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Probability 3.1 Sets

3.1 Sets

We begin our study of probability with some basic set theory. This will give us a rigorous framework on which to build. Set theory is, in itself, a large topic so we will adopt only the set-theoretic concepts that will be needed in our study of probability. Each of these concepts brings with it notation which is shorthand for an English phrase. In this respect, one can view our study of set notation as mastering a new language. Our format will be to introduce each piece of notation and discuss its use and underlying concept.

A set is a well-defined collection of objects. There are two ways to specify a set. Either

(1) list the contents (elements) of the set explicitly or (2) give a rule for membership to the set.

Three examples of a set formed by method (1) are:

{2, 4, 6, 8, D}, {-1, 0, 10, 15,17, 19}, {§,¨,©,ª}.

A set is defined strictly by its contents, not the manner in which they are presented. In particular, the order in which you list the elements does not matter. For instance, {1, 2, 3} is the same set as {3, 1, 2}. Moreover, multiple listings can be ignored. For instance, {1, 2, 3, 1, 1} is the same set as {3, 1, 2}. Each of these sets contains exactly three distinct elements.

Method (1) works fine for small sets but is not very practical (and sometimes impossible) for large sets. Consider the task of listing the contents of the set of all persons in the United States over the age of 40! In a case such as this, we use method (2) along with some special notation:

{persons in the U.S. | person is over age 40}.

“the set of all” “such that”

By convention, the expression to the left of the vertical bar is a global restriction on the elements of the set while the expression to the right of the vertical bar is a local restriction. For instance, consider the set, A = {integers x | x ³ 1}. The clause “integers x” is a global restriction which tells us that, out of all possible objects in the world (people, chairs, numbers, etc.), we have restricted our attention to integers (and therefore, numbers). The second clause, “x ³ 1”, is a more local restriction that tells us, out of the set of integers, we should consider only those that are greater than or equal to 1.Î - an element (member) of________________________________________________________________________________________

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If an object is in a set, then we say that it is an element or member of the set. We use the symbol Î to denote this relationship. For example, if A = {2, 4, 6}, then one writes 2 Î A which is to be read, “2 is an element of set A”. The symbol Ï denotes that an object is not an element of a set. For instance, 3 Ï {2, 4, 6}.

Í - contained in (or, subset of)

Let A and B be two sets. If every element in A is also an element of B, then we say that A is a subset of B or, equivalently, A is contained in B, which we denote by A Í B. For example, if A = {1, 2, 3, §} and B = {1, 2, 3, 4, 8, ©, §}, then A Í B since every element that appears in A also appears in B. We could also write {1, 2, 3, §} Í {1, 2, 3, 4, 8, ©, §}. Note that B is not a subset of A, denoted B Ë A, because there are elements of B that are not elements of A. For instance, 8 Î B and yet 8 Ï A.

= - equals

We say that two sets A and B are equal, written A = B, whenever both A Í B and B Í A, that is, whenever the sets contain exactly the same elements. Because the same set can be described many different ways, it may not be immediately obvious that two sets are equal. For instance, if A = {1, 2, 3, 5, 6, 2, 1, 8} and B = {8, 6, 2, 5, 3, 1}, then A = B. To verify this, we observe that every element in A is also in B (that is, A Í B) and that every element in B is also in A (that is, B Í A). Note that the repetition of the numbers 1 and 2 in set A can be ignored.

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A

24

6

3

Figure 1: 2 Î A, 3ÏA

A B

13

2§

48

©

Figure 2: A Í B


È - union

The union of two sets A and B is a new set, denoted A È B, whose elements belong to A or B (or both). Another way to view this is that A È B is the set of elements that belong to at least one of the sets A or B. You can form the union of two sets by merging them and then crossing out any repeats. For example, if A = {1, 2, 3, §} and B = {3, 4, 5, 7}, then we can form A È B as follows:

Strictly speaking, we do not need to strike the second 3 as we have done above. Listing an element twice is analogous to listing a person twice on a guest list - it’s not wrong but may lead to confusion.

Note that whenever x Î A, then x Î (A È B). Why? If x is in A, then it is an element of at least one of the two sets, namely, A. Similarly, whenever x Î B, then x Î (A È B). Ç - intersection

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A B

1

2

§ 3

4

57

Figure 3: The shaded region is A È B

A È B = {1, 2, 3, §} È {3, 4, 5, 7} = {1, 2, 3, §, 3, 4, 5, 7} = {1, 2, 3, §, 4, 5, 7}


The intersection of two sets A and B is a new set, denoted A Ç B, whose elements belong to both A and B. Phrased another way, A Ç B is the set of elements common to A and B. For instance, if A = {2, 3, 5, §} and B = {1, 9, 5, §}, then A Ç B = {5, §}. In a Venn diagram such as Figure 4, the intersection of two set is depicted as the overlap between two regions.

Note that, for any sets A and B, (A Ç B) Í A and (A Ç B) Í B. Moreover, whenever A Í B, then we know that A Ç B = A. Why? If A is contained in B, then every element in A is contained in B, hence, common to both A and B. Moreover, A Ç B can’t have any elements outside of A because A Ç B is the set of all elements in both A and B.

{ } - the empty set

The empty set, denoted by empty set brackets, { }, is the set that contains no elements. (Alternative notation is the Greek letter, Æ.) The empty set is a legitimate set to which the usual operations such as union and intersection can be applied.

There are a number of advantages to giving formal recognition to the empty set. For instance, if two sets A and B have no elements in common, then their intersection is still a well-defined set. That is, A Ç B = { }. Here are some properties of the empty set:

Property 1: A È { } = A, for any set A Property 2: A Ç { } = { }, for any set A

Property 3: { } Í A, for any set AProperty 4: the number of elements in { }is zero

Property 1 holds because the empty set is adding no new elements to set A. Property 2 holds because no set can have an element in common with the empty set. Property 3 holds because

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A B

2

3

5§

1

9

Figure 4: The shaded region is A Ç B

empty set

2

3 5

§

1

4

7 12

6¨

Figure 5: The empty set, { }.


logically, the only way this containment could fail is if the empty set contained an element not in A. But this cannot be, so, we are forced to say that the empty set is a subset of every set. Property 4 holds by definition of the empty set.

U - the universal set

The universal set, denoted by the symbol U, is the set of all objects to be considered in a given context. The purpose of declaring a universal set is to limit the scope of the objects that can be used to form sets. For instance, if we wish to limit set elements to positive integers, then we can declare that U = {1, 2, 3, 4, …}. Under this restriction, we can form sets such as {39, 12, 5} or {1, 4, 877, 11} but a set such as {-1, 5, 6, §} is meaningless.

It is customary in Venn diagrams, such as Figure 6, for the universal set to be represented by the contents of the outermost box.

Like the empty set, the universal set has some peculiar properties:

Property 1: U È A = U, for any set A Property 2: U Ç A = A, for any set A

Property 1 holds because the universal set contains all elements - nothing new is contributed by the set A. Property 2 holds because A is a subset of U so the elements they have in common are precisely those in A. Usually, the set operations of union and intersection can be performed without explicit knowledge of the universal set. For this reason, it is often left unspecified. However, the set operation we are about to introduce requires explicit knowledge of the universal set.

A C - the complement of set A

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A B

39

12

11

1

4

U= positive integers

5 87

Figure 6: The Universal Set.


AC denotes the complement of a given set, A. 1 This is the set of all elements that are not in the set A but are in the universal set. In Figure 7, the complement of A is depicted by the area outside set A (the shaded region). One way to form the complement of a set is to start with the universal set and remove all elements of set A from the universal set. The complement of set A. For example, if

U = { 1, 2, 3, 4, … } and A = {1, 2},

then AC = {3, 4, 5, …}. One way to see this is to imagine removing all elements from U that are in A, namely, 1 and 2.

It is important to note that the complement of a set is dependent upon the universal set that has been declared. For instance, in the above, we had that

A’ = {1, 2} and AC = {3, 4, 5, …}, when U = { 1, 2, 3, 4, … }.

But if the universal set were changed to U = {ª, 1, 2, 3, 4, …}, then we would have that

A = {1, 2} and AC = {ª, 3, 4, 5, …}.

In a Venn diagram, the complement of a set A is depicted by shading the region outside of the set A. See Figure 7.

n(A) - number of distinct elements in set A

Given a finite set A, we use n(A) to denote the number of distinct elements of set A. For instance, if A = {-2, -1, 0, 1, 3, 4, 5, 12}, then n(A) = 8, because there are 8 distinct elements in A. We emphasize the word “distinct” because an element may be represented more than once in the description of a set. This is analogous to listing a person twice on a guest list. For instance, one might debate whether the following set contains two elements or four: B = {1, 1, 1, 2}. But it is indisputable that B contains two distinct elements. That is, n(B) = 2.

2k = The number of subsets of a set with k-many distinct elements 1 Many textbooks use the notation A¢ instead of Ac to denote the complement of A. We have adopted the Ac notation because the ¢ symbol can be easily missed. ________________________________________________________________________________________

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ACA

Figure 7: The Complement of Set A.


If a finite set A has k-many distinct elements, then the number of distinct subsets of A is given by 2k. For example, if A = {1, 2, 3}, then A has 23 = 8 subsets. We can verify this by writing out all of the subsets of A. A systematic approach for doing this is to first list all subsets with no elements (there is only one, the empty set), then all subsets with exactly one element, then all subsets with exactly two elements, and, finally, all subsets with exactly three elements (there is only one, the set A itself):

{ }, {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}, {1, 2, 3}

It is impractical to list all the subsets of even a modest-size set such as {1, 2, …, 10}. This set has 210 = 1024 many subsets!

The Inclusion-Exclusion Principle:

The inclusion-exclusion principle states that the number of elements in the set A È B is the number of elements in A plus the number of elements in B minus the number of elements they have in common. It serves as a reminder that the size of the union of two sets might be less than the sum of the two sizes. For example, suppose that there are 50 students registered for an English class and 20 students registered for a math class and two students registered for both classes. If we bring all the students together in one room (union the two classes), there will be 50 + 20 - 2 = 68 students in the room. A common mistake is to ignore the intersection and conclude that there would be a total of 50 + 20 = 70 students. This counts twice those registered for both classes (i.e., those in the intersection).

The inclusion-exclusion principle (formula) can be used on a strictly algebraic level. For instance, consider the sets A = {1, 2, 3, 4} and B = {3, 4, 5, 6}. These sets have in common the elements 3 and 4. That is, A Ç B = {3, 4}. So, the number of elements in A È B is given by

n(A È B) = n(A) + n(B) - n(A Ç B) = 4 + 4 - 2 = 6.

Even when the intersection of two sets is unknown, the inclusion-exclusion principle can be used to establish an upper bound on the number of elements in the union of two sets. For instance, if there are 35 elements in set A and 40 elements in set B, then the number of

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The Inclusion-Exclusion Principle

n(A È B) = n(A) + n(B) - (A Ç B)


elements in A È B is at most 35 + 40 = 75. Once we know the number of elements that A and B have in common, then we can subtract this number from 75 to get the number of elements in A È B . For instance, if A and B have 7 elements in common, then the number of elements in A È B is 75 - 7 = 68.

Example 1: A computer program generates statistics from a database of flight information. Two of the statistics it searches for are the arrival time and departure time of each flight. A flight is considered to have bad data if either of these pieces of information is lacking. Otherwise, it is considered to have good data. The program outputs the following information:

105 = number of flights for which no arrival time was reported120 = number of flights for which no departure time was reported 35 = number of flights for which neither a departure time nor arrival time was reported680 = total number of flights

Does the program output sufficient information to deduce how many flights had bad data? If so, how many were there? What about the number of flights that had good data?

Solution: There are two ways to solve this problem. One is with the inclusion-exclusion and the other is using a Venn diagram. We will present both methods.

Method 1 (inculsion-exclusion principle):Let

A = set of all flights with no arrival timeD = set of all flights with no arrival time

Note that

A Ç D = set of all flights with no arrival time and no departure time.

We wish to find the n(AÈD), that is, the number of flights with no arrival time or no departure time (or both). We make the following substitutions into the inclusion-exclusion formula:

n(A È D) = n(A) + n(D) - n(A Ç D) = 120 + 105 - 35 = 190

Thus, we have enough information to conclude that there were 190 flights with bad data.

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Now we compute the number of flights with good data. This is simply the total number of flights minus the number of flights with bad data, that is, 680 - 190 = 490.

Method 2 Venn Diagram:As before, we establish the following sets:

A= set of all flights with no arrival timeD= set of all flights with no departure time

Note that

A Ç D = set of all flights with no arrival time and no departure time.

We have been given the following information: n(A) = 120 n(D)= 105 and n(A Ç D) = 35. We form a two-circle Venn diagram, as the figure below, and fill the number 35 into the center, to indicate the size of A Ç D.

