Applied Statistics and Probability for Engineers Sixth Edition Douglas C. Montgomery George C. Runger Chapter 2 Probability Copyright 2014 John Wiley & Sons, Inc. All rights reserved. 2 Probability CHAPTER OUTLINE 2-1 Sample Spaces and Events 2-1.1 Random Experiments 2-1.2 Sample Spaces 2-1.3 Events 2-1.4 Counting Techniques

2-2 Interpretations and Axioms of Probability 2-3 Addition Rules 2-4 Conditional Probability 2-5 Multiplication and Total Probability Rules 2-6 Independence 2-7 Bayes Theorem 2-8 Random Variables Chapter 2 Title and Outline 2 Copyright 2014 John Wiley & Sons, Inc. All rights reserved. Learning Objectives for Chapter 2 After careful study of this chapter, you should be able to do the following: 1. 2.

3. 4. 5. 6. Understand and describe sample spaces and events Interpret probabilities and calculate probabilities of events Use permutations and combinations to count outcomes Calculate the probabilities of joint events Interpret and calculate conditional probabilities Determine independence and use independence to calculate probabilities 7. Understand Bayes theorem and when to use it 8. Understand random variables Sec 2-1.1 Random Experiments Copyright 2014 John Wiley & Sons, Inc. All rights reserved. 3 Random Experiment

An experiment is a procedure that is carried out under controlled conditions, and executed to discover an unknown result. An experiment that results in different outcomes even when repeated in the same manner every time is a random experiment. Sec 2-1.1 Random Experiments Copyright 2014 John Wiley & Sons, Inc. All rights reserved. 4 Sample Spaces The set of all possible outcomes of a random experiment is called the sample space, S. S is discrete if it consists of a finite or countable infinite set of outcomes. S is continuous if it contains an interval

of real numbers. Sec 2-1.2 Sample Spaces 5 Copyright 2014 John Wiley & Sons, Inc. All rights reserved. Example 2-1: Defining Sample Spaces Randomly select a camera and record the recycle time of a flash. S = R+ = {x | x > 0}, the positive real numbers. Suppose it is known that all recycle times are between 1.5 and 5 seconds. Then S = {x | 1.5 < x < 5} is continuous. It is known that the recycle time has only three values(low, medium or high). Then S = {low, medium, high} is discrete. Does the camera conform to minimum recycle time specifications? S = {yes, no} is discrete. Sec 2-1.2 Sample Spaces

6 Copyright 2014 John Wiley & Sons, Inc. All rights reserved. Sample Space Defined By A Tree Diagram Example 2-2: Messages are classified as on-time(o) or late(l). Classify the next 3 messages. S = {ooo, ool, olo, oll, loo, lol, llo, lll} Sec 2-1.2 Sample Spaces 7 Copyright 2014 John Wiley & Sons, Inc. All rights reserved. Events are Sets of Outcomes An event (E) is a subset of the sample space of a random experiment. Event combinations The Union of two events consists of all outcomes that are contained in one event or the other, denoted as E1 E2.

The Intersection of two events consists of all outcomes that are contained in one event and the other, denoted as E1 E2. The Complement of an event is the set of outcomes in the sample space that are not contained in the event, denoted as E. Sec 2-1.3 Events 8 Copyright 2014 John Wiley & Sons, Inc. All rights reserved. Example 2-3 Discrete Events Suppose that the recycle times of two cameras are recorded. Consider only whether or not the cameras conform to the manufacturing specifications. We abbreviate yes and no as y and n. The sample space is S = {yy, yn, ny, nn}. Suppose, E1 denotes an event that at least one camera conforms to specifications, then E1 = {yy, yn, ny} Suppose, E2 denotes an event that no camera conforms to

specifications, then E2 = {nn} Suppose, E3 denotes an event that at least one camera does not conform. then E3 = {yn, ny, nn}, Then E1 E3 = S Then E1 E3 = {yn, ny} Then E1 = {nn} Sec 2-1.3 Events 9 Copyright 2014 John Wiley & Sons, Inc. All rights reserved. Example 2-4 Continuous Events Measurements of the thickness of a part are modeled with the sample space: S = R+. Let E1 = {x | 10 x < 12}, Let E2 = {x | 11 < x < 15}

