lecture 02 iub mat 212

Lecture Notes on Probability and Statistics

Independent University, Bangladesh

Lecture Notes on MAT 212 Probability & Statistics for Science & Engineering

Lecture 2 Probability Introduction We will now proceed to begin our formal discussion on the subject topic of our course. Normally there are two path along which we can proceed. In the first path, we can discuss descriptive statistics, and then we discuss probability, and show the relationship between them. In the second way we be begin with the concepts of probability, and then move to descriptive statistics as it would appear naturally. We would choose the second path, as that would give us better mathematical insight into both probability as well as statistics. So we try to answer the first question: what is probability? Before we answer this question, we look into some background by looking at some examples, especially from science and engineering. Take a simple example of turning on the light. If you turn the switch on, are you absolutely sure that the light will come on? The answer is obviously ‘no’. There are many things that can go wrong. The switch itself can be defective; there can be no electricity; the connecting wires can be out of order; the bulb itself can be burnt (if it is a fluorescent light, more things can be wrong)! No matter how little are the chances, all these events can happen! Therefore, there is a chance if we turn the lights on, it would not come on. Study of probability helps us to understand these ‘chances’ and helps us reduce the possibilities of non-events. Let us consider another example. You turn on your computer. What are the chances that it will operate correctly? There is a chance – no matter how small – that the computer will not operate correctly. This example is perhaps a little more critical than our first example. Consider further the instance of a computer controlling the take-off of an airplane. Can you afford the computer to become in-operational during take-off? Can you leave the take-off of an airplane to a ‘chance’ of the computer not-working? We mentioned about a quality control system in a factory. We accept the fact that a production line would manufacture defective items. The objective of the quality control system is to stop the defective item from being shipped out. If all goes well, a defective item would be stopped by the quality control system. What might also

20 Probability

Independent University, Bangladesh Lecture Notes on Probability and Statistics

happen is (i) a defective item passes through inspection; (ii) a non-defective item is stopped by inspection. There are possibilities that these events might happen. Study of probability helps us to design effective quality control systems. We discuss one last example before we start the discussion on our topic. This time we discuss a queuing system. We take the example of a telephone switching system. An important question here is what should be the capacity of the system. This depends upon call arriving rates, and length of calls. Both the call arrival rates and the length of calls have a pattern; hence probabilistic. Therefore, the capacity of the system depends upon both these probabilistic systems. Preliminary concepts, simple events To evaluate probability, we define three terms: experiment, space, and event. Definition: An experiment is a process that leads to a predictable, or unpredictable outcome. Definition: Space is defined as all possible outcome of an experiment. Definition: An event is defined as one particular outcome of an experiment. Example 2.1 We would like to investigate the rolling a 6-faced dice. Experiment: Rolling of a 6-faced dice Space: {1, 2, 3, 4, 5, 6} Event A: 6 is not observed.

Example 2.2 Experiment : flipping of two coins Space : {hh, ht, th, tt} Event B: at least one head is observed

With probability, we determine ‘What is the likelihood that an event would happen’. We denote this by p(A), p(B), etc. There are two methods to computer probability. The first method, called the relative frequency approach, assumes that the experiment has earlier been conducted for a large number of times. In this case the probability is defined as

conductedwasexperimentthetimesofNo.occuredeventtimesofNo.)( AAp =

Example 2.3

Probability 21



1000 items were inspected from a production line, and 17 were found defective. If an item is inspected at random, what is the probability of finding the item defective? We have conducted the ‘experiment of inspection 1000 times, and we observed the event 17 times. Therefore

p(defective item) = 17.0100017

=

Example 2.4 A team standing at a street intersection counts the types of automobiles passing through the intersection. They find that there were 32 buses, 64 auto rickshaws, 45 cars, 28 taxicabs, and 43 motorbikes. If an automobile passes through the intersection, what is the probability that the automobile is (i) a car, (ii) motorbike? In total there were 212 automobiles. Therefore

2123.021245

onintersectithethroughpassingsautomobileofno.totalonintersectithethroughpassingcarsofNo.)car( ===p

2028.021243

onintersectithethroughpassingsautomobileofno.totalonintersectithethroughpassingmotorbikesofNo.)motorbike( ===p

If all the possible outcome of an experiment is equally likely, then the probability of an event can be given by the classical formula.

proceedcanexperimentthewaysofNo.occurcaneventwaysofNo.)( AAp =

Example 2.5 A fair dice is rolled. Find the probability of (i) getting a 3; (ii) getting a number greater than 3. The space for this problem is S = {1, 2, 3, 4, 5, 6}. Let us define event A: the top face is a 3. Event A can happen only if the top face is a 3; i.e. only one way. In terms of set theory, we can write A = {1}. The experiment of rolling a dice has six possible

outcomes given by the space S. Therefore 61

)()()( ==

SnAnAp .

For the second problem, define the event B: the top face is greater than 3. This event can happen in three ways: if the top face is a 4, 5, or 6. So B = {4, 5, 6}. Therefore

5.063

)()()( ===

SnBnBp .

The interpretation of the first of these probabilities is that of the number of times the dice will be rolled, 1/6th of those rolls are expected to result in the top face being 3; similarly, the second result means that half of the rolls are expected to result in the top face being greater than 3. It should be understood that these probabilities would work for ‘large’ number of rolls only. Figure 1 shows a simulation of roll-dice for the first event. We observe as the number of rolls become, the probability of the event reaches 1/6.

22 Probability


Example 2.6 A fair coin is tossed twice. The space S = {hh, ht, th, tt}. Let us find the probabilities of the following events. A: at least one head is observed.

43},,{)( ==

SthhthhAp

B: No head is observed

41}{)( ==

SttBp

C: only one head is observed

21

42},{)( ===

SthhtCp

Example 2.7 A class in probability theory consists of 6 men and 4 women. An exam is given and the students are ranked according to their performance. Assuming that no two students obtain the same score, (a) how many different rankings are possible? (b) If all rankings are considered equally likely, what is the probability that women receive the top 4 scores? (a) Because each ranking corresponds to a particular ordered arrangement of the

10 people, we see the answer to this part is 10! = 3,628,800. (b) Because there are 4! possible rankings of the women among themselves and 6!

possible rankings of the men among themselves, it follows from the basic principle that there are (6!)(4!) = (720)(24)=17,280 possible rankings in which the women receive the top 4 scores. Hence, the desired probability is

2101

789101234

!10!4!6

=⋅⋅⋅⋅⋅⋅

= .

Example 2.8 Let us consider an example of two rolls of a fair dice. The space for this experiment is

=

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

S

Let us try to find the probability of the following events. A: the two rolls are same. From S, we can observe A = {(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)}. Therefore,

61

366

)()()( ===

SnAnAp

B: the second roll is 1

Probability 23



B = {(1,1), (2,1), (3,1), (4,1), (5,1), (6,1)}. Therefore, 61

366

)()()( ===

SnBnBp

C: The sum of the two roll is 8

C = {(6,2), (5,3), (4,4), (3,5), (2,6)}. Therefore, 365

)()()( ==

SnCnCp

D : the sum of the two rolls is at least 10 D = {(4,6), (5,5), (5,6), (6,4), (6,5), (6,6)}. Therefore,

61

366

)()()( ===

SnDnDp

Example 2.9 A committee of size 5 is to be selected from a group of 6 men and 9 women. If the selection is made randomly, what is the probability that the committee consists of 3 men and 2 women?

Let us assume that “randomly selected” means that each of the

5

15 possible

combinations is equally likely to be selected. Hence, since there are

36

possible

choices of 3 men and

29

possible choices of 2 women, it follows that the desired

probability is given by 1001240

515

29

36

=

.

Example 2.10 If n people are present in a room, what is the probability that no two of them celebrate their birthday on the same day of the year? How large need n be so that this probability is less than ½? Because each person can celebrate his or her birthday on any one of 365 days, there are a total of (365)n possible outcomes. (We are ignoring the possibility of someone having been born on February 29.) Furthermore, there are (365)(364)(363)・ (365 – n + 1) possible outcomes that result in no two of the people having the same birthday. This is so because the first person could have any one of 365 birthdays, the next person any of the remaining 364 days, the next any of the remaining 363, and so on. Hence, assuming that each outcome is equally likely, we see that the desired probability is

n

n365

)1365)...(363(364)(365( +−

It is a rather surprising fact that when n ≥ 23, this probability is less than ½. That is, if there are 23 or more people in a room, then the probability that at least two of

24 Probability


them have the same birthday exceeds ½. Many people are initially surprised by this result, since 23 seems so small in relation to 365, the number of days of the year.

