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Chapter 2 Probability

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**LEARNING OBJECTIVES Understand and 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 probabilities

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**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}

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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 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} S B A S

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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 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 S A

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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 Any number of sets, A1, A2, A3,… are mutually exclusive if and only if Ai∩Aj = Ø for i≠j. A1 ∩A2 = Ø A2 ∩ A3 = Ø A1 ∩ A3= Ø A2 ∩ A4 = Ø A1 ∩ A4 = Ø A3 ∩ A4 = Ø S S A1 A2 A3 A4

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**Example:1 For an experiment, let Determine: AB AC AB AC A´ {AC}´**

B = {3,4,5,6}, and C = {1,3,5} Determine: AB AC AB AC A´ {AC}´ Solution: AB={0,1,2,3,4,5,6} AC={0,1,2,3,4,5} AB={3,4} AC={1,3} A´={5,6} {AC}´={6}

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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) AB d) AB 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}

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Example:3 In an injection-molding operation, several characteristics of each 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) BC c) A´B d) AB Solution

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**Class Problem Solution:**

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 =…

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**Interpreting Probabilities**

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

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**Axioms of Probability For any event A, P (A) 0 P(S)=1**

The chance of occurring should be at least 0 P(S)=1 The maximum possible probability is assigned to S Let A1, A2, A3,…, An,… be a finite or infinite sequence of mutually exclusive events. Then P(A1∪A2∪A3…) = P (A1) + P(A2) + P(A3)+…=∑ P(Ai) Example If an experiment has the three possible and mutually exclusive outcomes A, B, and C, check in each case whether the assignment of probabilities is permissible: P(A) = 1/3, P(B)= 1/3, and P(C) = 1/3 P(A) = 0.64, P(B)= 0.38, and P(C) = -.02 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

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Class Problem The sample space of a random experiment is {a, b, c, d, e} with probabilities 0.1, 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(AB) = e) P(AB) =

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**Rules of Probability Complement Rule**

The probability of impossible events is 0: P(Ø) =0 Complement rule: P(A´)= 1- P(A) Proof From axioms 3 for finite case, let k=2, A1=A and A2=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) S A´ A

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**Addition Rule For any two events A and B**

P(A1∪A2)=P(A1)+P(A2) – P(A1∩A2) By Venn diagram The probability of a union of more than two events P(A1∪A2∪A3)= P(A1)+ P(A2)+ P(A3) +P(A1∩A2∩A3) -P(A1∩A2) -P(A2∩A3) -P(A1∩A3)

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Example: 1 If P(A) =0.3, P(B)=0.2, and P(AB) =0.1, determine the following probabilities a) P(A´) b) P(AB) c) P(A´B) d) P(AB´) e) P[(AB)´] f) P(A´B) Solution a) P(A') = 1- P(A) = 0.7 b) P(AB) = P(A) + P(B) - P(AB) = = 0.4 c) P(A´B)+ P(AB) = P(B). Therefore, P(A´B)= = 0.1 d) P(A) = P(AB) + P(AB´) Therefore, P(AB´) = = 0.2 e) P((AB) ') =1 - P(AB)= = 0.6 f) P(A´B)= P(A') + P(B) - P(A´ B)= = 0.8 from part c

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Example:2 Denote the six events 1,2,3,4,5, and 6 associated with tossing a six-sided die once by E1, E2, E3, E4, E5, and E6 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(E1)=P(E3)=P(E5)=… P(E2)=P(E4)=P(E6)=… Define A={outcome is even}= E2 E4 E6 P(A)= P(E2)+P(E4)+P(E6)=… B={outcome ≤3}= E1E2E3 P(B)= P(E1)+P(E2)+P(E3)=…

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**Class Problem Solution**

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 If a disk is selected at random, what is the probability that its scratch resistance is high and its shock resistance is high? b) If a disk is selected at random, what is the probability that its scratch resistance is high or its shock resistance is high? 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(AB) = … b) P(AB) = P(A) + P(B) - P(AB) = … c) Because (AB) does not equal Ø , A and B…

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**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)=…

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**Counting Techniques Example N= n1 n2 =3*3=9**

Ability to count number of elements in the sample space without listing actually each element The Product Rule for Ordered Pairs If the first object of an ordered pair can be selected in n1 ways, and for each of these n1 ways the second object of the pair can be selected in n2 ways, then the number of pairs is n1 n2 Example A homeowner requires two types of contractors, plumbing and electrical 3 plumbing contractors 3 electrical contractors How many possible ways of choosing the two types of contractors? N= n1 n2 =3*3=9

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**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= n1 n2 n3 … nk Example E1 E2 E3 P1 E1 P2 E2 P3 E3 E1 E2 E3

