engineering statistics and probability chapter 3

Chapter 3 Random Variables and Probability Distributions

Exercises

3.1 Classify the following random variables as dis-crete or continuous:

X: the number of automobile accidents per yearin Virginia.

Y: the length of time to play 18 holes of golf.

the amount of milk produced yearly by a par-

ticular

N: the number of eggs laid each month by a hen.

P: the number of building permits issued eachmonth in a certain city.

the weight of grain produced per acre.

3.7 The total number of hours, measured in units of100 hours, that a family runs a vacuum cleaner over a period of one year is a continuous random variable X that has the density function

1 ,

x, 1 < x < 2,

elsewhere.

Find the probability that over a period of one year, a family runs their vacuum cleaner

(a) less than 120 hours;

(b) between 50 and 100 hours,

3.2 An overseas shipment of 5 foreign automobilescontains 2 that have slight paint blemishes. If anagency receives 3 of these automobiles at random, listthe elements of the sample space S using the letters B and for blemished and respectively;then to each sample point assign a value x of the ran-dom variable X representing the number of automo-biles purchased by the agency with paint blemishes.

3.3 Let W be a random variable giving the numberof heads minus the number of tails in three tosses of a coin. List the elements of the sample space S for thethree tosses of the coin and to each sample point assigna value of W.

3.4 A coin is flipped until 3 heads in succession oc-cur. List only those elements of the sample space thatrequire 6 or less tosses. Is this a discrete sample space?Explain.

3.5 Determine the value c so that each of the follow-ing functions can serve as a probability distribution ofthe discrete random variable X:(a) f(x) = c(x2 + 4), for a: =

(b)f(x)

3.6 The shelf life, in days, for bottles of a certainprescribed medicine is a random variable having thedensity function

elsewhere.

Find the probability that a bottle of this medicine willhave a shell life of

(a) at least 200 days;

(b) anywhere from 80 to 120 days.

3.8 Find the probability distribution of the randomvariable W in Exercise 3.3. assuming that the coin isbiased so that a head is twice as likely to occur as a tail.

The proportion of people who respond to a certainmail-order solicitation is a continuous random variableX that has the function

0 < x < l ,

0, elsewhere.

(a) Show that P(0 < X < 1) =

(b) Find the probability that more than 1/4 but fewer

than 1/2 of the people contacted will respond to

this type of solicitation.

Find a formula for the probability distributionthe random variable X representing the outcome whena single die is rolled once.

3.11 A shipment of 7 television sets contains 2 de-fective sets. A hotel makes a random purchase of; 3 of the sets. If x is the number of defective sets pur-chased by the hotel, find the probability distributionof X. Express the results graphically as a probabilityhistogram.

3.12 An firm offers its customers munici-pal bonds that mature after varying numbers of years.Given that the cumulative distribution function of T,the number of years to maturity for a randomly se-lected bond, is,

F(t) =

t< 1,

1 < t < 3,

3 < t <5 ,

5 < t < 7, t>7,

Exercises 89

find

( a ) P ( T =

(b) P(T > 3);

(c) P(1.4 < T < 6).

3.13 The probability distribution of A, the numberof imperfections per 10 meters of a synthetic fabric incontinuous rolls of uniform width, is given by

0 3 4

0.41 0.16 0.05 0.01

Construct the cumulative distribution function of X.

3.14 The waiting time, in hours, between successivespeeders spotted by a radar unit is a continuous ran-dom variable with distribution function

x < 0,

x > 0.

Find the probability of waiting less than 12between successive speeders

(a) using the cumulative distribution function of X;(b) using the probability density function of

3.15 Find the cumulative distribution function of therandom variable X representing the number of defec-tives in Exercise 3.11. Then using F(x), find

(a) P(X = 1);

(b) P(0 < X < 2).

Construct a graph of the cumulative distributionfunction of Exercise 3.15.

3.17 A continuous random variable X that can as-

sume values between x = 1 and x 3 has a density

function given by f(x) = 1/2.

(a) Show that the area under the curve is equal to

(b) Find P(2 < X < 2.5).

(c) Find P(X < 1.6).

3.18 A continuous random variable X that can as-sume values between x = 2 and x = 5 has a densityfunction given by = 2(1 + Find

(a) P(X <(b) P(3 < X < 4).

3.19 For the density function of Exercise 3.17, findF(x). Use it to evaluate P(2 < X < 2.5).

3.20 For the density function of Exercise 3.18, findF(x), and use it to evaluate P(3 < X < 4).

3.21 Consider the density

< X < 1,

elsewhere.

(a) Evaluate k.(b) F(x) and use it to evaluate

P(0.3 < X < 0.6).

3.22 Three cards are drawn in succession from a deckwithout replacement. Find the probability distributionfor the number of spades.

3.23 Find the cumulative distribution function of the

random variable W in Exercise 3.8. Using find

(a) P(W > 0);

(b) P ( - l < W < 3 ) .

Find the probability distribution for the numberof jazz CDs when 4 CDs are selected at random froma collection consisting of 5 jazz CDs, 2 classical CDs,and 3 rock CDs. Express your results by means of a formula.

3.25 From a box containing 4 dimes and 2 nickels,3 coins are selected at random without replacement.Find the probability distribution for the total T of the3 coins. Express the probability distribution graphi-cally as a probability histogram.

3.26 From a box containing 4 black balls and 2 greenballs, 3 balls are drawn in succession, each ball beingreplaced in the box before the next draw is made. Findthe probability for the number of greenballs.

3.27 The time to failure in hours of an importantpiece of electronic equipment used in a manufacturedDVD player the density function

/ (2000 •a:/2000), x > 0,

x < 0 .

(a) Find F(x). (b) Determine the probability that the component (and

thus the DVD player) lasts more than 1000 hoursbefore the component needs to be replaced.

(c) Determine the probability that the component failsbefore 2000 hours.

3.28 A cereal manufacturer is aware that the weightof the product in the box varies slightly from boxto box. In fact, considerable historical data has al-lowed the determination of the density function thatdescribes the probability structure for the (in

90 Chapter 3 Random Variables and Probability

ounces). In fact, letting X be the random variable in ounces, the density function can be described

as

23.75 < x < 26.25,

elsewhere.

(a) Verify that this is a valid density

(b) Determine the probability that the weight issmaller than 24 ounces.

(c) The company desires that the weight exceeding 26ounces is an extremely rare occurrence. What is theprobability that this "rare occurrence" does actu-ally occur?

3.29 An important factor in solid fuel is theparticle size distribution. Significant problems occur ifthe particle sizes are too large. From production datain the past, it has been determined that the particlesize (in micrometers) distribution is characterized by

/ ( e)

> 1,

elsewhere.

(a) Verify that this is a valid density function.

(b) Evaluate F(x). (c) What is the probability that a random particle

from the manufactured fuel exceeds 4 micrometers?

3.30 Measurements of scientific systems alwayssubject to variation, some more than others. Thereare many structures for measurement error and statis-ticians spend a great deal of time modeling these errors.Suppose the measurement error X of a certain physicalquantity is decided by the density function

/(• x) { fc(30.

x2), - 1 < < 1,

elsewhere.

(a) Determine k that f(x) a valid density func-

tion.

(b) Find the probability that a random error in mea-

surement is less than 1/2.

(c) For this particular measurement, is undesirableif the magnitude of the error (i.e., |a:|), exceeds 0.8.What is the probability that this occurs?

3.31 Based on extensive testing, it is determined bythe manufacturer of a washing machine that the timeY (in years) before a major repair is required is char-acterized by the probability density function

elsewhere.

(a) Critics would certainly consider the product a bar-gain if it is unlikely to require a major repair beforethe sixth year. Comment on this by determiningP(Y > 6).