However, every pair of individuals has probability 3651

365365

2 = of having the same

birthday, and in a group of 23 people there are 253223

=

different pairs of

individuals. Looked at this way, the result no longer seems so surprising.

Exercise 2.1 A digital scale is used that provides weights to the nearest gram.

(a) What is the sample space for this experiment? Let A denote the event that a weight exceeds 11 grams, let B denote the event that a weight is less than or equal to 15 grams, and let C denote the event that a weight is greater than or equal to 8 grams and less than 12 grams. Describe the following events.

(b) A ∪ B (c) A ∩ B (d) A′ (e) A ∩ B ∩ C (f) (A ∪ C)′ (g) A ∩ B ∩ C (h) B′ ∩ C (i) A ∪ B ∩ C

2.2 In an injection-molding operation, the length and width, denoted as X and Y,

respectively, of each molded part are evaluated. Let A denote the event of 48 < X < 52 centimeters B denote the event of 9 <Y < 11 centimeters C denote the event that a critical length meets customer requirements.

Construct a Venn diagram that includes these events. Shade the areas that represent the following: (a) A (b) A ∩ B (c) A′ ∪ B (d) A ∪ B (e) If these events were mutually exclusive, how successful would this

production operation be? Would the process produce parts with X = 50 centimeters and Y = 10 centimeters?

2.3 Four bits are transmitted over a digital communications channel. Each bit is

either distorted or received without distortion. Let Ai denote the event that the ith bit is distorted, i = 1, 2, 3, 4. (a) Describe the sample space for this experiment. (b) Are the Ai’s mutually exclusive? Describe the outcomes in each of the following events: (c) A1 (d) A1′ (e) A1 ∩ A2 ∩ A3 ∩ A4 (f) (A1 ∩ A2) ∪ (A3 ∩ A4)

2.4 A sample of three calculators is selected from a manufacturing line, and each

calculator is classified as either defective or acceptable. Let A, B, and C denote

Probability 25



the events that the first, second, and third calculators respectively, are defective. Describe each of the following events: (a) A (b) B (c) A ∩ B (d) B ∪ C

2.5 A wireless garage door opener has a code determined by the up or down setting

of 12 switches. How many outcomes are in the sample space of possible codes? 2.6 Disks of polycarbonate plastic from a supplier are analyzed for scratch and

shock resistance. The results from 100 disks are summarized below: shock resistance

high low scratch high 70 9 resistance low 16 5

Let A denote the event that a disk has high shock resistance, and let B denote the event that a disk has high scratch resistance. Determine the number of disks in A ∩ B, A′ and A ∪ B.

2.7 Samples of a cast aluminum part are classified on the basis of surface finish (in

microinches) and edge finish. The results of 100 parts are summarized as follows:

edge finish excellent good

surface excellent 80 2 finish good 10 8

(a) Let A denote the event that a sample has excellent surface finish, and let B denote the event that a sample has excellent edge finish. Determine the number of samples in A′ ∩ B, B and A ∪ B.

(b) Assume that each of two samples is to be classified on the basis of surface finish, either excellent or good, edge finish, either excellent or good. Use a tree diagram to represent the possible outcomes of this experiment.

2.8 Samples of emissions from three suppliers are classified for conformance to air-

quality specifications. The results from 100 samples are summarized as follows: conforms yes no

1 22 8 Supplier 2 25 5

3 30 10 Let A denote the event that a sample is from supplier 1, and let B denote the event that a sample conforms to specifications. Determine the number of samples in A′ ∩ B, B and A ∪ B.

2.9 The rise time of a reactor is measured in minutes (and fractions of minutes). Let

the sample space be positive, real numbers. Define the events A and B as follows:

A = {x | x < 72.5} and

B = {x | x > 52.5} Describe each of the following events:

26 Probability


(a) A′ (b) B′ (c) A ∩ B (d) A ∪ B

2.10 A sample of two items is selected without replacement from a batch. Describe

the (ordered) sample space for each of the following batches: (a) The batch contains the items {a, b, c, d}. (b) The batch contains the items {a, b, c, d, e, f, g}. (c) The batch contains 4 defective items and 20 good items. (d) The batch contains 1 defective item and 20 good items.

2.11 A sample of two printed circuit boards is selected without replacement from a

batch. Describe the (ordered) sample space for each of the following batches: (a) The batch contains 90 boards that are not defective, 8 boards with minor

defects, and 2 boards with major defects. (b) The batch contains 90 boards that are not defective, 8 boards with minor

defects, and 1 board with major defects. 2.12 Counts of the Web pages provided by each of two computer servers in a selected

hour of the day are recorded. Let A denote the event that at least 10 pages are provided by server 1 and let B denote the event that at least 20 pages are provided by server 2. (a) Describe the sample space for the numbers of pages for two servers

graphically. Show each of the following events on the sample space graph: (b) A (c) B (d) A ∩ B (e) A ∪ B

2.13 The rise time of a reactor is measured in minutes (and fractions of minutes). Let

the sample space for the rise time of each batch be positive, real numbers. Consider the rise times of two batches. Let A denote the event that the rise time of batch 1 is less than 72.5 minutes, and let B denote the event that the rise time of batch 2 is greater than 52.5 minutes. Describe the sample space for the rise time of two batches graphically and show each of the following events on a two dimensional plot: (a) A (b) B′ (c) A ∩ B (d) A ∪ B

2.14 Each of the possible five outcomes of a random experiment is equally likely.

The sample space is {a, b, c, d, e}. Let A denote the event {a, b, c}, and let B denote the event {c, d, e}. Determine the following: (a) p(A) (b) p(B) (c) p(A′) (d) p(A ∪ B) (e) p(A ∩ B)

2.15 A part selected for testing is equally likely to have been produced on any one of

six cutting tools. (a) What is the sample space? (b) What is the probability that the part is from tool 1? (c) What is the probability that the part is from tool 3 or tool 5? (d) What is the probability that the part is not from tool 4?

Probability 27



2.16 An injection-molded part is equally likely to be obtained from any one of the eight cavities on a mold. (a) What is the sample space? (b) What is the probability a part is from cavity 1 or 2? (c) What is the probability that a part is neither from cavity 3 nor 4?

2.17 Orders for a computer are summarized by the optional features that are

requested as follows: proportion of orders

no optional features 0.3 one optional feature 0.5 more than one optional feature 0.2

(a) What is the probability that an order requests at least one optional feature? (b) What is the probability that an order does not request more than one optional

feature? 2.18 If the last digit of a weight measurement is equally likely to be any of the digits

0 through 9, (a) What is the probability that the last digit is 0? (b) What is the probability that the last digit is greater than or equal to 5?

2.19 A sample preparation for a chemical measurement is completed correctly by

25% of the lab technicians, completed with a minor error by 70%, and completed with a major error by 5%. (a) If a technician is selected randomly to complete the preparation, what is the

probability it is completed without error? (b) What is the probability that it is completed with either a minor or a major

error? 2.20 A credit card contains 16 digits between 0 and 9. However, only 100 million

numbers are valid. If a number is entered randomly, what is the probability that it is a valid number?

2.21 Suppose your vehicle is licensed in a state that issues license plates that consist

of three digits (between 0 and 9) followed by three letters (between A and Z). If a license number is selected randomly, what is the probability that yours is the one selected?

