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Fall 2013 Probability TABLE OF CONTEN

LIHORNE.COM

STAT 230Probability

Dr. Jiwei Zhao Fall 2013 University of Waterloo

Last Revision: December 4, 2013

Table of Contents

1 Introduction to Probability

1.1 Definitions of Probability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.2 Solutions to Problems on Chapter 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

2 Mathematical Probability Models

2.1 Sample Spaces and Probability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2 Solutions to Problems on Chapter 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

3 Probability - Counting Techniques

3.1 Counting Arguments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2 Review. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.3 Tutorial 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.4 Solutions to Problems on Chapter 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

4 Conditional

4.1 General Methods . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2 Rules for Unions of Events. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.3 Intersections of Events and Independence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.4 Conditional Probability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.5 Multiplication and Partition Rules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.6 Tutorial 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

4.7 Problems on Chapter 4. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

5 Probability Distributions

5.1 Solutions to Problems on Chapter 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.2 Tutorial 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

6 Chapter 7

6.1 Tutorial 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.2 Solutions to Problems on Chapter 7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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Fall 2013 Probability TABLE OF CONTEN

7 Chapter 8

7.1 Solutions to Problems on Chapter 8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.2 Tutorial 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

8 Continuous Probability Distributions

8.1 Solutions to Problems on Chapter 9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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Fall 2013 Probability 1 INTRODUCTION TO PROBABILI

Abstract

These notes are intended as a resource for myself; past, present, or future students of this course, and anyone

interested in the material. The goal is to provide an end-to-end resource that covers all material discussed in the

course displayed in an organized manner. If you spot any errors or would like to contribute, please contact me

directly.

1 Introduction to Probability

Why is it important to learn about Probability and Statistics? There are many mistakes made currently in industry, for example, watch the Ted Talk on "How juries are fooled by statistics.". For example, consider the cof Sally Clark. For Wednesday, understand what a sample space is, an event is, and a probability distribution

1.1 Definitions of Probability

Definition 1.1 (Randomness). Uncertainty or randomness (i.e. variability of results) is usually due to somixture of at least two factors including:

1. Variability in populations consisting of animate or inanimate objects (e.g., people vary in size, weight, blotype etc.)

2. Variability in processes or phenomena (e.g., the random selection of 6 numbers from 49 in a lottery draw clead to a very large number of different outcomes)

Definition 1.2 (sample space). We refer to the set of all possible distinct outcomes to a random experiment as sample space (usually denoted by S). Groups or sets of outcomes of possible interest, subsets of the sample spwe will call events.

Then we might define probability in three different ways:

Definition 1.3 (Probability (Classic), Relative Frequency, Subjective Probability). The classical definition: T

probabilityof some event isnumber of ways the event can occur

number of outcomes in S

provided all points in the sample spaceSare equally likely. For example, when a die is rolled the probabilitygetting a 2 is 16 because one of the six faces is a 2.

The relative frequency definition: The probability of an event is the (limiting) proportion (or fraction) of timthe event occurs in a very long series of repetitions of an experiment or process. For example, this definition cobe used to argue that the probability of getting a 2 from a rolled die is 16 .

The subjective probability definition: The probability of an event is a measure of how sure the person mak

the statement is that the event will happen. For example, after considering all available data, a weather forecasmight say that the probability of rain today is 30% or 0.3.

There are problems with all three of these definitions. The first being that "equally likely" is not defined, the secbeing unfeasible, and third being hard to agree on. Thus we create a mathematical model with a set of axiomsbetter understand probability.

1

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Fall 2013 Probability 2 MATHEMATICAL PROBABILITY MODE

Definition 1.4 (probability model).

a sample space of all possible outcomes of a random experiment is defined. a set of events, subsets of the sample space to which we can assign probabilities, is defined.

a mechanism for assigning probabilities (numbers between 0 and 1) to events is specified.1.2 Solutions to Problems on Chapter 1

1.1 The chance of head or tails; repetetive dice throwing in grade six; stock forecasts.

1.2 Claim on car insurance would be subjective based on someones basic opinion, or relative considering hoften other people have claims. A meltdown at nucelear power plant would be subjective since it doehappen often, isnt one of a set of probabilities, and is based on opinion or skepticism. A persons birthdaclassical since there are 12 possibilities and April is 1 of 12.

1.3 Lottery draws are classical in that a ticket is one of several million, thus there is a value for P(x) calculausing the formula above. Small businesses may use probability, or decision, trees to save money and tiwhen deciding upon which areas of their company require an extensive review due to fraud or human erThere is subjective probability in understanding disease transmission, as well relative frequency since lotdata and samples can be taken. Public opinion polls can be subjectively evaluated, for example a presedenelection (a type of publiv opinion poll) typically has forecasts by experts in politics and media.

1.4 Positions of small particles in space involve quantum physics, for example a particle must follow HeisenberUncertainty Principle. The velocity of an object dropped from the leaning tower of Pisa does not dependanything (if not acted on by another force). Values of stocks are not deterministic at all (though people trythink of them as such). Currency changes value steadily but it is not deterministic.

2 Mathematical Probability Models

2.1 Sample Spaces and Probability

Consider some repeatable phenomenon with certain events or outcomes labelled A1, A2, , An are needed. We this phenomenon an experiment and repetitions of it as a trial. The probability of an event ifA is 0 P(A)Definition 2.1 (Sample Space). A sample space S is a set of distinct outcomes for an experiment or procwith the property that in a single trial, one and only one of these outcomes occurs.

The outcomes that make up the sample space may sometimes be called sample pointsor just points on occas

Example 2.1. Roll a 6-sided die, and define the events ai= top face is i, for i = 1, 2, 3, 4, 5, 6. Then we can ta sample space as S=

{a1, a2, a3, a4, a5, a6

}. Rather than defining this sample space we could have also defin

S= {E, O}where E is the event that an even number comes up, and O the event that an off number turnsBoth sample spaces satisfy the definition.

However, we try to choose sample spaces that are indivisible, so the first is preferred.

Definition 2.2. A discrete sample space is one that consists of a finite or countably infinite set of simple evenRecall that a countably infinite sequence is one that can be put in one-one correspondance with positive intege

Definition 2.3. A non-discrete sample space is one that is not discrete.

2

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Fall 2013 Probability 2 MATHEMATICAL PROBABILITY MODE

Example 2.2. The sample space S= Z is discrete,S= Ris not, but S= {1, 2, 3} is.Definition 2.4. An event in a discrete sample space is a subset A S. If the event is indivisible so it contaonly one point, e.g. A1= {a1}we call it a simple event. An event A made up of two or more simple events suas A = {a1, a2} is called a compound event.

We will simplifyP({a1})to mean P(a1).Definition 2.5. Let S ={a1, a2, a3, } be a discrete sample space. Then probabilities P(ai) are numbattached to the ais(i= 1, 2, 3, )such that the following two conditions hold:

(1) 0 P(ai) 1(2)

i P(ai) = 1

The above function P(ai) is called a probability distribution on S.

Definition 2.6. The probability P(A) of an event A is the sum of the probabilities for all the simple events tmake up A or P(A) = aA P(a).Probability theory does not choose values for ai but rather defines probability to be restricted to the above tmathematical rules. All else is experiment to mimic the real world values.

Example 2.3. For a dice we would assign P(ai)equal to 16 for S= {a1, a2, a3, a4, a5, a6}. So,P(E) =P(a2, a4, a6

3 16 = 12 .The typical approach to a given problem of these types is the following:

(1) Specify a sample space S.

(2) Assign numerical probabilities to the simple events in S.

(3) For any compound eventA, find P(A)by adding the probabilities of all the simple events that make up

Example 2.4. Draw 1 card from a standard well-shuffled deck. Find the probability of finding a club.

Solution Let S ={ spade, heart, diamond, club}. S then has four sample points and one of them is clso P(club) = 14 . We could also have defined S ={ all cards} where a club event is A ={ all clubs} aP(A) =

P(one club) = 152 + + 152 = 1352 = 14 .

Definition 2.7. The odds in favour of an event A is the probability the event occurs divided by the probabilitdoes not occur or P(A)1P(A) . The odds against the event is the reciprocal.

When treating objects in an experiment as distinguishable leads to a different answer from treating them as identithe points in the sample space for identical objects are usually not "equally likely" in terms of their long run relat

frequencies. It is generally safer to pretend objects can be distinguished even when they cant be, in order to equally likely sample points.

3

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Fall 2013 Probability3 PROBABILITY - COUNTING TECHNIQU

2.2 Solutions to Problems on Chapter 2

2.1 a) S= {a1b1, a1b2, a1b3, a1b4, a2b1, a2b2, a2b3, a2b4, a3b1, a3b2, a3b3, a3b4, a4b1, a4b2, a4b3, a4b4}b) The probability that both students ask the same prof is P(A) whereA=

4i=1 aibi. P(A) =

116+

1

116 +

116 =

14 .

2.2 a) S= {H H H , H H T , H T H , T H H , H T T , T T H , T H T , T T T }b) P(T T) =P({H T T , T T H }) = 18+ 18 = 14

2.3 S= {(1, 2), (1, 3), (1, 4), (1, 5), (2, 1), (2, 3), (2, 4), (2, 5), (3, 1), (3, 2), (3, 4), (3, 5), (4, 1), (4, 2), (4, 3), (4, 5), (5(5, 3), (5, 4)} then the probability of of finding numbers that differ by 1 is P({(1, 2), (2, 1), (2, 3), (3, 2), (3, 4), ((5, 4)}) = 410 .

