Chapter 2 Probability LEARNING OBJECTIVES Understand describe sample

Chapter 2 Probability

LEARNING OBJECTIVES • Understand describe sample spaces and events • Interpret probabilities and use probabilities of outcomes to calculate probabilities of events in discrete sample spaces • Calculate the probabilities of joint events such as unions and intersections from the probabilities of individual events • Interpret and calculate conditional probabilities of events • Determine the independence of events and use independence to calculate probabilities • Use Bayes’ theorem to calculate conditional

Sample Space and Events • An experiment is any action or process that generates observations • The sample space of an experiment, denoted S, is the set of all possible outcomes, or sample points. • Example: Toss a fair coin 3 times in a row – The sample space has 8 sample points. – S={HHH, THH, HHT, THT, HTH, TTH, HTT, TTT} • An event is a subset of the sample space S • Example: look at 3 different events of previous example – The event of 3 heads, • A= {HHH} – The event of 2 heads, • B={HHT, HTH, THH} – The event that the last toss is a head, • C={HHH, HTH, THH, TTH}

Set Relations • Suppose S is the universal set, with two subsets, A and B • A set, A, is a subset of B if all elements of A belong to B, A⊂B S B • The union of two events A and B, denoted by A∪B, and read “A or B, ” is the event consisting of all elements that are either in A, in B, or in both • Or the union A∪B = { x | x ∈ A or x ∈ B} A S A B

Set Relations-Cont. • The intersection of two • events A and B, denoted by A∩B, and read “A and B, ” is the event consisting of all elements that are in both • The intersection A∩B={x | x ∈A and x ∈B} is the subset of S which contains all elements that are in both A & B S A B The complement of an event A, denoted by A´, is the set of all elements in S that are not contained in A S A´ A

Set Relations-Cont. • Sets A and B are mutually exclusive or disjoint, if and only if A∩B = Ø, or events A & B have no elements in common S A B • Any number of sets, A 1, A 2, A 3, … are mutually exclusive if and only if Ai∩Aj = Ø for i≠j. • A 1 ∩A 2 = Ø A 2 ∩ A 3 = Ø • A 1 ∩ A 3= Ø A 2 ∩ A 4 = Ø • A 1 ∩ A 4 = Ø A 3 ∩ A 4 = Ø S A 1 A 2 A 3 A 4

Example: 1 • For an experiment, let – A = {0, 1, 2, 3, 4}, – B = {3, 4, 5, 6}, and – C = {1, 3, 5} • • Determine: A B A C A´ {A C}´ • • Solution: A B={0, 1, 2, 3, 4, 5, 6} A C={0, 1, 2, 3, 4, 5} A B={3, 4} A C={1, 3} A´={5, 6} {A C}´={6}

Example: 2 • The rise time of a reactor is measured in minutes (and fractions of times). Let the sample space positive, real numbers. Define the events A and B as follows: A={x | x<72. 5} and B={x | x>52. 5} • Describe each of the following events. • a) A´ • b) B´ • c) A B • d) A B • • Solution: a) A = {x | x 72. 5} b) B = {x | x 52. 5} c) A B = {x | 52. 5 < x < 72. 5} • d) A B = {x | x > 0}

Example: 3 • In an injection-molding operation, several characteristics of each • Solution molded part are evaluated • Let A denote the event that a part meets customer shrinkage requirements, B denote the event that a part meets customer color requirements, and C denote the event that a critical length meets customer requirements • a) Construct a Venn diagram that includes these events and indicate the region in the diagram in which a part meets all customer requirements. Shade the areas that represent the following • b) B C • c) A´ B • d) A B

Class Problem • Disks of polycarbonate plastic from a supplier are analyzed for scratch resistance and shock resistance. The results from 100 disks are summarized below. Shock Resistance High low Scratch High 70 9 Resistance low 16 5 • Let A denote the event that a disk has high shock resistance, and let B denote the event that a disk has high scratch resistance. Determine the number of disks in A∩B, A´, and A∪B • Solution: Number of samples in A∩B=… Number of samples in A' = … Number of samples in A∪B =…

