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Jubail University College
MATH 117 Statistical Methods for Management I
Chapter Four
Introduction: In the real world events cannot be predicted with total certainty. Probability is a method to
measure the likelihood of the occurrence of an event. Probability is a number between 0 and 1 inclusive. It
can be represented as a decimal such as 0.6, fraction such as 6/10 or percentage such as 60%.
Approaches of Assigning Probabilities:
There are three major approaches of assigning probabilities.
1. Classical Approach: Is based on the assumption that the outcomes of an experiment are equally
likely to happen, and the events are mutually exclusive and collectively exhaustive.
P(X)=
Number of ways it can happen
Total number of possible outcomes
Example 1: In the experiment of flipping a coin; the possible outcomes are head “H” or tail “T”.
S= {H, T}, if X= {H} then P(X) = 1/2.
Example 2: The experiment of rolling a die S= {1, 2, 3, 4, 5, 6}, E= {3, 4} then P(E) = 2/6
Example 3: Choosing a drink from Coke, Pepsi and Tea. The sample space S= {Coke, Pepsi, Tea}
E= {Coke, Pepsi} the P(E)= 2/3
In the classical approach, the probability is obtained a prior without actually doing an experiment.
2. Relative Frequency Approach:
Based on the accumulated historical or experimental data and the
assumption that what happened in the past will hold under the same conditions.
P(X)=
Number of times an event occurred
Total number of opportunities for the event to occur
For example if 100 students took the final exam in Statistics I this semester and 9 failed. The
probability that a student is taking Statistics I will fail is 9/100=0.09
3. Subjective Approach: Is based on personal experience, intuitive judgment and expertise. No two
people may agree on a particular subjective probability, it depends on the person state of mind.
For example medical doctors sometimes assign subjective probability to the length of life expectancy
for a person who has cancer. Weather forecasting is another example of subjective probability.
Ms. Ghaida Barghouthi, JUC, Semester 332
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Important Terms:
Experiment: Any Process that is either observed or measured that can produce one single outcome.
Such as tossing a die, or rolling a coin, selecting a marker from three red, green and yellow.
Sample point: One possible outcome such as {H}, {2}.
Event: One or more possible outcome of an experiment. If the experiment is tossing a die, then E={2},
F={3}, G={2,4}, E=[1,2,3,4,5,6} are all events of the sample space S=[1,2,3,4,5,6}.
Elementary Event: Those type of events that cannot be broken into other events. For example the event {3}
is an elementary event, while {2, 4, 6} is not elementary.
Sample Space: Is a complete set of all possible outcomes of an experiment. Some examples are:
a) Flipping one coin, the sample space consists of 2 possible outcomes S= {H, T}
b) Rolling one die, the sample space consists of 6 possible outcomes, S= {1, 2, 3, 4, 5, 6}
c) Flipping two coins, the sample space consists of 4 possible outcomes S = HH, HT,TH,TT 
d) Three coins, the sample space S  HHH , HHT , HTH ,THH ,TTH ,THT , HTT ,TTT .
e) Rolling two distinguishable dice the sample space consists of 36 possible outcomes,
(1,1),(1, 2)...(1,6),(2,1),(2, 2)...(2,6),(3,1),(3, 2)...(3,6) 
S=

(4,1),(6, 2)...(4,6),(5,1),(5, 2)...(5,6),(6,1),(6, 2)...(6,6) 
Try to write the sample space for rolling 3 dice? How many sample point you will have?
Mutually Exclusive Events: Those events that cannot happen at the same together. In the experiment of
rolling a die one time S= {1, 2, 3, 4, 5, 6}. The event that the number is even is E= {2, 4, 6}, the event the
number is odd is O= {1, 3, 5} and let event C= {1, 4, 5, 6}. The two events E and O cannot happen at the
same time so they are mutually exclusive. The probability of two mutually exclusive events occurring at the
same time is zero. The events E and C are not mutually exclusive.
The Complementary event: The complement of an event A, denoted by A , consists of all the outcomes that
do not belong to A. The event A occurs whenever A does not occur. The complement of the events E= {2,
4, 6} is the event O= {1, 3, 5} if the sample space is S= {1, 2, 3, 4, 5, 6}.
Events can be illustrated graphically by Venn diagrams as shown below:
Mutually Exclusive Events
O
E
2
6
4
1 3
5
Not Mutually Exclusive Events
C
E
2
4
6
The complement of A
A
1
5
1
4
A
2
6
3
5
Independent Events: If one happened the occurrence of the other is not affected. If you toss a coin, the
Probability of getting a head the first time is 0.5. The probability that you get a head the second time given
that the first time was a head will be 0.5.
Ms. Ghaida Barghouthi, JUC, Semester 332
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Dependent Events: If the events are not independent means they are dependent. This happens when
selection form a finite set of object without replacement.
Selection with replacement: When selection is done with replacement, the selected object is put back before
the next object is selected. With replacement the events are independent. Consider selecting a letter
followed by another from the set X  A , B , C  with replacement.
The result can be: AA , AB , AC , BA , BB , BC , CA , CB , CC 
1
1 1 1
P  BB  = or P  BB  = P  B  P  B  = * =
9
3 3 9
Selection without replacement: When selection is done without replacement, the selected object will not be
put back before the next object is selected. Without replacement the events are dependent. Consider
selecting a letter followed by another from the set X  A , B , C  without replacement.
The result can be: S  AB , AC , BA , BC , CA , CB 
Rules of Probability:
For any event A of the sample space S = {e1 , e 2 , ....e n }
1
P{e i } =
n
A = {e 1 , e 2 , ....e k }, k  n
k
P(A) = P(e 1 ) + P(e 2 ) + ....P(e k ) =
n
0 ≤ P(A) ≤ 1
n
P{e i } = 1,