2.4 a) AB is for As letter in Bs envelope.S= {(W W , X X , Y Y , Z Z ), (WW,XX,Y Z,ZY ), (W W , X Z , Y Y , Z X ), (W Z , X X , Y Y , Z W ),(W W , X Y , Y X , Z Z ), (W Y , X X , Y W , Z Z ), (WX,XW,Y Y ,ZZ), (WW,XY ,Y Z,ZX),(WW,XZ,Y X,ZY ), (W Y , X X , Y Z , Z W ), (WZ,XX,Y W,ZY ), (W X , X Y , Y Y , Z Z ),

(W Y , X Z , Y Y , Z W ), (WX,XY ,Y W,ZZ), (W Y , X W , Y X , Z Z ), (WY ,XZ,Y W,ZX),(W Y , X W , Y Z , Z X ), (W Y , X Z , Y X , Z W ), (W X , X Y , Y Z , Z W ), (WX,XZ,Y W,ZY ),(WX,XW,Y Z,ZY ), (WZ,XW,Y X,ZY ), (W Z , X Y , Y W , Z X ), (W Z , X Y , Y X , Z W )}

c) 14 ,38 ,

14 , 0

2.5 a)

c)

2.6 P(test positive has disease) = 181000 ,P(has the disease) = 2100

TODO : Finish these problems

3 Probability - Counting Techniques

Definition 3.1. A uniform distribution over the set{a1, , an} is when the probability of each event isThat is, each event has equal probability. If a compound event A contains r points, then P(A) = rn .

3.1 Counting Arguments

There are two useful rules for counting.

1. The Addition Rule: Suppose we can do job 1 in p ways and job 2 in qways. Then we can do either joORjob 2, but not both, in p + qways.

2. The Multiplication Rule: Suppose we can d job 1 in p ways and, for each of these ways, we can do job 2qways. Then we can do both job 1 ANDjob 2 in p qways.

This interpretation using AND and OR occur throughout probability. For example, if we are picking two numbfrom{1, 2, 3, 4, 5}with uniform distribution. Then the probability of having at least one even can be found likewe need the first to be even AND the second odd OR the first odd AND the second even. Then,

P(one number is even) =2 3 + 3 2

5 5 =12

25

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Note that the phrases at random and uniformly are often used to mean that all of the points in a sample spare equally likely.

Example 3.1. Suppose a course has 4 sections with no limit on how many can enrol in each section. Three studeeach pick a section at random. The probability that they are all in the same section is

14

14

14

4 = 116 . Tprobablity that they all end up in different sections is

14

14

14

12 = 316 . The probability that nobody sele

section 1 is 34 34 34 = 2764Remark 3.1. In many problems the sample space is a set of arrangements or sequences. These are capermutations. A key step in the argument is to be sure to undrstand what it is that you are counting. Ihelpful to invent a notation for the outcomes in the sample space and the events of interest.

Example 3.2. Suppose the first six letters are arranged at random to form a six-letter word (an arrangement) -must use each letter once only. The sample space

S= {abcdef, abcdf e, , fedcba}

has a large number of outcomes and, because we formed the word "at random", we assign the same probablity

each. To count the number of words in S, count the number of ways that we can construct such a word, each wcorresponds to a unique word. For each word we follow a pattern of having six choices, then five, and so until uwere left with one. Thus there are 6 5 4 3 2 1 ways of constructing words; 720 ways.Now consider events such as A; the second letter is eorf. So there are now 5 2 4 3 2 1 = 240 ways.

P(A) =number of outcomes in A

number of outcomes in S =

240

720=

1

3

Generally, there are n!arrangements of lengthn using each symbol once and only once, and (nr +1)!arrangemeof length r using each symbol at most once. This product is denoted n(r) ("n to r factors").

There is an approximation to n! called Stirlings formula whcih states

n! nnen2n as n Example 3.3. A pin number of length 4 is formed by randomly selecting with replacement 4 digits from the{0, 1, 2, , 9]}.

A: The probability that the pin number is even is the probability that the last digit is even.

P(A) =5 103

104 =

1

2

B: The probability that the pin only has even digits is

P(B) =

54

104

C: That all digits are unique is

P(C) =10(4)

104 =

10 9 8 7104

= 63

125

D: That the pin number contains at least one 1 is

P(D) = 1 94

104 =

3439

10000

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Definition 3.2 (complement). For some general event A, the complement ofA denoted by A is the set ofoutcomes in Swhich are not in A. It is often easier toc ount outcomes in the complement rather than in the evitself.

Example 3.4. Suppose now that were selecting pins again, but this time without replacement.

A: The probability that the pin number is even is the probability that the last digit is even.

P(A) =5 9 8 7

10(4) =

1

2

B: The probability that the pin only has even digits is

P(B) =5 4 3 2

10(4) =

1

42

C: That the pin number contains 1 is

P(D) = 1

9(4)

10(4)

=2

5In some problems, the outcomes in the sample space are subsets of a fixed sized. Here we look at counting susubsets. Suppose we want to randomly select a subset of 3 digits from the set of digits 0 - 9. Let us now howe

consider order to be irrelevant, and numbers cannot be picked twice. Then we end up with

m=10(3)

3! = 120

arrangements of digits.

Definition 3.3. We use the combinatorial symboln

r

to denote the number of subsets of size r that can be selec

from a set ofn objects. If we denote mto be the number of subsets of size r that can be selected from nthi

thenm r! =n(r)

and so nr

=

n(r)

r!

Example 3.5. Suppose a box contains 10 balls of which 3 are red, 4 are white and 3 are green. A sample of 4 bais selected at random without replacement. Find the probablity of the events

E: the sample contains 2 red balls

F: the sample contains 2 red, 1 white, and 1 green ball

G: the sample contains 2 or more red balls

For simplicity, define 0,1,2 being red, 3,4,5,6 being white, and 7,8,9 being green. Then we construct a unifoprobability modelS= {{0, 1, 2, 3}, . . . ,{6, 7, 8, 9}}

Now for E, we take the number of ways we can take 2 red balls of the three AND take 2 other-coloured balls frthe other seven

P(E) =

32

72

104

= 310

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ForFit is the same idea, we multiply the number of ways of finding each ball. For G we take the sum of the wof finding 2 red balls and 3 red balls.

P(G) =

32

72

104

+ 3371104

Here are some important properties of the combinatorial function n chooser:

1. n(r) = n!(nr)! =n(n 1)(r1) forr 1

2.nr

= n!r!(nr)! =

n(r)

r!

3.n

r

= nnr

for allr= 0, 1, . . . , n.

4. If we define0! = 1, then the formulas above make sense forn0

=nn

= 1

5. (1 + x)n =n0

+n1

x +

n2

x2 + + nrxn (binomial theorem)

Example 3.6. Suppose the letters of the word STATISTICS are arranged at random. Find the probablity of the evG that the arrangement begins with and ends with S. The sample space is S= {SSSTTTIIAC,SSSTTTIICA,First consider the length of this set (the number of ways of spelling STATISTICS)

10

3

7

3

4

2

2

1

1

1

=n

Now the number of ways of spelling it with two Ss fixed at either end83

52

31

21

11

103

=kThe quotient kn =

115 is our answer to the probability that the words start and end with S.

Remark 3.2. The number of arrangements when some symbols are alikegeneraly, for n symbols of t

i {1, . . . , k} withki=1 ni=n, then the number of arrangements using all of the symbols isn

n1

n n1n2

n n1 n2n3

nknk

=

n!

n1!n2! nn!Example 3.7. Suppose we make a random arrangement of length 3 using letters from the set {1,b,c,d,e,f,g,h,ithen what is the probability of the event B that the letters are in alphabetic order if (a) the letters are selecwithout replacement, and (b) if the letters are selected with replacement.There are10(3) ways of picking 3 letter words from this set, and

103

ways of picking them in alphabetical or

thus for part (a) we simply take the quotient of the two.

P(B) = 10

3 10(3)With replacement implies three cases, one for each number of identical letters in a sequence of 3 letters. In casthere are 3 identical letters and there are obviously only 10 ways of achieving this, for case 2 thee are 2 identletters and so there are 6 possibilites of arrangements, but only two are identical, so we multiply the choices of tletters by two. The third case is the same as part (a). Thus

P(B) =10 +

102

+103

103

=11

50

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Note. While n(r) only has a physical interpretation when n and r are positive integers with n r, it still meaning when n is not a positive integer, as long as r is a non-negative integer. In general, we can den(r) =n(n 1) (n r+ 1). For example:

(2)(3) = (2)(2 1)(2 2) = 24

1.3(2) = (1.3)(1.3 1) = 0.39Note that in order for

n0

=nn

= 1we must define

n(0) = n!

(n 0)! = 1 and 0! = 1.

Alson

r

loses its physical meaning when nis not a non-negative r but we can use

n

r

=

n(r)

r!

to define it when n is not a positive integer but r is. For example,12

3

=

12

(3)3!

= 1

16

Also, when n and r are non-negative integers and r > n notice thatn

r

= n

(r)

r! = n(n1)(0)

r! = 0

3.2 Review of Useful Series and Sums

Recall the following series and sums.

