3
occurs. Find P (A ∪ B) and P (A ∩ B).

A = {2, 4, 6} → A  B = {2,3, 4, 6} → P ( A  B ) = 7 9

B = {3, 6} → A  B = {6} → P( A  B) = 2 9
Additive Rules
➢ If A and B are any two events, then
S

P ( A  B ) = P ( A) + P ( B ) − P ( A  B ) A A B B

A B
➢ If A and B are mutually exclusive, then
S

P( A  B) = P( A) + P ( B )
A B

A B
➢ For three events A, B, and C,
P( A  B  C ) = P( A) + P( B) + P(C ) −
P( A  B) − P( A  C ) − P( B  C ) + P( A  B  C )
? Can you prove using
Venn diagram?
➢ P (A ∩ B′ ) = P (A) – P (A ∩ B)
Additive Rules
Example: The probability of John to be hired by company A is 0.8, and
the probability that he gets an offer from company B is 0.6. If, on the
other hand he believes that the probability that he will get offers from
both companies is 0.5, what is the probability that he will get at least
one offer from these two companies?

P ( A  B ) = P ( A) + P ( B ) − P ( A  B ) = 0.8 + 0.6 − 0.5 = 0.9

Example: What is the probability of getting a total of 7 or 11 when a

pair of fair dice are tossed?
Let A be the event that 7 occurs and B the event that 11 comes up.
The events A and B are mutually exclusive, since a total of 7 and 11
cannot both occur on the same toss. Therefore,
6 2 2
P( A  B) = P( A) + P ( B ) = + =
36 36 9
**This result could also have been obtained by counting the total number of
points for the event A ∪ B, namely 8, and writing
P(A ∪ B) = 8/36 = 2/9
Additive Rules
If A and A ’ are complementary events, means A ∩ A ’ =  and
A ∪ A ’ = S, then
P( A) + P( A) = 1

Example: The probabilities that an automobile mechanic will service

3, 4, 5, 6, 7, or 8 or more cars on any given workday are, respectively,
0.12, 0.19, 0.28, 0.24, 0.10, and 0.07. What is the probability that he
will service at least 5 cars on his next day at work?
Let E be the event that at least 5 cars are serviced, then
E ’ is the event that fewer than 5 cars are serviced.

P ( E ) = 0.12 + 0.19 = 0.31

P( E ) = 1 − P( E ) = 1 − 0.31 = 0.69
A Probability Table

Probabilities and joint probabilities for two events A and B are

summarized in this table:

B B

A P(A  B) P(A  B ) P(A)

A P( A  B) P(A  B ) P(A)

P(B) P( B ) P(S) = 1.0

 Addition Rule Example

Consider a standard deck of 52 cards, with four suits:

Let event A = card is an Ace

Let event B = card is from a red suit
 Addition Rule Example
(continued)

P(Red U Ace) = P(Red) + P(Ace) - P(Red ∩ Ace)

= 26/52 + 4/52 - 2/52 = 28/52

Don’t count
the two red
aces twice!
Color
Type Red Black Total
Ace 2 2 4
Non-Ace 24 24 48
Total 26 26 52
Conditional Probability

➢ The probability of an event B occurring when it is known that some

event A has occurred is called a conditional probability.
➢ It is denoted by symbol P (B|A), usually read “the probability that B
occurs given that A occurs” or simply “the probability of B, given
A.”
➢ The probability P(B|A) can be seen as an “updating” of P (B) based
on the knowledge that even A has occurred.
➢ The conditional probability of B, given A, denoted by P (B |A), is
defined by
P( A  B)
P( B A) = if P ( A)  0
P( A)
Conditional Probability

• A conditional probability is the probability of

one event, given that another event has
occurred:
The conditional
P(A  B) probability of A
P(A | B) = given that B has
P(B) occurred

The conditional
P(A  B) probability of B
P(B | A) = given that A has
P(A) occurred
Conditional Example Probability

◼ Of the cars on a used car lot, 70% have air

conditioning (AC) and 40% have a CD player
(CD). 20% of the cars have both.

• What is the probability that a car has a CD

player, given that it has AC ?

i.e., we want to find P(CD | AC)

Conditional Probability Example
(continued)

◼ Of the cars on a used car lot, 70% have air conditioning (AC) and 40%
have a CD player (CD).
20% of the cars have both.

CD No CD Total

AC .2 .5 .7
No AC .2 .1 .3
Total .4 .6 1.0

P(CD  AC) .2
P(CD | AC) = = = .2857
P(AC) .7
Conditional Probability Example
(continued)

◼ Given AC, we only consider the top row (70% of the cars). Of
these, 20% have a CD player. 20% of 70% is 28.57%.

