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1

Solutions to Chapter 1

1. The in t ent of this rather vague problem is to get you to compare the two notions, probability

as intuition and relative frequency theory. There are many possible answers to how to

make the statement "Ralph is probably guilty of theft" have a numerical value in the relative

frequency theory. First step is to deﬁne a repeatable experiment along with its outcomes.

The favorable outcome in this case would be ’guilty.’ Repeating this experiment a large

number of times would then give the desired probability in a relative frequency sense. We

thus see that it may entail a lot of work to attach an objective numerical value to such a

subjective statement, if in fact it can be done at all.

One possible approach would be to look through courthouse statistics for cases similar to

Ralph’s, similar both in terms of the case itself and the defendant. If we found a suﬃciently

large number of these cases, ten at least, we could then form the probability =

,where

is the number of favorable (guilty) verdicts, and is the total n u mber of found cases.

Here we eﬀectively assume that the judge and jury are omniscient.

Another possibility is to ﬁnd a large number of people with personalities and backgrounds

similar to Ralph’s, and to expose them to a very similar situation in which theft is possible.

The fraction of these people that then steal in relation to the total number of people, would

then give an objective meaning to the phrase "Ralph is probably guilty of theft."

2. Note that → 3, but 3 6→ ,i.e., implies 3 but not the other way around. Thus if we

turn over card 2 and ﬁnd a 3. So what? It was never stated that

a 3 → . Likewise, with

card 3. On the other hand, if we turn over card 4 and ﬁnd a , then the rule is violated.

Hence, we must turn over card 4 and card 1, of course.

3.Firststephereistodecidewhichkindofprobability to use. Since no probabilities are

explicitly given, it is reasonable to assume that all numbers are equally likely. Eﬀectively

we assume that the wheel is “fair." This then allows us to use the classical theory along

with the axiomatic theory to solve this problem. Now we must ﬁnd the corresponding prob-

ability model. We are told in the problem statement that the experiment is “spinning the

wheel." We identify the pointed-to numbers as the outcomes .Thesamplespaceisthus

Ω = {1 2 3 4 5 6 7 8 9}The total nu mber of outcomes is then 9. The probability of each

elemental event {} is then taken as [{}] , =19, as in the classical theory. We are also

told in the problem statement that the contestant wins if an even number shows. The set of

even numbers in Ω is {2 4 6 8} We can write this event as a disjoint union of four singleton

(atomic) events

{2 4 6 8} = {2} ∪ {4} ∪ {6} ∪

{8}

Now we can apply axiom 3 of probability to write

[{2 4 6 8}]= [{2}]+ [{4}]+[{6}]+[{8}]

=

1

9

+

1

9

+

1

9

+

1

9

=

4

9

We have seen that some ’reasonable’ assumptions are necessary to transform the given word

problem into something that exactly corresponds to a probability model. It turns out that

this is a general problem for such word problems, i.e. problems given in natural English.

2

4. The experiment in volves ﬂipping a fair coin 3 times. The outcome of each coin toss is either

a head or a tail. Therefore, the sample space of the combined experiment that contains all

the possible outcomes of the 3 tosses, is given by

Ω = {}

Since all the coins are fair, all the outcomes of the experiment are equally lik ely. The proba-

bility of each singleton event, i.e. an event with a single outcome, is then

1

8

. We are interested

in ﬁnding the probability of the event , whic h is the event of obtaining 2 heads and 1 tail.

There are 3 favorable outcomes for this event given by = {}. Therefore,

[]= [{} ∪ {} ∪ {}]=[]+[]+ []=

3

8

.Notethat

weareabletowritetheprobabilityoftheevent as the sum of probability of the singleton

events (from Axiom 3) because the singlteon events of any experiment are mutually exclusive.

Why?

5. The experiment contains drawing two balls (with replacement) from an urn containing balls

numbered 1, 2,and3. The sample space of the experiment is given by

Ω = {11 12 13 21 22 23 31 32 33}

The event of drawing a ball twice is said to occur when one of the outcomes 11, 22,or33

occurs. Therefore, the event of dra wing 2 equal balls ,isgivenby = {11 22 33} and

[]= [{11}]+ [{22}]+ [{33}] Since the balls are drawn at random, it can assumed that

drawing each ball is equally likely. Therefore, the singleton events, or equivalently outcomes

of the experiment, are equally likely. Hence, []=3(

1

9

)=

1

3

.