(b) What is the probability that a major repair occursin the first year?

3.32 The proportion of the budgets for a certain typeof industrial company that is allotted to environmentaland pollution control is coming under scrutiny. A datacollection project determines that the distribution ofthese proportions is given by

elsewhere.

(a) Verify that the above is a valid density.

(b) What is the probability that a company chosen atrandom expends less than 10% of its budget on en-vironmental and pollution controls?

(c) What is the probability that a company selected atrandom spends more than 50% on environmentaland pollution control?

3.33 Suppose a special type of small data processingfirm is so specialized that some have difficulty makinga profit in their first year of operation. The pdf thatcharacterizes the proportion Y that make a profit isgiven by

0 < y < 1,

elsewhere.

(a) What is the value of k that renders the above a valid density function?

(b) Find probability that at most 50% of the firmsmake a profit in the first year.

(c) Find the probability at least 80% of the firmsmake a profit in the first year.

3.34 Magnetron are produced from an auto-mated assembly line. A sampling plan is used periodi-cally to assess quality on the lengths of the tubes. Thismeasurement is subject to uncertainty. It is thoughtthat the probability that a random tube meets lengthspecification is 0.99. A sampling plan is used in whichthe lengths of 5 random tubes are measured.

(a) Show that the probability function of the num-ber out of 5 that meet length is givenby the following discrete probability function

for y = 0.1,2,3,4,5.

3.4 Joint Probability Distributions 91

(b) Suppose random selections made off the line section.and 3 are outside specifications. Use above that only 2 carseither to or refute the conjecture that theprobability is 0.99 that a single tube meets , , . .,cations assignment, the equipment

working, the density function of the observed outcome,X is

3.35 Suppose it is known from large amounts of his-torical data that X, the of cars that arrive ata specific intersection during a 20 second time period, _ 2(1 — x), 0 < x < 1,is characterized by the following discrete probability )function

/(x) = x = 0,1.2, . . . . (a) Calculate < 1/3). (b) What is the probability that X will exceed 0.5?

(a) Find the probability that in a specific (c) Given that X > 0.5, what is the probability thattime period, more than 8 cars arrive at the inter- X will be less than 0.75?

3.4 Joint Probability Distributions

Our study of random variables and their probability distributions in the preced-

ing sections is restricted to one-dimensional sample spaces, in that we recorded

outcomes of an experiment as values assumed by a single random variable. There

will be situations, however, where we may find it desirable to record the simulta-

neous outcomes of several random variables. For example, we might measure the

amount of precipitate P and volume V of gas released from a controlled chemical

experiment, giving rise to a two-dimensional sample space consisting of the out-

comes (p, or we might be interested in the hardness H and tensile strength T of cold-drawn copper resulting in the outcomes (h, t). In a study to determine the

likelihood of success in college, based on high school data, we might use a three-

dimensional sample space and record for each individual his or her aptitude test

score, high school rank in class, and average at the end of the freshman

year in college.

If X and Y are two discrete random variables, the probability distribution for

their simultaneous occurrence can be represented by a function with values f(x,y) for any pair of values (x, y) within the range of the random variables X and Y. It

is customary to refer to this function as the probability distribution of

X and Y. Hence, in the discrete case,

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

that is, the values f(x, y) give the probability that outcomes x and y occur at the

same time. For example, if a television set is to be serviced and X represents the

age to the nearest year of the set and Y represents the number of defective tubes

in the set, then is the probability that the television set is 5 years old and

needs 3 new tubes.

Exercises 101

type. Indeed many of these are reflected in exercises in bo th Chapters 2 and 3.

When repeated observations are binary in na ture (e.g., "defective or

not," "survive or not," "allergic or not") with observations 0 or 1, the distribution

covering this situation is called the b i n o m i a l d i s t r i b u t i o n and the probability

function is known and will be demonstrated in its generality in Chapter 5. Ex-

3.34 in Section 3.3 and Review Exercise 3.82 are examples and there are

others tha t the reader should recognize. The scenario of a continuous distribution

in "time to failure" as in Review Exercise 3.71 or Exercise 3.27 on page 89 often

suggests distribution type called the e x p o n e n t i a l d i s t r ibut ion . These types of

illustrations are merely two of many so-called s t andard distributions tha t are used

extensively in real world problems because the scientific scenario that gives rise to

each of them is recognizable and occurs often in practice. Chapters 5 and 6 cover

many of these with some underlying theory concerning their use.

A second of transition to material in future chapters deals with the

notion of p o p u l a t i o n p a r a m e t e r s or d i s t r i b u t i o n a l p a r a m e t e r s . Recall in

Chapter 1 we discussed the need to use data to provide information about these

parameters . We went to lengths in discussing notion of a m e a n and variance

and provided a vision for the concepts in the context of a population. Indeed the

population mean and variance are easily found from the probability function for

the discrete case or probability density function for the continuous case. These pa-

rameters and their importance in the solution of many types of real world problems

will provide much of the material in Chapters 8 through 17.

Exercises

3.37 Determine the values of c so that the follow-ing functions represent joint probability distributionsof the random variables A" and Y:(a) f(x, y) — cxy, for x = 1, 2, 3; y = 1,

(b) f(x, = c\x - y\, for = y

3.38 If the joint probability distribution of X and Y is given by

f(x, y) ,.. o, 1, y = (1, 1, 2,

find

(a) P(X <2,Y = 1);

(b) P(X > 2,Y < 1); (c) P(X > Y); (d) P(X + Y = 4).

3.39 From a sack of fruit containing 3 oranges, 2 ap-ples, and 3 bananas, a random of 4 pieces offruit is selected. If X is the number of oranges Y is the number of apples in the sample:, find

(a) the joint probability distribution of A' and

(b) P[(X, Y) € .4], where A is the region that is given

by {(x,y) +

3.40 A privately owned liquor store operates both a drive-in facility and a walk-in facility. On a randomlyselected day, let and respectively, be the propor-tions of the time that the drive-in and walk-in facilitiesare in use, and suppose that the density functionof these random variables is

+ 2y), 1, 0 < y < 1,

elsewhere.

(a) Find the marginal density of X.(b) Find the marginal density of Y.(c) Find the probability that the drive-in facility is

busy less than one-half of the time.

3.41 A candy distributes boxes of choco-lates with a mixture of creams, toffees, and cordials.Suppose the weight of each box is 1 kilogram, butthe individual weights of the creams, toffees, and cor-dials vary from box to box. For a selectedbox, let X and Y represent the weights of the creamsand the toffees, respectively, and suppose the joint

102 Chapter 3 Random Variables and Probability Distributions

density of these variables is

< 1,

elsewhere.

(a) Find the probability that in a given box the cordials

account for more than 1/2 of the weight.

(b) Find the marginal density for the weight of the

creams.

(c) Find the probability that the weight of the toffees

in a box is less than 1/8 of a kilogram if it is known

that creams constitute 3/4 of the weight.

3.42 Let X and Y denote the lengths of life, in years,of two components in an electronic system. If the jointdensity function of these variables is

findP(0<X< 1 \Y = 2).

x > 0, y > elsewhere.

0,

3.43 Let X denote the reaction time, in seconds, toa certain stimulus and Y denote the temperatureat which a certain reaction starts to take place. Sup-pose that two random variables X and Y have the jointdensity

Find

(a) P(0

(b) P(X X < and

<Y).

1,

elsewhere.

4 — — 2 );

3.44 Each rear tire on an experimental airplane issupposed to be filled to a pressure of 40 pound persquare inch (psi). Let X denote the actual air pressurefor the right tire and Y denote the actual air pressurefor the left tire. Suppose that X and Y are randomvariables with the joint density

30 < < 50;

30 < y < 50,

elsewhere.