CD No CD Total

AC .2 .5 .7
No AC .2 .1 .3
Total .4 .6 1.0

P(CD  AC) .2
P(CD | AC) = = = .2857
P(AC) .7
Multiplication Rule

• Multiplication rule for two events A and B:

P(A  B) = P(A | B) P(B)

• also
P(A  B) = P(B | A) P(A)
Multiplication Rule Example
P(Red ∩ Ace) = P(Red| Ace)P(Ace)

 2  4  2
=    =
 4  52  52
number of cards that are red and ace 2
= =
total number of cards 52

Color
Type Red Black Total
Ace 2 2 4
Non-Ace 24 24 48
Total 26 26 52
Conditional Probability

Example: If a fair dice is tossed once, what is the probability of

getting a 6, given that the number you got is an even number?

A = {2, 4, 6} → P ( A) = 3 6
B = {6} → P( B) = 1 6

A  B = {6} → P( A  B) = 1 6
P( A  B) 1 6 1
P( B A) = = =
P( A) 36 3
 Example: The probability that a regularly scheduled flight departs
on time is P (D) = 0.83; the probability that it arrives on time is
P (A) = 0.82; and the probability that it departs and arrives on time
is P(D ∩ A) = 0.78. Find the probability that a plane
(a) arrives on time given that it departed on time,
(b) departed on time given that it has arrived on time, and
(c) arrives on time given that it did not depart on time

P( D  A) 0.78
(a) P( A D) = = = 0.94
P( D) 0.83 A S
D
A D
P( A  D) 0.78
(b) P( D A) = = = 0.95 D’
P( A) 0.82 A  D’

P( D  A) 0.82 − 0.78
(c) P( A D) = = = 0.24
P( D) 1 − 0.83
Statistical Independence

• Two events are statistically independent if

and only if:
P(A  B) = P(A) P(B)
– Events A and B are independent when the probability of one event is
not affected by the other event

• If A and B are independent, then

P(A | B) = P(A) if P(B)>0

P(B | A) = P(B) if P(A)>0

Statistical Independence Example

◼ Of the cars on a used car lot, 70% have air conditioning (AC) and 40%
have a CD player (CD).
20% of the cars have both.

CD No CD Total

AC .2 .5 .7
No AC .2 .1 .3
Total .4 .6 1.0

◼ Are the events AC and CD statistically independent?

Statistical Independence Example
(continued)

CD No CD Total

AC .2 .5 .7
No AC .2 .1 .3
Total .4 .6 1.0
P(AC ∩ CD) = 0.2

P(AC) = 0.7
P(AC)P(CD) = (0.7)(0.4) = 0.28
P(CD) = 0.4
P(AC ∩ CD) = 0.2 ≠ P(AC)P(CD) = 0.28
So the two events are not statistically independent
Bivariate Probabilities

Outcomes for bivariate events:

B1 B2 ... Bk

A1 P(A1B1) P(A1B2) ... P(A1Bk)

A2 P(A2B1) P(A2B2) ... P(A2Bk)

. . . . .
. . . . .
. . . . .

Ah P(AhB1) P(AhB2) ... P(AhBk)

 Joint and
Marginal Probabilities
• The probability of a joint event, A ∩ B:
number of outcomes satisfying A and B
P(A  B) =
total number of elementary outcomes

• Computing a marginal probability:

P(A) = P(A  B1 ) + P(A  B 2 ) +  + P(A  Bk )

• Where B1, B2, …, Bk are k mutually exclusive and

collectively exhaustive events
Marginal Probability Example

P(Ace)

2 2 4
= P(Ace  Red) + P(Ace  Black) = + =
52 52 52

Color
Type Red Black Total
Ace 2 2 4
Non-Ace 24 24 48
Total 26 26 52
Ch. 2-
 Using a Tree Diagram
.2
Given AC or no .7 P(AC ∩ CD) = .2
AC:

P(AC ∩ CD) = .5
.5
.7
All
Cars
.2
.3 P(AC ∩ CD) = .2

.1 P(AC ∩ CD) = .1
.3 Ch. 2-
Independent Events:
Basic Concept…
➢ Two events A and B are independent if and only if

P( B A) = P( B) or P( A B ) = P ( A)

Otherwise, A and B are dependent.

Multiplicative Rules
➢ If in an experiment the events A and B can both occur, then
P( A  B) = P( A)  P( B A)

Since A  B and B  A are equivalent, it follows that

P( B  A) = P( B)  P( A B)

➢ Two events A and B are independent if and only if

P( A  B ) = P ( A)  P ( B )

Suppose that we have a fuse box containing 20 fuses, of which 5 are

defective. If 2 fuses are selected at random and removed from the box in
succession without replacement, what is the probability that both fuses
are defective?

