1. STATISTICS AND PROBABILITY IN CIVIL ENGINEERING
TS4512
Doddy Prayogo, Ph.D.

2. 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. 3.2. Basic idea
Combining events
Events are constructed by combining simpler events.
* The Union
* The Intersection
* The complement

5. A B

6. A B
A U B
A ∩ B

10. 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.
The events A and B are
mutually exclusive A B

12. 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:
Classical probability
Relative frequency probability
Subjective probability

13. 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.

14. Example:
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 Ok
Too thick
Too short OK
Too long 10 38 2 3
900 25 5 4 13

15. P(too short and too thick) ? P(too short or too thick) ?
Length
Diameter
Too thin Ok
Too thick
Too short OK
Too long 10 38 2 3
900 25 5 4 13

16. Solution:
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 =18.
P(too short) = 18/1000 = 0.018

17. Example:
If a rods is sampled at random, what is the probability that it is either too short and too thick?
Solution:
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 =35. P(too short or too thick) = 35/1000=0.035.
P(too thick) = 22/1000
P(too short and too thick)=5/1000
P(too short or to thick) =P(too short)+P(too thick)-P(too short and too thick) = 18/ /1000 – 5/1000 = 35/1000

19. Example:
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.
What is the probability that a randomly chosen can has a flaw?
What is the probability that it has no flaw.

20. Solution:
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)
P(flaw on side or flaw on top)= P(flaw on side) + P(flaw on top)- P(flaw on side and flaw on top)= = 0.04.
To find the probability that a can has no flaw =P(no flaw)=1-P(flaw) = = 0.96.

21. 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

22. 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)

24. Example 3.4.2:
Suppose that two letters are to be selected from A,B,C,D, and E and arranged in order. How many permutations are possible?
Solution:
The number of permutations, with n = 5, and k = 2 is as follows:

25. 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.