(a) Find k.(b) P(30 < X < 40 and 40 < Y < 50).

(c) Find the probability that both tires are underfilled.

3.45 Let X denote the diameter of an armored elec-tric cable and denote the diameter of the ceramicmold that makes the cable. Both X and Y are scaled

so that they range between 0 and 1. Suppose that X and Y have the joint density

1 0, elsewhere.

Find P(X + Y> 1/2).

3.46 Referring to Exercise 3.38, find

(a) the marginal distribution of X;(b) the marginal distribution of

3.47 The amount of kerosene, in thousands of liters,in a tank at the beginning of any day is a randomamount Y from which a random amount X is sold dur-ing that day. Suppose that the tank is not resuppliedduring the day so that x < y, and assume that thejoint density function of these variables is

elsewhere.

< 1 ,

(a) Determine if X and Y are independent.

(b) Find P ( l / 4 < X < 1/2 | Y = 3/4).

3.48 Referring to Exercise 3.39, find

(a) f(y\2) for all values of y;(b) P(Y 0 | X = 2).

3.49 Let X denote the number of times a certain nu-merical control machine will malfunction: 1, 2, or 3 times on any given day. Let Y denote the number oftimes a technician is called on an emergency call. Theirjoint probability distribution is given as

1

y3

1

0.050.05

0

x2

0.050.10.2

3

0.10.350.1

(a) Evaluate the marginal distribution of

(b) Evaluate the marginal distribution of

(c) Find P(Y = 3 | X =

Suppose that X and Y have the following jointprobability distribution:

xf(x,y) 2 4

0.150.300.15

(a) Find the marginal distribution of A.

(b) Find the marginal distribution of Y.

Review Exercises 103

3.51 Consider an experiment that consists of 2 rollsof a balanced die. If X is the number of 4s and Y isthe number of obtained in the 2 rolls of the die, find

(a) the joint probability distribution of and

(b) P[(X, Y) € where A is the region {(x,y) \ 2x + <

3.52 Let X denote the: number of heads and Y the of heads minus the number of tails when 3

coins are tossed. Find the joint probability distribu-tion of X and

3.53 Three cards are drawn withoutfrom the 12 face cards (jacks, queens, and kings) ofan ordinary deck of 52 playing cards. Let X be thenumber of kings selected and Y the number of jacks.Find

(a) the joint probability distribution of X and Y;(b) P[(X,Y) e where A is the region given by

{(x,y) \x + y>2}.

3.54 A coin is tossed twice. Z denote the number

of heads on the first toss and W the total number of

heads on the 2 tosses. If the coin is unbalanced and a

head has a 40% chance of occurring, find

(a) the probability distribution of W and Z;(b) the marginal distribution of

(c) the marginal distribution of Z; the probability that at least 1 head occurs.

3.55 Given the joint density function

f(x, y) elsewhere,

find P ( l < Y < 3 | X =

3.56 Determine whether the two random variables ofExercise 3.49 are dependent or independent.


3.58 The joint function of the random vari-ables X and Y is

y) 0 < x < 1, 0 < y < 1

elsewhere.x,

(a) Show that X and are not independent.

(b) Find P(X > 0.3 | Y = 0.5).

3.59 Let X, Y, and Z have the joint probability den-sity function

z) 0 , '

1; 0< z <2, elsewhere.

(a) Find k.(b)


3.61 Determine whether the two random variables ofExercise 3.44 are or

3.62 The joint probability density function of the ran- variables X, and Z is

Find

0 1; 0< z < 3,

elsewhere.

(a) the joint marginal density function of Y and Z:(b) the marginal density of Y;(c) KZ<2);

= i , 2 = 2).

Review Exercises

3.63 A tobacco company produces blends of tobaccowith each blend containing various proportions ofTurkish, domestic, and other tobaccos. The propor-tions of Turkish and domestic in a blend are randomvariables with joint density function (X = Turkish andY = domestic)

0 < < 1: x + < 1,

elsewhere.

(a) Find the probability that in a given box the Turkishtobacco accounts for over half the blend.

(b) Find the marginal density function for the propor-tion of the domestic tobacco.

(c) Find the probability that the proportion of Turk-ish tobacco is less than 1/8 if it is known that theblend contains 3/4 domestic tobacco.

3.64 An insurance company offers its policyholders a

104 Chapter 3 Random. Variables and Probability Distributions

number of different premium payment options. For a randomly selected policyholder, let X be the number ofmonths between successive payments. The cumulativedistribution function of X is

F(x) = {

0,

0.4, if 1 < x < 3,

0.6, if 3 < x < 5,

0.8, if 5 < x < 7, 1.0, if x > 7.

(a) What is the probability mass function of X?(b) Compute P(4 < X < 7).

3.65 Two electronic components of a missile systemwork in harmony for the success of the total system.Let X and Y denote the life in hours of the two com-ponents. The joint density of X and Y is

-{ 0,

(a) Give the marginal density functions for both ran-dom variables.

(b) What is the probability that both components willexceed 2 hours?

3.66 A service facility operates with two service lines.On a randomly selected day, let X be the proportion oftime that the first line is in use whereas Y is the pro-portion of time that the second line is in use. Supposethat the joint probability density function for (A, V) is

fix +

elsewhere.

(a) Compute the probability that neither line is busymore than half the time.

(b) Find the probability that the first line is busy morethan 75% of time.

3.67 Let the number of phone calls received by a switchboard during a 5-minute interval be a randomvariable X with probability function

for x =

(a) Determine the probability that X equals 0, 1, 2, 3,4, 5, and 6.

(b) Graph the probability function for these val-ues x.

(c) Determine the cumulative distribution function forthese values of X.

3.68 Consider the random variables X and Y withjoint density function

fix, y) x + y, 0 < x,y < 1,

elsewhere.

(a) Find the marginal distributions of X and

(b) >0.5) .

3.69 An industrial process manufactures items thatcan be classified as either defective or not defective.The probability that an item is defective is 0.1. Anexperiment is conducted in which 5 items are drawnrandomly from the process. Let the random variable X be the number of defectives in this sample of 5. Whatis the probability mass function of X?

3.70 Consider the following joint probabilityfunction of the random variables X and Y:

f{x,y) 1 < x < 3, < y < 2,

elsewhere.

(a) Find the marginal density functions of X and Y.(b) Are X and Y independent?

(c) Find P(X > 2).

3.71 The life span in hours of an electrical compo-nent is a random variable with cumulative distributionfunction

F(x) {;-x > 0,

eleswhere.

(a) Determine its probability density function.

(b) Determine the probability that the life span of sucha component will 70 hours.

Pairs of pants are being produced by a particu-lar outlet facility. The pants are by a groupof 10 workers. The workers inspect pairs of pants takenrandomly from the production Each inspector isassigned a number from 1 through 10. A buyer selectsa pair of pants for purchase. Let the random variableX be the inspector number.

(a) Give a reasonable probability mass function for X.(b) Plot the cumulative distribution function for X.

3.73 The shelf life of a product is a random variablethat is related to consumer acceptance. It turns outthat the shelf life Y in days of a certain type of bakeryproduct has a density function

fiv) 0 < y < oc,

elsewhere.

Review Exercises 105

What fraction of the of this stocked to-day would you to be sellable 3 days from now?

Passenger congestion is a service problem in air-ports. Trains are installed within the airport to reducethe With the use of the train, the time X that it takes in to travel from the main termi-nal to a particular concourse has density function

fix) 0 < <

elsewhere.

(a) Show the pdf above is a valid density function.

(b) Find the probability that the time it takes a pas-senger to travel from the main terminal to the con-course will not exceed 7 minutes.