2.22 A message can follow different paths through servers on a network. The senders

message can go to one of five servers for the first step, each of them can send to five servers at the second step, each of which can send to four servers at the third step, and then the message goes to the recipients server. (a) How many paths are possible? (b) If all paths are equally likely, what is the probability that a message passes

through the first of four servers at the third step? 2.23 Disks of polycarbonate plastic from a supplier are analyzed for scratch and

shock resistance. The results from 100 disks are summarized as follows: shock resistance high low

28 Probability


scratch high 70 9 resistance low 16 5

Let A denote the event that a disk has high shock resistance, and let B denote the event that a disk has high scratch resistance. If a disk is selected at random, determine the following probabilities: (a) p(A) (b) p(B) (c) p(A′) (d) p(A ∪ B) (e) p(A ∩ B) (f) p(A′ ∩ B)


microinches) and edge finish. The results of 100 parts are summarized as follows:

edge finish excellent good


Let A denote the event that a sample has excellent surface finish, and let B denote the event that a sample has excellent length. If a part is selected at random, determine the following probabilities: (a) p(A) (b) p(B) (c) p(A′) (d) p(A ∪ B) (e) p(A ∩ B) (f) p(A′ ∩ B)


quality specifications. The results from 100 samples are summarized as follows: conforms

yes no 1 22 8

Supplier 2 25 5 3 30 10

Let A denote the event that a sample is from supplier 1, and let B denote the event that a sample conforms to specifications. If a sample is selected at random, determine the following probabilities: (a) p(A) (b) p(B) (c) p(A′) (d) p(A ∪ B) (e) p(A ∩ B) (f) p(A′ ∩ B)

Compound events So far we have been looking into simple events. In practical applications, especially in engineering application, it is often necessary to find probabilities of compound events. Compound events are two, or more, events happening together. We may be looking into two situations: anyone of the events would take place; or all the events would take place simultaneously. For example, for the Example 2.7, we may define the events:

A: two rolls are same B: the sum of two rolls is at least 8

Probability 29



Now we may want to find the probability of either A or B happening, or A and B happening. In terms of set theory, we are looking into A ∪ B and A ∩ B respectively. To see this working in real life, we discussed the case of anti-lock brake system in the first lecture. We had learnt that if the brake pedal is pressed, and the car is moving, and the wheels are locked, the anti-lock system would activate. Now let us define the events:

A: the brakes are applied B: the car is moving C: the wheels are locked

For the anti-lock system to activate itself we are looking for the event A ∩ B ∩ C. We consider another simple example of someone trying to make a call from a cell phone. Let us assume further that the call is being made to a number of another operator. The number is dialed and is received by the nearest Base Tower Station (BTS) of the originating operator. The call is passed to the central server for verification of the user. If it is found that the user is clear, the call is passed to the host operator. The host operator receives the host number, and performs verification. If the number is found clear, the call is passed to the nearest BTS. Finally the call is passed to the receiver. All these events are critical events. Let us define the events

A: not enough channels is available on the originating BTS B: not enough channels is available from BTS to central server C: not sufficient credit is available on you cell phone account D: not enough channels is available between the two operators E: the host number is not valid, or untraceable F: not enough channels is available to the nearest BTS of the receiving number G: not enough channels is available to handle the call on the BTS

The call will fail to go through for A ∪ B ∪ C ∪ D ∪ E ∪ F ∪ G. We learn the mechanics of compound event through an example. Example 2.11 Let us consider an example of two rolls of a fair dice. The space for this experiment is

=

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

S

We define two events A: two rolls are same B: the sum of two rolls is at least 8 We want to find p(A ∪ B) and p(A ∩ B).

30 Probability


From the space S, we find A = {(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)}, and B = {(2,6), (3,5), (3,6), (4,4), (4,5), (4,6), (5,3), (5,4), (5,5), (5,6), (6,2), (6,3),

(6,4), (6,5), (6,6)} From these two, we can obtain A ∪ B = {(1,1), (2,2), (3,3), (2,6), (3,5), (3,6), (4,4), (4,5), (4,6), (5,3), (5,4),

(5,5), (5,6), (6,2), (6,3), (6,4), (6,5), (6,6)}, and A ∩ B = {(4,4), (5,5), (6,6)} Therefore, we can find

21

3618

)()()( ==

∪=∪

SnBAnBAp

121

363

)()()( ==

∩=∩

SnBAnBAp

We learn a more formal method for finding p(A ∪ B). We have learnt in equation (1) in Lecture 2

n(A ∪ B) = n(A) + n(B) − n(A ∩ B)

Therefore

)()()()(

)()()(

SnBAnBnAn

SnBAnBAp ∩−+

=∪

=∪

)()(

)()(

)()(

SnBAn

SnBn

SnAn ∩

−+=

= p(A) + p(B) − p(A ∩ B) (1)

From Example 2.8, we obtain 366)( =Ap ,

3615)( =Bp , and

363)( =∩ BAp .

Combining all these, we obtain

3615

363

3615

366)( =−+=∪ BAp

If the events A and B are disjoint, or mutually exclusive; meaning if one happens, other would not happen, then A ∩ B = ∅, and p(A ∩ B) = 0. For example, in the last problem, if two events are defined as

A: first roll is 1 B: the sum of the two rolls is at least 9

Then A ∩ B = ∅, and p(A ∩ B) = 0. Therefore, for mutually exclusive events

p(A ∪ B) = p(A) + p(B) (2)

Exercise 2.26 If p(A) = 0.3, p(B) = 0.2 and p(A ∩ B) = 0.1, determine the following

probabilities:

Probability 31



(a) p(A′) (b) p(A ∪ B) (c) p(A′ ∩ B) (d) p(A ∩ B′) (e) p[(A ∪ B)]’ (f ) p(A′ ∪ B)

2.27 If A, B, and C are mutually exclusive events with p(A) = 0.3, p(B) = 0.2 and p(C)

= 0.1and determine the following probabilities: (a) p(A ∪ B ∪ C) (b) p(A ∩ B ∩ C) (c) p(A ∩ B) (d) p(A ∪ B ∩ C) (e) p(A′ ∩ B′ ∩ C′)

2.28 Disks of polycarbonate plastic from a supplier are analyzed for scratch and

shock resistance. The results from 100 disks are summarized as follows: shock resistance


(a) If a disk is selected at random, what is the probability that its scratch resistance is high and its shock resistance is high?

(b) If a disk is selected at random, what is the probability that its scratch resistance is high or its shock resistance is high?

(c) Consider the event that a disk has high scratch resistance and the event that a disk has high shock resistance. Are these two events mutually exclusive?

2.29 The analysis of shafts for a compressor is summarized by conformance to

specifications. roundness conforms

yes no surface finish yes 345 5 conforms no 12 8

(a) If a shaft is selected at random, what is the probability that the shaft conforms to surface finish requirements?

(b) What is the probability that the selected shaft conforms to surface finish requirements or to roundness requirements?

(c) What is the probability that the selected shaft either conforms to surface finish requirements or does not conform to roundness requirements?

(d) What is the probability that the selected shaft conforms to both surface finish and roundness requirements?

2.30 Cooking oil is produced in two main varieties: mono and polyunsaturated. Two

common sources of cooking oil are corn and canola. The following table shows the number of bottles of these oils at a supermarket:

type of oil canola corn

type of mono 7 13 unsaturation poly 93 77

(a) If a bottle of oil is selected at random, what is the probability that it belongs to the polyunsaturated category?

(b) What is the probability that the chosen bottle is monounsaturated canola oil?

32 Probability


2.31 The shafts in Exercise 2.29 are further classified in terms of the machine tool that was used for manufacturing the shaft.

Tool 1 roundness conforms


Tool 2 roundness conforms


(a) If a shaft is selected at random, what is the probability that the shaft conforms to surface finish requirements or to roundness requirements or is from Tool 1?

(b) If a shaft is selected at random, what is the probability that the shaft conforms to surface finish requirements or does not conform to roundness requirements or is from Tool 2?

(c) If a shaft is selected at random, what is the probability that the shaft conforms to both surface finish and roundness requirements or the shaft is from Tool 2?

(d) If a shaft is selected at random, what is the probability that the shaft conforms to surface finish requirements or the shaft is from Tool 2?