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Permutations Any ordered sequence of k objects taken from a set of n distinct objects is called a permutation of size k of the objects The number of permutations of size k that can be formed from the n objects is denoted by Pk,n Obtained from the general product rule Pk,n=n(n-1)(n-2)…(n-k+2)(n-k+1) Using factorial notation Example Consider the set {A,B,C,D,E} consisting of 5 elements Number of permutations of size 3? By taking 5 letters three at a time P5,3 = 5!/(5-3)! = 60 Class Problem Three awards will be given for a class of 25 graduate students If each student can receive at most one award How many possible selections? P25,3=…

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**Combinations 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 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

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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 D3={exactly 3 of the 6 cars chosen are domestic} P(D3)=… P(D3D4 D5 D6) =…

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**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. 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

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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

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**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 P(A|B)=… P(A|B C)=… P(A| reads at least one) = P(A|ABC)=… P(AB |C)=…

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**Class Problem Solution P(A) =… P(B) = … P(A/B)=… P(B/A)=…**

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. P(A) P(B) P(A/B) P(B/A) Solution P(A) =… P(B) = … P(A/B)=… P(B/A)=…

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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

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**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)(1/2)=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? P(C)P(D)=(0.65)(0.4)=0.260.24

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**Class Problem Solution: P(A) = … P(B) = 79/100**

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

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**Total Probability Rule**

Let A1, A2, A3,…, An be a collection of mutually exclusive events with known probabilities which partition S Consider an event B with known conditional probabilities How to use P(Ai) and P(B/Ai) to calculate P(B) Easily achieved by B=(A1∩B)∪… ∪(An∩B) Events Ai∩B are mutually exclusive P(B)=P(A1∩B)+… P(An∩B) =P(A1)P(B/A1)+…+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 A1 A2 An B

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Example In a certain assembly plant, three machines, B1, B2, and B3, 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 Select a finished product randomly What is the probability that it is defective? Let A: the product is defective B1=the product is made by machine B1 B2=the product is made by machine B2 B3=the product is made by machine B3 P(A)=P(B1)P(A/B1)+…+P(Bn)P(A/Bn) P(B1)P(A/B1)=(0.3)(0.02)=0.006 P(B2)P(A/B2)=(0.45)(0.03)=0.0135 P(B3)P(A/B3)=(0.25)(0.02)=0.005 P(A)= =0.0245

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**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(A1), …, P(An) are the prior probabilities of events A1, …, 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

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Example In a certain assembly plant, three machines, B1, B2, and B3, 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 B3)? Let A: the product is defective B1=the product is made by machine B1 B2=the product is made by machine B2 B3=the product is made by machine B3 Instead of asking P(A), we want to find P(Bi/A) Using the Bayes’ formula P(B3/A)=0.005/ = 10/49

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**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 B1=the product is made by machine B1 B2=the product is made by machine B2 B3=the product is made by machine B3 P(B1)=0.02 P(B2)=0.03 P(B3)=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(B1/A)=0.006/ = P(B2/A)=0.0135/ = P(B3/A)=0.005/ = 10/49

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**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 What is the probability that a selected buyer has a TV that will need repair while under warranty? If a customer returns a TV brand, what is the probability that is a brand 1? A brand 2? A brand 3?

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Solution

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**Tree Diagram P(B/A1) P(A1)=P(BA1)=0.125 P(B/A1)=0.25 Repair No repair**

Brand 1 P(B/A2) P(A2)=P(BA2)=0.06 Repair P(B/A2)=0.2 P(A2)=0.3 Brand 2 No repair P(A3)=0.3 P(B'/A2)=0.8 Brand 3 Repair P(B/A3) P(A3)=P(BA3)=0.02 P(B/A3)=0.1 No repair P(B'/A3)=0.90 P(B)=P[(brand1 and repair) or (brand 2 and repair) or (brand 3 and repair)] =0.205 P(B)= P(BA1)+ P(BA2)+ P(BA3)= =0.205 P(A1/B)=… P(A2/B)=… P(A3/B)=…

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Class Problem An assembly plant receives its voltage regulators from three different suppliers, 60% from supplier B1, 30% from supplier B2, and 10% from supplier B3. If 95% of the voltage regulators from B1, 80% of those from B2, and 65% of those from B3 perform according to specifications, what is the probability that a particular voltage regulator which is known to perform according to specifications came from supplier B3? If A denotes the event a voltage regulator received by the plant performs according to specifications and B1, B2, B3 are the events that it comes from the respective suppliers, then

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COUNTING AND PROBABILITY

COUNTING AND PROBABILITY

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