3.75 in the batch of final product of a chemical process often reflect a serious problem. Fromconsiderable plant data gathered, it is known that theproportion Y of impurities in a batch has a densityfunction given by

0 < y < 1,

elsewhere.

(a) Verify that the above is a valid density function.

(b) A batch is not sellable and then not-acceptable if the percentage of impurities exceeds60%. With the current quality of the process, whatis the percentage of batches that are not accept-able?

3.76 The time Z in minutes between calls to an elec-trical supply system has the probability density func-tion

0 < z < oo,

elsewhere.

(a) What is the probability that there are no callswithin a 20-minute time interval?

(b) What is the probability that the first call comeswithin 10 minutes of opening?

A chemical system that results from a chemicalreaction has two important components among othersin a blend. The joint distribution describing the pro-portion and of these two components is givenby

, , / 2 , 0 < < < 1,

(0, elsewhere.

(a) Give the marginal distribution of

(b) Give the marginal distribution of

(c) is the probability that component propor-tions produce the results < 0.2 and > 0.5?

(d) Give the distribution

3.78 Consider the situation of Review Exercise 3.77.But suppose the joint distribution of the two propor-tions is given by

0 < x X2 < < 1,

here.

(a) Give the marginal distribution of the pro-portion and verify that it is valid density func-tion.

(b) What is the probability that proportion is lessthan 0.5 given that is 0.7?

3.79 Consider the random variables X and Y thatrepresent the number of vehicles that arrive at 2 sep-arate street corners during a certain period.These street corners arc fairly close together so it is im-portant that traffic engineers deal with them jointly ifnecessary. The joint distribution of X and Y is knownto be

- - •

for x = and y =

(a) Are the two random variables X and Y indepen-dent? Explain why or why not.

(b) What is the probability that during the time pe-riod in question less than 4 vehicles arrive at thetwo street corners?

The behavior of series of components play a hugerole in scientific and engineering reliability problems.The reliability of the entire system is certainly no bet-ter than the weakest component in the series. In a series system, the components operate independentlyof each other. In a particular system containing threecomponents the probability of meeting specification forcomponents 1, 2, and 3, respectively, are 0.99, and0.92. What is the probability that the entire systemworks?

3.81 Another type of system that is employed en-gineering work is a group of parallel components or a parallel system. In this more conservative approach,the probability that the system operates is larger thanthe probability that any component operates. The sys-tem fails only when systems fail. Consider a situa-tion in which there are 4 independent components in a parallel system with probability of operation given by

Component 1: 0.95; Component 2: 0.94:

Component 3: 0.90; Component 4:

What is the probability that the system does not fail?

106 Chapter S Random Variables and Probability Distributions

3.82 Consider a system of components in which does have a redundancy built in such that it does not five components, each of which fail if 3 out of the 5 components are operational. What

an operational probability of 0.92. system is the probability that the total system is operational?

3.5 Potential Misconceptions and Hazards;

Relationship to Material in Other Chapters

In future chapters it will become apparent that probability distributions represent

the: structure through which probabilities that are computed aid in the evalua-

tion and understanding of a process. For example, in Review Exercise 3.67, the

probability distribution that quantifies the probability of a heavy load during cer-

tain time periods can be very useful in planning for any changes in the system.

Review Exercise 3.71 describes a scenario in which the life span of an electronic:

component is studied. Knowledge of the probability structure for the component

will contribute significantly toward an understanding of the reliability of a large

system of which the component is a part. In addition, an understanding of the

general nature of probability distributions will enhance the understanding of the

concept of a P-value which was introduced briefly in Chapter 1 and will play a

major role: beginning in Chapter 10 and extending throughout the balance of the

text.

Chapters 4, 5, and 6 depend heavily on the material in this chapter. In Chapter

4 we discuss the meaning of pa rame te r s in probability distributions.

These important parameters quantify notions of central tendency variabil-

ity in a system. In fact, knowledge of these quantities themselves, quite apart

from the complete distribution, can provide into the nature of the system.

Chapters 5 and 6 will deal with engineering, biological, or general scientific scenar-

ios identify special types of For example, the structure of the

probability function in Review Exercise 3.67 will easily be identified under certain

assumptions discussed in Chapter 5. The same holds for the scenario of Review

Exercise 3.71. This is a special type of t ime to failure problem for which the

probability density function will be Chapter 6.

As far as potential hazards with the use of material in this chapter, the

ing" to the reader is not to read more into the material than is evident. The

general nature of the probability distribution for a specific scientific: phenomenon

is not obvious from what is learned in this chapter. The purpose of this chapter is

to learn how to manipulate a probability distribution, not to learn how to identify

a specific type. Chapters 5 and 6 go a long way toward identification to

general nature of the scientific system.

Chapter 3

Random Variables and Probability

Distributions

3.1 Discrete; continuous; continuous; discrete; discrete; continuous.

3.2 A table of sample space and assigned values of the random variable is shown next.

Sample Space x

NNNNNBNBNBNNNBBBNBBBNBBB

01112223

3.3 A table of sample space and assigned values of the random variable is shown next.

Sample Space w

HHHHHTHTHTHHHTTTHTTTHTTT

3111−1−1−1−3

3.4 S = {HHH, THHH, HTHHH, TTHHH, TTTHHH, HTTHHH, THTHHH,HHTHHH, . . .}; The sample space is discrete containing as many elements as thereare positive integers.

29


3.5 (a) c = 1/30 since 1 =3∑

x=0

c(x2 + 4) = 30c.

(b) c = 1/10 since

1 =

2∑

x=0

c

(

2

x

)(

3

3− x

)

= c

[(

2

0

)(

3

3

)

+

(

2

1

)(

3

2

)

+

(

2

2

)(

3

1

)]

= 10c.

3.6 (a) P (X > 200) =∫∞

20020000

(x+100)3dx = − 10000

(x+100)2

∣

∣

∣

∞

200= 1

9.

(b) P (80 < X < 200) =∫ 120

8020000

(x+100)3dx = − 10000

(x+100)2

∣

∣

∣

120

80= 1000

9801= 0.1020.

3.7 (a) P (X < 1.2) =∫ 1

0x dx +

∫ 1.2

1(2− x) dx = x2

2

∣

∣

∣

1

0+

(

2x− x2

2

)∣

∣

∣

1.2

1= 0.68.

(b) P (0.5 < X < 1) =∫ 1

0.5x dx = x2

2

∣

∣

∣

1

0.5= 0.375.

3.8 Referring to the sample space in Exercise 3.3 and making use of the fact that P (H) =2/3 and P (T ) = 1/3, we haveP (W = −3) = P (TTT ) = (1/3)3 = 1/27;P (W = −1) = P (HTT ) + P (THT ) + P (TTH) = 3(2/3)(1/3)2 = 2/9;P (W = 1) = P (HHT ) + P (HTH) + P (THH) = 3(2/3)2(1/3) = 2/9;P (W = 3) = P (HHH) = (2/3)3 = 8/27;The probability distribution for W is then

w −3 −1 1 3P (W = w) 1/27 2/9 2/9 8/27

3.9 (a) P (0 < X < 1) =∫ 1

02(x+2)

5dx = (x+2)2

5

∣

∣

∣

1

0= 1.

(b) P (1/4 < X < 1/2) =∫ 1/2

1/42(x+2)

5dx = (x+2)2

5

∣

∣

∣

1/2

1/4= 19/80.

3.10 The die can land in 6 different ways each with probability 1/6. Therefore, f(x) = 16,

for x = 1, 2, . . . , 6.

3.11 We can select x defective sets from 2, and 3− x good sets from 5 in(

2x

)(

53−x

)

ways. A

random selection of 3 from 7 sets can be made in(

73

)

ways. Therefore,

f(x) =

(

2x

)(

53−x

)

(

73

) , x = 0, 1, 2.