Our goal is find the size of the union of sets A and B. First, we must find the size of the two regions indicated by question marks (the sum of these two numbers plus 35 is the size of the union). The question mark on the left, for example, indicates the number of flights in set A but not in set D, that is, the number of flights that had no arrival time but did have a departure time.

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A D

35

A Ç D

? ?


Since there were 120 flights with no arrival time and 35 of those had no departure time, there must have been 120 - 35 = 85 flights with no arrival time but with a departure time. We fill the number 85 into the appropriate region. (A common mistake is to put the number 120 in place of the number 85. But that would imply that there are a total of 120 + 35 = 155 flights in set A.)

Similarly, there were 105 flights with no departure time, 35 of which had no departure time, so, there must have been 105 - 35 = 70 flights with no departure time but with an arrival time. We fill the number 70 into the region inside D but outside A.

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A D

35

A Ç D

85

A D

35

A Ç D

85 70


Now, we have enough information to conclude that the size of the union of sets A and D is 85 + 35 + 70 = 190. In other words, there were 190 flights with bad data, meaning, no arrival time or no departure time or both.

Next, we compute the number of flights with good data. These flights are represented by the region outside of both circles. The total number of flights in the box is 680. We have already filled 85 + 35 + 70 = 190 flights into the diagram and there is only one remaining region so the corresponding number of flights must be 680 - 190 = 490.

Lastly, we double-check our arithmetic. The total of the A circle should be 120; indeed, 85 + 35 = 120. The total of the D circle should be 105; indeed, 35 + 70 = 105. And the total of all number in the diagram should be 680; indeed, 85 + 35 + 70 + 490 = 680.

n

Complex Set ExpressionsNo one of the set-theoretic concepts we have learned so far is particularly difficult. This does not mean that everything we do with them is simple. Indeed, a great many high-level mathematical proofs consists of showing that one set is contained in another or, perhaps, equal to it. As set notation begins to compound on set notation, the task of unraveling its meaning grows more difficult. Analogously, it would take even someone who is quite good at arithmetic quite a while to compute an expression such as this in their head: ((23+15)*3) -14/7+1/0.09).

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A D

35

A Ç D

85 70

490


The trick to unraveling complex set expressions is to proceed methodically and work from the inside of parentheses outward. For instance, to compute the set C = ((A Ç B) - D) È B, we would first compute AÇB, then subtract D from it, then union the result with B.

Example: (complex set expression)Let

A = B = C = D = { }

Solution: We start with the innermost parentheses.

Abuses of Set NotationAs we stated at the beginning of this section, mastering the use of set notation is like mastering a new language: at first, there are bound to be some grammatical errors. Below, we have listed some of the more common abuses of set notation.

Abuse 1: {3, 4} Î {3, 4, 5, 6}Note that “Î” is a relation that exists between an element and a set, while “Í” is a relation that exists between two sets. Therefore, the statement {2, 3} Î {2, 3, 4} is meaningless. A correct statement would be {2, 3} Í (2, 3, 4}. Similarly, the statement 3 Í {3, 4, 5} is meaningless, while a correct statement would be either {3} Í {3, 4, 5} or 3 Î {3, 4, 5}.

Abuse 2: {1, 2, 3, ...}There are times when a set is defined by a hybrid of the two methods we have described, namely, listing it contents explicitly or by giving a rule for membership. For instance, the set of all integers bigger or equal to 1 is sometimes written as {1, 2, 3, …}. The dots are a mathematical way of saying “and so on”, meaning, “the pattern continues forever”. Use of this notation relies heavily on the fact the pattern that the writer had in mind is clearly understood by the reader. This type of notation should be avoided, in general, because of possible multiple interpretations of the pattern. After a few more elements of the set {1, 2, 3, …}are given, it could turn out (though it is not likely) that the pattern is actually this: (1, 2, 3, 11, 12, 13, 21, 22, 23, 31, 32, 33, …}.

Abuse 3: { } = 0Recall that the empty set is the set that contains no elements. But the empty set is not “nothing”; it is still a well-defined set. If you think of sets as containers that hold objects, then the empty set is the container with no objects inside. It’s hard to tell what is intended by the abusive statement { } = 0. A similar but correct statement is that the number of ________________________________________________________________________________________

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elements in the empty set is 0, that is, if A = { }, then n(A) = 0. Another correct statement is { } = Æ, because the Greek letter Æ is alternative notation for the empty set.

Abuse 4: AÇBÈCThis is an example of an ambiguity, meaning, a situation in which there is more than one possible interpretation. At least one set of parentheses is required in this expression because, it is not clear whether it stands for the intersection of A and B followed by union with C, or whether it stands for the union of B with C, followed by intersection with A. These two interpretations would be denoted (AÇB)ÈC and AÇ(BÈC), respectively. An arithmetic analogy would be the expression 2*3+7, which could be interpreted as (2*3)+7 = 13, or interpreted as 2*(3+7) = 20.

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Problems

### For each pair of sets, state whether or not they are equal. If they are not, give an example of an element that is in one set that is not in the other. (a) A = the set of all drinkers who drink Budweiser, B = the set of all beer drinkers who drink Budweiser(b) A = {integers x | x > 1}, B = {1, 2, 3, …}

(c) A BRST UVWRST UVW12

35

67

48

610

1214

12

, , , , , ,

(d) A = the set of all persons over age 65, B = the set of all retired persons(e) A = the set of all persons with a life-threatening illness, B= the set of all persons with cancer(f) A = the set with exactly 0 elements, B = the empty set

Thinker: The set of all sets. Judy. Define numbers.

###(a) What are the two ways to define a set?(b) If A = B, is A Í B?(c) If A = B, is B Í A?(d) If A = B, what is A Ç B?(e) If A = { }, how many elements are in A?(f) If A Í B, what is A Ç B?

### Basketball teams.

Determine whether or not each of the following statements is true.

1. 24 Î {x | x ³ 12.5}2. 24.999 Î {x | x ³ 25}3. {Persons in the world} Í {Persons in the U.S.}4. {Persons in U.S.} Í {Persons in the world}5. {Persons in the U.S.} is not a subset of {Persons in Wyoming}6. {Persons in the U.S.} Ç {Persons in Kansas} = {Persons in U.S. | Persons in Kansas}7. p Î {x | x > 2 and x £ 5} [note: p » 3.14]8. p Î {x | x > 4}9. {2, 2.5, 3, 5} Ç {2, 2.5, 3, 5} = {2, 2.5, 3, 5}10. {1, 1, 1, 1} Í {1}11. {2, 2.5, 3, 5} Ç {2, 2.5, 3, 5} = Æ

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For Problems 12-21, determine if each statement is true for all choices of A and U.

12. A È { } = { }13. A È { }= A14. A È U = A15. A Ç U = A16. A Ç { } = { }17. A È AC = U18. A Ç AC = { }19. UC = { }20. { }C = U21. { } Î U

Let U = {x Î integers | x ³ -1}, A = {0, 2, 4, 6, ...}; B = {-1, 1, 3, 5, 7, ...} and C = {-1, 0}Compute the following:

22. A È B23. (A È B) È C24. C Ç (A Ç BC)25. C Ç (A Ç BC)C

26. Determine whether or not each of the following scenarios is possible. If so, give an example. If not, say why not. (a) There are two sets, A and B, such that n(A) = n(B) and yet A ¹ B.(b) There are two sets, A and B (possibly equal), such that A Ç B = { }, where A = B.(c) There are three sets, A, B and C such that A È B = A È C and A Ç B ¹ A Ç C. Hint: draw a Venn diagram.

Let U = {x | x ³ 0}, A = {x | x ³ 0 and x £ 2}, B = {x | x ³ 2 and x £ 3} and C = {x | x ³ 5}.[Note that these sets contain much more than integers; each one is infinite.] Compute the following:

27. (A Ç B) È C28. (A Ç B)C È C29. AC È {1}30. [ (B Ç { }) È U ] Ç CC

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Let A = {a, b, c, d}, B = {a, b, c}, C = {b, a, c} and D = {a, b, c, d, e}. Determine whether or not each statement is true:

31. A Í B32. A Í D33. d Î B34. c Î D35. e Î A36. B = C37. { } Í B38. B Í A39. B Í B40. B È C = A41. B Ç A = C42. Find the number of subsets of a set A, given that:(a) n(A) = 3 (b) n(A) = 5 (c) n(A) = 1 (d) n(A) = 0 (e) n(A) = 2

Use the inclusion-exclusion principle (formula) to make the following calculations. [Hint: The formula has four variables so if you are given any three of them, you can solve for the fourth.]

43. Find n(A È B), if n(A) =3, n(B) = 6 , and n(A Ç B) = 2.44. Find n(A Ç B), if n(A) = 25, n(B) = 10, and n(A È B) = 30.45. Find n(B), if n(A) = 8, n(A È B) = 14, and n(A Ç B) = 4.

46. A college campus survey revealed that there are 1200 Pepsi drinkers, 1400 Coke drinkers, and that 200 persons on campus drank both. How many persons on campus drink Coke or Pepsi (or both)?

47. A small county has 3200 voters registered as Democrats, 2100 voters registered as Republicans, and 500 registered to other (or no) affiliations. If we formed a list of all the registered voters in the county, how many voters would be on the list (assume no voter is registered twice).

48. In a certain town, there are 10,000 Chevy owners and 12,000 Ford owners and 6,000 owners of other types of cars. How many car owners are there in town?

49. A survey was taken of shoppers leaving an indoor mall. It was revealed that 300 persons had made a purchase at Sears, 200 had made a purchase at Hecht’s and that 150 had made a purchase at both Sears and Hecht’s. How many shoppers made a purchase from at least one of the two stores?

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50. Statistics at a stockbroker’s office revealed that 250 clients own CD’s, 360 own bonds, and 75 clients own both. How many clients own at least one of the two? How many only own bonds?

Thinkers

51. Well-defined Sets. In order for a set to be well-defined, the criteria for membership must be clear, unambiguous, and non-contradictory. Explain why each of the following “sets” is not well-defined. Specifically, state whether or not the membership for criteria is unclear, ambiguous (has more than one possible, but clear, interpretation) , or circular.

A = the set of all tall persons.B = the set of all fast computers.C = the set of all integers between 0 and 1. [Hint: think about the word, “between”.]D = {1, 2, 4, …}.E = the set of all elements inside set E.F = the set of all elements that are not inside set F. [Can such a set exist?]

52. Even Numbers. Consider the set of all even integers. Is this set infinite? Write a short paragraph to convince a classmate that it is (or isn’t). What about the set of odd integers?

53. Prime Numbers. A positive integer is prime if the only (positive) integers that divide into it evenly are 1 and itself. (The number 1 is declared not to be prime.) For example, 7 is prime because it is evenly divisible only by 1 and itself. The integer 4 is not prime, because it is divisible by 2. Consider A = the set of all prime numbers. This set is well-defined, because the test for membership has a clear, testable outcome for any given integer. (For any given positive integer, n, check the divisibility into n of all positive integers less than n.) Is this set infinite? How can you be sure (that it is or isn’t)?

54. Inclusion-Exclusion Principle. This principle reminds us that the size of the union of two sets is not necessarily equal to the sum of the sizes of the two sets. In other words, given two sets, A and B, the following conclusion could be wrong: n(A È B) = n(A) + n(B). The formula for the inclusion-exclusion principle reminds us that, when computing n(A È B), we need to subtract n(A Ç B) from the quantity n(A) + n(B). When is it valid to conclude that n(A È B) = n(A) + n(B)? Conversely, if you know that n(A È B) = n(A) + n(B), what can you conclude about the intersection of A and B?

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55. Tautologies. A statement that is always true is called a tautology. Explain why each of the following statements is true, no matter what universal set is declared.

Statement 1: { }C = U. Statement 2: UC = { }

56. The Lion’s Paradox2. It is said that, somewhere in Europe, outside of a medieval church, there sits the statue of a lion with a gaping mouth. Below the lion is a Latin inscription that when translated reads, “he who puts his hand in the lion’s mouth and utters a lie will not be able to remove his hand”. Suppose that you were to place your hand in the lion’s mouth and say, “I cannot remove my hand from the lion’s mouth”. Do you speak the truth?

57. Russell’s Paradox. In 1901, a British Philosopher, Bertrand Russell, posed a set-theoretic paradox now called, Russell’s Paradox. One version of Russell’s paradox supposes that there is a town in which all the men are clean shaven and the barber shaves every man who does not shave himself. The question is, who shaves the barber? This paradox challenged the notion that in mathematics, we are entitled to form sets based on any properties we wish. In particular, it showed that, unless we impose some restrictions, a paradox will always result if we work within a logical setting in which we freely allow sets to contain other sets. (The resulting paradox results from considering the set of all sets that do not contain themselves.) Here is an updated version of Russell’s paradox for you to think about: Every web site can have a link to any other web site. Some web sites have a link to themselves (usually, in the form of a home page button). Now imagine that there is a master web site that has a link to all web sites that do not have links to themselves. The question is this: does the master web site have a link to itself?