Interpreting Probabilities • The assignment of a • Example: Toss a fair coin weight between 0 and 1 to three times in a row indicate the likelihood of – The probability of getting 3 heads P(A)= 1/8 the occurrence of an event. – The probability of getting 2 • The probability of an heads P(B)=3/8 event is defined in terms – The probability that the last of an experiment and a toss is a head P(C) = 4/8 sample space. =1/2

Axioms of Probability 1. For any event A, P (A) 0 • Example • The chance of occurring • If an experiment has the three should be at least 0 possible and mutually 2. P(S)=1 exclusive outcomes A, B, and • The maximum possible C, check in each case whether probability is assigned to S the assignment of probabilities 3. Let A 1, A 2, A 3, …, An, … be is permissible: a finite or infinite sequence • P(A) = 1/3, P(B)= 1/3, and of mutually exclusive events. P(C) = 1/3 Then • P(A) = 0. 64, P(B)= 0. 38, and P(A 1∪A 2∪A 3…) = P (A 1) + P(C) = -. 02 P(A 2) + P(A 3)+…=∑ P(Ai) • P(A) =0. 35, P(B)=0. 52, and P(C) =0. 26 • P(A) =0. 57, P(B)=0. 24, and P(C) =0. 19

Class Problem • The sample space of a random experiment is {a, b, c, d, e} with probabilities 0. 1, 0. 2, 0. 4, and 0. 2, respectively. Let A denote the event {a, b, c}, and let B denote the event {c, d, e}. Determine the following • a) P(A) • b) P(B) • c) P(A') • d) P(A∪B) • e) P(A∩B) • • • Solution: a) P(A) = b) P(B) = c) P(A') = d) P(A B) = e) P(A B) =

Rules of Probability Complement Rule • The probability of impossible events is 0: P(Ø) =0 • Complement rule: P(A´)= 1 - P(A) S A´ A • Proof • From axioms 3 for finite case, let k=2, A 1=A and A 2=A' • By definition, A∪A' =S while A and A' are mutually exclusive • 1=P(S)=P(A∪A')=P(A)+ P(A' ) • P(A´)= 1 - P(A)

Addition Rule • For any two events A and B P(A 1∪A 2)=P(A 1)+P(A 2) – P(A 1∩A 2) • By Venn diagram • The probability of a union of more than two events P(A 1∪A 2∪A 3)= P(A 1)+ P(A 2)+ P(A 3) +P(A 1∩A 2∩A 3) -P(A 1∩A 2) -P(A 2∩A 3) -P(A 1∩A 3)

Example: 1 • If P(A) =0. 3, P(B)=0. 2, and P(A B) =0. 1, determine the following probabilities • • • a) P(A´) b) P(A B) c) P(A´ B) d) P(A B´) e) P[(A B)´] f) P(A´ B) • Solution • a) P(A') = 1 - P(A) = 0. 7 • b) P(A B) = P(A) + P(B) - P(A B) = 0. 3+0. 2 - 0. 1 = 0. 4 • c) P(A´ B)+ P(A B) = P(B). Therefore, P(A´ B)= 0. 2 - 0. 1 = 0. 1 • d) P(A) = P(A B) + P(A B´) Therefore, P(A B´) = 0. 3 - 0. 1 = 0. 2 • e) P((A B) ') =1 - P(A B)= 1 - 0. 4 = 0. 6 • f) P(A´ B)= P(A') + P(B) - P(A´ B)= 0. 7 + 0. 2 - 0. 1 = 0. 8 from part c

Example: 2 • Denote the six events 1, 2, 3, 4, 5, and 6 associated with tossing a six-sided die once by E 1, E 2, E 3, E 4, E 5, and E 6 • Suppose the die is constructed so that any of the three even outcomes is twice as likely to occur as any of the three odd outcomes • Determine P(A) where the event A is even • Determine P(B) where the event B is less than or equal to 3 • • Solution: P(E 1)=P(E 3)=P(E 5)=… P(E 2)=P(E 4)=P(E 6)=… Define A={outcome is even}= E 2 E 4 E 6 • P(A)= P(E 2)+P(E 4)+P(E 6)=… • B={outcome ≤ 3}= E 1 E 2 E 3 • P(B)= P(E 1)+P(E 2)+P(E 3)=…