In tabular form

x 0 1 2f(x) 2/7 4/7 1/7

Solutions for Exercises in Chapter 3 31

The following is a probability histogram:

1 2 3

xx

f(x)

f(x)

1/7

2/7

3/7

4/7

3.12 (a) P (T = 5) = F (5)− F (4) = 3/4− 1/2 = 1/4.

(b) P (T > 3) = 1− F (3) = 1− 1/2 = 1/2.

(c) P (1.4 < T < 6) = F (6)− F (1.4) = 3/4− 1/4 = 1/2.

3.13 The c.d.f. of X is

F (x) =

0, for x < 0,

0.41, for 0 ≤ x < 1,

0.78, for 1 ≤ x < 2,

0.94, for 2 ≤ x < 3,

0.99, for 3 ≤ x < 4,

1, for x ≥ 4.

3.14 (a) P (X < 0.2) = F (0.2) = 1− e−1.6 = 0.7981;

(b) f(x) = F ′(x) = 8e−8x. Therefore, P (X < 0.2) = 8∫ 0.2

0e−8x dx = −e−8x|0.2

0 =0.7981.


F (x) =

0, for x < 0,

2/7, for 0 ≤ x < 1,

6/7, for 1 ≤ x < 2,

1, for x ≥ 2.

(a) P (X = 1) = P (X ≤ 1)− P (X ≤ 0) = 6/7− 2/7 = 4/7;

(b) P (0 < X ≤ 2) = P (X ≤ 2)− P (X ≤ 0) = 1− 2/7 = 5/7.


3.16 A graph of the c.d.f. is shown next.

F(x)

F(x)

xx

1/7

2/7

3/7

4/7

5/7

6/7

1

0 1 2

3.17 (a) Area =∫ 3

1(1/2) dx = x

2

∣

∣

3

1= 1.

(b) P (2 < X < 2.5)∫ 2.5

2(1/2) dx = x

2

∣

∣

2.5

2= 1

4.

(c) P (X ≤ 1.6) =∫ 1.6

1(1/2) dx = x

2

∣

∣

1.6

1= 0.3.

3.18 (a) P (X < 4) =∫ 4

22(1+x)

27dx = (1+x)2

27

∣

∣

∣

4

2= 16/27.

(b) P (3 ≤ X < 4) =∫ 4

32(1+x)

27dx = (1+x)2

27

∣

∣

∣

4

3= 1/3.

3.19 F (x) =∫ x

1(1/2) dt = x−1

2,

P (2 < X < 2.5) = F (2.5)− F (2) = 1.52− 1

2= 1

4.

3.20 F (x) = 227

∫ x

2(1 + t) dt = 2

27

(

t + t2

2

)∣

∣

∣

x

2= (x+4)(x−2)

27,

P (3 ≤ X < 4) = F (4)− F (3) = (8)(2)27

− (7)(1)27

= 13.

3.21 (a) 1 = k∫ 1

0

√x dx = 2k

3x3/2

∣

∣

1

0= 2k

3. Therefore, k = 3

2.

(b) F (x) = 32

∫ x

0

√t dt = t3/2

∣

∣

x

0= x3/2.

P (0.3 < X < 0.6) = F (0.6)− F (0.3) = (0.6)3/2 − (0.3)3/2 = 0.3004.

3.22 Denote by X the number of spades int he three draws. Let S and N stand for a spadeand not a spade, respectively. ThenP (X = 0) = P (NNN) = (39/52)(38/51)(37/50) = 703/1700,P (X = 1) = P (SNN) + P (NSN) + P (NNS) = 3(13/52)(39/51)(38/50) = 741/1700,P (X = 3) = P (SSS) = (13/52)(12/51)(11/50) = 11/850, andP (X = 2) = 1− 703/1700− 741/1700− 11/850 = 117/850.The probability mass function for X is then

x 0 1 2 3f(x) 703/1700 741/1700 117/850 11/850



F (x) =

0, for w < −3,

1/27, for − 3 ≤ w < −1,

7/27, for − 1 ≤ w < 1,

19/27, for 1 ≤ w < 3,

1, for w ≥ 3,

(a) P (W > 0 = 1− P (W ≤ 0) = 1− 7/27 = 20/27.

(b) P (−1 ≤W < 3) = F (2)− F (−3) = 19/27− 1/27 = 2/3.

3.24 There are(

104

)

ways of selecting any 4 CDs from 10. We can select x jazz CDs from 5

and 4− x from the remaining CDs in(

5x

)(

54−x

)

ways. Hence

f(x) =

(

5x

)(

54−x

)

(

104

) , x = 0, 1, 2, 3, 4.

3.25 Let T be the total value of the three coins. Let D and N stand for a dime and nickel,respectively. Since we are selecting without replacement, the sample space containingelements for which t = 20, 25, and 30 cents corresponding to the selecting of 2 nickels

and 1 dime, 1 nickel and 2 dimes, and 3 dimes. Therefore, P (T = 20) =(2

2)(4

1)

(6

3)

= 15,

P (T = 25) =(2

1)(4

2)

(6

3)

= 35,

P (T = 30) =(4

3)

(6

3)

= 15,

and the probability distribution in tabular form is

t 20 25 30P (T = t) 1/5 3/5 1/5

As a probability histogram

20 25 30

xx

f(x)

f(x)

1/5

2/5

3/5


3.26 Denote by X the number of green balls in the three draws. Let G and B stand for thecolors of green and black, respectively.

Simple Event x P (X = x)BBBGBBBGBBBGBGGGBGGGBGGG

01112223

(2/3)3 = 8/27(1/3)(2/3)2 = 4/27(1/3)(2/3)2 = 4/27(1/3)(2/3)2 = 4/27(1/3)2(2/3) = 2/27(1/3)2(2/3) = 2/27(1/3)2(2/3) = 2/27(1/3)3 = 1/27

The probability mass function for X is then

x 0 1 2 3P (X = x) 8/27 4/9 2/9 1/27

3.27 (a) For x ≥ 0, F (x) =∫ x

01

2000exp(−t/2000) dt = − exp(−t/2000)|x0

= 1− exp(−x/2000). So

F (x) =

{

0, x < 0,

1− exp(−x/2000), x ≥ 0.

(b) P (X > 1000) = 1− F (1000) = 1− [1− exp(−1000/2000)] = 0.6065.

(c) P (X < 2000) = F (2000) = 1− exp(−2000/2000) = 0.6321.

3.28 (a) f(x) ≥ 0 and∫ 26.25

23.7525

dx = 25t∣

∣

26.25

23.75= 2.5

2.5= 1.

(b) P (X < 24) =∫ 24

23.7525

dx = 25(24− 23.75) = 0.1.

(c) P (X > 26) =∫ 26.25

2625

dx = 25(26.25− 26) = 0.1. It is not extremely rare.

3.29 (a) f(x) ≥ 0 and∫∞

13x−4 dx = −3 x−3

3

∣

∣

∣

∞

1= 1. So, this is a density function.

(b) For x ≥ 1, F (x) =∫ x

13t−4 dt = 1− x−3. So,

F (x) =

{

0, x < 1,

1− x−3, x ≥ 1.

(c) P (X > 4) = 1− F (4) = 4−3 = 0.0156.

3.30 (a) 1 = k∫ 1

−1(3− x2) dx = k

(

3x− x3

3

)∣

∣

∣

1

−1= 16

3k. So, k = 3

16.


(b) For −1 ≤ x < 1, F (x) = 316

∫ x

−1(3− t2) dt =

(

3t− 13t3)∣

∣

x

−1= 1

2+ 9

16x− x3

16.