Conditional probability We introduce an important concept in this section. We have observed that in the calculation of the probability, the assessment of the population is very important. We would like to be as realistic as possible in assessing the population. This means that if any additional information is available about the population, that information must be included into the calculation. To see how this works, let us consider the vents given in Example 2.8. We had defined two events A: two rolls are same B: the sum of two rolls is at least 8 We found p(A) and p(B). For both these events the space was considered to be S. Now, suppose we want to find the probability of the event A, but it is also stated that the event B has already occurred. The situation is that a fair dice has already been rolled twice, and we are told that the sum of the two dice is at least 8; now we want to find the probability that both the rolls show the same number. This event is shown symbolically as p(A | B), and is read as ‘probability of A, given B’. We will still use the classical formula to find the probability, but what may change here is that with the availability of the additional information about the occurrence of B, the space may change. As in this example, we have been told that the event B has already happened. Therefore, our space is now

B = {(2,6), (3,5), (3,6), (4,4), (4,5), (4,6), (5,3), (5,4), (5,5), (5,6), (6,2), (6,3), (6,4), (6,5), (6,6)}

Now looking for the event A within B gives us A = {(4,4), (5,5), (6,6)}

Probability 33



Therefore

153

)()|()|( ==

BnBAnBAp

To formalize this, we observe

)()(

)(/)()(/)(

)()(

)()|()|(

BpBAp

SnBnSnBAn

BnBAn

BnBAnBAp ∩

=∩

=∩

==

Therefore

)()()|(

BpBApBAp ∩

=

or p(A ∩ B) = p(A | B) · p(B) (3) Example 2.12 Cold solder and wrong components are the two most common type of production defects on printed circuit boards (PCBs) In a batch of 364 PCBs it was found that 17 have cold solder defects, 14 have wring component defects, and 8 have both cold solder and wrong component defects. If a PCB is chosen at random, find the probability that if a PCB is found to have a cold solder defect, it would also have wrong component defect. Let us define the events A: PCB has cold solder defect B: PCB has wrong component defect We have been asked to find p(B | A). From equation (1), we can write

)(

)()|(Ap

ABpABp ∩=

From the given numbers, we have364

8)( =∩ ABp , and36417)( =Ap . Therefore

178

364/17364/8)|( ==ABp

Example 2.13 Let us consider the same example of rolling a fair dice twice, as we considered in Example 2.8. In this example let us define the events A: the first roll shows a 1 B: The second roll also shows a 1 We would like to find p(B | A) From equation (1), we have

)(

)()|(Ap

ABpABp ∩=

From the space given in Example 2.8, we can write

61

6/136/1

)()()|( ==

∩=

ApABpABp

34 Probability


We notice an interesting feature of Example 2.10. From the space of the problem, we

observe p(B) = 61 . We have already seen that find p(B | A) =

61 . This means that for

the second roll, the additional information about the occurrence of A has no influence. Therefore, in this case, A and B are regarded as independent events. Definition: If the occurrence of the event A has no influence on the occurrence of the event B, the events A and B are regarded as independent events. In this case p(A | B) = p(A). For independent events, equation (2) can be simplified further as

p(A ∩ B) = p(A)·p(B) (4)

Some important axioms We have seen several examples of probability of events. We have learnt the underlying assumption for probability that all events will have a space, and if the experiment is repeated large number of times. From a purely mathematical viewpoint, we will suppose that for each event E of an experiment having a sample space S there is a number, denoted by p(E), that is in accord with the following three axioms. Axiom 1. p(E) ≥ 0, and p(E) ≤ 1 Axiom 2. p(S) = 1 Axiom 3. For any sequence of mutually exclusive events E1, E2, . . . (that is events for which EiEj = ∅ when i ≠ j)

∑==

=

n

ii

n

ii EpEp

11

)(U

Axiom 4. For any sequence of independent events E1, E2, . . .

∏==

=

n

ii

n

ii EEp

11I

An important extension of axiom 3 is that for any event E, the event itself and the complimentary event Ec are mutually exclusive. So we can write

p(S) = p(E) + p(Ec) = 1

or p(E) = 1 − p(Ec) (5)

This equation is useful in many places. Example 2.14 A box contains 8 red, 10 green, 6 white, and 4 blue marbles. Two marbles are drawn, without replacement, one after another from the box. Find the probabilities of the following events. a) A: two blue marbles are drawn A = {bb}

Probability 35



The draw of first and the second draws are independent. But after the first draw, since the marble is not put back into the box, the total number reduces. Therefore

p(A) = p(first blue) · p(second blue) = 631

273

284

=

b) B: a green and a white marble are drawn B: {gw, wg} The event {gw} and {wg} are mutually exclusive. Therefore

p(B) = p(gw) + p(wg) = 6310

2710

286

276

2810

=+

c) C: no white marbles are drawn The events can be listed as rg, rb, gb, . . . This would make the list immensely

unmanageable. We would rather handle this event in a more effective way. If we assume that w indicate the event of ‘drawing of a white marble’, then wc would indicate the event of ‘not drawing a white marble’. Therefore, our event is

C = {wcwc}

p(C) = 1811

2721

2822

=

d) D: At least one blue marble is drawn Once again we can list the events as br, rb, bg, gb, . . . Once again it would be

much easier to find the probability of the complimentary event Dc, the use that to find the probability of the event D.

p(D) = 1 − p(Dc) = 1 − p(no blue marbles are drawn)

= 1 − p(bcbc) = 1 − 6317

2723

2824

=

The same problem can also be solved by listing the event D as D = {bbc, bcb, bb} Therefore

p(D) = p(bbc) + p(bcb) + p(bb) = 6317

273

284

274

2824

2724

284

=++

e) E: At most one white marble is drawn The event is E = {wwc, wcw, wcwc}. Therefore

p(E) = p(wwc) + p(wcw) + p(wcwc) = 126121

2721

2822

276

2822

2722

286

=++

f) F: first marble is not green The event is F = {gcg, gcgc}. Therefore

p(F) = p(gcg) + p(gcgc) = 149

2717

2818

2710

2818

=+

g) G: second marble is white The event is G = {wcw, ww}. Therefore

p(G) = p(wcw) + p(ww) = 143

275

286

276

2822

=+

h) H: the first marble is either red or green and the second marble is either green or blue. H = {(r ∪ g) ∩ (g ∪ b)}

p(H) = {p(r) + p(g)}·{p(g) + p(b)} = 31

274

2710

2810

288

=

+

+

Example 2.15

36 Probability


A bin contains 5 defective (that immediately fail when put in use), 10 partially defective (that fail after a couple of hours of use), and 25 acceptable transistors. A transistor is chosen at random from the bin and put into use. If it does not immediately fail, what is the probability it is acceptable? Since the transistor did not immediately fail, we know that it is not one of the 5 defectives and so the desired probability is:

( ) ( )( )

( )( defectivenot

acceptabledefectivenot

defectivenotacceptabledefectivenot|acceptablep

pp

pp =∩

=

where the last equality follows since the transistor will be both acceptable and not defective if it is acceptable. Hence, assuming that each of the 40 transistors is equally likely to be chosen, we obtain that

( )75

40/3540/25defectivenot|acceptable ==p

It should be noted that we could also have derived this probability by working directly with the reduced sample space. That is, since we know that the chosen transistor is not defective, the problem reduces to computing the probability that a transistor, chosen at random from a bin containing 25 acceptable and 10 partially

defective transistors, is acceptable. This is clearly equal to 3525 .

Example 2.16 The organization that Jones works for is running a father–son dinner for those employees having at least one son. Each of these employees is invited to attend along with his youngest son. If Jones is known to have two children, what is the conditional probability that they are both boys given that he is invited to the dinner? Assume that the sample space S is given by S = {(b, b), (b, g), (g, b), (g, g)} and all outcomes are equally likely [(b, g) means, for instance, that the younger child is a boy and the older child is a girl]. The knowledge that Jones has been invited to the dinner is equivalent to knowing that he has at least one son. Hence, letting B denote the event that both children are boys, and A the event that at least one of them is a boy, we have that the desired probability p(B | A) is given by

( ) ( )( )

[ ][ ] 3

14/34/1

),(),,(),,(),(| ===

∩=

bggbbbpbbp

ApBApABp

Many readers incorrectly reason that the conditional probability of two boys given at least one is 1/2, as opposed to the correct 1/3, since they reason that the Jones child not attending the dinner is equally likely to be a boy or a girl. Their mistake, however, is in assuming that these two possibilities are equally likely. Remember that initially there were four equally likely outcomes. Now the information that at least one child is a boy is equivalent to knowing that the outcome is not (g, g). Hence we are left with the three equally likely outcomes (b, b), (b, g), (g, b), thus showing that the Jones child not attending the dinner is twice as likely to be a girl as a boy.