6. Let

1

2

6

represent the six balls. Each outcome will be represented by the two balls

that were drawn. In the ﬁrst experiment, the balls are drawn without replacement; hence,

the two balls drawn cannot have the same index. Then the sample space containing all the

outcomes is given by

Ω

1

= {

1

2

1

3

1

4

1

5

1

6

2

1

2

3

2

4

2

5

2

6

3

1

3

2

3

4

3

5

3

6

4

1

4

2

4

3

4

5

4

6

5

1

5

2

5

3

5

4

5

6

6

1

6

2

6

3

6

4

6

5

}

This can be written compactly as

Ω

1

= {

()

|1 ≤ ≤ 6 1 ≤ ≤ 66= }

If the ﬁrst ball is replaced before the second draw, then in addition to the outcomes in the

earlier part, there are outcomes where both the two balls drawn are the same. The sample

space for the new experiment is given by

Ω

2

= Ω

1

∪ {

1

1

2

2

3

3

4

4

5

5

6

6

}

This can also be written as Ω

2

= {

()

|1 ≤ ≤ 6 1 ≤ ≤ 6}.

3

7. Let

be the height of the man and

be the height of the woman. Each outcome of the

experiment can be expressed as a two-tuple (

).Thus

(a) The sample space Ω is the set of all possible pairs of heights for the man and woman.

This is given as

Ω = {(

):

0

0}

(b) The event , which is a subset of Ω is given by

= {(

):

0

0

}

8. The word problem describes the physical experiment of drawing numbered balls from an urn.

We need to ﬁnd a corresponding mathematical model. First we form an appropriate event

space with meaningful outcomes. Here the physical experiment is ’draw ball from urn,’ so

the outcome in words is ’particular labeled ball drawn ,’ which we can identify with its label.

So we select as outcome in our mathematical model, the number on the drawn ball’s face,

i.e. the particular label. The outcomes are thus the integers 1,2,3,4,5,6,7,8, 9, and 10. The

sample space is then Ω = {1 2 3 4 5 6 7 8 9 10} and is the set of all ten outcomes. We

are told that is ’the event of drawing a ball numbered no greater than 5.’ Thus we deﬁne

in our event ﬁeld = {1 2 3 4 5} The other event speciﬁed in the word problem is ’the

event of drawing a ball greater than 3 but less than 9.’ In our mathematical event ﬁeld this

corresponds to = {4 5 6 7 8} Having constructed our sample space with indicated events,

we can use elementary set theory to determine the following answers:

= {6 7 8 9 10}

= {1 2 3 9 10}

= {4 5}∪ = {1 2 3 4 5 6 7 8}

= {1 2 3}

= {6 7 8}

∪

= {1 2 3 6 7 8 9 10}

(

) ∪ (

)={1 2 3 6 7 8} () ∪ (

)={4 5 9 10}

( ∪ )

= {9 10} ()

= {1 2 3 6 7 8 9 10}

The last part of the problem asks us to ’express these events in words.’ Since we have a

mathematical model, we should really more precisely ask what each of these events corresponds

to in words.Weknowofcoursethat corresponds to ’drawing a ball numbered no greater

than 5.’ We can thus loosely write = {

0

drawing a ball numbered no greater than 5’},

although in our mathematical model is just the set of integers {1 2 3 4 5}.Sowhen

we write ={ ’drawing a ball numbered no greater than 5’}, what we really mean is that

the event in our mathematica l model corresponds to the physical event ’drawing a ball

numbered no greater than 5’ mentioned in the word problem. With this caveat in mind, we

canthenwrite:

= {

0

drawing a ball greater than 5

0

}

= {

0

drawing a ball not in the range 4-8 inclusive

0

}

= {

0

drawing a ball greater than 3 and no greater than 5

0

}

etc.

9. The sample space containing four equally likely outcomes is giv en by Ω = {

1

2

3

4

}.Two

events = {

1

2

} and = {

2

3

} are given. The required events can be easily obtained

by observation.

4

= set of outcomes in and not in = {

1

}.

= set of outcomes in and not in = {

3

}.

