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STATISTICS AND PROBABILITY
IN CIVIL ENGINEERING
TS4512
Doddy Prayogo, Ph.D.
1
3. Probability Methods
 3.1. Introduction
• Probability is now not only use in physical world, but
also an extensive branch of mathematics.
• A random experiment is a process leading to two or
more possible outcomes, without knowing exactly
which outcome will occur.
• The set of all possible outcomes of an experiment is
call the sample space for the experiment (symbol S)
• The possible outcomes of a random experiment are
called the basic outcomes, and the set of all basic
outcomes is called the sample space.
• A subset of sample space is called an event.
3.2. Basic idea
 3.2.1. Combining events
Events are constructed by combining simpler
events.
* The Union
* The Intersection
* The complement
 The Union of two events A and B, denoted A ∪ 𝐵, is
the set of outcomes that belong either to A, to B, or to
both. In words, A∪B means A or B. Thus the event
A ∪ 𝐵 occurs whenever either A or B (or both)
occurs.
 The Intersection of two events A and B, denoted A
∩ B, is the set of outcomes that belong both to A and
to B. In words, A ∩ B means A and B. Thus the
event A ∩ B occurs whenever both A and B occur.
 The complement of an event A, denoted Ac , is the
set of outcomes that do not belong to A. In words, Ac
means “not A”. Thus the event Ac occurs whenever
A does not occur.
A
A
B
A
B
B
A
B
A
AUB
B
A ∩B
A
B
B∩ 𝐴𝐶
Example:
An electrical engineer has on hand two boxes of resistors,
which four resistors in each box. The resistors in the first
box are labeled 10 Ω (ohms), but in fact their resistances
are 9,10,11,and 12Ω. The resistors in the second box are
labeled 20Ω, but in fact their resistances are 18.19.20. and
21 Ω. The engineer chooses one resistor from each box
and determines the resistance of each.
Let A be the event that the first resistor has a resistance
greater than 10, let B the event that the second resistor has
a resistance less than 19, and let C be the event that the
sum of the resistances is equal to 28. Find a sample space
for this experiment, and specify the subsets corresponding
to the events A, B, and C.
Solution:
The sample space S = {(9,18), (9,19), (9,20), (9,21),
(10,18), (10,19), (10,20), (10,21), (11,18), (11,19),
(11,20), (11,21), (12,18), (12,19), (12,20), (12,21)}.
The events A, B, and C are given by
A= {(11,18), (11,19), (11,20), (11,21), (12,18), (12,19),
(12,20), (12,21)}.
B = {(9,18), (10, 18), (11, 18), (12, 18)}.
C = {(9,19), (10, 18)}.
Example:
Refer to example above, find B ∪ C and A ∩ B C
Solution:
The event B ∪ C containts all the outcomes that belong
either to B or to C, or to both.
B ∪ C = {(9,18), (10, 18), (11, 18), (12, 18), (9, 19)}.
The event B C containts those outcomes in the sample
space that do not belong to B. It follows that the event
A ∩ B C containts the outcomes that belong to A and do
not belong to B.Therefore
A ∩ B C = {(11,19), (11,20), (11,21), (12,19), (12,20),
(12,21)}.
 3.2.2. Mutually exclusive events
• There are some events that can never occur together. For example,
it is impossible that a coin can come up both heads and tails.
• The events A and B are said to be mutually exclusive if they have no
outcomes in common.
• More generally, a collection of events A1, A2, A3, ....., An, is said to
be mutually exclusive if no two of them have any outcomes in
commons.
•
A
B
The events A and B are
mutually exclusive
Example:
Refer to example above. If the experiment is performed,
is it possible for event A and B both to occur? How about
B and C? A and C? Which pair of events is mutually
exclusive?
Solution:
If the outcome is (11,18) or (12,18), then events A and B
both occur. If the outcome is (10, 18), then both B and C
occur. It is impossible for A and C both to occur, because
these events are mutually exclusive, having no outcomes
in common.
 3.3. Probabilities
• Each event in a sample space has a probability of occuring.
• Probability is measured over the range from 0 (will not occur) to 1
(certain to occur).
• The expression P(A) denotes the probability that the event A
occurs.