So, P(

X < 12

)

= 12−

(

916

) (

12

)

− 116

(

12

)3= 99

128.

(c) P (|X| < 0.8) = P (X < −0.8) + P (X > 0.8) = F (−0.8) + 1− F (0.8)= 1 +

(

12− 9

160.8 + 1

160.83

)

−(

12

+ 916

0.8− 116

0.83)

= 0.164.

3.31 (a) For y ≥ 0, F (y) = 14

∫ y

0e−t/4 dy = 1− ey/4. So, P (Y > 6) = e−6/4 = 0.2231. This

probability certainly cannot be considered as “unlikely.”

(b) P (Y ≤ 1) = 1− e−1/4 = 0.2212, which is not so small either.

3.32 (a) f(y) ≥ 0 and∫ 1

05(1− y)4 dy = − (1− y)5|10 = 1. So, this is a density function.

(b) P (Y < 0.1) = − (1− y)5|0.10 = 1− (1− 0.1)5 = 0.4095.

(c) P (Y > 0.5) = (1− 0.5)5 = 0.03125.

3.33 (a) Using integral by parts and setting 1 = k∫ 1

0y4(1− y)3 dy, we obtain k = 280.

(b) For 0 ≤ y < 1, F (y) = 56y5(1 − Y )3 + 28y6(1 − y)2 + 8y7(1 − y) + y8. So,P (Y ≤ 0.5) = 0.3633.

(c) Using the cdf in (b), P (Y > 0.8) = 0.0563.

3.34 (a) The event Y = y means that among 5 selected, exactly y tubes meet the spec-ification (M) and 5 − y (M ′) does not. The probability for one combination ofsuch a situation is (0.99)y(1 − 0.99)5−y if we assume independence among thetubes. Since there are 5!

y!(5−y)!permutations of getting y Ms and 5 − y M ′s, the

probability of this event (Y = y) would be what it is specified in the problem.

(b) Three out of 5 is outside of specification means that Y = 2. P (Y = 2) = 9.8×10−6

which is extremely small. So, the conjecture is false.

3.35 (a) P (X > 8) = 1− P (X ≤ 8) =8∑

x=0

e−6 6x

x!= e−6

(

60

0!+ 61

1!+ · · ·+ 68

8!

)

= 0.1528.

(b) P (X = 2) = e−6 62

2!= 0.0446.

3.36 For 0 < x < 1, F (x) = 2∫ x

0(1− t) dt = − (1− t)2|x0 = 1− (1− x)2.

(a) P (X ≤ 1/3) = 1− (1− 1/3)2 = 5/9.

(b) P (X > 0.5) = (1− 1/2)2 = 1/4.

(c) P (X < 0.75 | X ≥ 0.5) = P (0.5≤X<0.75)P (X≥0.5)

= (1−0.5)2−(1−0.75)2

(1−0.5)2= 3

4.

3.37 (a)3∑

x=0

3∑

y=0

f(x, y) = c3∑

x=0

3∑

y=0

xy = 36c = 1. Hence c = 1/36.

(b)∑

x

∑

y

f(x, y) = c∑

x

∑

y

|x− y| = 15c = 1. Hence c = 1/15.

3.38 The joint probability distribution of (X, Y ) is


xf(x, y) 0 1 2 3

0 0 1/30 2/30 3/30y 1 1/30 2/30 3/30 4/30

2 2/30 3/30 4/30 5/30

(a) P (X ≤ 2, Y = 1) = f(0, 1) + f(1, 1) + f(2, 1) = 1/30 + 2/30 + 3/30 = 1/5.

(b) P (X > 2, Y ≤ 1) = f(3, 0) + f(3, 1) = 3/30 + 4/30 = 7/30.

(c) P (X > Y ) = f(1, 0) + f(2, 0) + f(3, 0) + f(2, 1) + f(3, 1) + f(3, 2)= 1/30 + 2/30 + 3/30 + 3/30 + 4/30 + 5/30 = 3/5.

(d) P (X + Y = 4) = f(2, 2) + f(3, 1) = 4/30 + 4/30 = 4/15.

3.39 (a) We can select x oranges from 3, y apples from 2, and 4− x − y bananas from 3in

(

3x

)(

2y

)(

34−x−y

)

ways. A random selection of 4 pieces of fruit can be made in(

84

)

ways. Therefore,

f(x, y) =

(

3x

)(

2y

)(

34−x−y

)

(

84

) , x = 0, 1, 2, 3; y = 0, 1, 2; 1 ≤ x + y ≤ 4.

(b) P [(X, Y ) ∈ A] = P (X + Y ≤ 2) = f(1, 0) + f(2, 0) + f(0, 1) + f(1, 1) + f(0, 2)= 3/70 + 9/70 + 2/70 + 18/70 + 3/70 = 1/2.

3.40 (a) g(x) = 23

∫ 1

0(x + 2y) dy = 2

3(x + 1), for 0 ≤ x ≤ 1.

(b) h(y) = 23

∫ 1

0(x + 2y) dy = 1

3(1 + 4y), for 0 ≤ y ≤ 1.

(c) P (X < 1/2) = 23

∫ 1/2

0(x + 1) dx = 5

12.

3.41 (a) P (X + Y ≤ 1/2) =∫ 1/2

0

∫ 1/2−y

024xy dx dy = 12

∫ 1/2

0

(

12− y

)2y dy = 1

16.

(b) g(x) =∫ 1−x

024xy dy = 12x(1− x)2, for 0 ≤ x < 1.

(c) f(y|x) = 24xy12x(1−x)2

= 2y(1−x)2

, for 0 ≤ y ≤ 1− x.

Therefore, P (Y < 1/8 | X = 3/4) = 32∫ 1/8

0y dy = 1/4.

3.42 Since h(y) = e−y∫∞

0e−x dx = e−y, for y > 0, then f(x|y) = f(x, y)/h(y) = e−x, for

x > 0. So, P (0 < X < 1 | Y = 2) =∫ 1

0e−x dx = 0.6321.

3.43 (a) P (0 ≤ X ≤ 1/2, 1/4 ≤ Y ≤ 1/2) =∫ 1/2

0

∫ 1/2

1/44xy dy dx = 3/8

∫ 1/2

0x dx = 3/64.

(b) P (X < Y ) =∫ 1

0

∫ y

04xy dx dy = 2

∫ 1

0y3 dy = 1/2.

3.44 (a) 1 = k∫ 50

30

∫ 50

30(x2 + y2) dx dy = k(50 − 30)

(

∫ 50

30x2 dx +

∫ 50

30y2 dy

)

= 392k3· 104.

So, k = 3392· 10−4.


(b) P (30 ≤ X ≤ 40, 40 ≤ Y ≤ 50) = 3392· 10−4

∫ 40

30

∫ 50

40(x2 + y2) dy dx

= 3392· 10−3(

∫ 40

30x2 dx +

∫ 50

40y2 dy) = 3

392· 10−3

(

403−303

3+ 503−403

3

)

= 49196

.

(c) P (30 ≤ X ≤ 40, 30 ≤ Y ≤ 40) = 3392· 10−4

∫ 40

30

∫ 40

30(x2 + y2) dx dy

= 2 3392· 10−4(40− 30)

∫ 40

30x2 dx = 3

196· 10−3 403−303

3= 37

196.

3.45 P (X + Y > 1/2) = 1− P (X + Y < 1/2) = 1−∫ 1/4

0

∫ 1/2−x

x1y

dy dx

= 1−∫ 1/4

0

[

ln(

12− x

)

− ln x]

dx = 1 +[(

12− x

)

ln(

12− x

)

− x ln x]∣

∣

1/4

0

= 1 + 14ln

(

14

)

= 0.6534.