1. Geometric Series

a + ar+ ar2 + + arn1 = a(1 rn

)1 r forr = 1

If|r|

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provided that this series is convergent. In this case, with f(a) = (1 + a)n, f(0) =n, f(0) =n(n 1) af(r)(0) =n(r). Substituting,

f(a) = 1 +n

1a +

n(n 1)2!

a2 + + n(r)

r! ar + =

x=0

n

x

ax

It is not hard to show that this converges whenever|a|

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Consider a box model where we have 12 slots and 10 objects to fill the the slots with. So,

P(A) =1010

1210

P(B) =

12(10)

1210

P(C) = 12

1210

P(D) =

121

102

111

83

10(5)

1210

Consider now a new case E= "5 in one month and remaining 5 in another month."

P(E) =

121

105

111

55

1210

not correct, we need to divide by 2

We take a different approach, reserving two months to begin with, so

P(E) =

122

105

55

1210

Example 3.9. Four numbers are randomly selected from{0, 1, 2, 3, 4, 5, 6, 7, 8, 9}without replacement and forsequence.

A = "The sequence is a 4-digit number"

B = "It is a 4-digit even number"

C = "It is a 4-digit even number bigger than 4000"

Our solution,P(A) =

91

9(3)

10(4) =

9

10

P(B) =

51

51

8(2) +

41

41

8(2)

10(4)

P(C) =

31

51

8(2) +

31

41

8(2)

10(4)

Example 3.10. Roll a regular die 4 times. A= "The sum of the 4 numbers appeared is 10"

S= {(1, 1, 1, 1), (1, 1, 1, 2), . . . , (6, 6, 6, 6)}

andn = 64. So,A= {(2, 3, 3, 2), (1, 4, 1, 4), . . .}

The number of ways to have 4 numbers add up to 10, each number is from {1, 2, 3, 4, 5, 6} is the same as the numof ways to put 10 identical balls into 4 boxes, and each box has at least one ball, at most 6 balls.

P(A) =

93

464

10

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Another way to solve this problem is to consider (x + x2 + x3 + x4 + x5 + x6)4, we need the coefficient in fronx10in the expansion. We can find this using binomial theorem and it describes the same thing. So,

x4(1 + x + x2 + x3 + x4 + x5)4 =x4

1 x61 x

4=x4(1 x6)4(1 x)4


3.1 (a) P(a) =(41)6

(5)

7(4) = 47

(b) P(b) = 55(5)

7(6)

(c) P(c) = (4+3+2+1)5(4)

7(6)

3.2(a) i. P = (n1)r

nr

(a) ii. P = n(r)

nr

3.3 (a) P(a) = 6(4)

64 = 5

18

(b) P(b) = 6(62)

64 = 572 (6 destinations for 6 spots split among groups of 2)

3.4 P(H) =(42)(

124 )(

367 )

(5213) (36 since 16 cards not applicable as "others")

3.5 (a) P(STATISTICS) = 110!3!3!2!1!1!1!

= 150400

(b) P("same letter at each end") = 8(73)(

42)(

21)+8(

73)(

42)(

21)+(

83)(

53)(

21)(

1)50400 =

745

3.6 (a) P(a) = 16 (take any 3 digits, the probability there are 3(3) arrangements, and only one is stric

increasing)

(b) P(b) =P(a) 10(3)103

3.7 P(A) = 365(r)

365r

3.8 (a) P(a) = 1n (not in order, just pick a single key out ofn)

(b) P(b) = 2n

3.9 P(A) = 1+3+5++(2n1)

(2n+13 )

3.10 (a) i. 610000 (6 ways to arrange this group, refer to Remark 3.2)

ii. 4(4)

10000= 24

10000, better odds to pick four non-repeating digits

(b) The operator loses money if the prize money exceeds $1 10000, that is, if there are more than 20pwinners. We need the probability that four digits are picked with no repeating digits. P(b) = 10

(4)

10000 .The amount of money the operator gives out is $500|S| whereSis the set of all possible rearrangemeof the four digits drawn.

3.11 (a) P(a) =(62)(

194 )

25 (note to self: remember you have 19 left after taking 6 away, not 23 kiddo)

(b) We could estimate there were 15 deer (extrapolate experiment three times to meet 6 tagged deer)

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Fall 2013 Probability 4 CONDITION

3.12 (a) P(a) = 1(496 )

(b) P(b) =(65)(

431 )

(496 )

(c) P(c) =(64)(

432 )

(496 )

(d) P(d) =(63)(

433 )

(496 )

3.13 (a) There are 2 Jacks, left, soP(a) = 1 (483 )

(503 )

(b) 1(452 )

(472 )

(c) (22)(

483 )

(505 ) =

(483 )(505 )

The binomial theorem states that

(1 + a)

n

=

n

x=0

nxaxSo,

n(1 + a)n1 =n

x=0

x

n

x

ax1 (Differentiate)

na(1 + a)n1 =n

x=0

x

n

x

ax

n

p

1p

1 +

p

1pn1

=n

x=0x

n

x

p

1px

a=

p1p

np=x=0

xn

x

pxpnx

4 Probability Rules and Conditional Probability

4.1 General Methods

Recall that a probability modelconsists of aSample Space S, a set of events or subsets of the sample spacewhich we can assign probabilities and a mechanism for assigning these probabilities. So the probability for soeventAis the sum of the probabilities of simple events in A. Thus we have these rules:

Rule 1P(S) = 1 Rule 2For any eventA, 0 P(A) 1. Rule 3IfA and B are two events with A B, P(A) P(B).

Now consider since I am not going to make venn diagrams right now a venn diagram with two circles A andinside a rectangle S. Then, S is the sample space and A and B are events. A B is everything in either cirA B is the intersection of both circles. As well, Ais everything outside ofA and is defined as the complemen

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Theorem 4.1 (De Morgans Laws).

A B=A B A B=A B

4.2 Rules for Unions of Events

In addition to the last two rules, we have the following

Rule 4a (probability of unions) P(A B) =P(A) + P(B) P(AB) Rule 4b (probability of union of three events)

P(A B C) =P(A) + P(B) + P(C) P(AB) P(AC) P(BC) + P(ABC)

Rule 4c (inclusion-exclusion principle)

P(A1 An) = i

P(Ai)i

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Example 4.2. If a fair coin is tossed 3 times, find the probablity that if at least 1 head occurs, then exactly 1 heoccurs.

LetA be the event that we obtain 1 head, and B be the event that at least one head occurs. We want P(A|B).

P(A|B) =P(AB)

P(B)

=3878

=3

7

4.5 Multiplication and Partition Rules

Product RuleLet A,B ,C,D, . . .be arbitary events in a sample space. Assume that P(A)> 0, P(AB)>andP(ABC)> 0. Then

P(AB) =P(A)P(B|A)P(ABC) =P(A)P(B|A)P(C|AB)

P(ABCD) =P(A)P(B|A)P(C|AB)P(D|ABC)

and so on.

Partition RuleLet A1, . . . , Ak be a partition of the sample space Sinto disjoint (mutually exclusive) evethat is

A1 A2 Ak = S. and Ai Aj = ifi =jLetB be an arbitary event in S. Then

P(B) =P(BA1) + P(BA2) + + P(BAk)=

ki=1

P(B|Ai)P(Ai)

Example 4.3. In an insurance portfolio 10% of the policy holders are in Class A1 (high risk), 40% are in Class(medium risk), and 50% are in Class A3 (low risk). The probability there is a claim on a Class A1 policy in a giyear is .10; similar probabilities for ClassesA2 andA3 are .05 and .02. Find the probability that if a claim is mait is made on a Class A1 policy.

Solution.

For a randomly selected policy, let B be the event that a policy has a claim, and Ai be the event that the policof class i fori= 1, 2, 3. We are asked to find P(A1|B). Note that,

P(A1|B) = P(A1B)P(B)

and thatP(B) =P(A1B) + P(A2B) + P(A3B).

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We are told thatP(A1) = 0.10, P(A2) = 0.40, P(A3) = 0.50

and thatP(B|A1) = 0.10, P(B|A2) = 0.05, P(B|A3) = 0.02.

Thus,

P(A1B) =P(A1)P(B|A1) = 0.01P(A2B) =P(A2)P(B|A2) = 0.02P(A3B) =P(A3)P(B|A3) = 0.01

ThereforeP(B) = 0.04 and P(A1|B) = 0.01/0.04 = 0.25.Theorem 4.3 (Bayes Theorem).

P(A|B) = P(B|A)P(A)P(B|A)P(A) + P(B|A)P(A)

4.6 Tutorial 2

Example 4.4. Independent games are to be played between teams X andY, with team Xhaving probabilityto win each game.

(i) A= "Team Xwins a best-of-seven series with a sweep" (4:0)

(ii) B= "Team Xwins a best of seven series."

Solution. LetBi = "Team Xwins game i" (1 i 7). For any game P(Bi) = 0.6. Note that B1, B2, . . . , B7 independent.