2 A paradox is statement that, in a certain context, can be neither true nor false.________________________________________________________________________________________

3-18


Solutions

1. True2. False3. False4. True5. True (there are people in the United States who do not live in Wyoming.)6. True7. True (p » 3.1415926536 which is bigger than 2 and less than 5)8. False9. True10. True, in fact these sets are equal (recall that a set A is a subset of B if all elements in A are elements of B) 11. False12. False, it is always true that A È Æ = A.13. True14. False, it is always true that A È U = U.15. True, Removing the contents of the empty set has no effect.16. True17. True18. True19. True. There are no elements that are not in U.20. True. Every element in U is not in the empty set. 21. Neither! This is an abuse of notation. The relation “Î“ relates elements and sets, not

sets between sets. However, it is true that Æ Ì U.22. U23. U24. {0}25. {-1}26. (a) Yes. For example, let A = {1} and B = {2}. (b) Yes. Let A = f and let B = f. Then A Ç B = f Ç f = f. (c) Yes. Let A = {1, 3, 4}, B = {3, 5} and C = {4, 5}.27. {x | x ³ 5 or x = 2}28. {x | x ³ 0 and x ¹ 2}29. {x | x = 1 or x > 2}30. {x | x ³ 0 and x < 5} ( or CC)31. False32. True33. False34. True________________________________________________________________________________________

3-19


35. False36. True37. True38. True39. True40. False41. True42. (a) 8 (b) 32 (c) 2 (d) 1 (e) 443. n(A È B) = 744. n(A Ç B) = 545. n(B) = 10 46. 2400 persons on campus drink Coke or Pepsi (or both).47. 5800 voters would be on the list.48. No way to tell! Some car owners may own more than one type of car. All we can say is that there are no more than 28,000 car owners in the town.49. 350 shoppers made a purchase from at least one of the two stores. 50. (a) 535 clients own a CD or bond. (b) 285 persons own only bonds.

________________________________________________________________________________________

3-20

Probability 3.2 The Basics of Probability

3.2 The Basics of Probability

We will be assigning numerical values, called probabilities, to events to help us assess their likelihood of happening. In order to do this rigorously, we need some basic definitions.

Definition: An experiment is an action followed by a recording of the outcome.

In order for an experiment to be well-defined, the nature of the recordings that one is willing to make must be specified in advance. The experiment “choose one person at random from a room containing three persons” is not well-defined because the nature of the recording has not been specified. One could record, for instance, the height of the person chosen, or last name, or hair color, etc. However, the experiment would be well-defined if we restated it as, “choose one person at random from a room containing 3 persons and record their first name”.

Definition: The sample space of a given experiment is the set of all possible outcomes (recordings) of the experiment. 3

Definition: An event is a subset of the sample space. Moreover, every subset of the sample space is an event.

Note that sample spaces and events are defined as sets. The sample space, denoted S, will play the role of the universal set in all of our work in probability. Given any event, E, we have that E Í S.

For any event E, there is an associated probability, denoted by P(E), which represents the likelihood of E occurring. We restrict P(E) to lie between 0 and 1 so that we can think of probability as a fraction (or percentage). If an event has probability 1 (100% chance), then the event will definitely occur. If an event E has probability 0 (0% chance), then the event will definitely not occur.

3 We will restrict our attention to finite sample spaces so that the number of elements in every event is finite.

for any event E,0 £ P(E) £ 1

P(S) = 1 and P({ }) = 0

Probability 3.3 Conditional Probability

Example 1: Consider the experiment in which one picks a person at random from a room containing Jack, Sue, and Sam, and announces the first name of the person chosen. Let E be the event of choosing a person whose name begins with the letter S.

Q1: What is the sample space for this experiment?

A1: The sample space is S = {Jack, Sue, Sam}, because these are all the possible outcomes of the experiment.

Q2: How can the event E be written in set notation?

A2: E = {Sue, Sam}.

Q3: Form another event for this experiment.

A3: Every subset of the sample space is an event. In particular, we choose to form the subset (event) F = {Jack, Sue}, which can be interpreted as the event that Sam is not chosen. Note that F = {Sam}C.

Q4: What is the probability of choosing Sue?

A4: Since there are 3 persons in the room and they are all equally likely to be chosen, the probability of choosing Sue is 1/3. n

In order to answer question 4 in the foregoing example, we counted the number of persons named Sue and divided by the number of persons. This demonstrates the key principle that we will employ, either directly or indirectly, in all of our computations:

The probability of an event E occurring is the number of elements in E divided by the number of elements (outcomes) in the sample space S, assuming that the outcomes in S are equally likely (and that S is finite).

This means that our calculations can be done by counting the number of elements in appropriate sets and forming appropriate ratios. However, there are two ways that complications can arise. First, one must know exactly what sets to form and be able to count their elements in a systematic manner. Many of the techniques we will learn are methods for


organized counting. The other complication is that the outcomes of a sample space may not be equally likely. In this case, our tactic will be to switch to an alternate sample space in which the outcomes are equally likely. We demonstrate this in the next example.

Example 2: Choose one person at random from a room containing 3 persons and record the first name of the person chosen. The room contains Jack Martin, Jack Jones, and Sue Sinclair.


A1: The sample space is S = {Jack, Sue} because these are the only possible outcomes (recordings) of the experiment. Note: The answer {Jack, Jack, Sue} is also correct because {Jack, Jack, Sue} = {Jack, Sue}.

Q2: What is the probability of the event E = {Jack} occurring?

A2: A faulty conclusion would be 1/2, based on the reasoning that there is one element in the set E = {Jack} and 2 elements in the sample space S = {Jack, Sue}. This incorrectly assumes that the outcomes in the sample space are equally likely. Since there are two persons named Jack, it is far more likely to choose someone named Jack than Sue. Our tactic is to consider the alternate sample space,

S2 = {Jack Martin, Jack Jones, Sue Sinclair.}.

The outcomes in this sample space are equally likely. Now we see that the probability of choosing a person named Jack is 2/3. [A similar approach would be to define the alternate sample space S3 = {Jack1, Jack2, Sue}]. n

Example 3: Roll a six-sided die and record the number showing on the top face. The sample space for this experiment is S = {1, 2, 3, 4, 5, 6}. Q: What are the probabilities of the following events?

A = the event of rolling an even number B = the event of rolling a number strictly greater than 4 C = the event of rolling a number that is even or strictly greater than 4D = the event of rolling a number that is both even and strictly greater than 4

A: We start by listing the four sets (events) explicitly. The first two are:


A = {2, 4, 6} and B = {5, 6}.

The word “or” can be translated as union (È) while the word “and” can be translated as intersection (Ç). Thus, C = A È B and D = A Ç B. Applying these set operations to A and B, we obtain

C = A È B = {2, 4, 6} È {5, 6} = {2, 4, 5, 6}

D = A Ç B = {2, 4, 6} Ç {5, 6} = {6}

For each event, the probability is the number of elements in the set divided by the number of elements in the sample space (each roll is equally likely). Thus,

P(A) = 3/6 P(B) = 2/6 P(C) = 4/6 P(D) = 1/6 n

Warning about notation: Only sets (or names of sets) should be placed inside the expression P( ). An expression such as P(1/5) is meaningless because 1/5 is not a set. Similarly, the expression P(Sam) is an abuse of notation. A correct expression would be P({Sam}). Also, keep in mind that for any event E the notation P(E) is a number between 0 and 1 as well as shorthand for the phrase, “the probability of event E occurring”.

Recall that given any set E, the notation EC stands for the complement of set E, that is, all elements that are not in the set E (but in the universal set). In probability, the sample space S serves as the universal set. This means that EC is the set of all elements that are not in E (but in the sample space S). The numbers P(E) and P(EC) bear the following relationship:

This means that if we already know, for instance, that the probability of event E happening is 1/3, then the probability of E not happening is 2/3 (these numbers must sum to one). Conversely, if we know, for instance, that the probability of event E not happening is 3/5, then the probability of E happening is 2/5. The most common use for this formula is for cases in which one wants to know P(E) but P(EC) is easier to calculate (or the reverse).

For any event E,

P(E) + P(EC) = 1


When computing probabilities, one must translate English into mathematics. In general, the word “or” can be translated into union (of sets) and the word “and” can be translated into intersection (of sets). For instance, if we want to know the probability of events E or F happening (or both), then we would translate this as P(E È F). The probability of the events E and F happening together would be translated as P(E Ç F). The following formula will aid us in the computation of unions of events:

You may have noticed that this formula has the same pattern as the inclusion-exclusion formula from the previous section. It serves much the same purpose, which is to prevent one from overestimating the size (probability) of the union of two sets.

The union formula can be thought of as a mechanism for guiding one through the computation of the number of elements in a union. One should use it whenever it’s too hard to directly count the number of elements in a union or when one wishes to verify reasoning in a solution that involves the word “or”. Whenever practical, we will compute probabilities both by “reasoning it out” and by employing the union formula. Each of these is a valuable skill and neither one alone serves all occasions.

Example 4: Flip a coin three times and record “heads” or “tails” for each successive flip, keeping track of the order in which they occur. Q1: What is the sample space for this experiment?

A1: S = {HHH, HHT, HTH, HTT, TTT, TTH, THT, THH}. Note that the outcomes in the sample space are equally likely.

Q2: Let E be the event of getting tails on the third flip and let F be the event of getting two (or more) tails in a row. Compute P(E), P(F), P(EC), and P(FC).

The union (“or”) formula

P(E È F) = P(E) + P(F) - P(E Ç F)


A2: In order to compute P(E), we will list the elements of set E. Since E is the event of getting tails on the third flip, we copy all outcomes from the sample space that end with the letter “T”:

E = {HHT, HTT, TTT, THT}.

Since the outcomes of the sample space are all equally likely, and since there are four elements in E and eight elements in S,

P(E) = 4/8 = 1/2 .

The shortcut for computing P(EC) is to recall that P(E) + P(EC) = 1 and that we already know P(E) = 1/2. Thus,

P(EC) = 1 - 1/2 = 1/2.

In order to compute P(F), we will list the elements of set F. We copy all outcomes of the sample space that have two or more consecutive “T’s”.

F = {HTT, TTT, TTH}

Thus, P(F) = the number of elements in F divided by the number of elements in S:

P(F) = 3/8

Since P(F) and P(FC) must sum to one,

P(FC) = 5/8 .

Q3: What is the meaning of the event E È F? Also, compute P(E È F).

A3 (reasoning): The meaning of the set E È F is the event of getting tails on the third flip or getting two or more tails in a row (or both). We select from S each outcome that meets at least one of the two properties: (1) tails on the third flip or (2) getting two or more tails in a row. Since S = {HHH, HHT, HTH, HTT, TTT, TTH, THT, THH},

E È F = {HHT, HTT, TTT, TTH, THT}

Since there are 5 elements in E È F and 8 in S,


P(E È F) = 5/8.

A3 (formulaic): We should get the same answer, P(E È F) = 5/8, using the union formula.

P(E È F) = P(E) + P(F) - P(E Ç F)

We have already computed P(E) and P(F) but, we have not computed P(E Ç F). We first write out E Ç F (all elements that are in both E and F):

E Ç F = {HTT, TTT}

Thus, P(E Ç F) = 2/8 because there are 2 elements in (E Ç F) and 8 in S. Substituting these numbers into the formula we obtain

P E F P E P F P E FÈ - Ç - -

48

38

28

4 3 28

58

. n

Definition: A probability distribution is a list of each outcome of a sample space and its corresponding probability.

We construct a probability distribution in the next example.

Example 5: Roll a pair of (six-sided) dice and record the sum of the numbers showing on the tops.

Q1: Write out the sample space for this experiment.

A1: The smallest possible sum is 1 + 1 = 2 and the biggest is 6 + 6 = 12. All integers between 2 and 12 are possible sums, so,

S = {2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}.

Q2: Are the outcomes in the sample space equally likely?

A2: No! There is only one way to roll a “2” (1 + 1), for instance, but there are many ways to roll a “7” (e.g., 1+ 6, 2 + 5, etc.). So these outcomes are not equally likely.

Q3: Write the probability distribution for the experiment.

A3: We need to fill out the following table:


Outcome 2 3 4 5 6 7 8 9 10 11 12probability

Since the outcomes in the sample space are not equally likely, we switch to an alternative sample space that distinguishes between the two dice. Let us imagine that one of the dice is blue and the other red. If we record the outcome (2, 3), for instance, then this will indicate that the blue die came up 2 and the red die came up 3 for a sum of 5.