3.46 (a) From the column totals of Exercise 3.38, we have

x 0 1 2 3g(x) 1/10 1/5 3/10 2/5

(b) From the row totals of Exercise 3.38, we have

y 0 1 2h(y) 1/5 1/3 7/15

3.47 (a) g(x) = 2∫ 1

xdy = 2(1− x) for 0 < x < 1;

h(y) = 2∫ y

0dx = 2y, for 0 < y < 1.

Since f(x, y) 6= g(x)h(y), X and Y are not independent.

(b) f(x|y) = f(x, y)/h(y) = 1/y, for 0 < x < y.

Therefore, P (1/4 < X < 1/2 | Y = 3/4) = 43

∫ 1/2

1/4dx = 1

3.

3.48 (a) g(2) =2∑

y=0

f(2, y) = f(2, 0) + f(2, 1) + f(2, 2) = 9/70 + 18/70 + 3/70 = 3/7. So,

f(y|2) = f(2, y)/g(2) = (7/3)f(2, y).f(0|2) = (7/3)f(2, 0) = (7/3)(9/70) = 3/10, f(1|2) = 3/5 and f(2|2) = 1/10. Intabular form,

y 0 1 2f(y|2) 3/10 3/5 1/10

(b) P (Y = 0 | X = 2) = f(0|2) = 3/10.

3.49 (a)x 1 2 3

g(x) 0.10 0.35 0.55

(b)y 1 2 3

h(y) 0.20 0.50 0.30

(c) P (Y = 3 | X = 2) = 0.20.05+0.10+0.20

= 0.5714.


3.50

xf(x, y) 2 4 h(y)

1 0.10 0.15 0.25y 3 0.20 0.30 0.50

5 0.10 0.15 0.25g(x) 0.40 0.60

(a)x 2 4

g(x) 0.40 0.60

(b)y 1 3 5

h(y) 0.25 0.50 0.25

3.51 (a) Let X be the number of 4’s and Y be the number of 5’s. The sample spaceconsists of 36 elements each with probability 1/36 of the form (m, n) wherem is the outcome of the first roll of the die and n is the value obtained onthe second roll. The joint probability distribution f(x, y) is defined for x =0, 1, 2 and y = 0, 1, 2 with 0 ≤ x + y ≤ 2. To find f(0, 1), for example,consider the event A of obtaining zero 4’s and one 5 in the 2 rolls. ThenA = {(1, 5), (2, 5), (3, 5), (6, 5), (5, 1), (5, 2), (5, 3), (5, 6)}, so f(0, 1) = 8/36 = 2/9.In a like manner we find f(0, 0) = 16/36 = 4/9, f(0, 2) = 1/36, f(1, 0) = 2/9,f(2, 0) = 1/36, and f(1, 1) = 1/18.

(b) P [(X, Y ) ∈ A] = P (2X + Y < 3) = f(0, 0) + f(0, 1) + f(0, 2) + f(1, 0) =4/9 + 1/9 + 1/36 + 2/9 = 11/12.

3.52 A tabular form of the experiment can be established as,

Sample Space x y

HHHHHTHTHTHHHTTTHTTTHTTT

32221110

3111

−1−1−1−3

So, the joint probability distribution is,

xf(x, y) 0 1 2 3

−3 1/8y −1 3/8

1 3/83 1/8


3.53 (a) If (x, y) represents the selection of x kings and y jacks in 3 draws, we must havex = 0, 1, 2, 3; y = 0, 1, 2, 3; and 0 ≤ x + y ≤ 3. Therefore, (1, 2) represents theselection of 1 king and 2 jacks which will occur with probability

f(1, 2) =

(

41

)(

42

)

(

123

) =6

55.

Proceeding in a similar fashion for the other possibilities, we arrive at the follow-ing joint probability distribution:

xf(x, y) 0 1 2 3

0 1/55 6/55 6/55 1/55y 1 6/55 16/55 6/55

2 6/55 6/553 1/55

(b) P [(X, Y ) ∈ A] = P (X +Y ≥ 2) = 1−P (X +Y < 2) = 1− 1/55− 6/55− 6/55 =42/55.

3.54 (a) P (H) = 0.4, P (T ) = 0.6, and S = {HH, HT, TH, TT}. Let (W, Z) representa typical outcome of the experiment. The particular outcome (1, 0) indicating atotal of 1 head and no heads on the first toss corresponds to the event TH . There-fore, f(1, 0) = P (W = 1, Z = 0) = P (TH) = P (T )P (H) = (0.6)(0.4) = 0.24.Similar calculations for the outcomes (0, 0), (1, 1), and (2, 1) lead to the followingjoint probability distribution:

wf(w, z) 0 1 2

z 0 0.36 0.241 0.24 0.16

(b) Summing the columns, the marginal distribution of W is

w 0 1 2g(w) 0.36 0.48 0.16

(c) Summing the rows, the marginal distribution of Z is

z 0 1h(z) 0.60 0.40

(d) P (W ≥ 1) = f(1, 0) + f(1, 1) + f(2, 1) = 0.24 + 0.24 + 0.16 = 0.64.


3.55 g(x) = 18

∫ 4

2(6− x− y) dy = 3−x

4, for 0 < x < 2.

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

= 6−x−y2(3−x)

, for 2 < y < 4,

and P (1 < Y < 3 | X = 1) = 14

∫ 3

2(5− y) dy = 5

8.

3.56 Since f(1, 1) 6= g(1)h(1), the variables are not independent.

3.57 X and Y are independent since f(x, y) = g(x)h(y) for all (x, y).

3.58 (a) h(y) = 6∫ 1−y

0x dx = 3(1− y)2, for 0 < y < 1. Since f(x|y) = f(x,y)

h(y)= 2x

(1−y)2, for

0 < x < 1− y, involves the variable y, X and Y are not independent.

(b) P (X > 0.3 | Y = 0.5) = 8∫ 0.5

0.3x dx = 0.64.

3.59 (a) 1 = k∫ 1

0

∫ 1

0

∫ 2

0xy2z dx dy dz = 2k

∫ 1

0

∫ 1

0y2z dy dz = 2k

3

∫ 1

0z dz = k

3. So, k = 3.

(b) P(

X < 14, Y > 1

2, 1 < Z < 2

)

= 3∫ 1/4

0

∫ 1

1/2

∫ 2

1xy2z dx dy dz = 9

2

∫ 1/4

0

∫ 1

1/2y2z dy dz

= 2116

∫ 1/4

0z dz = 21

512.

3.60 g(x) = 4∫ 1

0xy dy = 2x, for 0 < x < 1; h(y) = 4

∫ 1

0xy dx = 2y, for 0 < y < 1. Since

f(x, y) = g(x)h(y) for all (x, y), X and Y are independent.

3.61 g(x) = k∫ 50

30(x2 + y2) dy = k

(

x2y + y3

3

)∣

∣

∣

50

30= k

(

20x2 + 98,0003

)

, and

h(y) = k(

20y2 + 98,0003

)

.Since f(x, y) 6= g(x)h(y), X and Y are not independent.

3.62 (a) g(y, z) = 49

∫ 1

0xyz2 dx = 2

9yz2, for 0 < y < 1 and 0 < z < 3.

(b) h(y) = 29

∫ 3

0yz2 dz = 2y, for 0 < y < 1.

(c) P(

14

< X < 12, Y > 1

3, Z < 2

)

= 49

∫ 2

1

∫ 1

1/3

∫ 1/2

1/4xyz2 dx dy dz = 7

162.

(d) Since f(x|y, z) = f(x,y,z)g(y,z)

= 2x, for 0 < x < 1, P(

0 < X < 12| Y = 1

4, Z = 2

)

=

2∫ 1/2

0x dx = 1

4.