Then E1 E2 = {x | 10 x < 15} Then E1 E2 = {x | 11 < x < 12} Then E1 = {x | 0 < x < 10 or x 12} Then E1 E2 = {x | 12 x < 15} Sec 2-1.3 Events 10 Copyright 2014 John Wiley & Sons, Inc. All rights reserved. Venn Diagrams Events A & B contain their respective outcomes. The shaded regions indicate the event relation of each diagram. Sec 2-1.3 Events 11 Copyright 2014 John Wiley & Sons, Inc. All rights reserved. Mutually Exclusive Events Events A and B are mutually exclusive because they

share no common outcomes. The occurrence of one event precludes the occurrence of the other. Symbolically, A B = Sec 2-1.3 Events 12 Copyright 2014 John Wiley & Sons, Inc. All rights reserved. Mutually Exclusive Events - Laws Commutative law (event order is unimportant): A B = B A and A B = B A Distributive law (like in algebra): (A B) C = (A C) (B C) (A B) C = (A C) (B C) Associative law (like in algebra): (A B) C = A (B C) (A B) C = A (B C)

Sec 2-1.3 Events 13 Copyright 2014 John Wiley & Sons, Inc. All rights reserved. Mutually Exclusive Events - Laws DeMorgans law: (A B) = A B The complement of the union is the intersection of the complements. (A B) = A B The complement of the intersection is the union of the complements. Complement law: (A) = A. Sec 2-1.3 Events 14 Copyright 2014 John Wiley & Sons, Inc. All rights reserved. Counting Techniques

There are three special rules, or counting techniques, used to determine the number of outcomes in events. They are : 1. Multiplication rule 2. Permutation rule 3. Combination rule Each has its special purpose that must be applied properly the right tool for the right job. Sec 2-1.4 Counting Techniques Copyright 2014 John Wiley & Sons, Inc. All rights reserved. 15 Counting Multiplication Rule Multiplication rule: Let an operation consist of k steps and there are n1 ways of completing step 1, n2 ways of completing step 2, and

nk ways of completing step k. Then, the total number of ways to perform k steps is: n1 n2 nk Sec 2-1.4 Counting Techniques Copyright 2014 John Wiley & Sons, Inc. All rights reserved. 16 Example 2-5 - Web Site Design In the design for a website, we can choose to use among: 4 colors, 3 fonts, and 3 positions for an image. How many designs are possible? Answer via the multiplication rule: 4 3 3 = 36 Sec 2-1.4 Counting Techniques

Copyright 2014 John Wiley & Sons, Inc. All rights reserved. 17 Counting Permutation Rule A permutation is a unique sequence of distinct items. If S = {a, b, c}, then there are 6 permutations Namely: abc, acb, bac, bca, cab, cba (order matters) Number of permutations for a set of n items is n! n! = n(n-1)(n-2)21 7! = 7654321 = 5,040 = FACT(7) in Excel By definition: 0! = 1 Sec 2-1.4 Counting Techniques

Copyright 2014 John Wiley & Sons, Inc. All rights reserved. 18 CountingSubset Permutations and an example For a sequence of r items from a set of n items: n! P n(n 1)(n 2)...(n r 1) (n r )! n r Example 2-6: Printed Circuit Board 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 designs are possible? Answer: Order is important, so use the permutation formula with n = 8, r = 4. 8!

8 7 6 5 4! 8 P4 8 7 6 5 1,680 (8 4)! 4! Sec 2-1.4 Counting Techniques Copyright 2014 John Wiley & Sons, Inc. All rights reserved. 19 Counting - Similar Item Permutations Used for counting the sequences when some items are identical. The number of permutations of: n = n1 + n2 + + nr items of which n1, n2, ., nr are identical. is calculated as:

n! n1 ! n2 ! ... nr ! Sec 2-1.4 Counting Techniques Copyright 2014 John Wiley & Sons, Inc. All rights reserved. 20 Example 2-7: Hospital Schedule In a hospital, a operating room needs to schedule three knee surgeries and two hip surgeries in a day. The knee surgery is denoted as k and the hip as h. How many sequences are there? Since there are 2 identical hip surgeries and 3 identical knee surgeries, then 5! 5 4 3! 10 2!3! 2 1 3!