i=1
P(S) = 1
P(  ) = 0
P( A ) + P(A) = 1  P( A )  1  P(A) or P(A) = 1 - P( A )
General Computation Rules
Addition: When two or more events will happen at the same time, and the events A, B are not mutually
exclusive, then:
P(A or B) = P(A) + P(B) - P(A and B)
Or P(A ∪ B) = P(A) + P(B) - P(A  B)
Example 1:
P(A) = 0.3, P( B ) = 0.55, P(A and B) = 0.15, Find :
a) P(A or B)
P(A or B) = P(A) + P(B) - P(A and B)
P(B) = 1 - P( B ) = 1 - 0.55 = 0.45
P(A or B) = 0.3 + 0.45 - 0.15 = 0. 6
b) P( A or B ),(the Complement of P(A or B) )
P(A o r B =) 1 - P (A or B) = 1 - 0.6 = 0.4
c) P( A and B ),(the Complement of P(A and B))
P( A and B ) = 1 - P(A and B) = 1 - 0.15 = 0.85
Ms. Ghaida Barghouthi, JUC, Semester 332
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Example 2: If 70% of students in a Statistics I course at JUC are taking Arabic Writing and 55% are taking
Professional Ethics. If 45% are taking both Arabic and Professional Ethics, find:
a) The probability that a randomly selected student is taking Arabic or Professional Ethics.
b) The probability that a randomly selected student is taking neither Arabic nor Professional Ethics.
c) The probability that a randomly selected student is not taking both Arabic and Professional Ethics.
d) The probability that a randomly selected student is taking one course only.
e) The probability that a randomly selected student is taking at least one course.
f) The probability that a randomly selected student is taking at most one course.
Let A: student is taking Arabic, E: Student is taking Ethics
S
Given that P(A)=0.7, P(E)=0.55, P(A and E)=0.45
A
E
a)
b)
c)
d)
e)
f)
P(A or E)=P(A)+P(E)-P(A and E)=0.70+0.55-0.45=0.80
0.25 0.45
P(neither)=1-P(A or E)=1-0.8=0.20
P(Not both)=1-P(A and E)=1-0.45=0.55
0.20
P(Exactly one )=P( A or E)-P(A and E)=0.8-0.45=0.35
P( at least one)=P(A or E)=0.80 as above
At most one means: taking (Arabic or Ethics but not both) or (not taking at all).
As shown in the graph =0.25+0.10+0.20
Or P (At most one) = 1-P (taking both courses) = 1-0.45=0.55
0.10
Special Case of Addition Rule: When two or more events will happen at the same time, and the events
are mutually exclusive, then:
P(A and B) = P(A  B)  
P(A or B) = P(A ∪ B) = P(A) + P(B)
Example 1: The distribution of the blood type of 100 employees of a company is shown in the table below.
a) What is the probability that a randomly selected employee is of blood type A or B?
b) What is the probability that a randomly selected employee is of blood type A and male?
c) What is the probability that a randomly selected employee is of blood type A or male?
d) What is the probability that a randomly selected employee is of blood type male or female
Define the following events:
A: the person of blood type A.
B: the person of blood type B
M: the person is male
F: the person is female
Blood Type
A
AB
B
O
Total
Male
20
16
28
2
66
Female
5
14
7
8
34
Total
25
30
35
10
100
a) Events A, B are mutually exclusive, because a person cannot have two blood types.
P(A or B) = P(A) + P(B) =
25 35
60
+
=
= 0.6
100 100 100
b) P(A and M) = 20  0.20
100
Events F, M are mutually exclusive, because a person is either a male or a female not both.
Ms. Ghaida Barghouthi, JUC, Semester 332
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Conditional probability: When two events are dependent, the conditional probability of A given that B is
known to have occurred is written as P(B / A)
P(A and B)
;P(B) > 0
P(B)
P(A ∩ B)
or P(A / B) =
; P(B) > 0
P ( B)
P(A and B) = P(A / B).P(B)
P(A and B) = P(B / A).P(A)
P(A / B) =
Example 1: Consider the sample space S = {1, 2,3,4, 5,6,7,8,9}, A = {2,4,1}, E = {2,4,6,8}
3 1
=
9 3
4
P(E) =
9
P(A) =
P(A and E) = P({2,4}) =
P(A / E) =
2
9
2 1
=
4 2
If we apply the formula P(A / E) =
P(E / A) =
P(A and E)
P(E)

