BFC 34303
CIVIL ENGINEERING STATISTICS
Chapter 2
Probability
Faculty of Civil and Environmental Engineering
Universiti Tun Hussein Onn Malaysia
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What is a ‘probability’?
A probability is a number (between zero and one) that describes the
chance or likelihood that something will happen.
The following three key words are used in the study of probability:
Experiment
• A process that
leads to the
occurrence of one
and only one of
several possible
observations
Outcome
• A particular result
of an experiment
Event
• A collection of one
or more outcomes
of an experiment
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1
Example: Rolling a die
Experiment
Basic Outcomes
Possible Events
Rolling a die
1
2
3
4
5
6
• Observe an even
number
• Observe an odd
number
• Observe a number
less than 3
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Sample Space
Sample space is the set of all basic or possible outcomes. It is denoted by
the letter S.
Example: What is the size of the sample space when two
dice are rolled?
Sample space, S = { (1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,2), (2,3),
(2,4), (2,5), (2,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6),
(4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (5,1), (5,2), (5,3),
(5,4), (5,5), (5,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6) }
n(S) = 36
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2
Event
An event is the subset of basic outcomes from the sample space.
Example: Observing odd numbers when rolling a die.
The basic outcomes are 1, 2, 3, 4, 5 and 6
Thus, the sample space, S = { 1, 2, 3, 4, 5, 6 }
If A is the event of observing odd numbers, then A = { 1, 3, 5 }
A
2
4
6
1
3
5
5
Mutually Exclusive
If only one of several events can occur at one time, we refer to the events
as mutually exclusive. In other words, the occurrence of one event
means that none of the others can occur at the same time.
Example:
In a die tossing experiment, the event ‘getting an even number’ and the
event ‘getting an odd number’ are mutually exclusive. They can never
happen together.
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3
In a die tossing experiment, let A = getting an odd number and B = getting
an even number.
A = { 1, 3, 5 }
A
B = { 2, 4, 6 }
B
1
5
3
2
4
6
A∩B=∅
 Empty set because the two
events do not intersect
 A and B are mutually exclusive
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Collectively Exhaustive
If an experiment has a set of events that includes every possible outcome,
then the set of events is collectively exhaustive.
Example:
In a die tossing experiment, every outcome (1, 2, 3, 4, 5, 6) will be either
even or odd. So the set is collectively exhaustive.
Note:
If the set of events is collectively exhaustive and the events are mutually
exclusive, the sum of the probabilities equals to 1.
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4
In a die tossing experiment, let A = getting an odd number and B = getting
an even number.
A = { 1, 3, 5 }
A
B
1
5
4
2
B = { 2, 4, 6 }
6
3
A∪B=S
 A and B are collectively exhaustive
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Example:
In a die tossing experiment, let C = getting an odd number and D = getting
a multiple of 3. Are C and D mutually exclusive? Are C and D collectively
exhaustive?
C = { 1, 3, 5 }
2
C
D = { 3, 6 }
D
1 5
3
6
C∩D={3}≠∅
4
 Not mutually exclusive
C ∪ D = { 1, 3, 5, 6 } ≠ S
 Not collectively exhaustive
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5
Probability of an Event
The probability of an event A happening is denoted by P(A), where
P(A) =
Number of possible outcomes in A
Number of possible outcomes in S
=
n(A)
n(S)
P(A) satisfies the following conditions:
0 < P(A) < 1
If P(A) = 1, event A is a sure event.
If P(A) = 0, event A is an impossible event.
The complement of an event A is denoted by A’ or A. This means that the
event A does not happen.
P(A’) = 1 – P(A)
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Example:
Two dice are tossed. What is the probability of getting the sum of 6? Also,
what is the probability of not getting the sum of 6?