To construct our sample space, we start by listing all the ways that the sum of the die could be 2. There is only one: (1, 1). Next, we list all the ways to get a sum of 3: (1, 2) or (2, 1). Continuing in this fashion, we obtain the complete sample space which is the second column in the table below.

sum rolls

2 (1, 1)

3 (1, 2) (2, 1)

4 (1, 3) (2, 2) (3, 1)

5 (1, 4) (2, 3) (3, 2) (4, 1)

6 (1, 5) (2, 4) (3, 3) (4, 2) (5, 1)

7 (1, 6) (2, 5) (3, 4) (4, 3) (5, 2) (6, 1)

8 (2, 6) (3, 5) (4, 4) (5, 3) (6, 2)

9 (3, 6) (4, 5) (5, 4) (6, 3)

10 (4, 6) (5, 5) (6, 4)

11 (5, 6) (6, 5)

12 (6, 6)

The total number of outcomes in the (alternate) sample space is 36 and each outcome is equally likely. So, the probability of rolling a 2 is 1/36, the probability of rolling a 3 is 2/36, and so on. Our final probability distribution is: Outcome 2 3 4 5 6 7 8 9 10 11 12probability 1/36 2/36 3/36 4/36 5/36 6/36 5/36 4/36 3/36 2/36 1/36

n


Problems

For problems 1 and 2, consider the experiment in which a card is drawn at random from an ordinary deck of 52 playing cards, and then the card is announced.

1. Write the sample space for the experiment.

2. What is the probability of drawing:(a) the queen of spades? (b) the jack of hearts?(c) a face card?(d) a numbered card?(e) a red card?(f) a diamond?(g) a numbered card between six and ten (inclusively)?(h) a face card or a club?(i) a red or black card?(j) a queen or a face card?(k) a card that is a black jack?(l) a card that is a jack or black?(m) a face card or a numbered card strictly bigger than a 5?(n) a card other than a queen?(o) a card that is not a heart?

For problems 3 and 4, consider the experiment in which a person randomly chooses one month of the year and records the month.

3. What is the sample space for this experiment?

4. What is the probability of choosing:(a) a summer month (June, July or August)?(b) a month beginning with the letter J?(c) a month ending with the letter y?(d) a month that does not end in the letter y?(e) a month other than June?(f) a month beginning with the letter J or a month ending with y?(g) a month that is neither March, April nor May? (h) the month with your birthday in it?

5. Suppose that there are 22 students in a room sitting in seats numbered 1 through 22 and you pick one student at random and announce his/her first name. Is this enough information


to write down the sample space for this experiment? Exactly what information would you need to compute the probability of choosing a person named Suzanne?

For problems 6 and 7, consider the game in which you toss a coin three times in a row and write down whether it landed “heads” or “tails” each time. (A typical outcome would be “HHT”).

6. Write down the sample space S for this experiment. Compute n(S) ?

7. What is the probability that in three consecutive tosses you would get (a) HHT ?(b) all tails?(c) all heads?(d) a head on the first toss? (e) at least two tails?(f) exactly one tail?(g) at least one head or all tails?(h) exactly one tail or at least 2 heads?(i) one tail and 2 heads?(j) all tails and one head?

For problems 8 and 9, consider the game in which you roll a pair of ordinary (six-sided) dice and record the sum of the numbers showing on top (as in a board game).

8. Write down the sample space for this experiment; below each outcome in the sample space, write down the probability of each individual outcome.

9. What is the probability of rolling(a) a five?(b) an even number?(c) a number bigger than (or equal to) six?(d) a prime number?(e) a three or a nine?(f) a seven or an eleven?(g) an even number or a number greater than 7?(h) an odd number and a number less than 7?(i) a number that is not a 10?

For problems 10 and 11, consider the experiment in which you draw a marble at random from a bag that contains 2 red marbles, a green marble and three yellow marbles, and you record its color.


10. Write down the sample space for this experiment. Are all of the outcomes equally likely?

11. What is the probability of drawing(a) a red marble?(b) a green marble?(c) a marble that is not yellow?(d) a red or yellow marble?


Solutions

1. {A§, 2§, 3§, 4§, … , 10§, J§, Q§, K§ A¨, 2¨, 3¨, 4¨, … , 10¨, J¨, Q¨, K¨ A©, 2©, 3©, 4©, … , 10©, J©, Q©, K© Aª, 2ª, 3ª, 4ª, … , 10ª, Jª, Qª, Kª} [52 cards in all]

2. (a) 1/52 (b) 1/52 (c) 12/52 = 3/13 (d) 40/52 = 10/13 (e) 26/52 = 1/2 (f) 13/52 = 1/4 (g) 20/52 = 5/13 (h) 22/52 (i) 52/52 (j) 12/52 (k) 2/52 = 1/26 (l) 28/52 (m) 32/52 (n) 48/52 (o) 39/52

3. {January, February, March, April, May, June, July, August, September, October, November, December}

4. (a) 3/12 (b) 3/12 (c) 4/12 (d) 8/12 (e) 11/12 (f) 5/12 (g) 9/12 (h) 1/12.

5. No, we do not know the first names of the students. To calculate this probability, we need to know how many persons named Suzanne there are in the room.

6. {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}; n(S) = 8.

7. (a) 1/8 (b) 1/8 (c) 1/8 (d) This is the event {HHH, HHT, HTH, HTT}, so the probability is 4/8 = 1/2 (e) This is the event {HTT, THT, TTH, TTT}, so the probability is 4/8 = 1/2. (f) 3/8 (g) 8/8 = 1 (h) 4/8 (i) 3/8 (j) 0/8 = 0

8. S = {2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}

Outcome 2 3 4 5 6 7 8 9 10 11 12Probabilit

y1/36 2/36 3/36 4/36 5/36 6/36 5/36 4/36 3/36 2/36 1/36

9. (a) 4/36 (b) 18/36 = 1/2 (c) 26/36 = 13/18 (d) 15/36 (2 is prime!) (e) 6/36 = 1/6 (f) 8/36 = 2/9 (g) 30/36 (h) 6/36 = 1/6 (i) 33/36.

10. (a) S = {red, green, yellow} (b) No, picking a yellow is more likely than picking a green.

11. (a) 2/6 = 1/3 (b) 1/6 (c) 3/6 = 1/2 (d) 5/6.

Probability 3.4 Mutually Exclusive and Independent Events

3.3 Conditional Probability

In this section, we will be answering questions of the form, “what is the probability of event E happening, given that event F has occurred”? This is an example of a conditional probability, meaning, a probabilistic situation in which it is known, or can be assumed, that certain events have transpired. Having extra knowledge will usually change the assessment of the likelihood of an event happening. For instance, suppose that you were asked whether or not your favorite football team will win the Super Bowl this year and that you think their chances are quite good. Once you find out that four of their best players are out for the season with injuries, this will probably reduce your estimate.

We will be using two techniques to compute a conditional probability. One is to “reason” it out and the other is to apply the conditional probability formula, which we will introduce shortly. For most situations that we will see, either method will work. Whenever practical, we will demonstrate both methods. You may find one method consistently preferable over the other. However, there will be situations in which only one of the methods will work, so, you should be proficient with each of them.

The reasoning method is really just organized counting. The key is to know what to count. As an example, consider the experiment in which one picks a card form an ordinary deck of 52 playing cards and announces the card. The sample space for this experiment is

S = {Aª, 2ª, 3ª, 4ª, 5ª, 6ª, 7ª, 8ª, 9ª, 10ª, Jª, Qª, Kª A§, 2§, 3§, 4§, 5§, 6§, 7§, 8§, 9§, 10§, J§, Q§, K§ A¨, 2¨, 3¨, 4¨, 5¨, 6¨, 7¨, 8¨, 9¨, 10¨, J¨, Q¨, K¨ A©, 2©, 3©, 4©, 5©, 6©, 7©, 8©, 9©, 10©, J©, Q©, K©},

where J, Q, K denote Jack, Queen, King, respectively. The suits spades (ª) and clubs (§) are black while the suits diamonds (¨) and hearts (©) are red.

The probability of picking a red ace is 2/52, because there are two red aces (A¨ and A©) out of a total of 52 equally-likely cards (13 in each suit). But what is the probability of picking a red ace, given that a diamond has been chosen?

Since we know that the card chosen is a diamond, it must be one of these 13 cards:

{ A¨, 2¨, 3¨, 4¨, 5¨, 6¨, 7¨, 8¨, 9¨, 10¨, J¨, Q¨, K¨}

There is only one red ace (A¨) in this set of cards and they are all equally likely. So, the probability of picking a red ace, given that a diamond has been chosen, is 1/13.

Probability 3.1 - 3.4 Problems/Solutions

The key to solving this problem was to use the conditional information (the fact that a diamond had been chosen) to effectively reduce the sample space from all 52 cards to the set of 13 diamonds, then to count the number of red aces in this set (there was only one). Sometimes, this shrinking and counting process cannot be done so easily. The purpose of the conditional probability formula, given below, is to break up the calculation of a conditional probability into a series of smaller, more manageable calculations.

Conditional Probability Formula

P E FP E F

P F|

Ç

whenever P(F) ¹ 0

The expression P( E | F ) stands for “the probability that event E occurs given that event F has occurred”. Thus, the conditional probability formula can be read as, “the probability that event E occurs given that F has occurred is the probability that E and F occur simultaneously divided by the probability that F occurs”.

Let’s repeat our conditional probability calculation using the formula to see that it gives the same result as before (1/13). We wish to compute the probability of picking a red ace, given that a diamond has been chosen. We denote,

E = event of picking a red ace and F = event of picking a diamond.

We wish to compute the numerical expression P( E | F ) using the formula

P E FP E F

P F|

Ç.

We will need to compute the numerator and denominator of the fraction. In order to compute the numerator, P( E Ç F ), we must count the number of cards in E Ç F and divide by 52. First, we explicitly write out the contents of the sets E and F:


E = {A¨, A©} and F = { A¨, 2¨, 3¨, …, 10¨, J¨, Q¨, K¨}.

These sets have only the ace of diamonds in common. That is,

E Ç F = { A¨}.

There is one element in E Ç F, so, P( E Ç F ) = 1/52. Next, we compute the denominator, P(F). Since there are 13 cards in F, P(F) = 13/52. In all, we conclude that

P E F

P E FP F

| Ç

152

1352

1 152

13 152

113

.

Indeed, we get the same answer as before. Note that the use of the formula required us to first count the number of red aces in the set of diamonds (1), then the number of diamonds (13). This is the same counting process we went through when we “reasoned” out the answer.

Tip: Numbers versus sets. In the above example, we were careful to distinguish numbers from sets. For instance, P(E Ç F ) was a number, while E Ç F was a set.

Example 1: Suppose that there are 21 students in a room, each with a distinct first name, and that one of the seven students in the front row is named Darlene. We pick a student at random and announce their first name.

Q1: What is the probability of choosing Darlene?

A1: There is only one Darlene out of a total of 21 persons, each of whom is equally-likely to be chosen. So the probability of choosing Darlene is 1/21.

Q2: What is the probability of choosing Darlene given that the person chosen sits in the front row?

A2 (reasoning): Once we know that the person chosen sits in the front row, it must be one of 7 persons. There is only one Darlene amongst those 7 so the probability of choosing Darlene given that the person chosen sits in the front row is 1/7.

B


A2 (formulaic): Let D be the event of choosing Darlene and let F be the event of choosing a person from the front row. We want to find P ( D | F ) using the conditional probability formula:

P D FP D F

P F|

Ç

We will need to compute P(D Ç F) and P(F). D Ç F is the event of choosing a person from the front row and choosing Darlene. There is exactly one person with this property (Darlene) out of a total of 21 persons, so, P(D Ç F) = 1/21. F is the event of choosing a person from the front row. There are 7 such persons out of a total of 21 persons, so, P(F) = 7/21. In all, we have that

P D F

P D FP F

| Ç

1

217

21

1 121

7 121

17

.

Q3: What is the probability of choosing someone in the front row given that Darlene is chosen?

A3 (reasoning): We have been told that the person chosen is Darlene. Since Darlene sits in the front row, the probability of choosing someone in the front row given that Darlene was chosen is 1 [i.e., this person chosen definitely sits in the front row].