(i) P(B1B2B3B4) =P(B1)P(B2)P(B3)P(B4) = 0.64

(ii) Di = "Team Xwins the series in i games" fori = 4, 5, 6, 7.P(B) =P(D4 D5 D6 D7) =P(D4) + P(D5) + P(D6) + P(D7). Now, P(D4) = 0.64

P(D5) =P(B1B2B3B4B5 B1B2B3B4B5 B1B2B3B4B5 B1B2B3B4B5)=P(B1B2B3B4B5) + + P(B1B2B3B4B5)= 4 0.4 0.64

Then P(B) = 0.64 + 410.4 0.64 + 520.42 0.64 + 630.43 0.64Example 4.5. There aren lettersL1, . . . , Ln and n envelopesE1, . . . , E nand Liand Eiare intended to be matcrandomly, put letters into evenlopes, one letter per envelope. Find the probability that

A= "At least one match" B= "No matches" C= "The 1st r letters and the r envelopes matche dbut none of the others match"

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D= "Exactly r matches"Solution. Ai= "Li matchesEi" fori = 1, 2, . . . , n. AreA1, . . . , An independent? No. AreAi andAj disjoint?

P(A) =P(A1 A2 An)

=

ni=1

P(Ai)i

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4.7 Problems on Chapter 4

4.1 UseDe Morgans Lawsfor the second last one.

4.2 P(A) = 10103 , P(B) = 10(3)

103 , P(C) = 93

103 , P(D) = 53

103 , P(E) = 253

103 .

4.3 We wantP(A|B) which is P(AB)

P(B) = P(A)P(AB)

1P(B) = P(A)P(B)(A|B)

1P(B)

4.4 (a) Its the probablity of not getting a 1 to the 8th power. (1 P(1))8(b) Its the probablity of not getting a 2 to the 8th power. (1 P(2))8(c) Same idea, but for both. (1 P(1) P(2))8(d) 1 P((a)) P((b)) + P((c))

4.5 P(A B) =P(A) + P(B) P(AB) =P(A) + P(B) P(A)P(B) (independent)4.6 Let A,B,C be the events that A, B,C get the right answer. Then we want to find P(C|X) where X

the event that two people get the right answer. The probability that two people get the right answe

P(C|X) = P(CX)

P(X) = P(ABC)

P(X) = P(A)P(B)P(C)

P(A)P(B)P(C)+P(A)P(B)P(C)+P(A)P(B)P(C)

4.7 Let Xbe the event that a customer pays with credit card then

(a)53

0.730.32

(b)42

0.730.32

4.9 LetD be the probability that one randomly selected person out of this population is diseased. Then P(Dthe union of the probbilities at given A, A is diseased, and given B , B is diseased, and given C, C is diseawhich is 0.041. Then the probability that at least one of ten are diseased is one minus the probability tpeople selected have no disease. Thus P("at least one diseased of 10") = 1 (1 0.041)10.

4.10 The probability that A sweeps is P(AHAHAAAA) = 0.72

0.52

, and in five is P(BAAHAAAAAAAHBAAAAAHAHBHAAAA AHAHAABHAA) = 0.175, probability it doesnt go to six games is the sum of these tplus the same idea for B winning in four or five.

4.12 look up

4.13 A="you win the game", and Bi= "the sum of the two numbers from rolling two dice" i= 2, 3, 4, . . . , 11, 12. want P(A) =

12i=2 P(Bi)P(A|Bi). How do we find Bi? Well consider P(B2) = 136 ,P(B3) = 236 . Now the tri

part is the conditional probability. For example, P(A|B7) = 1. P(A|B2) = 0, and P(A|B5) = P(B5)P(B5)+P(So,

P(A) = 0 + 0 + P(B4)

2

P(B4) + P(B7)+

P(B5)2

P(B5) + P(B7)+

P(B6)2

P(B6) + P(B7)+ P(B7) +

P(B8)2

P(B8) + P(B7)+

+ P(B9)

2

P(B9) + P(B7)+

P(B10)2

P(B10) + P(B7)+ P(B11) + 0 = 0.492929292

4.14 (a) The probability that a student says yes is the 0.2 times the chance that he or she was born in one of two months added to 0.8 times the chance they have ever cheated on an exam p. Thus

P((a)) = 1

30+

4p

5

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Fall 2013 Probability 5 PROBABILITY DISTRIBUTIO

(b) Supposex ofnstudents answer yes, then set P((a)) = xn . So,

p=5x 16n

4n

(c) The proportion of students that answer yes responding to question B is

4p5

P((a))=

4p16+ 4p

== 24p

1 + 24p

4.16 (a) 153515 =

3125

(b) 110110

810 is the minimum.

4.17 (a) P(A|Spam) = P(Spam)P(A|Spam)P(A) = 0.50.20.5(0.2+0.001)(b) P(A|NotSpam)

4.18 (a) 1P(A1|NotSpam)(PAn|NotSpam)

P(A1|Spam)P(An|Spam))

(b) Same as above, use1 P(A3).(c)

Note. Sorry to anyone using these notes that I havent created a better summary of 4 or anything for 5, Ill tryget back on track after the midterm.

Note. Good questions to ask:

Can I use the negation? De Morgans Laws, remember to take 1 - "neg". Is there a product I dont know? Use product rule. (P(AB) =P(A)(P(B|A)) Probabiliy of a union? Take sum of parts, less their various intersections. Mutually exclusive? Intersection is 0. Independent? Probability of product is product of probabilities of individual parts of initial product Conditional probaility? Turn into fraction of product over probability of given part. Do I know probabilities of various condition parts but not main part? Paritition rule; P(B) =P(B|A1 + P(B|An)

5 Probability Distributions


Quick tip, when the feature of independence is mentioned in the question, were probably talking about eitBernoulli trials (Binomial, Negative Binomial), or Poisson process. Hypergeometric is NOT independent.

5.1

5.2

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5.3

5.4 Let P(X=x) =p(1p)x, then the probability function on Ris a function on X4,

P(R= r) =P(X=r) + P(X= (r+ 4)) + P(X= (r+ 8)) + =p(1

p)r +p(1

p)r+4 +p(1

p)r+8 +

=p(1 p)r(1 +p(1 p)4 +p(1 p)8 + )

The sum inside the brackets is a geometric series similar to

z+ z4 + z8 + = z1 z4

So,

P(R= r) = p(1p)r1p(1 p)4

5.5 (a) Event Xthat our friend takes 10 tickets to have 4 winners with probability of each trial is 0.3 ha

negative binomial distribution. We want 6 failures before 4th success.

f(x) =

x + 4 1

x

(0.3)4(0.7)x

then (x= 6failures)P(X= 6) = 0.0800483796

(b) We have a binomial distribution, we wantx= 0. 15 events, probability of success trial is 1/9.

f(x) =

15

x

1

9x

8

9

15x= P(X= 0) = 0.170888

(c) Same idea, 60 cups not 15, 1 or 0 not just 0.

P(X 1) =

60

1

1

9

8

9

59+

60

0

8

9

60= 7.2488 103

5.6 (a) One minus probability this guy wins no prizes. Probability of a win is 500500000 = 0.001. Binomdistribution, ten trials, 0 successes.

P(x 1) = 1 P(X= 0) = 1

10

0

(0.001)0(1 0.001)10 = 9.955 103 0.010

(b) Same thing, 2000 tickets.

P(X 1) = 1 P(X= 0) = 1 (1 0.001)2000 = 0.8648

5.7 (a) Straightforward. 40150 = 415

(b) Hypergeometric distribution, men are successes, women are failures. 150 people, 74 men, 76 women,people being picked.

P(Y =y) =

74y

1507412y

15012

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(c) Plug in numbers.P(Y 2) =P(Y= 0) + P(Y= 1) + P(Y = 2) = 0.176

5.8 This is classic Poisson Distribution from Poisson Process. We have a rate, a timeline, a period and sonumber of events in that period. Both units in months so were good.

P(X 4) = 1 3

k=0

P(X=k) = 1 et 3

k=0

(t)k

k!

= 0.9989497

5.9 Fish are more interesting than bacteria, so we use Possion (translate: fish) process. We have a rate 120

(a) Classic. V is volume of sample.

P(X=x) =(V)2eV

x! = P(X= 2) =

120 10

2e(

12010)

2! = 0.7581633

(b) Exactly the same procedure, use 1 - probability of finding none, and 1 instead of 10.

P(X 1) = 0.488

(c) We use binomial distribution on poisson distribution. p is probability of no fish present in sample.

p=(10)0e10

0! =e10

P(Y =y) =

10

y

py(1p)10y =

10

y

(e10)y(1 e10)10y

(d)

p=

3

10 = = 0.125.10 (a) One minus probability of no claims,

P(X 1) = 1 P(X= 0) = 1 8100 1

0e(

8100

1)

0! = 0.07688365

(b) Probability of no claims for 20 is binomial ontop of above,

P(X=x) =

20

x

(0.07688365)x(1 0.07688365)20x

Setx = 0 for first part of this question. For second part, calculate P(X

2).

5.11 80% of months have no failures.

(a) Straightforward.

P(X= 5) =

7

5

0.850.22 = 0.2753

(b) Negative binomial.

P(X= 7) =

2 + 5 1

2

0.220.85 = 0.1966

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(c) So for this part we think about it in terms of Poisson Distribution. So P(X=x) is the probability ta month has xpower failures,

P(X= 0) = 0.8

from the question, and t= 1for one month, so

0.8 =(t)0

et

0! =et =e(1) = = 0.223

Then,

P(X >1) = 1 (P(X= 0) + P(X= 1)) = 1 (e0.223 + (0.223)e0.223) = 0.0215

5.12(a) First note thatlimN

rN =

rN

lim

N

f(x) = lim

N rx

NrnxNn

=

n

x

limN

r!(N r)!n!(r x)!(N r n + x)!N!