What is the set of sequences? {kkkhh, kkhkh, kkhhk, khkkh, khkhk, khhkk, hkkkh, hkkhk, hkhkk, hhkkk} Sec 2-1.4 Counting Techniques Copyright 2014 John Wiley & Sons, Inc. All rights reserved. 21 Counting Combination Rule A combination is a selection of r items from a set of n where order does not matter. If S = {a, b, c}, n =3, then If r = 3, there is 1 combination, namely: abc If r = 2, there are 3 combinations, namely ab, ac, and bc # of permutations # of combinations Since order does not matter with combinations, we are dividing the # of permutations by r!, where r! is the # of arrangements of r elements. n

n! C r r !(n r )! n r Sec 2-1.4 Counting Techniques Copyright 2014 John Wiley & Sons, Inc. All rights reserved. 22 Example 2-8: Sampling w/o Replacement-1 A bin of 50 parts contains 3 defectives and 47 non-defective parts. A sample of 6 parts is selected from the 50 without replacement. How many samples of size 6 contain 2 defective parts? First, how many ways are there for selecting 2 parts from the 3 defective parts? 3!

C23 In Excel: 2!1! 3 different ways 3 = COMBIN(3,2) Sec 2-1.4 Counting Techniques Copyright 2014 John Wiley & Sons, Inc. All rights reserved. 23 Example 2-8: Sampling w/o Replacement-2 Now, how many ways are there for selecting 4 parts from the 47 non-defective parts? C 47

4 47! 47 46 45 44 43! 178,365 different ways 4!43! 4 3 2 143! In Excel: 178,365 = COMBIN(47,4) Sec 2-1.4 Counting Techniques Copyright 2014 John Wiley & Sons, Inc. All rights reserved. 24 Example 2-8: Sampling w/o Replacement-3 Now, how many ways are there to obtain: 2 from 3 defectives, and 4 from 47 non-defectives?

C23C447 3 178,365 535,095 different ways In Excel: 535,095 = COMBIN(3,2)*COMBIN(47,4) Sec 2-1.4 Counting Techniques Copyright 2014 John Wiley & Sons, Inc. All rights reserved. 25 Probability Probability is the likelihood or chance that a particular outcome or event from a random experiment will occur. In this chapter, we consider only discrete (finite or countably infinite) sample spaces. Probability is a number in the [0,1] interval. A probability of: 1 means certainty 0 means impossibility

Sec 2-2 Interpretations & Axioms of Probability Copyright 2014 John Wiley & Sons, Inc. All rights reserved. 26 Types of Probability Subjective probability is a degree of belief. Example: There is a 50% chance that Ill study tonight. Relative frequency probability is based on how often an event occurs over a very large sample space. n( A) lim n n Example: Sec 2-2 Interpretations & Axioms of Probability Copyright 2014 John Wiley & Sons, Inc. All rights reserved. 27

Probability Based on Equally-Likely Outcomes Whenever a sample space consists of N possible outcomes that are equally likely, the probability of each outcome is 1/N. Example: In a batch of 100 diodes, 1 is laser diode. A diode is randomly selected from the batch. Random means each diode has an equal chance of being selected. The probability of choosing the laser diode is 1/100 or 0.01, because each outcome in the sample space is equally likely. Sec 2-2 Interpretations & Axioms of Probabilities Copyright 2014 John Wiley & Sons, Inc. All rights reserved. 28 Probability of an Event For a discrete sample space, the probability of an event E, denoted by P(E), equals the sum of the probabilities of the outcomes in E.