We know that n(S) = 36
Let X be the event of getting the sum of 6
So, X = { (1,5), (2,4), (3,3), (4,2), (5,1) }
n(X) = 5
Therefore, the probability of getting the sum of 6, P(X) = n(X) / n(S)
P(X) = 5 / 36 = 0.139
The probability of not getting the sum of 6
P(X’) = 1 – 0.139 = 0.861
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6
Rules of Probability
Rules of Addition
For two events A and B, if their probabilities are written as P(A) and P(B)
then the rules of addition can be written as follows:
Mutually Exclusive
Events
Non-mutually Exclusive
Events
P(A 𝑜𝑟 B) = P(A) + P(B)
P(A ∪ B) = P(A) + P(B)
P(A 𝑜𝑟 B) = P(A) + P(B) − P(A 𝑎𝑛𝑑 B)
P(A ∪ B) = P(A) + P(B) − P(A ∩ B)
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Rules of Multiplication
For two events A and B, if their probabilities are written as P(A) and P(B)
then the rules of multiplication can be written as follows:
Mutually Exclusive
Events
Independent Events
Dependent Events
P(A 𝑎𝑛𝑑 B) = 0
P(A ∩ B) = 0
P(A 𝑎𝑛𝑑 B) = P(A) × P(B)
P(A ∩ B) = P(A) × P(B)
P(A 𝑎𝑛𝑑 B) = P(A) × P(B|A)
P(A ∩ B) = P(A) × P(B|A)
P(B|A) means the probability event B occurs when event A has occurred.
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7
Example:
An integer is randomly selected from a set of integers, S = { 1, 2, 3, 4, 5,
6, 7, 8, 9, 10, 11, 12 }. Find the probability that the integer
(a)
is an even number and a factor of 10.
(b)
is an even number or a factor of 10.
Let E = even number and F = factor of 10
E = { 2, 4, 6, 8, 10, 12 }
n(E) = 6
F = { 1, 2, 5, 10 }
n(F) = 4
We know that n(S) = 12
Therefore, P(E) = 6 / 12 = 0.50 and P(F) = 4 / 12 = 0.33
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E and F are not mutually exclusive.
(a)
(b)
P(E ∩ F) =
P(E) × P(F)
=
0.50 × 0.33
=
0.17
P(E ∪ F) =
P(E) + P(F) − P(E ∩ F)
=
0.50 + 0.33 − 0.17
=
0.66
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8
Probability Tree Diagrams
We can construct a probability tree diagram to solve probability problems.
A probability tree diagram shows all the possible events. The first event is
represented by a dot. From the dot, branches are drawn to represent all
possible outcomes of the event. The probability of each outcome is written
on its branch.
P(G)
P(H)
P(G)
Outcome G
P(H)
Outcome H
P(G)
Outcome G
P(H)
Outcome H
Outcome G
Outcome H
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Multiply
P(G)
P(H)
P(G)
Outcome G
P(G) × P(G) = a
P(H)
Outcome H
P(G) × P(H) = b
P(G)
Outcome G
P(H) × P(G) = c
P(H)
Outcome H
P(H) × P(H) = d
Outcome G
Add
Outcome H
a + b + c + d = 1.0
When using the tree diagram, the following rules apply:
1. Multiply probabilities along the branch
2. Add probabilities down the column
3. All probabilities should total up to 1.0
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9
Example:
A bag contains 2 apples and 6 oranges. Jane picks a fruit at random from
the bag and returns it to the bag. She mixes the fruits in the bag and then
picks another fruit at random from the bag.
Calculate the probability of Jane picking
(a) the same fruit
(b) at least one apple
(c) an orange in her second pick
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Let A = apple and O = orange
P(A) = 2/8 = 0.25, P(O) = 6/8 = 0.75
0.25
0.75
0.25
A
0.25 × 0.25 = 0.0625
0.75
O
0.25 × 0.75 = 0.1875
0.25
O
0.75 × 0.25 = 0.1875
0.75
O
0.75 × 0.75 = 0.5625
A
O
Total = 1.0
20
10
(a)
Probability of Jane picking the same fruit
P(A ∩ A) + P(O ∩ O) = 0.0625 + 0.5625 = 0.625
(b)
Probability of Jane picking at least one apple
P(A ∩ A) + P(A ∩ O) + P(O ∩ A)
= 0.0625 + 0.1875 + 0.1875 = 0.4375
(c)
Probability of Jane picking an orange in her second pick
P(A ∩ O) + P(O ∩ O) = 0.1875 + 0.5625 = 0.75
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Conditional Probability
When an event occurs with the condition that another event has occurred,
then the event is a conditional event or dependent event.