=

n

x

limN

[r(r 1) (r x + 1)][(N r)(N r 1) (N r n + x + a)]N(N 1) (N n + 1)

Forx = 0, 1, 2, . . . , n. It must be th ecase thatr sincep = rN is fixed. We expect thatNr Nandn could be ignored as r, Ngo to infinity. Let q= 1p =

Nr . Then,

f(x) =

n

x

[r(r 1) (r x + a)][(q 1)r((q 1)r 1) ((q 1)r n + x + 1)]

qr(qr 1) (qr n + 1)

=n

x

1q

xq 1q

nx=

n

x

px(1p)nx

5.13 (a) Call this event X. We have a rate per hectare and measure which is n hectare plots, and were lookfor at least k of some item. So its Poisson process. We are looking for one minus the probability tall of the hectare plots have less than k spruce budworms. That probability is then-th power of probablility that a single hectare plot has less than k spruce budworms, which is the sum from i = 0i= k 1 of the poisson distribution over one hectare (so = 1 ) fori. Therefore we conclude,

P(X) = 1k1i=0

( 1)ie1i!

n

5.14 So were essentially told at tht beginning here that = 20, andp= 0.2.

(a) Probability of two sales in 5 calls is straightforward Binomial Distribution,5

2

0.22 0.83 = 0.2048

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(b) This is negative binomial distribution, 8 fails up to reach 2 successes.2 + 6 1

1

0.22 0.86 = 0.0734

(c) This situation occurs only if calls 4,5,6,7 are failures. Thus our probability is (

) the ways that the fi

three can yield one success, then () four failures, then ( ) one success, over all possibilities whertakes 8 trials for 2 successes (which we found in part b).

3

1

0.2 0.82

()

0.84()

0.2()

P((b)) =

0.03145728

0.0734 = 0.428

(d) This is Poisson, we have to convert minutes to hours, is per hour, so just use 14 for 15 minutes.

(14

)3e4

3! =

(14

20)3e204

3! = 0.14037

5.15 Collection ofN= 105objects,n = 35of typeA and r = 70of typeB . We choose objects without replacemuntil 8 of type B have been selected. Let Xbe the number of type Awithdrawn. This is Hypergeomedistribution. Think about it this way: how many ways can you choose x objects of type A, and also 7 objeof type B? (7 here since the last one is always of type B to get our 8th success) Well,

35x

707 . possibilities are

1057+x

because youre taking 7 of type B for sure, and then x of typeA. Now finally all

need is the probability that the last one is an 8. Well, thats simply the number of objects of type B left aour first 7 are taken, 63, over the total number less the x and 7 we already took.

f(x) =P(X=x) = 35x

707 1057+x

70 7

105 7 x5.16 (a) Basic Poisson questions, just convert seconds to hours.

540 112011

e(540 1120)

11! = 4.26438 103

and

1

10k=0

540 1120

ke(540

1120)

k!

= 6.668672 103

(b) This is binomial on the first part of part (a)20

2

(0.004264)2(1 0.004264)18 = 3.19877 103

(c) (i) Its not quite the same as part (b), this time were using negative binomial distribution to find probablity of only 11 successes in 1399 trials, then the 1400th being golden.

1399

11

(0.004264)11(1 0.004264)1388 0.004264

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(ii) Check page 90 of textbook for "use the Poisson distribution with = np as a close approximatto the binomial distributionBi(n, p) in processes for which nis large and p is small."

(0.004264 1399)11 e(0.0042641399)11!

0.004264 = 9.331 105

5.17 (a) The probability of a single sheet of glass with no bubbles is

(1.2 0.8)0e(1.20.8)0!

Therefore,

f(x) =P(X=x) =

n

x

(e0.96)x(1 e0.96)nx

(b)

1

2e0.8

ln(2)

0.8

0.866

5.18 4k2 = 1 = k = 0.5, then just draw the same little table for f(x), and add up f(3) andf(4).5.19 (a)

P(Y y) =P(Y =y) + P(Y =y + 1) + =p(1p)y +p(1 p)y+1 +

=p(1p)y

(1 + (1p) + (1 p)2

+ )=

p(1 p)y1 (1 p)

=p(1 p)y

p

= (1 p)y

Then,

P(Y s + t|Y s) = P(Y s + t)P(Y s)

=(1p)s+t

(1 p)s= (1p)t=P(Y t)

(b) Since p 1, and (1 p) is therefore less than one, the probablity function is decreasing. The mprobable value ofY is 0. P(Y= 0) =p.

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(c) Call this eventX,

P(X) =P(Y= 0) + P(Y= 3) + P(Y= 6) + =p(1 p)0 +p(1 p)3 +p(1 p)6 + =p(1 + (1

p)3 + (1

p)6 +

)

= p1 (1 p)3

(d)

P(R= r) =p(1p)r +p(1 p)r+3 +p(1p)r+6 + =p(1p)r(1 + (1p)3 + (1 p)6 + )

= p(1p)r1 (1 p)3

5.2 Tutorial 3

Recall the Poisson approximation to Binomial, for nlarge and p small, = np (so p = n). Thenn

x

px(1 p)nx

xe

x!

Example 5.1. For example, consider X Bi(100, 0.02)i. Find the probablity ofP(X= 0), P(X= 1), and P(X= 2)

P(X= 0) = 1000 (0.02)0(0.98)100 = 0.1326P(X= 1) =

100

1

(0.02)1(0.98)99 = 0.2707

P(X= 2) =

100

2

(0.02)2(0.98)98 = 0.2734

ii. Use Posson approximation to fnd probabilities in i. (= 100 0.02 = 2)

P(X= 0) =0e

0! = 0.135

P(X= 1) =1e

1! = 0.2707

P(X= 2) =2e

2! = 0.2707

Example 5.2. SayX NB(2, 0.02). Find probability P(X= 99)

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Use Poisson approx to find P(X= 99) Let

f(x) =P(X=x) =

x + k 1

x

pk(1p)x

Then

P(X= 99) =

99 + 2 1

99

0.022 0.9899

=

100

99

(0.02)2(0.98)99

=

100

1

(0.02)(0.98)99 0.02

=1

1!e 0.02

Example 5.3. Traffic accident occurs following a poisson process. It is known that 10% of days have no accide

Letx be the number of accidents in a day. Find distribution ofX. Find the probability that there are two accidents in a day. Find the probablity that there are four accidents in two days. Find the probablity that there are at least 2 accidents in a day. Find the probablity that, among a week of 7 days, there are exactly 3 days which have at least 2 accidents

day.

Then for i., X P oi(), and = t not available. However, we know that x= 0means no accidents, and the

P(X= 0) =e = 0.1 = = 2.3

So,X P oi(2.3).

For ii.,P(X= 2) = 2

2! e.

For iii., lety be the number of accidents in 2 days. Then y P oi(2.3 2) =P oi(4.6). So,

P(y= 4) =4.64

4! e4.6

For iv., Letxbe the number of accidents in a day,X P oi(2.3). Then

p= P(X 2)= 1 [f(0) + f(1)]= 1 [e + e]= 0.67

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For v., let y be the number of days among a week of seven days that have at least 2 accidents. Theny Bi(7, 0.P(Y= 3) =

73

0.6730.334.

For vi. Find the probability that we have to observe a total of 10 days to see the third day with greater thanequal to 2 accidents in the day. Let y be the days required to see 3rd day with 2 accidents.

P(y= 10) =9

2

(0.67)3(0.33)7

(could also use y the number of failures before 3rd success, and then y N B(3, 0.67))

Now vii, if there are a total of 13 accidents in a week of 7 days, whats the probablity that there are 6 accidentsthe first 3 days of the week?

Let y be the number of accidents in 7 days. Let x1 be the number of accidents in the first 3 days. Let x2 be number of accidents in the last 4 days. Then,

y

P oi() for = 2.3

7.

x1 P oi(1)for1= 2.3 3. x2 P oi(2)for2= 2.3 4.

Then

P(x= 6|y= 13) = P(x1= 6 y= 13)P(y= 13)

=P(x1= 6 x2= 7)

P(y= 13)

=

P(x1= 6)P(x2= 7)

P(y= 13)

=

61e1

6! 72e

2

7!13e

13!

=

13

6

3

7

647

7

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Fall 2013 Probability 6 CHAPTE

6 Chapter 7

6.1 Tutorial 4

(1) Moment generating function:M(t) =E(etx), t

(

h, h), h >0

IfX is discrete with probability function f(x),

M(t) =

x

etxf(x)

Then,

X Bi(n, p), M(t) = (pet + q)n, anyt X Geo(p), M(t) = p1qet ,t < ln(q) X P oi(), M(t) =e(et1), any t

(2) XandYare two discrete random variables with probability functions fx(x)and fy(y),MGFsthat areMandMy(t), t (h, h),h >0. Then,fx(x) =fy(x) for allx

if and only ifMx(t) =My(t) fort (h, h)

Two random variables have the same distribution if and only if they have the same moment generatfunction.

Example 6.1. IfXhas Moment Generating Function,

M(t) = 0.5et + 0.2e3t + 0.3e5t

find the probability function ofX

X f(x) : x 1 3 5f(x) 0.5 0.2 0.3

(3) Xn has probablity fnction fn(x), moment generating function Mn(x);Xhas probablity function f(x), moment generating function M(t)

IfMn(t) M(t), t (h, h)asn . Then,

fn(x) f(x)for all x, as n Example 6.2. Letp= n . Use the MGF technique to show that

n

x

px(1 p)nx

x

x!e

whenn .