The discrete sample space may be: A finite set of outcomes A countably infinite set of outcomes. Sec 2-2 Interpretations & Axioms of Probability Copyright 2014 John Wiley & Sons, Inc. All rights reserved. 29 Example 2-9: Probabilities of Events A random experiment has a sample space {a,b,c,d}. These outcomes are not equally-likely; their probabilities are: 0.1, 0.3, 0.5, 0.1. Let Event A = {a,b}, B = {b,c,d}, and C = {d} P(A) = 0.1 + 0.3 = 0.4 P(B) = 0.3 + 0.5 + 0.1 = 0.9 P(C) = 0.1 P(A ) = 0.6 and P(B ) = 0.1 and P(C ) = 0.9 Since event AB = {b}, then P(AB) = 0.3 Since event AB = {a,b,c,d}, then P(AB) = 1.0 Since event AC = {null}, then P(AC ) = 0

Sec 2-2 Interpretations & Axioms of Probability Copyright 2014 John Wiley & Sons, Inc. All rights reserved. 30 Axioms of Probability Probability is a number that is assigned to each member of a collection of events from a random experiment that satisfies the following properties: If S is the sample space and E is any event in the random experiment, 1. P(S) = 1 2. 0 P(E) 1 3. For any two events E1 and E2 with E1E2 = , P(E1E2) = P(E1) + P(E2) The axioms imply that: P() =0 and P(E ) = 1 P(E) If E1 is contained in E2, then P(E1) P(E2). Sec 2-2 Interpretations & Axioms of Probability Copyright 2014 John Wiley & Sons, Inc. All rights reserved.

31 Addition Rules Joint events are generated by applying basic set operations to individual events, specifically: Unions of events, A B Intersections of events, A B Complements of events, A Probabilities of joint events can often be determined from the probabilities of the individual events that comprise them. Sec 2-3 Addition Rules 32 Copyright 2014 John Wiley & Sons, Inc. All rights reserved. Example 2-10: Semiconductor Wafers A wafer is randomly selected from a batch that is classified by

contamination and location. Let H be the event of high concentrations of contaminants. Then P(H) = 358/940. Let C be the event of the wafer being located at the center of a sputtering tool. Then P(C) = 626/940. Location of Tool P(HC) = 112/940 Contamination Total Low High Total Center 514 112 626 Edge 68 246

314 582 358 940 P(HC) = P(H) + P(C) P(HC) = (358 + 626 112)/940 This is the addition rule. Sec 2-3 Addition Rules 33 Copyright 2014 John Wiley & Sons, Inc. All rights reserved. Probability of a Union For any two events A and B, the probability of union is given by: P( A B) P( A) P ( B ) P( A B ) If events A and B are mutually exclusive, then

P ( A B) , and therefore: P ( A B) P ( A) P ( B ) Sec 2-3 Addition Rules 34 Copyright 2014 John Wiley & Sons, Inc. All rights reserved. Addition Rule: 3 or More Events P ( A B C ) P ( A) P ( B) P (C ) P ( A B ) P ( A C ) P ( B C ) P ( A B C ) Note the alternating signs. If a collection of events Ei are pairwise mutually exclusive; that is Ei E j , for all i, j k Then : P ( E1 E 2 ... Ek ) P ( Ei ) i 1

Sec 2-3 Addition Rules 35 Copyright 2014 John Wiley & Sons, Inc. All rights reserved. Conditional Probability P(B | A) is the probability of event B occurring, given that event A has already occurred. A communications channel has an error rate of 1 per 1000 bits transmitted. Errors are rare, but do tend to occur in bursts. If a bit is in error, the probability that the next bit is also in error is greater than 1/1000. Sec 2-4 Conditional Probability Copyright 2014 John Wiley & Sons, Inc. All rights reserved. 36 Conditional Probability Rule

The conditional probability of an event B given an event A, denoted as P(B | A), is: P(B | A) = P(AB) / P(A) for P(A) > 0. From a relative frequency perspective of n equally likely outcomes: P(A) = (number of outcomes in A) / n P(AB) = (number of outcomes in AB) / n P(B | A) = number of outcomes in AB / number of outcomes in A Sec 2-4 Conditional Probability Copyright 2014 John Wiley & Sons, Inc. All rights reserved. 37 Example 2-11 There are 4 probabilities conditioned on flaws in the below table. Parts Classified Surface Flaws Yes (F ) No (F' )