Probability 37



Example 2.17 Ms. Perez figures that there is a 30 percent chance that her company will set up a branch office in Phoenix. If it does, she is 60 percent certain that she will be made manager of this new operation. What is the probability that Perez will be a Phoenix branch office manager? If we let B denote the event that the company sets up a branch office in Phoenix and M the event that Perez is made the Phoenix manager, then the desired probability is p(B ∩ M), which is obtained as follows:

p(B ∩ M) = p(B)p(M | B) =0.3×0.6 = 0.18

Hence, there is an 18 percent chance that Perez will be the Phoenix manager.

Exercise 2.32 Disks of polycarbonate plastic from a supplier are analyzed for scratch and

shock resistance. The results from 100 disks are summarized as follows: shock resistance


Let A denote the event that a disk has high shock resistance, and let B denote the event that a disk has high scratch resistance. Determine the following probabilities: (a) p(A) (b) p(B) (c) p(A | B) (d) p(B | A)


microinches) and length measurements. The results of 100 parts are summarized as follows:

length excellent good


Let A denote the event that a sample has excellent surface finish, and let B denote the event that a sample has excellent length. Determine: (a) p(A) (b) p(B) (c) p(A | B) (d) p(B | A) (e) If the selected part has excellent surface finish, what is the probability that

the length is excellent? (f) If the selected part has good length, what is the probability that the surface

finish is excellent?

38 Probability


2.34 The analysis of shafts for a compressor is summarized by conformance to specifications:

roundness conforms yes no

surface finish yes 345 5 conforms no 12 8

(a) If we know that a shaft conforms to roundness requirements, what is the probability that it conforms to surface finish requirements?

(b) If we know that a shaft does not conform to roundness requirements, what is the probability that it conforms to surface finish requirements?

2.35 The following table summarizes the analysis of samples of galvanized steel for

coating weight and surface roughness: coating weight high low

surface high 12 16 roughness low 88 34

(a) If the coating weight of a sample is high, what is the probability that the surface roughness is high?

(b) If the surface roughness of a sample is high, what is the probability that the coating weight is high?

(c) If the surface roughness of a sample is low, what is the probability that the coating weight is low?

2.36 A lot of 100 semiconductor chips contains 20 that are defective. Two are

selected randomly, without replacement, from the lot. (a) What is the probability that the first one selected is defective? (b) What is the probability that the second one selected is defective given that

the first one was defective? (c) What is the probability that both are defective? (d) How does the answer to part (b) change if chips selected were replaced prior

to the next selection? 2.37 A lot contains 15 castings from a local supplier and 25 castings from a supplier

in the next state. Two castings are selected randomly, without replacement, from the lot of 40. Let A be the event that the first casting selected is from the local supplier, and let B denote the event that the second casting is selected from the local supplier. Determine: (a) p(A) (b) p(B) (c) p(A ∪ B) (d) p(A ∩ B)

2.38 Continuation of Exercise 2.37. Suppose three castings are selected at random,

without replacement, from the lot of 40. In addition to the definitions of events A and B, let C denote the event that the third casting selected is from the local supplier. Determine: (a) p(A ∩ B ∩ C) (b) p(A ∩ B ∩ C′)

2.39 A batch of 500 containers for frozen orange juice contains 5 that are defective.

Two are selected, at random, without replacement from the batch.

Probability 39



(a) What is the probability that the second one selected is defective given that the first one was defective?

(b) What is the probability that both are defective? (c) What is the probability that both are acceptable?

2.40 Continuation of Exercise 2.39. Three containers are selected, at random, without

replacement, from the batch. (a) What is the probability that the third one selected is defective given that the

first and second one selected were defective? (b) What is the probability that the third one selected is defective given that the

first one selected was defective and the second one selected was okay? (c) What is the probability that all three are defective?

2.41 A maintenance firm has gathered the following information regarding the failure

mechanisms for air conditioning systems: evidence of gas leaks

yes no evidence of yes 55 17 electrical failure no 32 3

The units without evidence of gas leaks or electrical failure showed other types of failure. If this is a representative sample of AC failure, find the probability (a) That failure involves a gas leak (b) That there is evidence of electrical failure given that there was a gas leak (c) That there is evidence of a gas leak given that there is evidence of electrical

failure 2.42 The probability is 1% that an electrical connector that is kept dry fails during the

warranty period of a portable computer. If the connector is ever wet, the probability of a failure during the warranty period is 5%. If 90% of the connectors are kept dry and 10% are wet, what proportion of connectors fail during the warranty period?

2.43 Suppose 2% of cotton fabric rolls and 3% of nylon fabric rolls contain flaws. Of

the rolls used by a manufacturer, 70% are cotton and 30% are nylon. What is the probability that a randomly selected roll used by the manufacturer contains flaws?

2.44 In the manufacturing of a chemical adhesive, 3% of all batches have raw

materials from two different lots. This occurs when holding tanks are replenished and the remaining portion of a lot is insufficient to fill the tanks. Only 5% of batches with material from a single lot require reprocessing. However, the viscosity of batches consisting of two or more lots of material is more difficult to control, and 40% of such batches require additional processing to achieve the required viscosity. Let A denote the event that a batch is formed from two different lots, and let B denote the event that a lot requires additional processing. Determine the following probabilities: (a) p(A) (b) p(A′) (c) p(B | A) (d) p(B | A′) (e) p(A ∩ B) (f) p(A ∩ B′)

40 Probability


(g) p(B) 2.45 The edge roughness of slit paper products increases as knife blades wear. Only

1% of products slit with new blades have rough edges, 3% of products slit with blades of average sharpness exhibit roughness, and 5% of products slit with worn blades exhibit roughness. If 25% of the blades in manufacturing are new, 60% are of average sharpness, and 15% are worn, what is the proportion of products that exhibit edge roughness?

2.46 Samples of laboratory glass are in small, light packaging or heavy, large

packaging. Suppose that 2 and 1% of the sample shipped in small and large packages, respectively, break during transit. If 60% of the samples are shipped in large packages and 40% are shipped in small packages, what proportion of samples break during shipment?

2.47 Incoming calls to a customer service center are classified as complaints (75% of

call) or requests for information (25% of calls). Of the complaints, 40% deal with computer equipment that does not respond and 57% deal with incomplete software installation; and in the remaining 3% of complaints the user has improperly followed the installation instructions. The requests for information are evenly divided on technical questions (50%) and requests to purchase more products (50%). (a) What is the probability that an incoming call to the customer service center

will be from a customer who has not followed installation instructions properly?

(b) Find the probability that an incoming call is a request for purchasing more products.

2.48 Computer keyboard failures are due to faulty electrical connects (12%) or

mechanical defects (88%). Mechanical defects are related to loose keys (27%) or improper assembly (73%). Electrical connect defects are caused by defective wires (35%), improper connections (13%), or poorly welded wires (52%). (a) Find the probability that a failure is due to loose keys. (b) Find the probability that a failure is due to improperly connected or poorly

welded wires. 2.49 A batch of 25 injection-molded parts contains 5 that have suffered excessive

shrinkage. (a) If two parts are selected at random, and without replacement, what is the

probability that the second part selected is one with excessive shrinkage? (b) If three parts are selected at random, and without replacement, what is the

probability that the third part selected is one with excessive shrinkage? 2.50 A lot of 100 semiconductor chips contains 20 that are defective.

(a) Two are selected, at random, without replacement, from the lot. Determine the probability that the second chip selected is defective.

(b) Three are selected, at random, without replacement, from the lot. Determine the probability that all are defective.

Probability 41



2.51 Disks of polycarbonate plastic from a supplier are analyzed for scratch and shock resistance. The results from 100 disks are summarized as follows:

shock resistance high low

scratch high 70 9 resistance low 16 5

Let A denote the event that a disk has high shock resistance, and let B denote the event that a disk has high scratch resistance. Are events A and B independent?


microinches) and length measurements. The results of 100 parts are summarized as follows:

length excellent good


Let A denote the event that a sample has excellent surface finish, and let B denote the event that a sample has excellent length. Are events A and B independent?


quality specifications. The results from 100 samples are summarized as follows: conforms yes no

1 22 8 Supplier 2 25 5

3 30 10 Let A denote the event that a sample is from supplier 1, and let B denote the event that a sample conforms to specifications. (a) Are events A and B independent? (b) Determine p(B | A)

2.54 The probability that a lab specimen contains high levels of contamination is

0.10. Five samples are checked, and the samples are independent. (a) What is the probability that none contains high levels of contamination? (b) What is the probability that exactly one contains high levels of contamination? (c) What is the probability that at least one contains high levels of

contamination? 2.55 In a test of a printed circuit board using a random test pattern, an array of 10 bits

is equally likely to be 0 or 1. Assume the bits are independent. (a) What is the probability that all bits are 1s? (b) What is the probability that all bits are 0s? (c) What is the probability that exactly five bits are 1s and five bits are 0s?