Let A be the event that the first fuse is defective and B the event that the
second fuse is defective, then
5 4 1
P( A  B) = P( A)  P( B A) =  =
20 19 19
Multiplicative Rules

Example: One bag contains 4 white balls and 3 black balls, and a
second bag contains 3 white balls and 5 black balls. One ball is drawn
from the first bag and placed unseen in the second bag. What is the
probability that a ball now drawn from the second bag is black?

B1 : the drawing of a black ball from bag 1

B2 : the drawing of a black ball from bag 2
W1 : the drawing of a white ball from bag 1

P( B2 ) = P( B1  B2 ) + P (W1  B2 )
3 6 4 5 38
P( B2 ) = P( B1 )  P( B2 B1 ) + P(W1 )  P( B2 W1 ) =  +  =
7 9 7 9 63
P( B2 ) = P( B1  B2 ) + P (W1  B2 )
3 6 4 5 38
P( B2 ) = P( B1 )  P( B2 B1 ) + P(W1 )  P( B2 W1 ) =  +  =
7 9 7 9 63
Multiplicative Rules
Example: An electrical system consists of four components as
illustrated below. The system works if components A and B work and
either of the components C or D work. The reliability (probability of
working) of each component is also indicated. Find the probability
that
(a) the entire system works
(b) the component C does not work,
given that the entire system works
(c) the entire system works given that
the component C does not work.
Assume that four components work
independently.

(a) P( A  B  (C  D)) = P( A)  P( B)  ( P(C ) + P( D) − P(C  D))

= (0.9)(0.9)((0.8) + (0.8) − (0.8)(0.8))
= 0.7776
Continue… Example

(b) Find the probability that the component

C does not work, given that the entire
system works
P(system works while C not working)
P(C  system works) =
P(system works)
P( A  B  C   D) (0.9)(0.9)(1 − 0.8)(0.8)
= = = 0.1667
P(system works) 0.7776

(c) Find the probability that the entire system works given that the
component C does not work
P(C not working but system works )
P(system works C ) =
P(C )
P( A  B  C   D) (0.9)(0.9)(1 − 0.8)(0.8)
= = = 0.648
P(C ) (1 − 0.8)
Bayes’ Rule
➢Refer to the following figure.
A = ( E  A)  ( E   A)
P( A) = P  ( E  A)  ( E   A) 
= P ( E  A) + P ( E   A)
= P( E ) P( A E ) + P( E ) P( A E )

➢If the events B1, B2, ..., Bk constitute a

partition of the sample space S such
that P(Bi) = 0 for i = 1, 2, ..., k, then for
any event A of S,
k k
P( A) =  P( Bi  A) =  P( Bi )  P( A Bi )
i =1 i =1
Theorem 2.14
Bayes’ Rule

Example: A travel agent offers 4-day and 8-day trips around USA.
Based on long-range sales, the probability that a customer will book a
4-day trip is 0.75. Of those that book that trip, 60% also order the bus
pass. But only 30% of 8-day trip customers order the bus pass. A
randomly selected buyer purchases a bus pass and a round trip. What
is the probability that the trip she orders is a 4-day trip?
F : the customer books a 4-day round trip
E : the customer books an 8-day round trip
B : the customer orders a bus pass
P( B  F )
P( B) = P( B  F ) + P( B  E ) P( F B) =
P( B)
= P( F )  P( B F ) + P( E )  P( B E ) P( F )  P( B F )
=
= (0.75)(0.6) + (1 − 0.75)(0.3) P( B)
(0.75)(0.6)
= 0.525 =
0.525
= 0.857
Example: In a certain assembly plant, three machines, B1, B2, and B3,
make 30%, 45%, and 25%, respectively, of the products. It is known
from past experience that 2%, 3%, and 2% of the products made by
each machine, respectively, are defective. Now, suppose that a
finished product is randomly selected. What is the probability that it
is defective?

= (0.3)(0.02) + (0.45)(0.03) + (0.25)(0.02) = 0.0245

Example: With reference to the last example, if a product were
chosen randomly and found to be defective, what is the probability
that it was made by machine B3?

P(B3 )P( A | B3 )
P(B3 | A) =
P(B1 )P( A | B1 ) + P(B2 )P( A | B2 ) + P(B3 )P( A | B3 )
0.005 10
= =
0.006 + 0.0135 + 0.005 49

(0.25)(0.02)
= = 0.204
0.0245