A3 (formulaic): Let F be the event of choosing a person from the front row and let D be the event of choosing Darlene. We want to find P ( F | D ) in the conditional probability formula:

P F DP F D

P D|

Ç

We will need to compute P(F Ç D) and P(D). Since F Ç D is the same as D Ç F, we see that P(D Ç F) = P(F Ç D). In Q2, we computed that P(D Ç F) = 1/21. So, P(F Ç D) = 1/21. Next, we compute P(D). This is the probability of choosing Darlene, which is simply 1/21. In all, we have that


P F DP F D

P D|

Ç

121

121

1 n

Tip: When computing a conditional probability by reasoning it out (counting), start with the conditional fact. For instance, when computing “the probability of event A occurring, given that B has occurred”, use the fact that event B has occurred to reduce

the sample space. Then, compute the probability of event A happening in this reduced sample space. It is rarely productive to first compute “the probability of event A happening”, because the effect of event B has not been taken into account.

Example 2: Consider the experiment in which one draws a card at random from a deck of 52 playing cards and then announces the card.

Q1: What is the probability of picking an ace given that a black card is chosen?

A1 (reasoning): We start with the conditional information that a black card has been chosen. It must be one of the following 26 (black) cards:

Aª, 2ª, 3ª, 4ª, 5ª, 6ª, 7ª, 8ª, 9ª, 10ª, Jª, Qª, Kª A§, 2§, 3§, 4§, 5§, 6§, 7§, 8§, 9§, 10§, J§, Q§, K§

We have reduced our sample space to the set of black cards (by eliminating all of the red cards). Note that we have also eliminated the 2 red aces. We are left with two aces (Aª and A§) out of a total of 26 cards. So we conclude that the probability of picking an ace given that the card is black is 2/26.

A1 (formulaic): Let A be the event of picking an ace and let B be the event of picking a black card. We wish to compute P (A | B) in the conditional probability formula:

P A BP A B

P B|

Ç

We must first compute P( A Ç B ) and P(B). The event A Ç B is the event of choosing an ace and a black card. There are only two such cards (A§, and Aª), so, P ( A Ç B ) = 2 / 52. B is the event of picking a black card. There are 26 such cards, so, P(B) = 26/52. In all, we obtain

B

Probability of picking The Godfather and a drama = 1/8


P A B

P A BP B

| Ç

2

5226

52

2 152

26 152

226

113

.

n

Be careful not to “tinker” with the wording of a conditional probability question too much when trying to interpret it. In conditional probability, the phrase, “given that” is not interchangeable with “and”. For instance, suppose that you have eight different movie rentals to choose from. Three of them are dramas, and five of them are comedies. One of the dramas is the movie, The Godfather. You can’t decide which movie to watch tonight, so you choose one of these eight movies at random. What is the probability that the movie you choose is The Godfather AND is a drama? Since The Godfather is a drama and there is only one such movie out of a total of eight movies, the probability is 1/8.

Now, what is the probability that the movie you choose is The Godfather, GIVEN THAT you choose a drama? If you definitely choose a drama, it can only be one of three movies (the three dramas). The Godfather is just one movie out of these three, so the probability is 1/3.

The Godfather

Drama 2

Drama 3

Comedy 1

Comedy 2

Comedy 3

Comedy 4

Comedy 5

The Godfather

Drama 2

Drama 3

Probability of picking The Godfather AND a drama = 1/8


Probability of picking The Godfather, given that a drama is chosen= 1/3

Probability of picking The Godfather, given that a drama is chosen= 1/3

So, the probability that you choose the movie The Godfather AND a drama is not the same as the probability that you choose The Godfather, GIVEN THAT you choose a drama. The point is this: in conditional probability, the phrase “given that” is not interchangeable with word, “and”.

Probability 3.5 Expected Value

3.4 Mutually exclusive and Independent Events

Two events E and F are mutually exclusive if they rule each other out, meaning whenever E happens, F cannot happen and whenever F happens, E cannot happen. Sometimes, one can see that two events are mutually exclusive without performing any calculations. For instance, if E is the event of being in Chicago at a given time and F is the event of being in Boston at the same time, then E and F are mutually exclusive because a person cannot be in two places at the same time. When the events become more complicated, it can be difficult to determine by inspection whether or not two events are mutually exclusive. So, we will rely on the following mathematical definition:

Definition: Two events, E and F are mutually exclusive if and only if P( E Ç F ) = 0.

One can think of the above definition as a test. In order to apply the test to two sets, E and F, we will first compute P( E Ç F ), then check whether or not it equals zero. If it does, we will conclude that the sets are mutually exclusive. Otherwise, we will conclude that they are not.

A slightly different notion, called independence, is whether or not the occurrence of one event affects the occurrence of another event. Often, there is no intuition for whether or not two events are independent. But once again, we have a mathematical test to fall back on:

Definition: Given two events, E and F, consider the quantities P( E Ç F ) and P(E) ´ P(F). We say that events E and F are independent if and only if these two quantities are equal.

Another way to phrase this is that two events are independent if and only if the probability that they occur simultaneously, i.e., P( E Ç F ), is the same as the product, i.e., P(E) ´ P(F), of their probabilities.4 The above definition will be used in two ways:

(1) As a test. In order to answer the question of whether or not two events E and F are independent we will calculate, then compare, two numbers: P (E Ç F) and P(E) ´ P(F). If the two numbers are equal, then we will conclude that E and F are independent and if the two numbers are not equal, we will conclude that the events are not independent.

(2) As a formula. If we know in advance that two events, E and F, are independent then we can find the probability of them happening together by multiplying their probabilities.

4 An alternative definition of independence is as follows. Two events, E and F are independent if one of these cases holds: Case 1: P(E|F) = P(E) and P(F) ¹ 0 or, Case 2: P(F) = 0.

A

Warning: It’s important to realize that the terms independent and mutually exclusive are not synonymous. Example 1 demonstrates that it is possible to have two events that are independent but not mutually exclusive, while Example 2 demonstrates that

it is possible to have two events that are mutually exclusive but not independent.

Example 1: Consider the experiment in which one rolls a pair of dice and records the sum of the numbers on top. Let S be the sample space for this experiment and let E be the event of rolling a 2 or 3.

Q1: Are E and S mutually exclusive?

A1: By the definition (test) of mutually exclusive events, we need to check whether or not P( E Ç S ) = 0. First, we find the set E Ç S. Since

E = {2, 3} and S = {2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12},

we see that

E Ç S = E = {2, 3}.

The outcomes in S are not equally likely so we cannot simply count the number of elements in E and divide by the number of elements in S. Recall the probability distribution table for this experiment:

Outcome 2 3 4 5 6 7 8 9 10 11 12probability 1/36 2/36 3/36 4/36 5/36 6/36 5/36 4/36 3/36 2/36 1/36

We add the probabilities of 2 and 3 from the table. 5 That is,

P (E Ç S) = P ( {2, 3} ) = 136

236

336

.

Since P (E Ç S) = 3/36 is not equal to zero, we conclude that E and S are not mutually exclusive events.

Q2: Are E and S independent events?

A2: We turn to the definition of independent events. E and S are independent if and only if 5 Justification of the addition: Since {2} Ç {3} = { }, these events have empty intersection. The union formula tells us that we can add these probabilities directly. In general, the probabilities of any outcomes from the sample space can be added.

A

P ( E Ç S ) = P ( E ) ´ P ( S ).

In the previous answer, we computed that P (E Ç S) = P ( {2, 3} ) = 3/36. Also, we noted that P(E) = P (E Ç S) = 3/36. Since P(S) = 1 (this is true for any sample space), we see that

P (E Ç S) = 3

36

and

P(E) = P ( E ) ´ P ( S ) = 3

361 3

36´ .

Therefore, P (E Ç S) = P ( E ) ´ P ( S ) and these events are independent. n

Example 2: Consider the deck of cards experiment, in which one card is randomly chosen from a deck of 52 cards and announce the card. Consider the events D of picking a diamond, E of picking an eight, and F of picking a five.

Q1: Are the events D and E mutually exclusive?

A1: To answer this, we calculate P ( D Ç E ). Now D Ç E is the event of picking a card that is both a diamond and an eight simultaneously, i.e., D Ç E = {8¨}. So, P ( D Ç E ) = 1/52. Since 1/52 is not equal to zero, we conclude that the events D and E are not mutually exclusive.

Q2: Are the events E and F mutually exclusive?

A2: To answer this, we calculate P ( E Ç F ). The event E Ç F is the event of choosing a card which is a five and an eight simultaneously. However, there is no such card. This means that E Ç F = { }and P ( E Ç F ) = 0. We conclude that E and F are mutually exclusive.

Q3: Are the events D and E independent?

A3: To do this, we calculate P ( D Ç E ) , which we saw is 1/52. We then calculate

P ( D ) ´ P ( E ) =3

524

5252

27041

52´ .

This is, however, equal to 1/52 (verify this). So we conclude that the events D and E are independent.

A

Q4: Are the events E and F independent?

A4: We saw that P ( E Ç F ) = 0. Now

P E P F´ ´ 4

524

5216

2704.

Since these two numbers are not equal, we conclude that E and F are not independent. n

In the next example, we will use the fact that if we know in advance that two events are independent, then we can multiply their respective probabilities to get the probability that they will happen simultaneously.

Example 3: Main Street in Humphreyville has two traffic lights which operate independently. The first light is green 70% of the time, red 25% of the time and yellow 5% of the time. The second light is green 50% of the time, red 46% of the time and yellow 4% of the time. At a random time, the colors of both lights are recorded.


A1: One possible outcome is that both lights are green, which we denote GG. Another possible outcome is GR. Continuing in this fashion, we generate the entire sample space:

S = {GG, GR, GY, RG, RR, RY, YG, YR, YY}

Note that these outcomes are not equally likely, since it is more likely, for instance, that both lights are green than both being yellow (look at their probabilities).

Q2: What is the probability of both lights being yellow?

A2: Since these events are independent, the probability of them happening together is the product of their probabilities. The probability that the first light is yellow is 0.05 while the probability that the second light is yellow is 0.04. Multiplying these two numbers, we conclude that the probability that both lights are yellow is (0.05)(0.04) = 0.002.

Q3: What is the probability that at least one of the lights is not green?

A

A3: This is the same as the probability that at least one of the lights is red or yellow. Let E be the event that at least one of the lights is red or yellow. Recall the sample space for this experiment:

S = {GG, GR, GY, RG, RR, RY, YG, YR, YY}

Then E is every outcome in the sample space except for GG:

E = {GR, GY, RG, RR, RY, YG, YR, YY}.

We could compute the probability of every outcome in E then add them up but it would be simpler to compute the probability of the complement of E,

P(EC) = P({GG}).

Again, we use the fact that the lights operate independently. The probability that the first light is green is .70. The probability that the second light is green is .50. Therefore, the probability that both lights are green is P(EC) = (.70)(.50) = .35. Since P(EC) and P(E) must sum to 1, P(E) = 1 - .35 = .65. In all, we conclude that the probability that at least one of the lights is not green is .65. n

Working from a Table

Having data stored in tabular form can greatly facilitate the computation of probabilities. Often times, one can form the required ratios simply by selecting entries from the table. The trick is to know which numbers to select.

Example: The table below summarizes the ages and smoking habits of 1032 persons surveyed.

Age bracket Smoker Non-smoker Total18-23 30 63 9324-29 47 95 14230-35 46 99 14536-41 44 89 13342-53 73 130 203

over 53 68 248 316Total 308 724 1032

What is the probability that one person randomly chosen from the survey is

A

(a) a smoker?(b) a person between the ages of 30-35 (inclusively)?(c) at least 42 years old?(d) a person who is 24-29 years old or a smoker?(e) a person who is not over 53 years old?(f) Are the events of being a smoker and a non-smoker mutually exclusive?(g) Are the events of being a smoker and a non-smoker independent?(h) What is the probability of choosing someone in the age bracket 36-41, given that they are a non-smoker?

Solutions:(a) 308/1032; there are 308 smokers out of a group of 1032 (each of the 1032 are equally likely to be chosen).

(b) 145/1032; there are 46 + 99 = 145 persons between the ages of 30-35 (inclusively).

(c) 519/1032; there are 519 persons 42 or older. Note: 519 = (73 + 130 + 68 + 248 ).

(d) There are (47 + 95) = 142 persons between the ages of 24 and 29 [add up the numbers in this row]. There are 308 smokers [sum of the smoker column]. These two groups have 47 persons in common, so, the number of persons that are between ages 24 and 29 or a smoker is the number of persons in the union of these two groups which is 142 + 308 - 47 = 403, by the inclusion-exclusion principle. So, the probability is 403/1032.

(e) Let E be the event of not being over 53. Then EC is the event of being over 53. Since, P(E) + P(EC) = 1, and P(EC) = 316/1032, P(E) must equal 716/1032. (f) Let S = the event of choosing a smoker and let N = the event of choosing a non-smoker. By the definition of mutually exclusive events, we wish to compute P( S Ç N ) and check whether or not it equals zero. Since it is not possible to be both a smoker and a non-smoker, this set is empty and we see that P( S Ç N ) = P({ }) = 0. Our conclusion is that, yes, these events are mutually exclusive.