2.56 Eight cavities in an injection-molding tool produce plastic connectors that fall

into a common stream. A sample is chosen every several minutes. Assume that the samples are independent.

42 Probability


(a) What is the probability that five successive samples were all produced in cavity one of the mold?

(b) What is the probability that five successive samples were all produced in the same cavity of the mold?

(c) What is the probability that four out of five successive samples were produced in cavity one of the mold?

2.57 The following circuit operates if and only if there is a path of functional devices

from left to right. The probability that each device functions is as shown. Assume that the probability that a device is functional does not depend on whether or not other devices are functional. What is the probability that the circuit operates?

2.58 The following circuit operates if and only if there is a path of functional devices

from left to right. The probability each device functions is as shown. Assume that the probability that a device functions does not depend on whether or not other devices are functional. What is the probability that the circuit operates?

2.59 An optical storage device uses an error recovery procedure that requires an

immediate satisfactory readback of any written data. If the readback is not successful after three writing operations, that sector of the disk is eliminated as unacceptable for data storage. On an acceptable portion of the disk, the probability of a satisfactory readback is 0.98. Assume the readbacks are independent. What is the probability that an acceptable portion of the disk is eliminated as unacceptable for data storage?

2.60 A batch of 500 containers for frozen orange juice contains 5 that are defective.

Two are selected, at random, without replacement, from the batch. Let A and B denote the events that the first and second container selected is defective, respectively. (a) Are A and B independent events? (b) If the sampling were done with replacement, would A and B be independent?

0.9 0.8 0.7

0.95 0.95 0.95

0.9 0.8 0.7

0.95 0.95 0.95

Probability 43



Bayes’ theorem The general multiplication rules are useful in solving many problems in which ultimate outcome of an experiment depends on the outcomes of various intermediate stages. Suppose, for instance, that an assembly plant receives its voltage regulators from three suppliers, 60% from supplier B1, 30% from supplier B2, and 10% from supplier B3. Also suppose that 95% of regulators supplied by B1, 80% of those of B2, and 65% of those of B3 perform according to specifications. First of all, we would like to know the probability that any one of the voltage regulator received by the plant will work according to specification. If A denotes the event that a voltage regulator received by the plant works according to specifications, and B1, B2, and B3 that it comes from the respective suppliers, we can write

A = A ∩ (B1 ∪ B2 ∪ B3)

= (A ∩ B1) ∪ (A ∩ B2) ∪ (A ∩ B3)

The events B1, B2, and B3 are mutually exclusive events; therefore the event A ∩ B1, A ∩ B2, and A ∩ B3 must also be mutually exclusive. Therefore, we can write

p(A) = p(A ∩ B1) + p(A ∩ B2) + p(A ∩ B3)

Using equation (3), we can write

p(A) = p(B1) p(A | B1) + p(B2) p(A | B2) + p(B3) p(A | B3) (6)

Substituting the number from the problem into the above equation, we obtain

p(A) = (0.60)(0.95) + (0.30)(0.80) + (0.10)(0.65)

= 0.875

This is the probability that any one voltage regulator received by the given plant will perform according to specifications. We want to turn the table around and ask a question that if a voltage regulator is known to perform according to specifications, what is the probability that the regulator has come from a particular supplier. Symbolically, we want to determine p(B1 | A), p(B2 | A), and p(B3 | A). To do this, we first write

)()()|( 1

1 ApBApABp ∩

=

Using equations (3) and (6), we now write

)|()()|()()|()()|()()|(

332211

111 BApBpBApBpBApBp

BApBpABp⋅+⋅+⋅

⋅=

Similarly, for other quantities, we obtain

)|()()|()()|()()|()()|(

332211

222 BApBpBApBpBApBp

BApBpABp⋅+⋅+⋅

⋅=

)|()()|()()|()()|()(

)|(332211

333 BApBpBApBpBApBp

BApBpABp

⋅+⋅+⋅⋅

=

Substituting all the values of probabilities, we obtain

44 Probability


651.065.010.080.030.095.060.0

95.060.0)|( 1 =×+×+×

×=ABp

274.065.010.080.030.095.060.0

80.030.0)|( 2 =×+×+×

×=ABp

074.065.010.080.030.095.060.0

65.010.0)|( 3 =×+×+×

×=ABp

We should use care to interpret the above result. As an example if we consider the supplier B2, our input data tells us that 30% of the items are supplied by them, of which 80% work according to specification. The above result tells us that if a voltage regulator, that is working according to specifications, is chosen at random, there is a 27.4% possibility that the item has been delivered by B2. Similar interpretations can be made for other suppliers also. In many texts, p(B1 | A) is referred as a-priori probability, and p(A | B1) is referred to as a-posteriori probability. The method used to solve the preceding example can be easily generalized. For a particular experiment, let the space, S be

S = {A1, A2, . . . An}

The events A1, A2, etc are all mutually exclusive. Further, let B be an arbitrary event in the S. Then, we can write

B = B ∩ S = B ∩ (A1 ∪ A2 ∪ . . . ∪ An) = (B ∩ A1) ∪ (B ∩ A2) ∪ . . . ∪ (B ∩ An)

Since A1, A2 are all mutually exclusive events, for p(B), we can write

p(B) = p(B ∩ A1) + p(B ∩ A2) + . . . + p(B ∩ An)

Now using equation (3)

p(B) = p(B | A1) p(A2) + p(B | A2) p(A2) + . . . + p(B | An) p(An)

= ∑=

n

iii ApABp

1)()|(

Our target is to find p(Aj | B), which we write as

∑=

=∩

=∩

= n

iii

jjjjj

ApABp

ApABpBp

ABpBp

BApBAp

1

)()|(

)()|()(

)()(

)()|( (7)

Example 2.18 Four technicians regularly make repairs when breakdowns occur on an automated production line. Technician A services 20% of the breakdowns, makes an incomplete repair 1 time in 20; technician B services 60% of the breakdowns, makes an incomplete repair 1 in 10; technician C services 15% of the breakdowns, makes an incomplete repair 1 in 10; and technician D services 5% of the breakdowns, and make an incomplete repair 1 time in 20. If the supervisor detects an incomplete

Probability 45



repair, find the probability of that incomplete repair being done by each technician. Let us define E be the event that the repair was incomplete, T1 that the initial repair was done by A, T2 that the initial repair was done by B, T3 that the initial repair was done by C and T4 that initial repair was done by D. We will be using equation (7) for our problem. Therefore, let us find the denominator first.

)()|()()|(

)()|()()|()()|(

4433

22111

TpTEpTpTEp

TpTEpTpTEpTpTEpn

iii

++

+=∑=

= (0.20)(0.05) + (0.60)(0.10) + (0.15)(0.10) + (0.05)(0.05)

= 0.0875

Therefore

1143.00875.0

)05.0)(20.0(

)()|(

)()|()|(

1

111 ===

∑=

n

iii TpTEp

TpTEpETp

6857.00875.0

)10.0)(60.0(

)()|(

)()|()|(

1

222 ===

∑=

n

iii TpTEp

TpTEpETp

1714.00875.0

)10.0)(15.0(

)()|(

)()|()|(

1

333 ===

∑=

n

iii TpTEp

TpTEpETp

0286.00875.0

)05.0)(05.0(

)()|(

)()|()|(

1

444 ===

∑=

n

iii TpTEp

TpTEpETp

Example 2.19 A laboratory test is 99% effective in detecting a certain disease when it is, in fact, present. However, the test also yields a ‘false positive’ result for 1% of the healthy persons tested. If 0.5% of the population actually has the disease, what is the probability a person has the disease given that his test result is positive? Let D be the event that the tested person has the disease and E the event that his test results are positive. The desired probability p(D | E) is obtained by