Yes (D ) 10 18 No (D' ) 30 342 Total 40 360 Defective Total 28 372 400 P ( F ) 40 400 and P( D) 28 400 10 P ( D | F ) P( D F ) P ( F ) 400

10 40 40 400 30 P D ' | F P D ' F P F 400 40 400 18 P D | F ' P D F ' P F ' 400 P D ' | F ' P D ' F ' P F ' 342 400 30 40

360 400 18 360 360 400 342 360 38 Sec 2-4 Conditional Probability Copyright 2014 John Wiley & Sons, Inc. All rights reserved. Random Samples Random means each item is equally likely to be chosen. If more than one item is sampled, random means that every sampling outcome is equally likely.

2 items are taken from S = {a,b,c} without replacement. Ordered sample space: S = {ab,ac,bc,ba,ca,cb} Unordered sample space: S = {ab,ac,bc} Sec 2-4 Conditional Probability Copyright 2014 John Wiley & Sons, Inc. All rights reserved. 39 Example 2-12 : Sampling Without Enumeration A batch of 50 parts contains 10 made by Tool 1 and 40 made by Tool 2. If 2 parts are selected randomly*, a) What is the probability that the 2nd part came from Tool 2, given that the 1st part came from Tool 1? P(E1)= P(1st part came from Tool 1) = 10/50 P(E2 | E1) = P(2nd part came from Tool 2 given that 1st part came from Tool 1) = 40/49

b) What is the probability that the 1st part came from Tool 1 and the 2nd part came from Tool 2? P(E1E2) = P(1st part came from Tool 1 and 2nd part came from Tool 2) = (10/50)(40/49) = 8/49 *Selected randomly implies that at each step of the sample, the items remain in the batch are equally likely to be selected. Sec 2-4 Conditional Probability Copyright 2014 John Wiley & Sons, Inc. All rights reserved. 40 Multiplication Rule The conditional probability can be rewritten to generalize a multiplication rule. P(AB) = P(B|A)P(A) = P(A|B)P(B) The last expression is obtained by exchanging the roles of A and B.

Sec 2-5 Multiplication & Total Probability Rules Copyright 2014 John Wiley & Sons, Inc. All rights reserved. 41 Example 2-13: Machining Stages The probability that a part made in the 1st stage of a machining operation meets specifications is 0.90. The probability that it meets specifications in the 2nd stage, given that met specifications in the first stage is 0.95. What is the probability that both stages meet specifications? Let A and B denote the events that the part has met1st and 2nd stage specifications, respectively. P(AB) = P(B | A)P(A) = 0.950.90 = 0.855 Sec 2-5 Multiplication & Total Probability Rules Copyright 2014 John Wiley & Sons, Inc. All rights reserved. 42

Two Mutually Exclusive Subsets A and A are mutually exclusive. AB and A B are mutually exclusive B = (AB) (A B) Total Probability Rule For any two events A and B Sec 2-5 Multiplication & Total Probability Rules Copyright 2014 John Wiley & Sons, Inc. All rights reserved. 43 Example 2-14: Semiconductor Contamination Information about product failure based on chip manufacturing process contamination is given below. Find the probability of failure. Probability of Failure 0.1

0.005 Level of Probability Contamination of Level High Not High 0.2 0.8 Let F denote the event that the product fails. Let H denote the event that the chip is exposed to high contamination during manufacture. Then P(F | H) = 0.100 and P(H) = 0.2, so P(F H) = 0.02 P(F | H ) = 0.005 and P(H ) = 0.8, so P(F H ) = 0.004 P(F) = P(F H) + P(F H ) (Using Total Probability rule) = 0.020 + 0.004 = 0.024 Sec 2-5 Multiplication & Total Probability Rules Copyright 2014 John Wiley & Sons, Inc. All rights reserved.