(g) Again, let S = the event of choosing a smoker and let N = the event of choosing a non-smoker. By the definition of independent events, we need to check whether or not compute P( S Ç N ) = P(S) ´ P(N). We have already computed that P( S Ç N ) = 0. Since there are 308 smokers and 724 non-smokers, P(S) = 308/1032 and P(N) = 724/1032. Thus,

P(S) ´ P(N) = 308/1032 ´ 724/1032

A

We do not need to multiply these numbers to see that their product in non-zero. Since P( S Ç N ) = 0, we find that

P( S Ç N ) ¹ P(S) ´ P(N)

and, therefore, these events are not independent.

(h) (reasoning): We know that the person is a non-smoker so it must be one of 724 persons. In essence, we have circled the non-smoker column in the table and disregarded the rest of the table. This corresponds to shrinking the sample space. Out of these persons, there are 89 in the age range of 36-41, so the probability of picking someone who is 36-41 given that they are a non-smoker is 89/724.

(h) (formulaic):

PP

Pperson is36 - 41 | non - smoker

person is36 - 41 non smokernon smoker

Ç -

-

891032

7241032

89 11032

724 11032

89724

.

n


Problems

1. A traffic light on Elm St. is red 50% of the time, green 40% of the time and yellow 10% of the time. Consider the experiment in which a person (at a random time) records the color of the light.(a) What is the sample space for this experiment? (b) Are the outcomes equally likely?(c) What is the probability that the light is green?(d) What is the probability that the light is not yellow?(e) What is the probability that the light is green or not yellow?

2. Consider the experiment in which one draws at random from a deck of 52 playing cards.(a) What is the probability of drawing a red card or a jack?(b) What is the probability of drawing an ace or a black card?(c) What is the probability of drawing a card that is black or not an ace?(d) What is the probability of drawing a card that is black and not an ace?(e) What is the probability of drawing an ace given that a red card has been drawn?(f) What is the probability of drawing a numbered card between 3 and 6 (inclusively) given that a numbered card has been drawn?(g) What is the probability of drawing a face card given that a queen has been drawn? (h) What is the probability of drawing a queen given that a face card has been drawn?(i) Let F be the event of drawing a face card and J be the event of drawing a jack. Are F and J mutually exclusive? Are they independent?(j) Let N be the event of drawing a numbered card and F be the event of drawing a face card. Are these events mutually exclusive? Are they independent?

3. A survey was made of 125 students. Each of 125 students was asked their opinion on a television commercial for Batch Beer. The results are shown below.

Liked commercial Did not like commercialFreshman 7 9

Sophomore 11 32Junior 12 13Senior 33 8

One of the 125 persons is selected at random. Let J be the event of choosing a Junior and L be the event that the person chosen liked the commercial. (a) P(J Ç L) = ?


(b) P(J È L) = ?(c) P( J | L ) = ?(d) P( L | J ) = ?(e) P( J Ç LC) = ?(f) What is the probability of picking a person who did not like the commercial who is not a Junior?(g) What is the probability of picking someone who liked the commercial given that the person chosen was a Senior?

4. A room with 21 students has 2 English majors, 5 History majors and 1 person who is both a History and English major. One student is chosen at random.(a) What is the probability of choosing a person that is an English major or a History major but not both ?(b) What is the probability of choosing a History major given that an English major was chosen ?(c) What is the probability of not choosing a History major given that an English major was not chosen? (d) What is the probability of choosing a History major given that a non-History major was chosen?

5. Consider the experiment in which one rolls a blue six sided die and a red six-sided die and observes the number showing on the top of each of the faces. (a) What is the probability of rolling a sum of eight?(b) What is the probability of rolling a 5 on the red die?(c) What is the probability of rolling a sum of eight given that a 5 was rolled on the red die? (d) What is the probability of rolling a 5 on the red die given that the sum of the dice is eight?(e) What is the probability of rolling a sum of seven?(f) What is the probability of rolling a sum of seven given that a 5 was rolled on the red die?(g) In light of (e) and (f), do the chances of rolling a seven change when you are told that a 5 was rolled on the red die?

6. In a certain state lottery, there are 40 ping-pong balls, each numbered from 1 to 40. A vacuum cleaner-like device randomly picks four balls without replacement. A lottery ticket wins the grand prize of $1,000,000 if it has each of the numbers in the correct order in which they were chosen.

(a) How many elements are in the sample space? Note: the ticket 1, 4, 5, 24 is different from the ticket 24, 1, 4, 5.Suppose that you have the lottery ticket 13, 1, 36, 21. (b) What is the probability that you will win the grand prize?


(c) What is the probability that you will win the grand prize given that the first number chosen was a 13 and the second number chosen was a 1?

7. Consider the experiment in which a card is chosen at random from a deck of 52 playing cards. Let E be the event of picking an eight; let F be the event of choosing a club; let G be the event of picking the queen of clubs.(a) Are the events E and F mutually exclusive?(b) Are E and F independent?(c) Are the events F and G independent?(d) Are E and G independent?(e) Are E and G mutually exclusive?

8. A survey was made of 200 CEO’s (corporate executive officers). The results are shown below.

Had an MBA Did not have an MBAUnder 40 4 3

40-49 54 3550-59 45 22

60 or over 3 34

One of the executives is selected at random.

(a) Let M be the event that the CEO has an MBA and let F be the event that the CEO was age 40-49. Are the events M and F mutually exclusive?(b) Are the events M and F independent? (c) Are the events M and MC mutually exclusive?(d) Are the events M and MC independent?(e) What is the probability that the CEO chosen is 60 or over?(f) What is the probability that the CEO chosen is 60 or over, given that the CEO does not have an MBA?(g) Let A be the event A that the CEO is 60 or over and let B be the event that the CEO does not have an MBA. Are the events A and B independent?

9. Suppose that two traffic lights operate independently. The first traffic light is green 50% of the time, red 40% of the time, and yellow 10% of the time. The second traffic light is green 30% of the time, red 65% of the time, and yellow 5% of the time. Suppose that at a random time, the color of both of the traffic lights is recorded.(a) What is the sample space for this experiment?


(b) What is the probability that both lights are yellow simultaneously? (c) What is the probability that the first light is green and the second is yellow?(d) What is the probability that neither light is yellow?

10. Two life-support systems on a spacecraft operate independently. System 1 malfunctions with a probability of .001 and system 2 malfunctions with a probability of .0001. (a) What is the probability that both systems will malfunction? (b) What is the probability that neither system will malfunction?(c) What is the probability that system 1 malfunctions while system 2 does not?

11. Consider the experiment in which two cards are chosen from a deck of 52 cards without replacement. The first card picked is recorded and then the second card picked is recorded.(a) How many pairs can be chosen?Let P be the event of choosing two cards that are the same denomination (e.g. two fours).Let K be the event of choosing a king for the first card.Let J be the event of choosing the jack of diamonds for the second card.Let D be the event of choosing a diamonds for the first card.(b) Are P and K independent?(c) Are P and K mutually exclusive?(d) Are K and J independent?(e) Are K and J mutually exclusive?(f) Are J and D independent?

Thinkers

12. Mutually Exclusive vs. Independent Events. Review the definitions of mutually exclusive and independent events. Then answer the following questions.(a) Is it possible to have two events that are mutually exclusive but not independent? Why or why not? (b) Is it possible to have two events that are independent but mutually exclusive? Why or why not? (c) Is it true that an impossible event (probability 0) is independent of all other events?(d) Is it true that an impossible event (probability 0) is independent of all other events?

13. Equivalent Definitions? A proper definition of (test for) independent events should not depend upon the order in which the events are presented. In other words, if the test concludes that A and B are independent, then, when the test is applied to B and A, should reveal that B and A are also independent. Some text math books (incorrectly) define independent events as follows.

[Faulty] definition: Two events, A and B, are independent if (and only if) P(A | B) = P(A).


Construct an example of two events, A and B, with the property that P(A | B) = P(A) and yet P(B | A) ¹ P(B). Note that, by the faulty definition, we are forced to conclude that A and B are independent and that B and A are not independent.

14. In a sentence or two, explain why it is necessary to assume that P(B) is not equal to zero in order to consider P(A | B).

Web

15. The Monty Hall Problem. There was a television game show called “Let’s Make a Deal” in which, you, the contestant would be placed in the following dilemma. A prize is hidden behind exactly one of three doors; the other two contain a “zonk” (undesirable) prize. The object of the game is to select the door with the prize behind it. But, after you make your initial choice, the game show host, Monty Hall, will open one of the doors (that the prize is not behind) and gives you the opportunity to change your selection. Should you switch? A syndicated columnist (Marilyn vos Savant) was heavily criticized by the mathematical community when she published an answer to this question in which she stated that, yes, you should switch. Her reasoning was that the probability that you have picked the correct door started out at 1/3 (all equally likely doors) and remains 1/3 after the Monty opens one of the other doors. However, once Monty has opened a door, she argues, the probability that the prize is behind the door you did not select goes up to 2/3 and that, therefore, you should switch. Many (but not all) of the mathematicians argued that the prize is still equally likely to be behind either of the two remaining doors (probabilities 1/2, 1/2) so that it makes no difference if you switch or not. Who is right? One way to help you settle this debate is to simulate the game. Have one classmate play the role of Monty Hall and have other classmates take turns playing of the contestants. For each play of the game, record the contestant’s strategy (switch or stay) and whether or not it was successful. Note: You’ll find relevant discussions and automated simulations on the web. Do a search on “Monty Hall problem”. 6

16. Infinite Coin Toss. The infinite coin toss is the game in which you flip a coin infinitely many times and record the result. (Of course, in reality, you can’t complete even one trial of this experiment, but we can still think about it.) It is possible that you get all heads but, intuitively, it seems unlikely that this would happen. Write a convincing (English) argument

6 There are two key (implicit) assumptions about this game that you should be aware of. One is that Monty is obligated to open a zonk door and the other is that he knows which door the prize is behind. By the way, Monty Hall was later asked who was right. He replied that neither argument applies on his game show because, in fact, he is not obligated to open any of the doors!


that it is extremely unlikely to get all heads on the infinite coin toss. Be sure to back up your argument with some probability calculations. To get started, compute the probability of getting all heads on the 3-coin toss, then on the 4-coin toss, and so on, until you see a pattern.

17. Monkeys on Typewriters. Suppose that, at random, a monkey types characters one-by-one on a keyboard. For simplicity, assume that the only characters the monkey can type are lower-case letters and a space (27 characters, in all). Also, assume that each keystroke made by the monkey is independent of all others.(a) If the monkey types exactly 4 characters, what is the probability that he has typed the word “four”? (b) If the monkey types exactly 13 characters, what is the probability that he has typed the phrase, “fourscore and”? (c) The Gettysburg Address7 contains 1397 characters, including spaces but ignoring punctuation. If the monkey types exactly 1397 characters, what is the probability that the monkey types the Gettysburg Address? Confirm that your scientific calculator (hint: use exponents) gives an answer of 0. Based on your calculator’s result, can you conclude that the event that the monkey types the Gettysburg Address is impossible? Explain in a short paragraph.

7 The Gettysburg Address is a famous speech delivered by Abraham Lincoln in 1863 that begins with the phrase, “fourscore and seven years ago”. One of many websites that display the Gettysburg Address is http://www.pathfinder.com/photo/week/address.htm


Solutions

1. (a) S = {red, green, yellow} (b) No, the light is red with higher probability than it is yellow. (c) 40% = 40/100 = 2/5 (d) 90% = 90/100 = 9/10 (e) 90% = 9/10.

2. (a) 28/52 (b) 28/52 (c) 50/52 (d) 24/52 (e) 2/26 = 1/13 (f) 4/10 = 2/5 (an ace is considered numbered) (g) 4/4 = 1 (it is certain) (h) 4/12 = 1/3 (i) No they are not mutually exclusive because the set J Ç F = {J§, J¨, J©, Jª} is not empty and therefore, P(J Ç F) ¹ 0. No they are not independent because P(J) ¹ P( J | F ) [Note: P(J) = 4/52 and P(J | F) = 4/12]. (j) Yes these are mutually exclusive since P(N Ç F) = 0 [Note: N Ç F = { }]. No, they are not independent because P(N) ¹ P( N | F ). [Note: P(N) = 40/ 52 and P(N | F) = 0/12]

3. (a) J Ç L is the event of choosing a Junior who likes the commercial. There are exactly 12 such persons out of a total of 125, so, P( J Ç L ) = 12/125. (b) J È L is the event of choosing someone who is a Junior or liked the commercial (or both). The union (“or”) formula reveals:

P J L P J P L P J LÈ - Ç - 25

12563

12512125

76125

(c) The fact that the person liked the commercial reduces the sample space to 63 students. Of these 63, there are 12 who are juniors, so P ( J | L ) = 12/63.(d) Given that a junior was chosen, we reduce our sample space to the 25 juniors. There are 12 of these people who liked the commercial. So, P ( L | J ) = 12/25.(e) J Ç LC is the event of choosing someone who is a junior and did not like the commercial. There are 13 such persons out of a total of 125. So, P ( J Ç LC ) = 13/125.(f) There are 9 + 32 + 8 = 49 such persons. So the probability is 49/125. (g) We are told that a senior was picked, so we reduce our sample space down to the 41 seniors. Of these seniors, 33 liked the commercial. So, P ( L | S ) = 33/41.