2
9
4
9

2 1

4 2
2
3
P(A and E) 92 2
= 3=
P(A)
3
9
Example 2: Suppose 10 marbles in a bag, where 3 are defective. Two marbles are selected, one after
the other without replacement.
a) What is the probability that both are defective?
b) What is the probability that both are not defective?
Let A: the first is defective, B: the second is defective.
3
P(A ) =
10
2
P(B / A) =
9
P( both are defectives) = P(A and B) = P(B / A).P(A) =
P( A ) =
2 3
1
. =
9 10 15
7
10
P( B / A ) =
6
9
P( both are not defectives) = P( A and B ) = P( B / A ).P( A ) =
Ms. Ghaida Barghouthi, JUC, Semester 332
6 7
7
. =
9 10 15
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Example 3: Given P (A) =0.4, P(A/B)=0.5, P(B)=0.5 find:
a) P(A and B)
b) P(B/A)
c) Are the events A and B independent? Why?
a) P(A and B)=P(A/B).P(B)=0.5(0.5)=0.25
b) P(B/A)=P(A and B)/P(A)=0.25/0.4=0.625
c) No, they are not independent because P(A/B)=0.5≠P(A)=0.4
Or because P (B/A) = 0.625≠P (B) = 0.5
Or because P (A and B) = 0.25≠ P( A).P(B)=0.4*0.5=0.20
Special Case of Multiplicative Rule: When two or more events will happen at the same time,
and the events are independent, then:
P(A / B) = P(A); P(B) > 0
P(B / A) = P(B); P(A) > 0
P(A and B) = P(A).P(B)
Example 1: Consider the experiment of tossing a coin two times. What is the probability of getting head on
both times?
We can list the sample space S= {HH, HT, TH, TT}, therefore P ({HH}) = 1/4
Or since the two events are independent, P (H and H) = P (H).P (H) = (1/2) (1/2) =1/4
Example 2: What is the probability of rolling a die twice and getting two sixes?
By listing the sample space which consists of 36 points, P ({6, 6}) =1/36
(1,1),(1, 2)...(1,6),(2,1),(2, 2)...(2,6),(3,1),(3, 2)...(3,6) 
S=

(4,1),(6, 2)...(4,6),(5,1),(5, 2)...(5,6),(6,1),(6, 2)...(6,6) 
Or we can find the probability of getting (6, 6), knowing the fact that getting 6 on the first time is independent
of getting 6 in the second time. Therefore P ({6, 6}} =P ({6}) ·P ({6}) = 1/6 × 1/6 = 1/36
Example 3: Given A, B is two events with P (A) =0.6, P (B) =0.3 find P (A or B) if:
a) The events A and B are mutually exclusive.
b) The events A and B Independent.
c) If P(A/B)=0.25
a)
b)
Ms. Ghaida Barghouthi, JUC, Semester 332
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The Bayes' Rule (Theorem):
For any two events A, B, most of the time P(A/B) ≠P(B/A). Bayes’ rule provides a formula that allows us to
calculate one of the conditional probabilities if we know the other one.
Example 1: In a certain assembly plant, two machines A and B makes 30% and 70%, respectively of the
products. If it is known from past experience that 2%, 4% of the products made by each machine
respectively, are defectives. A product was selected randomly.
1) What is the probability that it is defective?
2) If a product was selected and found to be defective what is the probability that it was from
Machine A?
3) If a product was selected and found to be defective what is the probability that it was from
Machine B?
Machine-A (30%)
Define the events:
A: the product is made by machine.
B: the product is made by machine B.
D: the product is defective.
Also we are given the probabilities:
Machine-B (70%)
Defective
D∩A
D
Find:
Bayes’ Formula (Theorem)
For more than two events
D
………
Ms. Ghaida Barghouthi, JUC, Semester 332
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Example 2: Application using tree diagram.
Suppose two machines, A and B, produce identical parts. Machine A has probability 0.1 of producing a
defective each time, whereas Machine B has probability 0.4 of producing a defective. Each machine produces
one part. One of these parts is selected at random, tested, and found to be defective. What is the probability
that it was produced by Machine B? Probability tree diagrams depict events or sequences of events as
branches of a tree. Tree diagrams are useful for visualizing the conditional probabilities:
The probabilities at the end of each branch are the probability that events leading to that end will happen
simultaneously. The above tree diagram indicates that the probability of a part testing Good is 9/20 + 6/20 =
3/4, therefore the probability of Bad is 1/4. Thus, P(made by B | it is bad) = (4/20) / (1/4) = 4/5.
Now using the Bayes' Rule we are able to obtain useful information such as:
P(it is bad | made by B) = 1/4(4/5) / [1/4(4/5) + 3/4(2/5)] = 2/5.
Equivalently, using the above conditional probability, results in:
P(it is bad | made by B) = P(it is bad & made by B)/P(made by B) = (4/20)/(1/2) = 2/5.
Ms. Ghaida Barghouthi, JUC, Semester 332
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