44 Total Probability Rule (Multiple Events) A collection of sets E1, E2, Ek such that E1 E2 Ek = S is said to be exhaustive. Assume E1, E2, Ek are k mutually exclusive and exhaustive. ThenP( B) P( B E ) P( B E ) ... P( B E ) 1 2 k P( B | E1 ) P( E1 ) P( B | E2 ) P( E2 ) ... P( B | Ek ) P( Ek ) Sec 2-5 Multiplication & Total Probability Rules Copyright 2014 John Wiley & Sons, Inc. All rights reserved. 45

Example 2-15: Semiconductor Failures-1 Continuing the discussion of contamination during chip manufacture, find the probability of failure. Probability of Failure 0.100 0.010 0.001 Level of Contamination High Medium Low Sec 2-5 Multiplication & Total Probability Rules Copyright 2014 John Wiley & Sons, Inc. All rights reserved.

Probability of Level 0.2 0.3 0.5 46 Example 2-15: Semiconductor Failures-2 Let F denote the event that a chip fails Let H denote the event that a chip is exposed to high levels of contamination Let M denote the event that a chip is exposed to medium levels of contamination Let L denote the event that a chip is exposed to low levels of contamination. Using Total Probability Rule, P(F) = P(F | H)P(H) + P(F | M)P(M) + P(F | L)P(L) = (0.1)(0.2) + (0.01)(0.3) + (0.001)(0.5) = 0.0235 Sec 2-5 Multiplication & Total Probability Rules

Copyright 2014 John Wiley & Sons, Inc. All rights reserved. 47 Event Independence Two events are independent if any one of the following equivalent statements is true: 1. P(A | B) = P(A) 2. P(B | A) = P(B) 3. P(AB) = P(A)P(B) This means that occurrence of one event has no impact on the probability of occurrence of the other event. Sec 2-6 Independence 48 Copyright 2014 John Wiley & Sons, Inc. All rights reserved. Example 2-16: Flaws and Functions

Table 1 provides an example of 400 parts classified by surface flaws and as (functionally) defective. Suppose that the situation is different and follows Table 2. Let F denote the event that the part has surface flaws. Let D denote the event that the part is defective. The data shows whether the events are independent. TABLE 1 Parts Classified TABLE 2 Parts Classified (data chg'd) Surface Flaws Surface Flaws Defective Yes (F ) No (F' ) Total Defective Yes (F ) No (F' ) Total Yes (D ) 10 18 28 Yes (D ) 2 18 20 No (D' ) 30

342 372 No (D' ) 38 342 380 Total 40 360 400 Total 40 360 400 P (D |F ) = 10/40 = 0.25 P (D ) = 28/400 = 0.10 not same Events D & F are dependent P (D |F ) = 2/40 = 0.05 P (D ) = 20/400 = 0.05

same Events D & F are independent Sec 2-6 Independence 49 Copyright 2014 John Wiley & Sons, Inc. All rights reserved. Independence with Multiple Events The events E1, E2, , Ek are independent if and only if, for any subset of these events: P(Ei1Ei2 , Eik) = P(Ei1)P(Ei2)P(Eik) Sec 2-6 Independence 50 Copyright 2014 John Wiley & Sons, Inc. All rights reserved. Example 2-17: Semiconductor Wafers Assume the probability that a wafer contains a large particle of contamination is 0.01

and that the wafers are independent; that is, the probability that a wafer contains a large particle does not depend on the characteristics of any of the other wafers. If 15 wafers are analyzed, what is the probability that no large particles are found? Solution: Let Ei denote the event that the ith wafer contains no large particles, i = 1, 2, ,15. Then , P(Ei) = 0.99. The required probability is P(E1E2 E15). From the assumption of independence, P(E1E2 E15) = P(E1)P(E2)P(E15) = (0.99)15 = 0.86. Sec 2-6 Independence 51 Copyright 2014 John Wiley & Sons, Inc. All rights reserved. Bayes Theorem Thomas Bayes (1702-1761) was an English mathematician and Presbyterian minister.