)()|()()|(

)()|()|( cc DpDEpDpDEpDpDEpEDp

+=

3322.0)995.0)(01.0()005.0)(99.0(

)005.0)(99.0(=

+=

Thus only 33% of those persons whose test results are positive actually have the disease. Since most people are often surprised at this result, it is probably worthwhile to present a second argument which, though less rigorous than the foregoing, is

46 Probability


probably more revealing. We do that now. Since 0.5% of the population actually has the disease, it follows that, on the average, 1 person out of every 200 tested will have it. The test will correctly confirm that this person has the disease with a probability of 0.99. Thus on the average, out of every 200 person test, the test will correctly confirm that 0.99 person has the disease. On the other hand, out of 199 healthy people, the test will incorrectly state that 199 × 0.01 of these people have the disease. Hence for every 0.99 diseased person that the test correctly states ill, there are 1.99 healthy persons that the test incorrectly states ill. Hence the proportion of time that the test result is correct when it states that the person is ill is

3322.099.199.0

99.0=

+=

Example 2.20 Consider two urns. The first contains two white and seven black balls, and the second contains five white and six black balls. We flip a fair coin and then draw a ball from the first urn or the second urn depending on whether the outcome was heads or tails. What is the conditional probability that the outcome of the toss was heads given that a white ball was selected? Let W be the event that a white ball is drawn, and let H be the event that the coin comes up heads. The desired probability P(H|W) may be calculated as follows:

)()()|(

)()()|(

WpHpHWp

WpWHpWHp =

∩=

6722

21

115

21

92

21

92

)()|()()|()()|(

=+

=+

= cc HpHWpHpHWpHpHWp

Example 2.21 In answering a question on a multiple-choice test, a student either knows the answer or she guesses. Let p be the probability that she knows the answer and 1 − p the probability that she guesses. Assume that a student who guesses at the answer will be correct with probability 1/m, where m is the number of multiple-choice alternatives. What is the conditional probability that a student knew the answer to a question given that she answered it correctly? Let C and K denote, respectively, the events that the student answers the question correctly and the event that she actually knows the answer. To compute

)()()|(

CpCKpCKp ∩

=

we first note that ppKpKCpCKp =⋅==∩ 1)()|()(

Probability 47



To compute the probability that the student answers correctly, we condition on whether or not she knows the answer. That is,

p(C) = p(C | K )p(K )+p(C | Kc)p(Kc) = p + (1/m)(1 − p)

Hence, the desired probability is given by

pmmp

pm

p

pCKp)1(1)1(1

)|(−+

=−+

=

Thus, for example, if m = 5, p = 21 , then the probability that a student knew the

answer to a question she correctly answered is 65 .

Exercise 2.61 Software to detect fraud in consumer phone cards tracks the number of

metropolitan areas where calls originate each day. It is found that 1% of the legitimate users originate calls from two or more metropolitan areas in a single day. However, 30% of fraudulent users originate calls from two or more metropolitan areas in a single day. The proportion of fraudulent users is 0.01%. If the same user originates calls from two or more metropolitan areas in a single day, what is the probability that the user is fraudulent?

2.62 Semiconductor lasers used in optical storage products require higher power

levels for write operations than for read operations. High-power-level operations lower the useful life of the laser. Lasers in products used for backup of higher speed magnetic disks primarily write, and the probability that the useful life exceeds five years is 0.95. Lasers that are in products that are used for main storage spend approximately an equal amount of time reading and writing, and the probability that the useful life exceeds five years is 0.995. Now, 25% of the products from a manufacturer are used for backup and 75% of the products are used for main storage. Let A denote the event that a laser’s useful life exceeds five years, and let B denote the event that a laser is in a product that is used for backup. Determine the following: (a) p(B) (b) p(A | B) (c) p(A | B′) (d) p(A ∩ B) (e) p(A ∩ B′) (f) p(A) (g) What is the probability that the useful life of a laser exceeds five years? (h) What is the probability that a laser that failed before five years came from a

product used for backup? 2.63 Customers are used to evaluate preliminary product designs. In the past, 95% of

highly successful products received good reviews, 60% of moderately successful products received good reviews, and 10% of poor products received

48 Probability


good reviews. In addition, 40% of products have been highly successful, 35% have been moderately successful, and 25% have been poor products. (a) What is the probability that a product attains a good review? (b) If a new design attains a good review, what is the probability that it will be a

highly successful product? (c) If a product does not attain a good review, what is the probability that it will

be a highly successful product? 2.64 An inspector working for a manufacturing company has a 99% chance of

correctly identifying defective items and a 0.5% chance of incorrectly classifying a good item as defective. The company has evidence that its line produces 0.9% of nonconforming items. (a) What is the probability that an item selected for inspection is classified as

defective? (b) If an item selected at random is classified as non-defective, what is the

probability that it is indeed good? 2.65 A new analytical method to detect pollutants in water is being tested. This new

method of chemical analysis is important because, if adopted, it could be used to detect three different contaminants—organic pollutants, volatile solvents, and chlorinated compounds—instead of having to use a single test for each pollutant. The makers of the test claim that it can detect high levels of organic pollutants with 99.7% accuracy, volatile solvents with 99.95% accuracy, and chlorinated compounds with 89.7% accuracy. If a pollutant is not present, the test does not signal. Samples are prepared for the calibration of the test and 60% of them are contaminated with organic pollutants, 27% with volatile solvents, and 13% with traces of chlorinated compounds. A test sample is selected randomly. (a) What is the probability that the test will signal? (b) If the test signals, what is the probability that chlorinated compounds are

present.

Review exercise 2.66 A box contains three marbles — one red, one green, and one blue. Consider an

experiment that consists of taking one marble from the box, then replacing it in the box and drawing a second marble from the box. Describe the sample space. Repeat for the case in which the second marble is drawn without first replacing the first marble.

2.67 An experiment consists of tossing a coin three times. What is the sample space

of this experiment? Which event corresponds to the experiment resulting in more heads than tails?

2.68 Two dice are thrown. Let E be the event that the sum of the dice is odd, let F be

the event that the first die lands on 1, and let G be the event that the sum is 5. Describe the events E ∩ F, E ∪ F, F ∩ G, E ∩ Fc , E ∩ F ∩ G.

Probability 49



2.69 Let E, F, G be three events. Find expressions for the events that of E, F, G (a) only E occurs; (b) both E and G but not F occur; (c) at least one of the events occurs; (d) at least two of the events occur; (e) all three occur; (f) none of the events occurs; (g) at most one of them occurs; (h) at most two of them occur; (i) exactly two of them occur; (j) at most three of them occur. 2.70 Find simple expressions for the events (a) E ∪ Ec; (b) E ∩ Ec; (c) (E ∪ F)(E ∪ Fc); (d) (E ∪ F)(Ec ∪ F)(E ∪ Fc); (e) (E ∪ F)(F ∪ G). 2.71 Prove that (a) p(E ∩ Fc) = p(E ) − p(E ∩ F) (b) p(Ec ∩ Fc) = 1 − p(E) − p(F) + p(E ∩ F ) 2.72 Show that the probability that exactly one of the events E or F occurs is equal to

p(E) + p(F) – 2p(E ∩ F). 2.73 A group of 5 boys and 10 girls is lined up in random order — that is, each of the

15! permutations is assumed to be equally likely. (a) What is the probability that the person in the 4th position is a boy? (b) What about the person in the 12th position? (c) What is the probability that a particular boy is in the 3rd position? 2.74 A town contains 4 television repairmen. If 4 sets break down, what is the

probability that exactly 2 of the repairmen are called? What assumptions are you making?

2.75 A woman has n keys, of which one will open her door. If she tries the keys at

random, discarding those that do not work, what is the probability that she will open the door on her kth try? What if she does not discard previously tried keys?

2.76 A closet contains 8 pairs of shoes. If 4 shoes are randomly selected, what is the

probability that there will be (a) no complete pair and (b) exactly 1 complete pair?

2.77 Of three cards, one is painted red on both sides; one is painted black on both

sides; and one is painted red on one side and black on the other. A card is randomly chosen and placed on a table. If the side facing up is red, what is the probability that the other side is also red?