4. We can use the following Venn diagram to solve this problem.

E H

1 1 4

15


(a) There is 1 English major that is not a History major and 4 History majors that are not English majors so the probability is 5/21. (b) Since we are given that an English major was chosen, we reduce our sample space to the two English majors. Of these 2 persons, one is also a History major. So the probability of choosing a History major given that an English major was already chosen is ½. (c) Since we know that an English major was not chosen, we reduce our sample space to the 19 persons (21 - 2) that are not English majors. Of these 19, 4 are History majors, and 15 are not History majors. So, the probability is 15/19.(d) We know that a non-history major was chosen and there are 16 such persons. Of these 16, none are History majors, so this probability is 0/16 or simply 0.

5. (a) 5/36. (b) 6/36(c) We are given that a 5 was rolled on the red die. So we reduce our sample space to the 6 outcomes : (1,5) (2,5) (3,5) (4,5) (5,5) and (6,5) [here the first entry is what is rolled on the blue and the second entry is what is rolled on the red ]. Of these, there is only 1 way to roll a sum of eight (3,5) so the probability is 1/6. (d) We are given that the sum is an eight, this reduces our sample space to: (2,6) (3,5) (4,4) (5,3) and (6,2). There is only one way to roll a 5 on the red die with this reduced sample space, namely (5,3) so the probability is 1/5.(e) 6/36(f) We are given that a 5 was rolled on the red die, so our reduced sample space is: (1,5) (2,5) (3,5) (4,5) (5,5) and (6,5). There is one way to roll a sum of seven in this sample space, so the probability is 1/6.(g) The probability did not change. In particular, knowing that a 5 was rolled did not increase or decrease the chances of rolling a 7.

6. (a) There are 40 possible outcomes for the first ball. Having removed one of these, there are 39 possibilities for the second ball. Similarly, there are 38 possibilities for the third ball, and 37 for the fourth. So the number of total outcomes is 40´39´38´37 = 2,193,360.(b) 1 / 2,193,360(c) The ticket agrees with the numbers on the first two balls chosen so we can turn our attention to the last two balls. The third ball could be one of 38 balls (the 13 and the 1 have been removed). The fourth ball could be any one of the 37 remaining balls. Thus, there are 38 ´ 37 = 1406 possible outcomes for the remainder of the game. You are holding one lottery ticket so the probability of winning is 1/1406.

7. (a) E Ç F is the event of picking a card that is both an eight and a club. That is, E Ç F = {8§}. Since P( E Ç F ) = 1/52, which is not equal to zero, E and F are not mutually exclusive.(b) Yes, E and F are independent events. To see this, calculate


P E P F´ ´ 4

521352

527024

152

.

But P( E Ç F ) = 1/52 so P( E Ç F ) = P( E ) ´ P( F ).(c) The event F Ç G is the event of choosing a card that is a club and the queen of clubs which is the same as choosing the queen of clubs. Thus, F Ç G = {Q§}.

P( F Ç G ) = 1/52 P( F ) ´ P( G ) = 13/52 ´ 1/52 = 13/2704.

But 13/2704 does not equal 1/52, so these events are not independent. (d) E and G are not independent. E Ç G is the event of picking a card that is an eight and the queen of clubs. There is no such card, so, E Ç G = { } and P( E Ç F) = 0. P( E ) = 4/52, and P( G ) = 1/52, so we see that P( E ) ´ P( G ) = 4/2704 which is not equal to P( E Ç G ), which is zero. (e) As in (d), we see that P( E Ç G ) = 0, so yes E and G are mutually exclusive.

8. (a) M and F are not mutually exclusive. We calculate P( M Ç F ); M Ç F is the event that a CEO is chosen who has an MBA and is in the age range of 40-49. Since there are 54 such persons out of a total of 200 (see the table), P( M Ç F ) = 54/200. This is not equal to zero, so the events M and F are not mutually exclusive. (b) The events M and F are not independent. P( M Ç F ) = 54/200 = 0.27, while P( M ) ´ P( F ) = 106/200 ´ 89/200 = 0.23585. These numbers are not equal.(c) These events are mutually exclusive. M Ç MC is the event of choosing a CEO who has an MBA and does not have an MBA. Since no such person exists, we have that M Ç MC = { } and P( M Ç MC ) = 0. (d) P( M Ç MC) = 0, while P( M ) ´ P( MC ) = 106/200 ´ 94/200 = .2491, which is not equal to zero. We conclude that these events are not independent. (e) P( 60 or over ) = 37/200.(f) P( 60 or over | no MBA ) = 34/94, (reduce the sample space to the 94 CEO’s without an MBA).(g) P( A Ç B ) = 34/200 = .17; while P( A ) ´ P( B ) = 37/200 ´ 94/200 = .08695, so these events are not independent.

9. (a) S = {GG, GR, GY, RG, RR, RY, YG, YR, YY} (b) Note: The word “and” translates to intersection. Since the two events are independent, we can multiply their probabilities to compute the probability of them happening together.

P( light 1 yellow Ç light 2 yellow ) = P( light 1 yellow ) ´ P( light 2 yellow ) = .10 ´ .05 = .005.


(c) Note: The word “and” translates to intersection. Since the two events are independent, we can multiply their probabilities to compute the probability of them happening together.

P( light 1 green Ç light 2 yellow ) = P( light 1 green ) ´ P( light 2 yellow ) = .50 ´ .05 = .025.

(d) This is the event {GG, GR, RG, RR}.

P( {RR, GR, RG, RR} ) =P( {GG} ) + P( {GR} ) + P( {RG} ) + P( {RR} ) = .15 + .325 + .12 + .26 = 0.855.

10. Note: In (a) - (c), the word “and” translates to intersection. We can multiply probabilities because the events are independent. (a) P(system 1 fails Ç system 2 fails) = P(system 1 fails) ´ P(system 2 fails) =

.001 ´ .0001 = .000001.

(b) P(system 1 works Ç system 2 works) = P(system 1 works) ´ P (system 2 works) = .999 ´ .9999 = .9989001.

(c) P(system # 1 fails Ç system # 2 works) = P(system # 1 fails) ´ P (system # 2 works) = .001 ´ .9999 = .0009999.

11. (a) There are 52 possibilities for the first card. After picking one card, there are 51 possibilities left for the second card, so in all, there are 52 ´ 51 = 2652 possible outcomes in the sample space.

(b) We need to compute and then compare the probabilities P( P Ç K ) and P( P ) ´ P ( K ).

Compute P ( P Ç K ) : The event P Ç K is the event of getting a king as the first card and getting a pair. The only way to do this is to get a pair of kings. There are 12 pairs of kings, because any one of the four kings could be drawn the first time and any one of the three remaining kings could be drawn the second time. There are 2652 pairs that can be drawn, so,

P( P Ç K ) = 12/2652.

Compute P ( P ) ́ P ( K ) : We need to compute P(P) and P(K). The event P is the event of getting a pair of the same type of card. Given any card drawn on the first draw there are 3 ways to get a pair out of a total of 51 cards remaining. For


instance, if the seven of diamonds was drawn first, then in order to get a pair, the next card must be one of the three remaining sevens. Thus, P(P) = 3/51.

The event K is the event of getting a king on the first draw. There are 52 cards, of which 4 are kings, so P( K ) = 4/52.

Thus,

P P P K´ ´ 3

514

5212

2652

Compare: We conclude that P and K are independent because P( P Ç K ) = P( P ) ´ P( K ).

(c) From part (b) we know that P( P Ç K ) ¹ 0, so P and K are not mutually exclusive.

(d) We compute and compare P( K ) ´ P( J ) with P( K Ç J ).

Compute P ( K ) ́ P ( J ) :The event K is the event of getting a king on the first draw. There are 52 cards, of which 4 are kings, so P( K ) = 4/52.

Out of a total of 2652 pairs, there are 51 pairs that can be drawn in which the second card is the jack of diamonds. Why? If the second card is the jack of diamonds, then the first card is not the jack of diamonds, so it must be one of the 51 other cards. Therefore, P ( J ) = 51/2652 = 1/52.

Therefore,

P K P J´ ´ 4

521

524

2704.

Compute P ( K Ç J ) :K Ç J is the event that a king is chosen as the first card and a jack of diamonds is chosen as the second card. There are exactly four ways to do this

(Kª J¨), (K§ J¨), (K¨ J¨), (K© J¨)

out of a total of 2652. So P( K Ç J ) = 4/2652.

Compare:


P( K ) ´ P( J ) = 4/52 ´ 1/52 = 4/2704 » .0014792899.

P( K Ç J ) = 4/2652 » .0015082956.

Since these are numbers are not equal, we conclude that the events K and J are not independent.

(e) In (d), we saw that P( K Ç J ) is not zero, so K and J are not mutually exclusive.

(f) We compute and compare P( J ) ´ P( D ) with P( J Ç D ).

Compute P ( J ) ́ P ( D ) :Out of a total of 2652 pairs, there are 51 pairs that can be drawn in which the second card is the jack of diamonds. Why? If the second card is the jack of diamonds, then the first card is not the jack of diamonds, so it must be one of the 51 other cards. Therefore, P ( J ) = 51/2652 = 1/52.

Since there are 13 ways to draw a diamond for the first card out of a total of 52 cards, we see that P(D) = 13/52.

In all, P( J ) ´ P( D ) = 1/52 ´ 13/52 = 13 / 2704.

Compute P ( J Ç D ) :The event J Ç D the event of getting a diamond for the first card and the jack of diamonds for the second card. Since there are 12 ways to do this (why?), P( J Ç D ) = 12 / 2652.

Compare:P( J ) ´ P( D ) = 13 / 2704 = .004807692P( J Ç D ) = 12 / 2652 » .004524886

Since P(J Ç D) does not equal P( J ) ´ P(D), we conclude that J and D are not independent events.


3.5 Expected Value

In this section, we will learn how to compute and interpret a number called expected value. This number is a weighted average that is used to aid decision-making in probabilistic games and scenarios in which one stands to gain or lose.

In order to compute the expected value of a game or scenario, we need to know two things in advance: (1) the set of possible values (or payoffs) that can be obtained and (2) the probability of obtaining each value.8 Once this information is assembled into a table, it is easy to calculate expected value: one simply multiplies each probability by its respective value and then adds up the resulting terms. That is, given a (finite) set of values and their respective probabilities,

Value X1 X2 … Xn

Probability P(X1) P(X2) … P(Xn)

we have that

expected value = X1P(X1) + X2P(X2) + … + Xn P(Xn) .

Example 1: Consider a board game in which the roll of a single (six-sided) die determines the number of spaces to advance on the board. Then the expected value of rolling a die can be computed from the table below:

Value of a roll 1 2 3 4 5 6Probability 1/6 1/6 1/6 1/6 1/6 1/6

expected value =

616

615

614

613

612

611

=

66

65

64

63

62

61

8 In Probability and Statistics, the assignment of a value to each outcome in an experiment is called a random variable. When the number of values that the random variable can take on is finite, the random variable is called discrete.


= 1 2 3 4 5 6

6216

35 .

This means that the expected value of a single roll is 3.5 spaces. Of course, it is not possible to roll a “3.5” in the board game. The non-integer number is a consequence of the fact that expected value is arrived at by an averaging process. n

The next example shows how to interpret expected value and how to use it to make long range predictions.

Example 2: Consider the game in which a player rolls a six-sided die and wins the following amounts of money for the respective rolls:

Roll 1 2 3 4 5 6Value ($) -1 +3 -6 +2 -4 +5

(a) What is the expected value for this game? (b) What does the expected value say about the game?(c) How much money can a player expect to win or lose after 300 rolls?

Solution: (a) In order to compute the expected value, we need the probability of each value. Since there are six equally likely rolls, the probability of obtaining any one of them is 1/6. We add a probability row to the table given to us:

Roll 1 2 3 4 5 6Value ($) -1 +3 -6 +2 -4 +5Probability 1/6 1/6 1/6 1/6 1/6 1/6

The expected value is computed by multiplying each value by its probability and then summing the terms:

exp value = 615 +

614

612+

616

613 +

611

-

-

-

=- - - 1 3 6 2 4 5

6 = -

16

dollars per roll.