His idea was that we observe conditional probabilities through prior information. Bayes theorem states that, Sec 2-7 Bayes Theorem 52 Copyright 2014 John Wiley & Sons, Inc. All rights reserved. Example 2-18 The conditional probability that a high level of contamination was present when a failure occurred is to be determined. The information from Example 2-14 is summarized here. Probability of Failure Level of Contamination Probability of Level

0.1 0.005 High Not High 0.2 0.8 Solution: Let F denote the event that the product fails, and let H denote the event that the chip is exposed to high levels of contamination. The requested probability is P(F). P( H | F ) P ( F | H ) P ( H ) 0.10 0.20 0.83 P( F ) 0.024

P ( F ) P ( F | H ) P ( H ) P ( F | H ' ) P ( H ' ) 0.1 0.2 0.005 0.8 0.024 Sec 2-7 Bayes Theorem 53 Copyright 2014 John Wiley & Sons, Inc. All rights reserved. Bayes Theorem with Total Probability If E1, E2, Ek are k mutually exclusive and exhaustive events and B is any event, P B | E1 P E1 P E1 | B P B | E1 P E1 P B | E2 P E2 ... P B | Ek P Ek where P(B) > 0 Note : Numerator expression is always one of the terms in the sum of the denominator. Sec 2-7 Bayes Theorem

54 Copyright 2014 John Wiley & Sons, Inc. All rights reserved. Example 2-19: Bayesian Network A printer manufacturer obtained the following three types of printer failure probabilities. Hardware P(H) = 0.3, software P(S) = 0.6, and other P(O) = 0.1. Also, P(F | H) = 0.9, P(F | S) = 0.2, and P(F | O) = 0.5. If a failure occurs, determine if its most likely due to hardware, software, or other. P ( F ) P ( F | H ) P ( H ) P ( F | S ) P ( S ) P ( F | O ) P (O ) 0.9(0.1) 0.2(0.6) 0.5(0.3) 0.36 P ( F | H ) P ( H ) 0.9 0.1 P( H | F ) 0.250 P( F ) 0.36 P ( F | S ) P ( S ) 0.2 0.6 P(S | F )

0.333 P( F ) 0.36 P( F | O ) P (O ) 0.5 0.3 P (O | F ) 0.417 P( F ) 0.36 Note that the conditionals given failure add to 1. Because P(O | F) is largest, the most likely cause of the problem is in the other category. Sec 2-7 Bayes Theorem 55 Copyright 2014 John Wiley & Sons, Inc. All rights reserved. Random Variable and its Notation A variable that associates a number with the outcome of a random experiment is called a random variable. A random variable is a function that assigns a real number to each outcome in the sample space of a

random experiment. A random variable is denoted by an uppercase letter such as X. After the experiment is conducted, the measured value of the random variable is denoted by a lowercase letter such as x = 70 milliamperes. X and x are shown in italics, e.g., P(X = x). Sec 2-8 Random Variables 56 Copyright 2014 John Wiley & Sons, Inc. All rights reserved. Discrete & Continuous Random Variables A discrete random variable is a random variable with a finite or countably infinite range. Its values are obtained by counting. A continuous random variable is a random variable with an interval (either finite or infinite) of real numbers for its range. Its values are obtained by measuring.

Sec 2-8 Random Variables 57 Copyright 2014 John Wiley & Sons, Inc. All rights reserved. Examples of Discrete & Continuous Random Variables Discrete random variables: Number of scratches on a surface. Proportion of defective parts among 100 tested. Number of transmitted bits received in error. Number of common stock shares traded per day. Continuous random variables: Electrical current and voltage. Physical measurements, e.g., length, weight, time, temperature, pressure. Sec 2-8 Random Variables 58 Copyright 2014 John Wiley & Sons, Inc. All rights reserved.

Important Terms & Concepts of Chapter 2 Addition rule Random variable Axioms of probability Discrete Bayes theorem Continuous Combination Sample space Conditional probability Discrete

Equally likely outcomes Continuous Event Total probability rule Independence Tree diagram Multiplication rule Venn diagram Chapter 2 Summary Mutually exclusive events With

replacement Copyright 2014 John Wiley & Sons, Inc. All rights reserved. 59