Class Problem • Disks of polycarbonate plastic from a supplier are analyzed for scratch resistance and shock resistance. The results from 100 disks are summarized below. Shock Resistance High low Scratch High 70 9 Resistance low 16 5 a) If a disk is selected at random, what is the probability that its scratch resistance is high and its shock resistance is high? b) b) If a disk is selected at random, what is the probability that its scratch resistance is high or its shock resistance is high? c) Consider the event that a disk has high scratch resistance and the event that a disk has high shock resistance. Are these two events mutually exclusive? • Solution • Let A denote the event that a sample has high shock resistance and let B denote the event that a sample has high scratch resistance. a) P(A B) = … b) P(A B) = P(A) + P(B) - P(A B) = … c) Because (A B) does not equal Ø , A and B…

Equally Likely Outcomes • In an experiment consisting of N outcomes, it is reasonable to assign equal probabilities to all N sample events • So, p=1/N • Example • When two dice are rolled separately, there are N=36 outcomes, which are equally likely or P(Ei)=1/36 • Let A={sum of two numbers=7} • P(A)=…

Counting Techniques • Ability to count number of • Example elements in the sample space – A homeowner requires without listing actually each two types of contractors, element plumbing and electrical • The Product Rule for – 3 plumbing contractors Ordered Pairs – 3 electrical contractors • If the first object of an ordered – How many possible ways pair can be selected in n 1 ways, of choosing the two types and for each of these n 1 ways of contractors? the second object of the pair • N= n 1 n 2 =3*3=9 can be selected in n 2 ways, then the number of pairs is n 1 n 2

Tree Diagrams • Used to represent pictorially all the possibilities • Starting on the left side of the diagram, for each possible first element of a pair a straight-line segment emanates rightward • Construct another line segment emanating from the tip of the branch for each possible choice of a second element of the pair • A more general Product Rule • N= n 1 n 2 n 3 … nk • Example P 1 P 2 P 3 E 1 E 2 E 3

Permutations • Any ordered sequence of k objects • Example taken from a set of n distinct objects – Consider the set {A, B, C, D, E} consisting of 5 elements is called a permutation of size k of – Number of permutations of the objects size 3? • The number of permutations of size – By taking 5 letters three at a k that can be formed from the n time objects is denoted by Pk, n – P 5, 3 = 5!/(5 -3)! = 60 • Obtained from the general product • Class Problem rule – Three awards will be given for • Pk, n=n(n-1)(n-2)…(n-k+2)(n-k+1) a class of 25 graduate students – If each student can receive at • Using factorial notation most one award – How many possible selections? • P 25, 3=…

Combinations • Any unordered sequence of k objects taken from a set of n distinct objects is called combination of size k of the objects • The number of combinations of size k that can be formed from n distinct objects will be denoted by • Smaller than the number of permutations because the order is disregarded • Example: – Consider the set {A, B, C, D, E} consisting of 5 elements – Number of combinations of size 3? – There are six permutations of size 3 consisting of the elements A, B, and C (3!=6) – These six permutations are equivalent to the single combination {A, B, C} – So, 60/3! =10 • In general

Class Problem • Car dealer service 10 foreign cars and 15 domestic cars on a particular day • There are only 6 mechanics • If 6 cars are chosen at random, what is the probability that exactly 3 of the cars are domestic and the other cars are foreign? • What is the probability that at least 3 of the cars are domestic and the other cars are foreign? • Solution • Let D 3={exactly 3 of the 6 cars chosen are domestic} • P(D 3)=… • P(D 3 D 4 D 5 D 6) =…

Conditional Probability • Interested at the probability • of an event occurring conditional on the knowledge that another • event has occurred • Let A and B be events with P(A) 0 • The conditional probability of B given A is P(B|A)= P(A∩B)/P(A) • Note: P(B|A) is undefined if P(A) =0. A B Simple example concerns the situation in which two events are mutually exclusive P(B|A)= P(A∩B)/P(A)=0/P(B)=0 A B