50 Probability


2.78 A couple has 2 children. What is the probability that both are girls if the eldest is a girl?

2.79 Fifty-two percent of the students at a certain college are females. Five percent of

the students in this college are majoring in computer science. Two percent of the students are women majoring in computer science. If a student is selected at random, find the conditional probability that

(a) this student is female, given that the student is majoring in computer science; (b) this student is majoring in computer science, given that the student is female. 2.80 A total of 500 married working couples were polled about their annual salaries,

with the following information resulting. Husband

Wife Less than $25,000 More than $25,000

Less than $25,000 212 198

More than $25,000 36 54 Thus, for instance, in 36 of the couples the wife earned more and the husband earned less than $25,000. If one of the couples is randomly chosen, what is

(a) the probability that the husband earns less than $25,000; (b) the conditional probability that the wife earns more than $25,000 given

that the husband earns more than this amount; (c) the conditional probability that the wife earns more than $25,000 given

that the husband earns less than this amount? 2.81 There are two local factories that produce radios. Each radio produced at factory

A is defective with probability .05, whereas each one produced at factory B is defective with probability 0.01. Suppose you purchase two radios that were produced at the same factory, which is equally likely to have been either factory A or factory B. If the first radio that you check is defective, what is the conditional probability that the other one is also defective?

2.82 A red die, a blue die, and a yellow die (all six-sided) are rolled. We are

interested in the probability that the number appearing on the blue die is less than that appearing on the yellow die which is less than that appearing on the red die. (That is, if B (R) [Y] is the number appearing on the blue (red) [yellow] die, then we are interested in p(B < Y < R).)

(a) What is the probability that no two of the dice land on the same number? (b) Given that no two of the dice land on the same number, what is the

conditional probability that B < Y < R? (c) What is p(B < Y < R)? (d) If we regard the outcome of the experiment as the vector B, R, Y, how many

outcomes are there in the sample space? (e) Without using the answer to (c), determine the number of outcomes that

result in B < Y < R. (f ) Use the results of parts (d) and (e) to verify your answer to part (c). 2.83 You ask your neighbor to water a sickly plant while you are on vacation.

Without water it will die with probability 0.8; with water it will die with

Probability 51



probability 0.15. You are 90 percent certain that your neighbor will remember to water the plant.

(a What is the probability that the plant will be alive when you return? (b) If it is dead, what is the probability your neighbor forgot to water it? 2.84 Two balls, each equally likely to be colored either red or blue, are put in an urn.

At each stage one of the balls is randomly chosen, its color is noted, and it is then returned to the urn. If the first two balls chosen are colored red, what is the probability that

(a) both balls in the urn are colored red; (b) the next ball chosen will be red? 2.85 A total of 600 of the 1,000 people in a retirement community classify

themselves as Republicans, while the others classify themselves as Democrats. In a local election in which everyone voted, 60 Republicans voted for the Democratic candidate, and 50 Democrats voted for the Republican candidate. If a randomly chosen community member voted for the Republican, what is the probability that she or he is a Democrat?

2.86 Each of 2 balls is painted black or gold and then placed in an urn. Suppose that

each ball is colored black with probability 12, and that these events are independent.

(a) Suppose that you obtain information that the gold paint has been used (and thus at least one of the balls is painted gold). Compute the conditional probability that both balls are painted gold.

(b) Suppose, now, that the urn tips over and 1 ball falls out. It is painted gold. What is the probability that both balls are gold in this case? Explain.

2.87 Each of 2 cabinets identical in appearance has 2 drawers. Cabinet A contains a

silver coin in each drawer, and cabinet B contains a silver coin in one of its drawers and a gold coin in the other. A cabinet is randomly selected, one of its drawers is opened, and a silver coin is found. What is the probability that there is a silver coin in the other drawer?

2.88 Prostate cancer is the most common type of cancer found in males. As an

indicator of whether a male has prostate cancer, doctors often perform a test that measures the level of the PSA protein (prostate specific antigen) that is produced only by the prostate gland. Although higher PSA levels are indicative of cancer, the test is notoriously unreliable. Indeed, the probability that a noncancerous man will have an elevated PSA level is approximately .135, with this probability increasing to approximately .268 if the man does have cancer. If, based on other factors, a physician is 70 percent certain that a male has prostate cancer, what is the conditional probability that he has the cancer given that

(a) the test indicates an elevated PSA level; (b) the test does not indicate an elevated PSA level?

Repeat the preceding, this time assuming that the physician initially believes there is a 30 percent chance the man has prostate cancer.

2.89 Suppose that an insurance company classifies people into one of three classes —

good risks, average risks, and bad risks. Their records indicate that the

52 Probability


probabilities that good, average, and bad risk persons will be involved in an accident over a 1-year span are, respectively, 0.05, 0.15, and 0.30. If 20 percent of the population are “good risks,” 50 percent are “average risks,” and 30 percent are “bad risks,” what proportion of people have accidents in a fixed year? If policy holder A had no accidents in 1987, what is the probability that he or she is a good (average) risk?

2.90 A pair of fair dice is rolled. Let E denote the event that the sum of the dice is

equal to 7. (a) Show that E is independent of the event that the first die lands on 4. (b) Show that E is independent of the event that the second die lands on 3. 2.91 The probability of the closing of the ith relay in the circuits shown is given by pi

, i = 1, 2, 3, 4, 5. If all relays function independently, what is the probability that a current flows between A and B for the respective circuits?

2.92 An engineering system consisting of n components is said to be a k-out-of n

system (k ≤ n) if the system functions if and only if at least k of the n components function. Suppose that all components function independently of each other. (a) If the ith component functions with probability pi , i = 1, 2, 3, 4, compute the

probability that a 2-out-of-4 system functions. (b) Repeat (a) for a 3-out-of-5 system.

2.93 Five independent flips of a fair coin are made. Find the probability that (a) the first three flips are the same; (b) either the first three flips are the same, or the last three flips are the same; (c) there are at least two heads among the first three flips, and at least two tails

among the last three flips. 2.94 Suppose that n independent trials, each of which results in any of the outcomes

0, 1, or 2, with respective probabilities .3, .5, and .2, are performed. Find the probability that both outcome 1 and outcome 2 occur at least once. (Hint: Consider the complementary probability.)

A B1 2 3

4 5(a)

BA 1 2

3 4

5

(b)

BA

1 4

2 5

3

(c)

Probability 53



2.95 A parallel system functions whenever at least one of its components works. Consider a parallel system of n components, and suppose that each component independently works with probability 1/2. Find the conditional probability that component 1 works, given that the system is functioning.

2.96 A certain organism possesses a pair of each of 5 different genes (which we will

designate by the first 5 letters of the English alphabet). Each gene appears in 2 forms (which we designate by lowercase and capital letters). The capital letter will be assumed to be the dominant gene in the sense that if an organism possesses the gene pair xX, then it will outwardly have the appearance of the X gene. For instance, if X stands for brown eyes and x for blue eyes, then an individual having either gene pair XX or xX will have brown eyes, whereas one having gene pair xx will be blue-eyed. The characteristic appearance of an organism is called its phenotype, whereas its genetic constitution is called its genotype. (Thus 2 organisms with respective genotypes aA, bB, cc, dD, ee and AA, BB, cc, DD, ee would have different genotypes but the same phenotype.) In a mating between 2 organisms each one contributes, at random, one of its gene pairs of each type. The 5 contributions of an organism (one of each of the 5 types) are assumed to be independent and are also independent of the contributions of its mate. In a mating between organisms having genotypes aA, bB, cC, dD, eE, and aa, bB, cc, Dd, ee, what is the probability that the progeny will (1) phenotypically, (2) genotypically resemble (a) the first parent; (b) the second parent; (c) either parent; (d) neither parent?

2.97 Three prisoners are informed by their jailer that one of them has been chosen at

random to be executed, and the other two are to be freed. Prisoner A asks the jailer to tell him privately which of his fellow prisoners will be set free, claiming that there would be no harm in divulging this information because he already knows that at least one of the two will go free. The jailer refuses to answer this question, pointing out that if A knew which of his fellow prisoners were to be set free, then his own probability of being executed would rise from 13 to 12because he would then be one of two prisoners. What do you think of the jailer’s reasoning?

2.98 Although both my parents have brown eyes, I have blue eyes. What is the

probability that my sister has blue eyes?

lecture 02 iub mat 212

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