• P(A) is the proportion of times that event A would occur in the long
run, if the experiment were to be repeated over and over again.
• Three definition of prabability:
a. Classical probability
b. Relative frequency probability
c. Subjective probability
 3.3.1. Classical probability
• Classical probability is the proportion of times
that an event will occur, assuming that all
outcomes in a sample space are equally likely to
occur.
• The probability of an event A is P(A) = NA/N
where NA is the number of outcomes that satisfy
the condition of event A, and N is the total
number of outcomes in the sample space.
• One can develop probability from fundamental
reasoning about the process.
• Example:
o An extrusion die is used to produce aluminium rods.
Specification are given for the length and the diameter of the
rods. For each rods, the length is classified as too short, too
long, or ok. The diameter is classified as too thin, too thick, or
ok. In a population of 1000 rods, the number of rods in each
class is as follows:
Length
Diameter
Too thin
Too short
OK
Too long
10
38
2
Ok
3
900
25
Too thick
5
4
13
Length
Diameter
Too thin
Too short
OK
Too long
10
38
2
Ok
3
900
25
P(too short) ?
P(too short and too thick) ?
P(too short or too thick) ?
Too thick
5
4
13
• Solution:
o We can take of each of the 1000 rods as an
outcome in a sample space. Each of the 1000
outcomes is equally likely. We’ll solve the
problem by counting the number of outcomes
that correspond to the event. The number of
rods that are too short is 10+3+5=18.
P(too short) = 18/1000 = 0.018
• Example:
o If a rods is sampled at random, what is the probability that
it is either too short and too thick?
• Solution:
o Note that there are 5 rods that are booth too short and
toothick. Of the 1000 outcomes, the number that are either
too short or too thick is 10+3+5+4+13=35. P(too short or
too thick) = 35/1000=0.035.
o P(too thick) = 22/1000
o P(too short and too thick)=5/1000
o P(too short or to thick) =P(too short)+P(too thick)-P(too
short and too thick) = 18/1000 + 22/1000 – 5/1000 =
35/1000
*The axioms of probability:
1. Let S be a sample space. Then P(S) = 1
2. For any eventA, 0≤ P(A) ≤ 1
3. If A and B arwe mutually exclusive events,
then P(A∪ 𝑩) = P(A) + P(B).
More generally, if A1, A2, A3, .......An are
mutually exclusive events, then
P(A1∪A2∪......Un) = P(A1) = P(A2) + ......P(An)
*For any event A, P(𝑨𝑪 ) = 1- P(A)
Let ∅ the empty set, Then P(∅) = 0
*Let A and B are any events. Then
P(A∪B) = P(A) + P(B) – P(A∩B)
• Example:
o In a process that manufactures aluminium cans,
the probability that a can has a flaw on its side is
0.02, the probability that a can has a flaw on the
top is 0.03, and the probability that a can has a
flaw on both the side and the top is 0.01.
o What is the probability that a randomly chosen
can has a flaw?
o What is the probability that it has no flaw.
• Solution:
o We are given that P(flaw on side)=0.02, P(flaw on
top)=0.03, P(flaw on side and flaw on top)=0.01.
Now P (flaw) =P(flaw on side or flaw on top)
o P(flaw on side or flaw on top)= P(flaw on side) +
P(flaw on top)- P(flaw on side and flaw on
top)=0.02+0.03-0.01 = 0.04.
o To find the probability that a can has no flaw
=P(no flaw)=1-P(flaw) = 1-0.04 = 0.96.
 Permutations
• A permutation is an ordering of a collection of
objects.
• For any positive integer n, n!=n(n-1)(n-2) ... (3)
(2)(1). Also, we define 0!=1
• The number of permutations of n objects is n!
• The number of permutations of k objects chosen
from a group of n objects is
𝑃𝒌𝒏 =
𝒏!
𝒏−𝒌 !
• The total number of permutations of k objects
chosen from n,
is the number of possible
arrangements when k objects are to be
selected from a total of n and arranged in
order.
•
= n(n-1)(n-2)......(n-k+1)
The total number of permutations of k objects chosen
from n, 𝑃𝒌𝒏 is number of possible arrangements when k
objects are to be selected from a total of n and arranged in
order.