Examples • If the probability that a research project will be planned is 0. 80 and the probability that it will be planned and well executed is 0. 72, what is the probability that a research project, which is well planned, will also be well executed? • Solution • P(A/B) = P(A∩B)/P(B)=0. 72/0. 80=0. 90 • If the probability that a communication system will have high fidelity is 0. 81, and the probability that it will have high fidelity and high selectivity is 0. 18, what is the probability that a system with high fidelity will also have high selectivity? – A: A communication system which has high selectivity – B: A communication system which has high fidelity • Data: P(B)=0. 81, P(A∩B)=0. 18 • P(A/B)=P(A∩B)/P(B)= 0. 18/0. 81=2/9

Class Problem • A magazine publishes three columns entitled A, B, and C • Reading habits of a selected readers with respect to these columns Read regularly. A B C A∩B A∩C B∩C A∩B∩C Probability 0. 14 0. 23 0. 37 0. 08 0. 09 0. 13 0. 05 • P(A|B)=… • P(A|B C)=… • P(A| reads at least one) = P(A|A B C)=… • P(A B |C)=…

Class Problem • Disks of polycarbonate plastic from a supplier are analyzed for scratch resistance and shock resistance. The results from 100 disks are summarized below. Shock Resistance High low Scratch High 70 9 Resistance low 16 5 • Let A denote the event that a disk has high shock resistance, and let B denote the event that a disk has high scratch resistance. Determine the following probabilities. a) P(A) b) P(B) c) P(A/B) d) P(B/A) • a) b) c) d) Solution P(A) =… P(B) = … P(A/B)=… P(B/A)=…

Multiplication Rule • From the definition of conditional probability, we can write P(A∩B)= P(A) P(B/A) • Example: The supervisor of a construction group (20 workers) wants to get the opinion of 2 of them about new safety regulations. If 12 of them favor the new regulations and the other 8 are against it, what is the probability that both of them chosen by the supervisor will be against the new safety regulation? • Solution • P(A)=the first worker selected will be against the new safety regulation=8/20 • P(B/A)=the second worker selected will be against given that the first one is against=7/19 • P(A∩B)= P(A) P(B/A)= – (8/20)(7/19)=14/95

Independent Events • Two events A and B are independent if and only if P(B/A)=P(B) if P(A) 0 and P(A/B)=P(A) if P(B) 0 • P(B/A)=P(B) means the probability of event B remains the same whether or not event A is conditional upon • One event does not affect the probability of the another event • P(A∩B)=P(A)P(B/A)= P(A)P(B) • Example 1: – What is the probability of getting two heads in two flips of a balanced coin? • Solution – (1/2)=1/4 • Example 2: – If P(C)=0. 65, P(D)=0. 4, and P(C∩D)=0. 24, are the events C and D independent? • Solution – P(C)P(D)=(0. 65)(0. 4)=0. 26 0. 24

Class Problem • Disks of polycarbonate plastic from a supplier are analyzed for scratch resistance and shock resistance. The results from 100 disks are summarized below. Shock Resistance High low Scratch High 70 9 Resistance low 16 5 • Let A denote the event that a disk has high shock resistance, and let B denote the event that a disk has scratch resistance. Are events A and B independent? • Solution: – P(A∩B)= = … – P(A) = … – P(B) = 79/100

Total Probability Rule • Let A 1, A 2, A 3, …, An be a collection of mutually exclusive events with known probabilities which partition S • Consider an event B with known conditional probabilities A 1 A 2 An B • How to use P(Ai) and P(B/Ai) to calculate P(B) • Easily achieved by B=(A 1∩B)∪… ∪(An∩B) • Events Ai∩B are mutually exclusive P(B)=P(A 1∩B)+… P(An∩B) =P(A 1)P(B/A 1)+…+P(An)P(B/An) • Known as the law of total probability • Example – Suppose that P(A/B)=0. 2, P(A/B')=0. 3, and P(B)=0. 8. What is P(A)? • Solution

Example • In a certain assembly plant, • Let three machines, B 1, B 2, and – A: the product is defective B 3, make 30%, 45%, and – B 1=the product is made by machine B 1 25%, respectively, of the – B 2=the product is made by machine B 2 products – B 3=the product is made by machine B 3 • It is known from past • P(A)=P(B 1)P(A/B 1)+…+P(Bn)P(A/Bn) experience that 2%, 3%, and – P(B 1)P(A/B 1)=(0. 3)(0. 02)=0. 006 2% of the products made by – P(B 2)P(A/B 2)=(0. 45)(0. 03)=0. 0135 each machine, respectively, – P(B 3)P(A/B 3)=(0. 25)(0. 02)=0. 005 are defective • P(A)=0. 006+0. 0135+0. 005=0. 0245 • Select a finished product randomly • What is the probability that it is defective?