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Xn Bi(n, p) : Mn(t) = (pet + q)nX P oi() :M(t) =e(et1)Then

(pet + q)n = (pet + 1 p)n

= 1 + n et 1n=

1 + c

n

nx= (et 1)

And(n ) ec =e(et1)

(4) Ify= aX+ b for some constants a andb, then

My(t) =ebtMx(at)

special case:Ify= X+ b, thenMy(y) =e

btMx(t). Since by defintiion,

Mx(t) =E(etx)

My(t) =E(ety)

=E(et(aX+b))

=E(ebteatX)

=ebtE(eatX)

=ebtMx(at)

Question 7.7 from the book: Suppose thatn people take a blood test for a disease,w here each person has probabipof having the disease, independent of the other person. TO save time and money, blood samples from k people pooled and analyzed together. If none of the k persons has te disease then the test will be negative, but otherwiswill be positive. If the pooled test if positive then each of the k persons is tested seperately (so k + 1 tests are din that case).

(a) Let Xbe the number of tests required for a group ofk people, show that

E(X) =k + 1 k(1 p)k

LetXbe the number of tests required for a group ofk persons, then x 1 k+1

f(x) (1

p)k 1

(1

p)k

Th

E(x) = 1 (1 p)k + (k+ 1) (1 (1 p)k)=k + 1 k(1 p)k

(b) What is the expected number of tests required for n/k groups ofk people each? Ifp= 0.01, evaluate thiscaesk = 1, 5, 10.

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We have that

n

k E(x) = n

k

k+ 1 k(1 p)k

=n +

n

k n(1 p)k ()

Question 1: Show that() n kp + 1kQuestion 2: ()is minimized at k = p1/2

(c) Show that ifp is small, the expected number of tests in part (b) is approximately n(kp +k1), and is minimifork p1/2.

Sincepis small,

(1p)k = 1 kp + k(k 1)2!

p2 + 1 kp

Then,

n +n

k n(1 p)k n +n

k n(1 kp)

=n

1

k+ kp

=n

1

k

k

p

2+ 2

p

This is minimized when the squared term is 0, so

1k

= kp

k= p1/2

Question 7.11 from the book: Find the distributions that correspnd to the following moment-generating functio

(a) M(t) = 13et2 fort

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Theny= x + 1 :The total number of trials to the get the first success.

f(y) =P(Y =y) =qy1p, y= 1, 2

(b) M(t) =e2(et1) fort <

Question 7.13 from the book: Let Xbe a random variable in the set {0, 1, 2} with moments E(X) = 1 and E(X2) =

Note. When you have a random variable X f(x)we notice x 0 1 2f(x) p0 p1 p2

thenE(X) = 1 : 0 p0+ 1 p1+ 2 p2= 1

and

E(X2) =3

2: 02 p0+ 12 p1+ 22 p2 = 3

2

and

p0+p1+p2= 1


7.1 The mean multiplies eachx by its f(x), so

11

40 1 +

6x=2

x

2x=

11

40+

5

2= 2.775

The variance is then E((X 2.775)2)which is

V ar(X) = (1 2.775)2 11

40+

6x=2

(x

2.775)2

2x = 2.574375

7.2 E(X)is the expected winnings based on the expected number of tosses,g(X).

E(g(X)) = 0.5121 + 0.5222 + 0.5323 + 0.5424 + 0.5525 +x>5

0.5x(256) = 5 +25612

61 12

= 3

7.3 Expected cost per person if a population is tested for the disease using the inexpensive test followednecessary, by the expensive test. First letX = 1 if the initial test is negative, X = 2 if the inital tespositive. Let g(x) be the total cost of testing the person, then

E(g(X)) =g(1)f(1) + g(2)f(2)

Then

P(X= 2) =P(D) + P(R|D)P(D) = 0.02 + 0.05 (1 0.02) = 0.069

ThenE(g(X)) = 110 0.069 + 10 (1 0.069) = $16.90

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7.4 P(condition) =P(C) = 0.02. P(A|C) = 0.8andP(A|C) = 0.05. Also,P(B|C) = 1and P(B|C) = 1. Th(a) We want

150

i=1 P(C) = 3(b) Let g(X) be the expected total cost for the same Xas last time, then

P(X= 1) =P(A|C)P(C) + P(A|C)P(C) = 0.065

Thus,E(g(X)) = 101 0.065 + 1 (1 0.065) = 7.5

So for 2000 people, thats 7.5 2000 = $15000. Then the expected number of detected cases are ones where someone has the condition, and is verified by A, because if they are verified by A and dohave it, obviously the second test B will confirm they are alright.

2000 0.08 0.02 = 32

7.5 (a) Expected "winnings" if you bet a dollar is

E(X) =

18

37

1

19

37

1

multiply by ten for ten plays and we get$1037 .If you spend 10 on a single play, your expected winnings are

E(X) =18

37 10 19

37 (10)

reduce by the 10 you put in again and find its the same number from part (a).

(b) The probability that your winnings are positive if you bet one dollar ten times is the probability thyou win at least 6, 7, 8, 9, or 10 games

10x=6

10

x

18

37

x1937

10x

the other result is straightforward, the probability you win one game.

7.6 The probability of each can be calculated by multiplying the probability of getting it on a single wheel forthree wheels, thus

E(g(X)) = 20

2

10

6

10

2

10

+ 10

4

10

3

10

3

10

+ 5

4

10

1

10

5

10

which is $0.94.

7.7 (a) There are two cases, either no one has it or at least one person has it. Probability of no one having i(1p)k, in which case 1 test is done. The probability of at least one person having it is 1 (1p)

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which casek+ 1 tests are done.

E(X) = (1 p)k + (1 (1 p)k)(k+ 1)= (1 p)k + k+ 1 (1p)k(k+ 1)=k + 1

k(1

p)k

(b) Just multiply by the number of groups from part(a). n + nk + n(1p)k.(c) Since p is small,

(1p)k = 1 kp +k(k 1)2!

p2 + 1 kp

Then,

n +n

k n(1p)k n + n

k n(1 kp)

=n1

k+ kp

=n

1

k

k

p

2+ 2

p

This is minimized when the squared term is 0, so

1k

=

k

p

k= p1/2

7.8 The probability a radio is defective is 0.05. In a single carton ofnradios, we can find the expected valuethe amount that the manufacturer will pay to the reailer, it is

E(g(x)) = 59.50n 25 200 E(X2)

Find E(X2) using the fact that V ar(X) =E(X2) + E(X)2, and that for a binomial distribution V ar(Xnp(1 p)andE(X) =np. Plug in to the above equation and find max using methods from Calculus.

7.9 (a) By definition,

M(t) =E(etX)

= x0

etx

f(x)

=x0

etxp(1p)x

=px0

(et(1p))x

= p

1 et(1p)

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with the contraint thatet(1 p)< 1.(b)

E(X) =M(0)

= p

(1 (1 p))2(

1)(1

p)

=1p

p

E(X2) =M(0)

=(1p)2 +p(1p)

p2

V ar(X) =E(X2)E(X)2

=(1

p)2 +p(1

p)

p2 (1

p)p

p2

=

1p

p

2(c)

7.10 (a) IfXis the number of comparisons needed, then since in general E(X) =n

i=1 E(X|Ai)P(Ai), its cthat

Cn =n

i=1E(X|initial pivot if the ith smallest number)

1

n

(b) Once we have a pivot, we need to split the remaining n 1numbers into 2 groups (one comparison eaand then solve the two sub-problems; one of size i 1 and one of size n i, respectively. So,

E(X|initial pivot if the ith smallest number) =n 1 + Ci1+ Cn1The recursion comes from plugging in this result to the formula from part a.

Cn= n 1 + 2n

n1k=1

Ck, n= 2, 3, . . .

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(c) So,

Cn+1= n + 2

n + 1

nk=1

Ck n= 2, 3, . . .

(n + 1)Cn+1= (n + 1)n + 2

nk=1

Ck

= (n + 1)n + 2Cn+ 2n1k=1

Ck

= (n + 1)n + 2Cn+ n(Cn n + 1)= 2n + (n + 2)Cn n= 1, 2, . . .

(d) From part c,

limn

Cn+1n + 1

= limn

2n + (n + 2)Cn(n + 1)2

= limn

2n + nCn+ 2Cnn2 + 2n + 1

= limn

2n + n

n 1 + 2nn1

k=1Ck

+ 2

n 1 + 2n

n1k=1Ck

n2 + 2n + 1

= limn

2n +

n2 n + 2n1k=1Ck + 2n 2 + 4n n1k=1Ckn2 + 2n + 1

= limn

3n + n2 +

1 + 4nn1

k=1Ck

n2 + 2n + 1

7.11 Look above

7.12

M(t) =E(etx)

=b

x=a

etx 1

b a + 1

= 1

b a + 1b

x=aetx

= 1b a + 1 (e

at + e(a+1)t + + ebt))

= 1

b a + 1

e(b+1)t eat

et 1

Ifa= b thenM(t) =eat

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Ifb= a + 1 then

M(t) =1

2eat(1 + et)

7.13 First, x 0 1 2

f(x) p0 p1 p2Then we know that

E(X) = 1 : 0 p0+ 1 p1+ 2 p2= 1

and

E(X2) =3

2 : 02 p0+ 12 p1+ 22 p2= 3

2

andp0+ p1+p2= 1

Solving this system of equations gets us p0= 14 ,p1=

12 andp2=

14 . Then,

(a) M(t) = 0.25 + 0.5et + 0.25e2t

(b) M

(k)

(0) for k = 1, 2, 3, 4, 5, 6,

M(0) = 1, M(0) =3

2, M(3)(0) =

5

2, M(4)(0) =

9

2, M(5)(0) =

17

2 , M(6)(0) =

33

2

(c) See above

(d) For given values ofE(X) =1,E(X2) =2, there is a unique solution top0 +p1 +p2 = 1,p1 + 2p2 =

andp1+ 4p2= 2. Then,

1 1 1 10 1 2 10 1 4 2

1 0 0 1 + 22 3120 1 0 22 10 0 1 122 1

7 Chapter 8


8.1 (a) We check iff(x, y) = f1(x)f2(y) for all (x, y). Consider (0, 0). Then f(0, 0) = .15= f1(0)f2(0(0.15 + 0.35)(0.15 + 0.1 + 0.5).

