1. Chapter 2
Probability

2. 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

3. 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}

4. 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

5. 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

6. 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

7. Example:1 For an experiment, let Determine: AB AC AB AC A´ {AC}´
B = {3,4,5,6}, and
C = {1,3,5}
Determine:
AB
AC
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}

8. 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}

9. 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) BC
c) A´B
d) AB
Solution

10. 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 =…

11. 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

12. 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

13. 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(AB) =
e) P(AB) =

14. 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

15. 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)

16. 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) = = 0.4
c) P(A´B)+ P(AB) = P(B). Therefore, P(A´B)= = 0.1
d) P(A) = P(AB) + P(AB´) Therefore, P(AB´) = = 0.2
e) P((AB) ') =1 - P(AB)= = 0.6
f) P(A´B)= P(A') + P(B) - P(A´  B)= = 0.8 from part c

17. 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}= E1E2E3
P(B)= P(E1)+P(E2)+P(E3)=…

18. 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(AB) = …
b) P(AB) = P(A) + P(B) - P(AB) = …
c) Because (AB) does not equal Ø , A and B…

19. 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)=…

20. 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

21. 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

22. 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=…

23. 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

24. 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(D3D4 D5 D6) =…

25. 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

26. 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

27. 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|ABC)=…
P(AB |C)=…

28. 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)=…

29. 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

30. 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)=0.260.24

31. 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

32. 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

33. 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

34. 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

35. 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

36. 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

37. 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?

38. Solution

39. Tree Diagram P(B/A1) P(A1)=P(BA1)=0.125 P(B/A1)=0.25 Repair No repair
Brand 1
P(B/A2) P(A2)=P(BA2)=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(BA3)=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(BA1)+ P(BA2)+ P(BA3)= =0.205
P(A1/B)=…
P(A2/B)=…
P(A3/B)=…

40. 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