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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