# University of Technology, Building and Construction Engineering Department (Undergraduate study) PROBABILITY THEORY

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1 ENGIEERING STATISTICS (Lectures) University of Technology, Building and Construction Engineering Department (Undergraduate study) PROBABILITY THEORY Dr. Maan S. Hassan Lecturer: Azhar H. Mahdi

2 Probability Theory A random variable refers to a measurement or observation that cannot be known in advance. An experiment that can result in different outcomes, even though it is repeated in the same manner every time, is called a random experiment. Roman letter is used to represent a random variable, the most common letter being X. A lower case x is used to represent an observed value corresponding to the random variable X. So the notation X =x means that the observed value of X is x. The set of all possible outcomes or values of X we might observe is called the sample space. The set of all possible outcomes of a random experiment is called the sample space of the experiment. The sample space is denoted as S. EXAMPLE 1: Consider an experiment in which you select a plastic pipe, and measure its thickness. Sample space as simply the positive real line because a negative value for thickness cannot occur S= R + = { x >0 } If it is known that all connectors will be between 10 and 11 millimeters thick, the sample space could be S= { x 10 < x < 11 }

3 If the objective of the analysis is to consider only whether a particular part is low, medium, or high for thickness, the sample space might be taken to be the set of three outcomes: S = { low, medium, high } If the objective of the analysis is to consider only whether or not a particular part conforms to the manufacturing specifications, the sample space might be simplified to the set of two outcomes, S = { yes, no } that indicate whether or not the part conforms. A discrete random variable meaning that there are gaps between any value and the next possible value. A continuous random variable meaning that for any two outcomes, any value between these two values is possible. EXAMPLE 2: If two connectors are selected and measured, the sample space is depending on the objective of the study. If the objective of the analysis is to consider only whether or not the parts conform to the manufacturing specifications, either part may or may not conform. The sample space can be represented by the four outcomes:

4 S= {yy, yn, ny, nn } If we are only interested in the number of conforming parts in the sample, we might summarize the sample space as S= {0, 1, 2 } In random experiments in which items are selected from a batch, we will indicate whether or not a selected item is replaced before the next one is selected. For example, if the batch consists of three items {a, b, c} and our experiment is to select two items without replacement, the sample space can be represented as S without = { ab, ac, ba, bc, ca, cb } S with = { aa, ab, ac, ba, bb, bc, ca, cb, cc } Events: Often we are interested in a collection of related outcomes from a random experiment. An event is a subset of the sample space of a random experiment. Some of the basic set operations are summarized below in terms of events: The union of two events is the event that consists of all outcomes that are contained in either of the two events. We denote the union as E 1 UE 2. The intersection of two events is the event that consists of all outcomes that are contained in both of the two events. We denote the intersection as E 1 E 2. The complement of an event in a sample space is the set of outcomes in the sample space that are not in the event. We denote the component of the event E as É.

5 EXAMPLE 3: Consider the sample space S {yy, yn, ny, nn} in Example 2. Suppose that the set of all outcomes for which at least one part conforms is denoted as E 1. Then, E 1 = { yy, yn, ny } The event in which both parts do not conform, denoted as E 2, contains only the single outcome, E 2 {nn}. Other examples of events are E 3 = Ø, the null set, and E 4 =S, the sample space. If E 5 ={yn, ny, nn}, E 1 U E 5 = S E 1 E 5 = { yn, ny } É 1 = { nn } EXAMPLE 4: Measurements of the time needed to complete a chemical reaction might be modeled with the sample space S= R +, the set of positive real numbers. Let Then, Also, E 1 = { x 1 < 10} and E 2 = { x 1 < x < 118} E 1 U E 2 = { x 1 < 118} and E 1 E 2 = { x 3 < x < 10} É 1 = { x 10} and É 1 E 2 = { x 10 x < 118}

6 EXAMPLE 5: Samples of concrete surface are analyzed for abrasion resistance and impact strength. The results from 50 samples are summarized as follows: abrasion resistance impact strength High Low High 40 4 Low 1 5 Let A denote the event that a sample has high impact strength, Let B denote the event that a sample has high abrasion resistance. Determine the number of samples in A, Á, and AUB The event A consists of the 40 samples for which abrasion resistance and impact strength are high. The event Á consists of the 9 samples in which the impact strength is low. The event A U B consists of the 45 samples in which the abrasion resistance, impact strength, or both are high. Figure 1: Venn diagrams

7 Venn diagrams are often used to describe relationships between events and sets. Two events, denoted as E 1 and E 2, such that are said to be mutually exclusive. E 1 E 2 = Ø The two events in Fig. 1(b) are mutually exclusive, whereas the two events in Fig. 1(a) are not. Additional results involving events are summarized below. The de 1 E 2 E The distributive law for set operations implies that Table 1: Corresponding statements in set theory and probability Set theory Probability theory Probability is used to quantify the likelihood, or chance, that an outcome of a random experiment will occur. The chance of rain today is 30% is a statement that quantifies our feeling about the possibility of rain.