= set of outcomes in and = {

2

}.

∪ = set of outcomes in or in = {

1

2

3

}.

10. = ∪

Thiscanbeprovedusingthedistributivelawon

= Ω = ( ∪

)= ∪

∪ =(

) ∪ (

) ∪ () Here we ﬁrst write = ( ∪

) and = ( ∪

)

Then we can write

∪ =(( ∪

)) ∪ (( ∪

))

=( ∪

) ∪ ( ∪

)

= ∪

∪ ∪

= ∪

∪

using the above laws and formulas. Notice that the above two decompositions are into disjoint

sets. From the third axiom of probability, we know that the probability of union of disjoint

sets is the sum of the probabilities of the disjoin t sets. Therefore, we can add the probabilities

over the unions.

11. In a given random experiment there are four equal ly likely outcomes

1

2

3

and

4

Let

the event , {

1

2

}

[]= [{

1

2

}]= [{

1

}]+ [{

2

}]=

1

4

+

1

4

=

1

2

= {

3

4

}

[

]= [{

3

4

}]=[{

3

}]+[{

4

}]=

1

4

+

1

4

=

1

2

Note that we are told that the four outcomes are equally likely. This means that the four

singleton (atomic) events have equal probability. []=

1

2

=1− [

]=1−

1

2

12. (a) The three axioms of probability are given below

(a) [label=()]

(b) For any event , the probability of the even occuring is always non-negative.

[] ≥ 0

This ensures that probability is never negativ e.

(c) The probability of occurence of the sample space event Ω is one.

[Ω]=1

This ensures that probability of no event exceeds one. The ﬁrst two axioms ensures that

the probability is a quantity between 0 and 1,inclusive.

(d) For any two events that are disjoint, the probability of the union of the events is

the sum of the probabilities of the two events.

[ ∪ ]=[]+[] when =

This axiom tells us that the probability of any event can be obtained b y the sum disjoint

events that constitute the event.

5

(b) The even t ∪ can be obtained as the disjoint union of the three sets

.

Hence by applying the third axiom of probability, we obtain

[ ∪ ]= [ ∪ (

∪

)]

= []+ [

∪

]

= []+ [

]+ [

]

No w the event can be written as the disjoint union of and

(Axiom 3). Therefore

[]= []+ [

]=⇒ [

]= [] − []

Similarly

[]=[]+ [

]=⇒ [

]= [] − []

Therefore [ ∪ ]=[]+( [] − []) + ( [] − []) = []+[] − [].

13. We ﬁrst form our mathematical model by setting outcomes ς =(

1

2

) where

1

corresponds

to the label on the ﬁrst ball drawn, and

2

corresponds to the label on the second ball drawn.

We can also write the outcomes as strings ς =

1

2

The sample space Ω canthenbeidentiﬁed

with the 2-D array

11 12 13 14 15

21 22 23 24 25

31 32 33 34 35

41 42 43 44 45

51 52 53 54 55

There are thus 25 outcomes in the sample space. Now the word problem statement uses

the phrase ’at random’ to describe the drawing. This is a technical term that can be read

’equally likely.’ Thus all the elementary events {

1

2

} in our mathematical model must have

equal probability, i.e. [{

1

2

}]=125 Armed thusly we can attack the given problem as

follows. Deﬁne the event ={’sum of labels equals ﬁve’}, or precisely = {41 32 23 14}

Then we decompose this event into four singleton events as

= {41} ∪ {32} ∪ {23} ∪ {14}

Since diﬀeren t singleton events are disjoint, probability adds, and we have

[]=

1

25

+

1

25

+

1

25

+

1

25

=

4

25

"Dim" ignored that outcome is diﬀerent (distinguishable) from outcome . "Dense" talked

about the sums and correctly noted that there were nine of them. However, he incorrectly

assumed that each sum was equally likely. Looking at our sample space above, we can

see that the sum 2 has only one favorable outcome 11, while the sum 6 has ﬁve favorable

outcomes, just looking at the anti-diagonals of this matrix.

14. First we show ∩ ( ∪ ) ⊂ ( ∩ ) ∪ ( ∩ ).

Let ∈ ∩ ( ∪ ).

Then ∈ and ∈ ( ∪ ).

∈ and ∈ or ∈ .

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