(b) P(X > Y) =f(1, 0) + f(2, 0) + f(2, 1) = 0.3andP(X= 1|Y= 0) = f(1,0)f2(0) = 13 .

8.2 (a) Since X Poisson(1= 0.10) and Y Poisson(2= 0.5)we have

P(X=x) =

x1e1

x! P(Y =y) =

y2e2

y!

IfXandY are independent, then

P(X=x, Y =y) =x1e

1

x!

y2e2

y!

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parameterizing for P(T =t) =P(X+ Y =t) =

x P(X=x, Y =t x) then we have

P(T =t) =

x

x1e1

x!

tx2 e2

(t x)!

= 0.05te0.15x

1

x!(t x)! 0.01

0.005x

= 0.05te0.151

t!

x

t!

x!(t x)!

0.01

0.005

x

= 0.05te0.151

t!

x

t

x

0.01

0.005

x

=0.05te0.15

t!

1 +

0.1

0.05

t

=e0.150.15t

t!

Then we want P(t >1), its just

1 (P(T= 1) + P(T= 0)) 0.0102

(b) The mean and variance of a poisson distribution are both, so we get 0.15 for both.

8.4 Lets say the events areA,O,andT, andP(A) =p, P(O) =q, P(T) = 1p q.(a) We want P(X=x, Y =y). This is a case of negative binomial distribution, the first combinatorial te

is just the number of ways that we can arrange the x of type A, y of type O, and 9 of type Tbefore tenth T.

f(x, y) =(x + y+ 9)!

x!y!9! px

qy

(1p q)10

(b) The conditional probability function:

f(y|x) = f(x, y)f1(x)

=

(x+y+9)!x!y!9! p

xqy(1p q)10f1(x)

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Now,

f1(x) =y0

(x + y+ 9)!

x!y!9! pxqy(1p q)10

= y0

(x + y+ 9)!

x!y!9! pxqy(1

p

q)10

=px(1p q)10

x!9!

y0

(x + y+ 9)!

y! qy

=px(1p q)10

x!9! (x + 9)!

y0

x + y+ 9

y

qy

=px(1p q)10x+9x

(1 q)x+10y0

x + y+ 10 1

y

qy(1 q)x+10

= x + 9

x p

1 qx

1p q

1 q 10

Then

f(y|x) = f(x, y)f1(x)

=

(x+y+9)!x!y!9! p

xqy(1p q)10x+9x

p1q

x 1pq1q

10=

x + y+ 9

y

qy(1 q)x+10

8.7 On any draw, of the 5 yellow balls you choose x, and of the 3 remaining you choose 2 - x. Then, of remaining yellow balls you choose y xand then 2 + x y of the remaining.

f(x, y) =

5x

32x

5xyx

1+x2+xy

82

62

for part (b), to determine if they are not independent we need to find a case where f(x, y) =f1(x)f2(y). Nf1(0) = 0, f2(3) = 0 but f(0, 3) = 0.

8.8 Hypergeometric,

(a)

f(x, y) =

2x

1y

73xy

103

(b) Marginal probability functions are f1(x),f2(y), note that since

n

k

=

n 1k 1

+

n 1

k

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f1(x) =

2x

83x

103

f2(x) =

1y

93y

103

(c) Then P(X=Y) is f(x, x)

f(x, x) =f(0, 0) + f(1, 1) = 49

120

andP(X= 1|Y = 0) is f(1|0) = f(1,0)f2(0) = 12

8.9 (a) The marginal probability function ofX is

f1(x) =y0

k2x+y

x!y!

=2x

kx!

y

2y

y!

=2xk

x!

1 +

2

1!+

22

2!

=

2xk

x! e2

(b) Evaluatingk ... x

f1(x) = 1

x

k2x

x! e2 = 1

k= 1

e2

x2x

x!

k= e4

(c) Might as well check, note

f2(y) =k2y

y! e2

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by symmetry, then

f(x, y) =f1(x)f2(y)

= 2x

x!e22y

y!e2

= 2x+y

x!y!e4

=k2x+y

x!y!

=f(x, y)

so yeah, independent

(d) P(T =X+ Y) =f(t) is

xf(x, t x) =

x2x+tx

x!(t x)!e4

= 2t

e41t!

x

tx

=

2t

t!e4(1 + 1)t

= 22t

t!e4

8.10 (a) Let Yrepresent number of Type A eveners that occur in a 1-min period. Let Xbe the total numbeevents in a 1 minute period. Then, P(X=type A) =p and X Poisson().

P(Y =y) = x P(Y =y|X=x)P(X=x)=

x

f(y|x)f1(x)

=

x

x

y

py(1 p)xy e

x

x!

=epyy

x

x

y

((1 p))xy

x!

=e

y! pyy

x

((1p))xy(x y)!

= e

(p)y

y! e(1p)

=(p)yep

y!

(b) P(Y =y) = e0.05330(0.05330)y

y!

14

y=0

P(Y =y)

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8.11 (a) Multinomial distribution...

P(10 of each type) = 40!

10!10!10!10!

3

16

10 516

10 516

10 316

10(b)

40!

16!24!

8

16

40=

40

16

3 + 5

16

165 + 316

24(c)

P(10 of 1 | 16 of 1 and 2) =40!

10!6!24!

316

10 516

6 816

2440!

24!16!

816

16 816

24=

16

10

3

8

1058

6

8.12 (a)

E(X) =

x

xf(x)

= 0 .20 + + 8 .01=

44

25

(b) First,

f(y|x) =

x

y

0.5y0.5xy

So,

f(x, y) =f(x)f(y|x)

=f(x)

x

y

0.5y0.5xy

=f(x)

x

y

0.5x

Then,

fY(y) =8

x=y f(x, y)So finally,

E(y) =

y

yfY(y) =1

2E(x) = 0.88

8.13 (a) This is multinomial distribution,

f(x1, x2, x3, x4, x5, x6) = 10!

x1!x2!x3!x4!x5!x6!px11 p

x22 p

x33 p

x44 p

x55 p

x66

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(b)

P(X3 1|x1+ x2+ x3+ x4= 4) = 1= 1 P(X3= 0|1|x1+ x2+ x3+ x4= 4)

= 1

P(X3= 0, x1+ x2+ x3+ x4= 4, x5+ x6= 6)

P(x1+ x2+ x3+ x4= 4)

= 110!4!6!p

03(p1+p2+ p4)

4(p5+p6)6

10!4!6! (p1+ p2+p3+p4)

4(p5+p6)6

= 0.4602

(c)10 (0.1(500) + 0.05(500) + 0.05(700) + 0.15(2000) + 0.15(400) + 0.5(200)) = 5700

8.14 Proof.

E[g(X1, . . . , X n)] = x1,...,xn g(x1, . . . , xn)f(x1, . . . , xn)then

x1,...,xn

af(x1, . . . , xn)

x1,...,xn

g(x1, . . . , xn)f(x1, . . . , xn)

x1,...,xn

bf(x1, . . . , xn)

a

x1,...,xn

f(x1, . . . , xn)

x1,...,xn

g(x1, . . . , xn)f(x1, . . . , xn) b

x1,...,xn

f(x1, . . . , xn)

a

x1,...,xn

g(x1, . . . , xn)f(x1, . . . , xn) b

a E[g(X1, . . . , X n)] b

8.15

V ar(X 2Y) = (1)V ar(X) (2)2V ar(Y) + 2(1)(2)Cov(X, Y)= 13 + 4(34) 4(0.7)

13

34

= 207.86666666666666666666

8.16 (a) T Bin(n, p + q)(b) E(T) =n(p + q) because its binomially distributed, so also V ar(T) =n(p + q)(1 p q)(c)

Cov(X, Y) =E(XY) E(X)E(Y)

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not helpful. Try again:

V ar(T) =V ar(X+ Y)

=V ar(X) + V ar(Y) + 2Cov(X, Y)

n(p + q)(1 p q) =np(1p) + nq(1 q) + 2Cov(X, Y)Cov(X, Y) = npq

It should be negative since as one goes up, the other goes down.

8.17 (a) First find them forXandY.

E(X) =1

2(2) = 1, E(Y) =

1

2(2) = 1

V ar(X) =E(X)1

2=

1

2, V a r(Y) =

1

2

Then,

E(U) =E(X) + E(Y) = 2, E(V) = 0V(U) =V(X) + V(Y) = 1, V(V) = 1

(b)

Cov(U, V) =C ov(X+ Y, X Y)=C ov(X, X) + (1)Cov(X, Y) + Cov(Y, X) + (1)Cov(Y, Y)=V ar(X) V ar(Y)= 0

(c) Nah,U= 0and V = 1 is a counterexample.