Posterior Probabilities • How to use the probabilities P(Ai) and P(B/Ai) to calculate the probabilities P(Ai/B) • This is the revised probabilities of the events Ai conditional on the event B • Assume P(A 1), …, P(An) are the prior probabilities of events A 1, …, An • Observation of the event B provides some additional information to revise these prior probabilities • Called posterior probabilities and conditional on the event B • From the law of total probability • Known as Bayes’ Theorem

Example • In a certain assembly plant, three machines, B 1, B 2, and B 3, make 30%, 45%, and 25%, respectively, of the products • It is known from past experience that 2%, 3%, and 2% of the products made by each machine, respectively, are defective • Suppose a product is selected randomly and it is defective • What is the probability made by machine Bi (say B 3)? • Let – – A: the product is defective B 1=the product is made by machine B 1 B 2=the product is made by machine B 2 B 3=the product is made by machine B 3 • Instead of asking P(A), we want to find P(Bi/A) • Using the Bayes’ formula • P(B 3/A)=0. 005/0. 0245 = 10/49

Prior and Posterior Probabilities • Prior probabilities • – It is known from past experience that 2%, 3%, and 2% of the products made by each machine, respectively, are defective Past data – B 1=the product is made by machine B 1 – B 2=the product is made by machine B 2 – B 3=the product is made by machine B 3 • P(B 1)=0. 02 • P(B 2)=0. 03 • P(B 3)=0. 02 • Posterior probabilities – Revision of the prior probabilities conditional on the event A – A product is selected randomly and it is defective, what is the probability made by machine Bi? • P(B 1/A)=0. 006/0. 0245 = • P(B 2/A)=0. 0135/0. 0245 = • P(B 3/A)=0. 005/0. 0245 = 10/49

Class Problem • • A TV store sells 3 different brands of TVs Of its TV sales, 50% are brand 1, 30% are brand 2, and 20% are brand 3 • Known that 25% of brand 1 requires warranty repair work, whereas brand 2 and 3 are 20% and 10%, respectively 1. What is the probability that a selected buyer has a TV that will need repair while under warranty? 2. If a customer returns a TV brand, what is the probability that is a brand 1? A brand 2? A brand 3?

Solution

Tree Diagram 5 P(B/A 1) P(A 1)=P(B A 1)=0. 125 )=0. 2 A 1 / P(B 5 =0. 1 ) P(A 1 rand B P(A 2)=0. 3 P(A Brand 2 Bra 3 )=0. 3 nd 3 air Rep No repai r P(B'/A 1 )=0. 7 5 r Repai =0. 2 B/A 2) P( No re pair P(B'/ A 2 )=0. 8 Repair )=0. 1 A / 3 B ( P No re pair P(B'/ A 3 )= 0. 90 P(B/A 2) P(A 2)=P(B A 2)=0. 06 P(B/A 3) P(A 3)=P(B A 3)=0. 02 P(B)=P[(brand 1 and repair) or (brand 2 and repair) or (brand 3 and repair)] P(B)= P(B A 1)+ P(B A 2)+ P(B A 3)=0. 125+0. 06+0. 02 P(A 1/B)=… P(A 2/B)=… P(A 3/B)=… =0. 205

Class Problem • An assembly plant receives • If A denotes the event a its voltage regulators from voltage regulator received by three different suppliers, 60% the plant performs according to from supplier B 1, 30% from specifications and B 1, B 2, B 3 supplier B 2, and 10% from are the events that it comes supplier B 3. If 95% of the voltage regulators from B 1, from the respective suppliers, 80% of those from B 2, and then 65% of those from B 3 perform according to specifications, what is the probability that a particular voltage regulator which is known to perform according to specifications came from supplier B 3?