𝑃𝒌𝒏 = n(n-1)(n-2)....(n-k+1)
Multiplying and dividing by (n-k)(n-k-1)...((2)(1)=(n-k)!
Gives
𝒏 𝑛
𝑃𝒌 =
𝑃𝒌𝒏
=
𝑛−1 𝑛−2 ... 𝑛−𝑘+1 n−k+1 n−k n−k−1 … 2 (1)
n−k n−k−1 … 2 (1)
𝒏!
𝒏−𝒌 !
• Example 3.4.2:
o Suppose that two letters are to be selected from A,B,C,D, and
E and arranged in order. How many permutations are
possible?
o Solution:
The number of permutations, with n = 5, and k = 2 is as
follows:
𝑃𝒌𝒏 =
𝒏!
𝒏−𝒌 !
 𝑃𝟐𝟓 =
These are AB
BA
BD
DB
𝟓!
𝟓−𝟐 !
=20
AC AD AE BC
CA DA EA CB
BE CD CE DE
EB DC EC ED
 Combinations
• A combination is each distinct group of objects that can be
selected without regard to order.
• Suppose that we are interested in the number of different
ways that k objects can be selected from n (where no object
nay be chosen more than one) but are not concerned about
the order.
Number of combinations:
The number of combinations,
Or simply 𝐶𝑘𝑛 =
𝑛!
𝐶𝑘𝑛
=
𝑃𝑘𝑛
𝑘!
𝑘! 𝑛−𝑘 !
In some applications use this notation:
𝑛
𝑛!
𝑛
= 𝐶𝑘 =
𝑘! 𝑛−𝑘 !
𝑘
• Example 3.4.3 :
o A personnel officer has eight candidates to fill four similar
positions. Five candidates are men, and three are women. If, in
fact, every combination of candidates is equally likely to be
chosen, what is the probability that no women will be hired?
o Solution:
o First, the total number of combinations of four candidates chosen
from eight is as follows:
𝑛!
8!
𝐶𝑘𝑛 =
= 𝐶48 =
= 70
𝑘! 𝑛−𝑘 !
8! 8−4 !
Now, in order for no women to be hired, it follows that the four
successful candidates must come from the available five men. The
5!
number of such combinations is 𝐶45 =
= 5.
4! 5−4 !
Therefore, if at the outset each of the 70 possible combinations
was equally likely to be chosen, the probability that one of the five
all-male combinations would be selected is 5/70 = 1/14.
 Example 3.4.4 :
Suppose that John’s store now containts 10 Gateway computers, 5
CompaqcComputers, and 5 Acer computers. Melati enters the store
and wants to purchase 3 computers. The computers are selected
purely by chance from the shelf. Noe what is the probability that 2
Gateways computers and 1 Compaq computers are selected?
 Solution:
The total number of outcomes in the sample space
20!
20
N = 𝐶3 =
= 1140.
3! 20−3 !
The number of ways that we can select 2 Gateway computers from
10!
10
the 10 available is 𝐶2 =
= 40.
2! 10−2 !
Similarly, that the number of ways that we can select 1 Compaq
computer from the 5 available is 5 and therefore, the number of
CompaqcComputers, and 5 Acer computers. Melati enters the store
and wants to purchase 3 computers. The computers are selected
purely by chance from the shelf. Noe what is the probability that 2
Gateways computers and 1 Compaq computers are selected?
 Solution:
The total number of outcomes in the sample space
20!
N = 𝐶320 =
= 1140.
3! 20−3 !
The number of ways that we can select 2 Gateway computers from
10!
the 10 available is 𝐶210 =
= 40.
2! 10−2 !
Similarly, that the number of ways that we can select 1 Compaq
computer from the 5 available is 5 and therefore, the number of
outcomes that satisfyevent A is 𝑁𝐴 = 𝐶210 x 𝐶15 = 45 x 5 = 225.
Finally, the probability of A = [2 Gateway and 1 Compaq] is
𝑃𝐴 =
𝑁𝐴
𝑁
=
𝐶210 x 𝐶15
𝐶320
=
45 x 5
1140
= 0.197