8.18 (a) LetXi be whether or not question i was answered right (1 for yes 0 for no). Then,

(i)

Y =

100i=1

Xi 14

100

100i=1

Xi

(ii) E(Xi);V(Xi). LetAi = "know answer to question Qi"

P(Xi= 1) =P(Ai)P(Xi= 1|Ai) + P(Ai)P(Xi= 1|Ai)=pi

1 + (1

pi)

1

5=

4

5pi+

1

5

Then since X= 0 or 1,

E(X2) =E(X) = 0 f(0) + 1 f(1) =45

pi+1

5

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and then

V ar(X) =E(X2) E(X)2

=

4

5pi+

1

5

4

5pi+

1

5

2=12

25pi 16

25p2i + 425

So were set to find E(Y) and V ar(Y)

E(Y) =E

100i=1

Xi 14

100

100i=1

Xi

=100i=1

5

4E(Xi)

25

=5

4 100

i=1

4

5pi+

1

5 25=

100i=1

pi

as required. Next,

V ar(Y) =V ar

100i=1

Xi 14

100

100i=1

Xi

=V ar100

i=15

4Xi

=25

16

V ar(Xi)

= 3

4pi p2i +

1

4

=

pi(1 pi) +100

pi4

as required.

(b)

Y =

Xi

P(Xi= 1) =P(Ai)P(Xi= 1|Ai) =pithen

E(Xi) =pi

V ar(Xi) =E(X2i) E(Xi)2

=pi p2i=pi(1pi)

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so we have

E(Y) =E

Xi

=

pi

and

V ar(Y) =V ar

Xi

=

V ar(Xi)

=

pi(1 pi)

8.19

Cov(X+ Y, X Y) =C ov(X, X) Cov(Y, Y)=V ar(X) V ar(Y)= 1 2= 1

8.20 (a) We know that L = A + B+ C. So,

V ar(L) =V ar(A) + V ar(B) + V ar(C) = 0.62 + 0.82 + 0.72

so=

2 = 1.22.

(b) Do the same thing as before with the new standard deviation ofB , and take the quotient of the it wanswer from part (a).

8.21 X= total number of islands being cut off. Let

Xi=

1 island i being cut off0 otherwise

, i= 1, 2, 3, 4, 5

(i) X=X1+ X2+ X3+ X4+ X5. Now we want to find the expected value ofXi, the variance, and probability thatXi= 1. So,

E(Xi) =pi

V ar(Xi) =pi(1 pi)pi= P(Xi = 1)

howeverpi varies for i = 5.

P(X1= 1) =p3

P(X5= 1) =p4

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now, it also varies if you cut off two bridges at a time (since some bridges are shared)

Cov(Xi, Xj) =E(XiXj)E(Xi)E(Xj)E(X1X2) =P(X1= 1, X2= 1) =p

5

E(X1X5) =P(X1= 1, X5= 1) =p6

So first off,E(X) =E(X1+ X2+ X3+ X4+ X5) = 4p

3 +p4

so then

V ar(X) =V ar(X1+ X2+ X3+ X4+ X5)

=V ar(X1) + + V ar(X5) + 2i

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Note that Xi Bi(n, pi). Similarly, Xj Bi(n, pj). Also, Xi+ Xj Bi(n, pi+ pj). What about Xi|Xk =Xi|Xk =t Bi

n t, pi1pk

. So,

(a) Cov(Xi, Xj) =E(XiXj) E(Xi)E(Xj). But what is E(XiXj)? Lets approach it differently,

V ar(Xi+ Xj) = V ar(Xi) + V ar(Xj) + 2Cov(Xi, Xj). From this equation we know all three V arterms sithey are all binomially distributed. So,

V ar(Xi+ Xj) =V ar(Xi) + V ar(Xj) + 2Cov(Xi, Xj)

n(pi+pj)(1 pi pj) =npi(1 pi) + npj(1pj) + 2Cov(Xi, Xj)Cov(Xi, Xj) = npipj

Example 7.4. Randomly put n letters L1, L2, . . . , Ln into n envelopes E1, E2, . . . , E n, one letter per envelopeletterLi is in envelope Ei, its called a match. Let Xbe the total number of matches. Find E(X)andV ar(X

Method 1. Usef(x). Then,

E(X) =

x

xf(x)

This is hopelessly complicated. Method 2. Use indicator variables. Let

Xi=

1 letter Li matches Ei0 otherwise

, i= 1, 2, . . . , n

i. X=X1+ X2+ + Xnii. X1, . . . , X n are not independent

iii. Xi

Bernoulli (pi), sof(0) = 1

pi andf(1) =pi.

E(Xi) =pi

V ar(Xi) =pi(1pi)

pi= P(Xi= 1)

= 1

n

Note that

E(X) =E n

i=1 Xi =n

i=1 E(Xi) =n 1

n

= 1

Now,

V(X) =V

ni=1

Xi

=n

i=1

V(Xi) + 2i

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now before we continue, note that:

Cov(Xi, Xj) =E(XiXj) E(Xi)E(Xj)

E(XiXj) = x y xyf(x, y)=f(1, 1)

=P(Xi= 1, Xj = 1)

= 1

n(n 1)So,

V(X) =V

ni=1

Xi

=

ni=1

V(Xi) + 2i

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Fall 2013 Probability8 CONTINUOUS PROBABILITY DISTRIBUTIO

now, it also varies if you cut off two bridges at a time (since some bridges are shared)

Cov(Xi, Xj) =E(XiXj) E(Xi)E(Xj)E(X1X2) =P(X1= 1, X2= 2) =p

5

E(X1X5) =P(X1= 1, X5= 1) =p6

Example 7.6. Question 8.18 from the book. A multiple choice exam has 100 questions, each with 5 possanswers. One mark is awarded for a correct answer and 1/4 mark is deducted for an incorrect answe. A particustudent has probability pi of knowing the correct answer to the ith question, independent of other questions.

(a) Suppose that on a question where the student does not know the answer, he or she guesses randomly. Sh

that his or her total mark has mean

pi and variance

pi(1pi) + (100

pi)4 .

(b) Show that the total mark for a student who refrains from guessing also has mean

pi, but with variapi(1 pi). Compare the variances when all pis equal (i) .9, (ii) .5.

LetY = total mark and let

Xi=

1 Qi answered correctly0 otherwise

, i= 1, 2, . . . , 100

(i) y= 1100i=1 Xi 14 100100i=1(ii) E(Xi);V(Xi)

P(Xi= 1) = P(Ai)P(Xi = 1|Ai) + P(Ai)P(Xi= 1|Ai)=p1 1 + (1 pi) 1

5

LetAi = "know answer to question Qi"

8 Continuous Probability Distributions


9.1 Y = 43

X2

3. Plug in X= 0.6andX= 1.0to get the bounds 0.036 y 6 . Next,

F(x) =x 0.6

0.4

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so then

F(y) =P(Y y)

=P

4

3

X

2

3 y

=P

X 3

6y

=

3

6y 0.60.4

= 3

750y

8 3

2

which implies

f(y) = d

dyF(y)

= d

dy3

750y

8 3

2

=5

6

6

13

y23

9.2 (a) 11

f(x) = 1

11

k(1 x2)dx= 1

kx k3

x311

= 1

k=3

4

Then,

P(X x) = x1

3

4(1 x2)dx

=3

4x

1

4x3x1

=3

4

2

3+ x x

3

3

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(b)

F(c) F(c) = 0.953

4

2

3+ c x

3

3

3

4

2

3 c x

3

3

= 0.95

c= 0.811

9.3 (a)

E(X) =

x

xf(x)

=

12

04xdx +

112

4(1 x)dx

=1

2

2 =E(X2) E(X)2

=

x

x2f(x) 14

= 1

24

9.4

F(x) = x

20

then sinceX= 20Y

10,

F(y) =20y 10

20

so

f(y) = 1

9.5 (a) we just need + 1> 0. > 1.(b)

PX 12 = 12

0 ( + 1)x

=

1

2

+1

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E(X) =

10

xf(x)

= + 1

+ 2

(c)

F(x) =x+1

FT(t) =P

1

X t

= 1 P

1

t > X

= 1

1

t

+1so

f(t) = ( + 1)t(+2)

= + 1

t+2

9.6 (a) In this case= 25 .

(1 e(5))3 = (1 e2)3

(b)

P(exceeds 4

|exceeds 4) =

(1 (1 e(5))(1 (1 e(4)))

=e25

9.7 We know thatE(X) = 1000, therefore 1000 = 1 = = 11000 . So, the median lifetime x implies

F(x) = 0.5

So,

0.5 = 1 ex

x= 1

ln

1

2

= 693.1472

9.8 X N(0.65, 0.01). Thus,E(X) = 0.65 and V ar(X) = 0.01.(a) For each one just calculate the z -score, for example

P(X 80%) = 1 P(X

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(b) X is the average score in a group of 25 students. So X N(, 2/25)

P(X >70%) = 1 P(X

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We need

P(|X 28| < .26) =P(27.74< X b 9.5

2

= 0.9 = b= 6.94

9.12 (a)

P(A < l) =P

Z

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