8 A 0 probability indicates an outcome will not occur. A probability of 1 indicates an outcome will occur with certainty. Elements Fig. 2: Probability of the event E is the sum of the probabilities of the outcomes in E. For a discrete sample space, the probability of an event E, denoted as P(E), equals the sum of the probabilities of the outcomes in E. EXAMPLE 6: A random experiment can result in one of the outcomes {a, b, c, d} with probabilities 0.1, 0.3, 0.5, and 0.1, respectively. Let A denote the event {a, b}, B the event {b, c, d}, and C the event {d}.then, P(A)= = 0.4 P(B)= = 0.9 P(C) = 0.1 Also: P (Á)= 0.6, P(B )= 0.1, P(C ) = 0.9 P (A B)= 0.3 P (A U B)= 1 P (A )= 0

9 EXAMPLE 7: A visual inspection of a defects location on concrete element manufacturing process resulted in the following table: Number of defects Proportion of concrete element or more 0.1 a) If one element is selected randomly from this process to inspected, what is the probability that it contains no defects? The event that there is no defect in the inspected concrete elements, denoted as E 1, can be considered to be comprised of the single outcome, E 1 = {0}. Therefore, P(E 1 ) = 0.4 b) What is the probability that it contains 3 or more defects? Let the event that it contains 3 or more defects, denoted as E 2 P(E 2 ) = = 0.25 EXAMPLE 8: Suppose that a batch contains six parts with part numbers {a, b, c, d, e, f}. Suppose that two parts are selected without replacement. Let E denote the event that the part number of the first part selected is a. Then E can be written as E {ab, ac, ad, ae, af}. The sample space can be counted. It has 30 outcomes. If each outcome is equally likely, P(E) = 5/30 = 1/6

10 ADDITION RULES P( A U B ) = P( A ) + P( B ) - P( A EXAMPLE 8: The defects such as those described in Example 7 were further classified as either in the center or at the edge of the concrete elements, and by the degree of damage. The following table shows the proportion of defects in each category. What is the probability that a defect was either at the edge or that it contains four or more defects? Location in Concrete Element Surface Defects Center Edge Total Low High Total Let E 1 denote the event that a defect contains four or more defects, and let E 2 denote the event that a defect is at the edge. Defects Classified by Location and Degree Number of defects Center Edge Totals or more Totals

11 The requested probability is P (E 1 U E 2 ). Now, P (E 1 ) = 0.15 and P (E 2 ) = Also, from the table above, P (E 1 2 ) = 0.04 Therefore, P (E 1 U E 2 ) = = 0.39 What is the probability that concrete surface contains less than two defects (denoted as E 3 ) or that it is both at the edge and contains more than four defects (denoted as E 4 )? The requested probability is P (E 3 U E 4 ). Now P (E 3 ) = 0.6, and P (E 4 ) = Also, E 3 and E 4 are mutually exclusive. Therefore, P (E 3 4 ) = Ø and P (E 3 U E 4 ) = = 0.63 for the case of three events:

12 EXAMPLE 9: Let X denote the ph of a sample. Consider the event that X is greater than 6.5 but less than or equal to 7.8. This probability is the sum of any collection of mutually exclusive events with union equal to the same range for X. One example is: Another example is The best choice depends on the particular probabilities available.

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14 CONDITIONAL PROBABILITY In a manufacturing process, 10% of the parts contain visible surface flaws and 25% of the parts with surface flaws are (functionally) defective parts. However, only 5% of parts without surface flaws are defective parts. The probability of a defective part depends on our knowledge of the presence or absence of a surface flaw. Let D denote the event that a part is defective and let F denote the event that a part has a surface flaw. Then, the probability of D given, or assuming, that a part has a surface flaw as P(D. This notation is read as the conditional probability of D given F, and it is interpreted as the probability that a part is defective, given that the part has a surface flaw.

15 EXAMPLE 1: Table 1 below provides an example of 400 parts classified by surface flaws and as (functionally) defective. For this table the conditional probabilities match those discussed previously in this section. For example, of the parts with surface flaws (40 parts) the number defective is 10. Table 1: Parts Classified Therefore, and of the parts without surface flaws (360 parts) the number defective is 18. Therefore, Figure 1: Tree diagram for parts classified Therefore, P ( B can be interpreted as the relative frequency of event B among the trials that produce an outcome in event A.

16 EXAMPLE 2: Again consider the 400 parts in Table 1 above (example 1). From this table Note that in this example all four of the following probabilities are different: Here, P (D) and P (D ) are probabilities of the same event, but they are computed under two different states of knowledge. Similarly, P (F) and P (F The tree diagram in Fig. 1 can also be used to display conditional probabilities.

17 Permutations Another useful calculation is the number of ordered sequences of the elements of a set. Consider a set of elements, such as S {a, b, c}. A permutation of the elements is an ordered sequence of the elements. For example, abc, acb, bac, bca, cab, and cba are all of the permutations of the elements of S. In some situations, we are interested in the number of arrangements of only some of the elements of a set. The following result also follows from the multiplication rule. EXAMPLE 3: A printed circuit board has eight different locations in which a component can be placed. If four different components are to be placed on the board, how many different designs are possible? Each design consists of selecting a location from the eight locations for the first component, a location from the remaining seven for the second component, a location from the remaining six for the third component, and a location from the remaining five for the fourth component. Therefore,

18 Combinations Another counting problem of interest is the number of subsets of r elements that can be selected from a set of n elements. Here, order is not important. EXAMPLE 4: A printed circuit board has eight different locations in which a component can be placed. If five identical components are to be placed on the board, how many different designs are possible? Each design is a subset of the eight locations that are to contain the components. From the Equation above, the number of possible designs is The following example uses the multiplication rule in combination with the above equation to answer a more difficult, but common, question. EXAMPLE 5: A bin of 50 manufactured parts contains three defective parts and 47 non-defective parts. A sample of six parts is selected from the 50 parts. Selected parts are not replaced. That is, each part can only be selected once and the sample is a subset of the 50 parts. How many different samples are there of size six that contain exactly two defective parts?

19 A subset containing exactly two defective parts can be formed by first choosing the two defective parts from the three defective parts. Then, the second step is to select the remaining four parts from the 47 acceptable parts in the bin. The second step can be completed in Therefore, from the multiplication rule, the number of subsets of size six that contain exactly two defective items is 3 * 178,365 = 535,095 As an additional computation, the total number of different subsets of size six is found to be Therefore, the probability that a sample contains exactly two defective parts is

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