(b) The expected value tells us that, in the long run, a player can expect to lose an average of 1/6 dollars per roll. (That’s a little over 16 cents per roll.) This number should not be taken too literally. It is quite possible that a player will roll the positive payoff numbers (2,


4, and 6) often enough to come out ahead. This is a remote possibility, however, in general, this will tend to be a losing game for the player.

(c) After 300 rolls, a player’s average earnings should be fairly close to the expected value. The player should expect to lose about -$1/6 per roll, for a total of

300 rolls ´ (-$1/6 per roll) = -$50 dollars.

That is, a player can expect to lose about 50 dollars after 300 rounds of play. n

Example 3: A roulette wheel has a total of 38 numbers: 00, 0, 1, 2, …, 36. A player is allowed to bet $1 on any one of the numbers 1, 2, …, 36 and if that number comes up, he wins $35. However, if any other number comes up, he loses his dollar. What is the expected value of this game and how much can one expect to win or lose after 200 plays?

Solution: There are only two possible values in this game: win $35 or lose $1. No matter what number the player bets on, the probability of winning is 1/38 (there are 38 numbers on the wheel) and the probability of losing is 37/38 (these fractions must sum to one). Thus, we obtain the following table:

Outcome win loseRandom var +35 -1Prob 1/38 37/38

The expected value is:

exp value = (35)(1/38) + (-1)(37/38) = 35/38 - 37/38 = -2/38 dollars per play

This means that, on average, one would expect to lose about $0.053 dollars per play. That’s a little over five cents. If one played the game 200 times, one would expect to lose about (200)(2/38) = 10.53 dollars. n

Tip: The key to solving the previous problem was to start by listing the values (payoffs) that could be obtained in the game rather than the outcomes that generate them. One might have been tempted to start by filling into a table all possible outcomes on the

wheel (0, 00, 1, 2, 3, .., 36) and then their probabilities (1/38 for each). However, the value (payoff) of each number would depend upon the number that is bet, which is unknown.

B


Despite the fact that the ball could land on any one of 38 numbers, there were only two possible values: winning $35 or losing $1.

Part (c) of the next example highlights the fact that expected value might not be a good predictor for a small sampling.

Example 4: Inglow Company Inc. manufactures florescent light bulbs and ships them in packages of six. Their research department has collected statistics on the number of bulbs that are broken per package upon arrival to their customers. The results are tabulated below.

Broken bulbs per package of 6 0 1 2 3 4 5 6Probability .80 .02 .01 .01 .02 .06 .08

(a) On average, how many bulbs should Inglow expect to be broken during shipping?(b) If Inglow selects a package of bulbs at random upon arrival at the construction site, what is the most likely number of bulbs for them to see broken? (c) Inglow is about to ship 6000 bulbs to a customer at a construction site. How many bulbs should they expect to arrive unbroken?

Solution: (a) In this problem, the “value” of a package is the number of broken bulbs per package. (One could also solve the problem by making it the number of unbroken bulbs per package.) From the table above, we get

exp value = 0(.80) + 1(.02) + 2(.01) + 3(.01) + 4(.02) + 5(.06) + 6(.08) = 0 + .02 + .02 + .03 + .08 + .30 + .48 = .93

So, they should expect, on average, 0.93 broken bulbs per package.

(b) Based on the expected value, it is tempting to answer that they are most likely to see 0.93 broken bulbs. However, this expected value is an average. For any one package, the number of broken bulbs will be exactly one of 0, 1, 2, 3, 4, 5, or 6. Looking at the table of probabilities, we see that “0” has the highest probability associated with it (.80), so they should expect to see 0 broken bulbs.

(c) There are six bulbs per package and they are shipping 6000 bulbs. The total number of packages they are shipping is 6000/6 = 1000. Since the expected value is .93 broken bulbs per package, they should expect 1000(.93) = 930 bulbs to arrive broken out of a total of 6000 bulbs. Therefore, they should expect around 6000 - 930 = 5070 bulbs to arrive unbroken. n


Using Microsoft Excelâ to Analyze Expected Values

You will need one 6-sided die to perform this experiment.

Roll a six-sided die at least 100 times (this won’t take as long as you think!) and record the outcomes in a spreadsheet. [Suggestion: Do this with another person, one person roll and the other record data]. You will need to mimic the example below.

Example: Suppose that Sue rolls a 5, then a 4, then a 2, 1, 6, 2, 4, and a 3 . Then she would record this as follows.

A B C D E F1 Roll # Outcome2 1 53 2 44 3 25 4 16 5 67 6 28 7 49 8 3

We want to use the power of the spreadsheet to analyze this data. To do this, we would like to have a running average, i.e., after each roll we want to average all of the previous rolls.

If Sue wanted to do this, then in column C, she might keep a running total which adds up all of the previous rolls, so in column C2, she would put C2 = B2, and she would set C3 = C2 + B3. Now she would cut and paste this formula all of the way down column C. If she does this correctly, she will get:

A B C D E F1 Roll # Outcome running tot2 1 5 53 2 4 94 3 2 115 4 1 126 5 6 187 6 2 208 7 4 249 8 3 27


Of course, we want running averages, so in column D, she would divide by the number of rolls that have been added, or she would set D2 = C2/A2 and copy this formula down column D as below:

A B C D E F1 Roll # Outcome running total averages2 1 5 5 53 2 4 9 4.54 3 2 11 3.6666675 4 1 12 36 5 6 18 3.67 6 2 20 3.3333338 7 4 24 3.4285719 8 3 27 3.375

Your job is to mimic Sue’s experiment with 100 (or more rolls) and hand the resulting spreadsheet in with the following questions answered.

(a) What is the theoretical expected value for the experiment in which one 6-sided die is rolled?

(b) The expected average value can be used to predict the running total after 100 rolls, what is this expected total after 100 rolls?

(c) Did you actually get the number predicted in (b) as your 100th running total?

(d) If you ran this experiment over again, would you think that the results would vary dramatically?

(e) About how long does it take before the running averages start to “behave” ?


Problems

1. Calculate the expected value for the experiment in which one die is rolled and the number on the top face is recorded.

2. A company manufactures telephones in batches of 1000. The quality manager has computed the following table which shows the number of telephones that do not meet the required specifications:

Number of phones that do not meet specifications 0 1 2 3 4Probability 1/4 1/2 1/8 1/16 1/16

(a) What is the expected number of phones which will not meet specifications in each batch? (b) If you examined 500 batches of telephones (that’s 500,000 phones), how many defective phones would you expect to find? (c) If you examined one batch of phones, how many defective phones would you probably find?

3. A fire insurance company has determined that the probability that a fire strikes a random building on any given year is 1/10,000. If a fire strikes a building, they have estimated that they will pay $1,000,000 in damages. Suppose that the company charges $250 each year for fire insurance per building.(a) What is the expected gain (loss) that the company makes per year per building? [Hint: the sample space for each building is S = {fire, no fire}. ](b) If the company insures 300,000 customers, what is their expected profit (loss).(c) On any given year, for a fixed building, how much would you guess that the company would make (or lose) from that individual building?(d) Suppose that each of the company’s 300,000 insured buildings burned down one year. What is the probability of this happening?(e) If each of the company’s 300,000 insured buildings burned down, how much would the company lose?

4. Consider the following game in which you pay $20 to play. You roll a six-sided die with the numbers 2, 2, 2, 2, 32, and 62 on it and you win the dollar amount of the number that you roll, e.g., if you roll a “62” then you win $62 (net win is $42 = $62 - $20, since it cost $20 to play).(a) Calculate the expected winning (loss) for the game.(b) How much would you expect to win (lose) on a single roll?(c) After 1,000 plays of this game, how much would you expect to win (lose)?


5. Consider the roulette-like game in which a wheel with 42 numbers on it is spun and a marble (randomly) lands on one of the numbers. A player bets $20 on one of the numbers and wins four times that amount if the marble lands on that number. (a) What is the probability of winning on a single play? (b) What is the expected value for this game? (c) If you play this game 600 times, how much total would expect to win (lose)?(d) If you play the game one time, how much would you guess that you would win (lose)?

(e) For the player, is this a winning game or a losing game?

6. Consider the experiment in which a pair of six-sided dice are rolled and the sum of the numbers on the top faces are recorded. Calculate the expected value of this experiment.

7. You roll one standard six-sided die and record the square of the number on the top face. What is the expected value?

8. The following game has a $5.00 fee to play. Roll one six-sided die. If the number on the top face is even, you win $6.00, otherwise you lose $1.00. (a) What are the expected winnings (losses) of this game? (b) Is it a winning or losing game? (c) Play this game 100 times and record your findings in a spreadsheet. What were your average winnings (losses) per game? Was your average close to the expected value?

Web

9. The Pennsylvania State Lottery's Daily Number Game generates random digits by using a machine that mechanically selects one ping-pong ball out of a machine that has 10 balls labeled with the digits 0, 1, 2, …, 9. Prior to selection, these balls are tossed around inside the machine by an air-flow. Someone hypothesizes that the ball with number '8' comes up less than the other balls since the ball with the ball number '8' is slightly heavier than the other balls (since it takes more ink to draw an '8' than a '1'). Use the web to find the results of the Pennsylvania's Daily Number game over the past 6 months and see whether the number '8' is appearing approximately 1/10 of the time. In a paragraph or two, relay your findings. Justify them with mathematical calculations to support your conclusions.


Solutions

1. The outcomes together with their probabilities are given below:

Outcome 1 2 3 4 5 6Probability

1/6 1/6 1/6 1/6 1/6 1/6

Multiplying each outcome by its probability and adding we see that the expected average value is

5.3621

6654321

61 6

61 5

61 4

61 3

61 2

61 1

2. (a) The exp. value =

1875.11619

161 4

161 3

81 2

21 1

41 0

per batch

(b) 500 ´ 1.1875 = 593.75, so we would expect to find about 594 defective phones out of 500,000 phones. (c) If you examined one batch of phones, you probably would find one defective phone. (it has the highest probability).

3. (a) If there is a fire, the company pays $1,000,000 but they will get $250 to insure the building, so the company loses $999,750. If the building does not burn, the company earns $250. We compile this information below:

Fire No FireValue - $ 999,750 $250Probability 1/10,000 9999/10,000

So exp. value = (-999,750 ´ 1/10,000) + (250 ´ 9999/10,000) = 1500000/10000 = $150 per building.


(b) 300,000 ´ $150 = $45,000,000 in expected profits (per year).(c) Chances are that any particular building will not burn down, so the company will probably earn $250 on such a building.(d) The probability of this happening is (1/10,000)300,000 . This number is so small that most calculators will round it to zero. [In fact, if you wrote this number as a decimal, it would have three billion zeros to the right of the decimal, followed by a 1.] (e) The company would lose $999,750 ´ 300,000 = $299,925,000,000.

4. (a) Since it costs $20 to play, the possible (net) winnings are 2 - 20 = -18, 32 - 20 = 12 and 62 - 20 = 42. Tabulating these dollar amounts with their respective probabilities, we have:

Winnings -$18 $12 $42Probability 4/6 1/6 1/6

Therefore,

exp. value = -

-

618

6142

6112

6418 -$3.00 per roll.

(b) On any single roll, it is most likely that you would lose $18 (look at the probabilities).(c) After 1000 plays you expect to lose (on average) 1000 ´ 3.00 = $3000.00.

5. (a) 1/42(b)

Winnings $60 -$20Probability 1/42 41/42

So,

exp. value = 42

76042

8204260

424120

42160

-

-

» -$18.10 per game.

(c) On average you would expect to lose around 600 ´ (-$18.10) = -$10,860. (d) You expect to lose $20 because this outcome has the highest probability. Note: expected value is usually not a good estimator for one play.(e) This is a losing game for the player because it has a negative expected value.


6. The “value” received on each roll is simply the number itself. We tabulate the values with their respective probabilities:

Value 2 3 4 5 6 7 8 9 10 11 12Prob 1/36 2/36 3/36 4/36 5/36 6/36 5/36 4/36 3/36 2/36 1/36

To compute the expected value, we multiply each value by its probability and sum the resulting fractions:

736252

3612

3622

3630

3636

3640

3642

3630

3620

3612

366

362

7. Roll 1 2 3 4 5 6

Value 1 4 9 16 25 36Prob 1/6 1/6 1/6 1/6 1/6 1/6

exp. value =

691

6136

6125

6116

619

614

611

» 15.166666667 per roll.

8.Roll even odd

Net Win/Loss $1 -$6Probability 1/2 1/2

(a) The expected winnings = 25

216

211 -

-

= -$2.50 per game.

(b) This is a losing game (for the player) because the expected value is negative.

3hoff/joe/book/probability/pr… · web view1998/08/29 · these flights are represented by the...

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