3.63 g(x) = 24∫ 1−x

0xy dy = 12x(1− x)2, for 0 < x < 1.

(a) P (X ≥ 0.5) = 12∫ 1

0.5x(1− x)2 dx =

∫ 1

0.5(12x− 24x2 + 12x3) dx = 5

16= 0.3125.

(b) h(y) = 24∫ 1−y

0xy dx = 12y(1− y)2, for 0 < y < 1.

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

= 24xy12y(1−y)2

= 2x(1−y)2

, for 0 < x < 1− y.

So, P(

X < 18| Y = 3

4

)

=∫ 1/8

02x

1/16dx = 32

∫ 1/8

0= 0.25.

3.64 (a)x 1 3 5 7

f(x) 0.4 0.2 0.2 0.2

(b) P (4 < X ≤ 7) = P (X ≤ 7)− P (X ≤ 4) = F (7)− F (4) = 1− 0.6 = 0.4.


3.65 (a) g(x) =∫∞

0ye−y(1+x) dy = − 1

1+xye−y(1+x)

∣

∣

∞

0+ 1

1+x

∫∞

0e−y(1+x) dy

= − 1(1+x)2

e−y(1+x)∣

∣

∞

0

= 1(1+x)2

, for x > 0.

h(y) = ye−y∫∞

0e−yx dx = −e−y e−yx|∞0 = e−y, for y > 0.

(b) P (X ≥ 2, Y ≥ 2) =∫∞

2

∫∞

2ye−y(1+x) dx dy = −

∫∞

2e−y e−yx|∞2 dy =

∫∞

2e−3ydy

= − 13e−3y

∣

∣

∞

2= 1

3e6 .

3.66 (a) P(

X ≤ 12, Y ≤ 1

2

)

= 32

∫ 1/2

0

∫ 1/2

0(x2 + y2) dxdy = 3

2

∫ 1/2

0

(

x2y + y3

3

)∣

∣

∣

1/2

0dx

= 34

∫ 1/2

0

(

x2 + 112

)

dx = 116

.

(b) P(

X ≥ 34

)

= 32

∫ 1

3/4

(

x2 + 13

)

dx = 53128

.

3.67 (a)x 0 1 2 3 4 5 6

f(x) 0.1353 0.2707 0.2707 0.1804 0.0902 0.0361 0.0120

(b) A histogram is shown next.

1 2 3 4 5 6 7

xx

f(x)

f(x)

0.0

0.1

0.2

0.3

(c)x 0 1 2 3 4 5 6

F (x) 0.1353 0.4060 0.6767 0.8571 0.9473 0.9834 0.9954

3.68 (a) g(x) =∫ 1

0(x + y) dy = x + 1

2, for 0 < x < 1, and h(y) = y + 1

2for 0 < y < 1.

(b) P (X > 0.5, Y > 0.5) =∫ 1

0.5

∫ 1

0.5(x + y) dx dy =

∫ 1

0.5

(

x2

2+ xy

)∣

∣

∣

1

0.5dy

=∫ 1

0.5

[(

12

+ y)

−(

18

+ y2

)]

dy = 38.

3.69 f(x) =(

5x

)

(0.1)x(1− 0.1)5−x, for x = 0, 1, 2, 3, 4, 5.

3.70 (a) g(x) =∫ 2

1

(

3x−y9

)

dy = 3xy−y2/29

∣

∣

∣

2

1= x

3− 1

6, for 1 < x < 3, and

h(y) =∫ 3

1

(

3x−y9

)

dx = 43− 2

9y, for 1 < y < 2.

(b) No, since g(x)h(y) 6= f(x, y).

(c) P (X > 2) =∫ 3

2

(

x3− 1

6

)

dx =(

x2

6− x

6

)∣

∣

∣

3

2= 2

3.


3.71 (a) f(x) = ddx

F (x) = 150

e−x/50, for x > 0.

(b) P (X > 70) = 1− P (X ≤ 70) = 1− F (70) = 1− (1− e−70/50) = 0.2466.

3.72 (a) f(x) = 110

, for x = 1, 2, . . . , 10.

(b) A c.d.f. plot is shown next.F(x)

F(x)

xx

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

1 2 3 4 5 6 7 8 9 10

3.73 P (X ≥ 3) = 12

∫∞

3e−y/2 = e−3/2 = 0.2231.

3.74 (a) f(x) ≥ 0 and∫ 10

0110

dx = 1. This is a continuous uniform distribution.

(b) P (X ≤ 7) = 110

∫ 7

0dx = 0.7.

3.75 (a) f(y) ≥ 0 and∫ 1

0f(y) dy = 10

∫ 1

0(1− y)9 dy = − 10

10(1− y)10

∣

∣

1

0= 1.

(b) P (Y > 0.6) =∫ 1

0.6f(y) dy = − (1− y)10|10.6 = (1− 0.6)10 = 0.0001.

3.76 (a) P (Z > 20) = 110

∫∞

20e−z/10 dz = − e−z/10

∣

∣

∞

20= e−20/10 = 0.1353.

(b) P (Z ≤ 10) = − e−z/10∣

∣

10

0= 1− e−10/10 = 0.6321.

3.77 (a) g(x1) =∫ 1

x12 dx2 = 2(1− x1), for 0 < x1 < 1.

(b) h(x2) =∫ x2

02 dx1 = 2x2, for 0 < x2 < 1.

(c) P (X1 < 0.2, X2 > 0, 5) =∫ 1

0.5

∫ 0.2

02 dx1 dx2 = 2(1− 0.5)(0.2− 0) = 0.2.

(d) fX1|X2(x1|x2) = f(x1,x2)

h(x2)= 2

2x2

= 1x2

, for 0 < x1 < x2.

3.78 (a) fX1(x1) =

∫ x1

06x2 dx2 = 3x2

1, for 0 < x1 < 1. Apparently, fX1(x1) ≥ 0 and

∫ 1

0fX1

(x1) dx1 =∫ 1

03x2

1 dx1 = 1. So, fX1(x1) is a density function.

(b) fX2|X1(x2|x1) = f(x1,x2)

fX1(x1)

= 6x2

3x2

1

= 2x2

x2

1

, for 0 < x2 < x1.

So, P (X2 < 0.5 | X1 = 0.7) = 20.72

∫ 0.5

0x2 dx2 = 25

49.


3.79 (a) g(x) = 9(16)4y

∞∑

x=0

14x = 9

(16)4y1

1−1/4= 3

4· 14x , for x = 0, 1, 2, . . . ; similarly, h(y) = 3

4· 14y ,

for y = 0, 1, 2, . . . . Since f(x, y) = g(x)h(y), X and Y are independent.

(b) P (X + Y < 4) = f(0, 0) + f(0, 1) + f(0, 2) + f(0, 3) + f(1, 0) + f(1, 1) + f(1, 2) +f(2, 0) + f(2, 1) + f(3, 0) = 9

16

(

1 + 14

+ 142 + 1

43 + 14

+ 142 + 1

43 + 1r2 + 1

43 + 143

)

=916

(

1 + 24

+ 342 + 4

43

)

= 6364

.

3.80 P (the system works) = P (all components work) = (0.95)(0.99)(0.92) = 0.86526.

3.81 P (the system does not fail) = P (at least one of the components works)= 1−P (all components fail) = 1− (1−0.95)(1−0.94)(1−0.90)(1−0.97) = 0.999991.

3.82 Denote by X the number of components (out of 5) work.Then, P (the system is operational) = P (X ≥ 3) = P (X = 3) + P (X = 4) + P (X =5) =

(

53

)

(0.92)3(1− 0.92)2 +(

54

)

(0.92)4(1− 0.92) +(

55

)

(0.92)5 = 0.9955.

engineering statistics and probability chapter 3

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