The probability that an event will occur given that another event has
already occurred is called conditional probability.
For events A and B in sample space S, the probability of “A occurring
given that B has occurred” or simply mentioned as the probability of “A
given B” is denoted by P(A|B).
P(A|B) =
P(A ∩ B)
P(B)
where P(B) ≠ 0
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11
Example:
Rajesh travels to work by either route A or route B. The probability that he
chooses route A is 0.6. The probability that he is late for work if he uses
route A is 0.3 while the probability that he is late for work if he uses route
B is 0.2.
Given that Rajesh is not late for work, what is the probability he had
chosen route B?
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Let A = route A, B = route B and L = late for work.
0.6
0.4
0.3
L
0.6 × 0.3 = 0.18
0.7
L’
0.6 × 0.7 = 0.42
0.2
L
0.4 × 0.2 = 0.08
0.8
L’
0.4 × 0.8 = 0.32
A
B
Total = 1.0
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12
The probability of being late for work
P(L) = P(A ∩ L) + P(B ∩ L) = 0.18 + 0.08 = 0.26
The probability of not being late for work
P(L’) = P(A ∩ L’) + P(B ∩ L’) = 0.42 + 0.32 = 0.74
or P(L’) = 1 - P(L) = 1 – 0.26 = 0.74
The probability of choosing route B given that Rajesh is not late for work
P(B|L’) = P(B ∩ L’) =
P(L’)
0.32 = 0.43
0.74
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Note:
If A and B are independent events, it means that the outcome of one
event does not affect the outcome of the other.
Therefore:
P(A|B) = P(A)
P(B|A) = P(B)
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13
Question:
If P(A|B) is 0.2, what is P(B|A)?
Bayes’ Theorem
Bayes’ Theorem (formulated by Thomas Bayes) describes the probability
of an event, based on prior knowledge of conditions that might be related
to the event.
Bayes’ Theorem is useful when we have to “reverse the conditions” in a
problem.
P(A|B) =
P(B|A) × P(A)
P(B)
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Example:
From a survey conducted on employees in a company, it was found that
90% of the employees living more than 5 km from work travel to work by
car. Of the remaining employees, 50% travel to work by car. It is known
that 75% of employees live more than 5 km from work.
Determine
(a) the percentage of employees who travel to work by car.
(b) the probability an employee lives more than 5 km from work who
travels to work by car.
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D1 = lives more than 5 km from work
D2 = lives not more than 5 km from work
C = travels to work by car
We know the following:
P(D1) = 0.75  therefore, P(D2) = 0.25
P(C|D1) = 0.90
P(C|D2) = 0.50
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0.75
0.25
0.9
C
0.75 × 0.9 = 0.675
0.1
C’
0.75 × 0.1 = 0.075
0.5
C
0.25 × 0.5 = 0.125
0.5
C’
0.25 × 0.5 = 0.125
D1
D2
Total = 1.0
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(a)
Percentage of employees who travel to work by car
P(C) = P(D1) × P(C|D1) + P(D2) × P(C|D2)
= (0.75 × 0.9) + (0.25 × 0.5)
= 0.675 + 0.125
= 0.80
 80% of employees travel to work by car
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(b)
Probability an employee lives more than 5 km from work who
travels to work by car,
which is P(D1|C)
We know that P(C|D1) = 0.90
Using Bayes’ Theorem
P(D1|C) = P(C|D1) × P(D1) =
P(C)
0.90 × 0.75
0.80
= 0.844
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