251Solnh3 2/13/08 (Open This Document in 'Page Layout' View!)

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H. Introduction to Probability

1. Experiments and Probability

Text problems 4.1, 4.2.

2. The Venn Diagram and the Addition Rule.

Downing & Clark, pg. 96 (pg 85 in 3rd ed) Basics 1, Application 2, 13. H1 (H0A), Text problems 4.3, 4.4, 4.10, 4.11, 4.8!, 4.9!.[4.3, 4.4, 4.8*, 4.9*]., (4.3,4.4, 4.8*, 4.9*)., H4, H5 (H1, H2).

3. Conditional and Joint Probability, Bayes’ Rule.

Text 4.16a-c, 4.18 [4.14a-c, 4.16] (4.13a-c, 4.15). H2, H3 (H0B, H0C), D&C pg 113 (pg 103 in 3rd ed) 14 (Note error in text - 5/6 of the people in the city support Jones, 5/9 of the people in the country support Jones), 15, 16. H6 (H3).

4. Statistical Independence.

Text 4.16d, 4.22, 4.21!, 4.24, 4.30, 4.31, 4.33 [4.14d*, 4.19*, 4.20*, 4.22*, 4.28, 4.29, 4.31, 4.68*] (4.18*, 4.19*, 4.21*, 4.26, 4.27, 4.29). H8, H9 (H5, H6). pg. 97( pg. 85 in 3rd ed) Applications 4, 5, 8, 9, 10, 11, 44. H7(H4).

5. Review.

Section 4 is in this document.

Remember and are statistically independent if and that this implies that

Exercise 4.16d [4.14d in 9th] (Not in 8th edition):The text gave the following contingency table . We found the following probabilities: (a) =.3333; (b) =.3333 and (c) = .6667. Are the events and statistically independent?

Solution:We say thatthe events and statistically independent if P(A |B) =. Alternately, we can note that if the events and statistically independent.

Note that we found the following. This represents . Since we found P(A |B) = P(A |) = 0.3333, it is extremely obvious that the occurrence of the event does not in any way depend on whether event occurs, so they are statisticallyindependent. We can also note that every number on the inside of the joint probability table is the product of the numbers on the outside. We can also see that the second column is proportional to the first and that the second row is proportional to the first. These are all symptoms of independence.

Exercise 4.22 [4.19 in 8th] (Not in 8th edition):

The table reads:

Condition of Die
Quality / No Particles / Particles / Total
Good / 320 / 14 / 334
Bad / 80 / 36 / 116
Total / 400 / 50 / 450

The table represents the condition of the dies used and the acceptability of 450 wafers. ‘Particles’ means that particles were on the die that produced a given wafer. A wafer can be classified as ‘Good’ or ‘Bad.’ a) If a wafer is bad, what is the probability that it was produced from a die that had particles? b) If a wafer is good, what is the probability that it was produced from a die that had particles? c) Are the two events ‘Particles’ and ‘Good’ independent?

Solution:Define the following events:

“Yes, there are particles.”

“There are no particles.”

“The wafer is good.”

“The wafer is bad.”

If we use the events above, we get the table below. If we divide by 450, we get the second table.

According to the Instructor’s Solutions Manual(edited)

(a)P(had particles | bad) = 36/116 = 0.3103

(b)P(had particles | good) = 14/334 = 0.0419

(c)P(no particles | good ) = 320/334 = 0.9581

P(no particles) = 400/450 = 0.8889

Since P(no particles | good ) P(no particles), “a good wafer” and “a die with no particle” are not statistically independent.

Exercise 4.21 [Not in 8thor 9th]: 2000 community members were sampled with the following results:

Drives to Work / Homeowner / Renter / Total
Yes / 824 / 681 / 1505
No / 176 / 319 / 495
Total / 1000 / 1000 / 2000

Find a) the probability that someone who drives to work is a homeowner, b) the probability that a homeowner drives to work, c) the difference between a) and b) and d) whether driving to work and being a homeowner are independent.

Solution: This table was made into a joint probability table by dividing through by 2000.Use for homeowner, for renter and for “drives to work.”

So, for example . This is the probability that a respondent is both a homeowner and drives to work.

According to the Instructor’s Solutions Manual(edited)

4.21(a) P (a homeowner | drives to work) = 824/1505= 0.5475 or

(b) P (drives to work | a homeowner) = 824/1000 = 0.8240 or

(d) Since P (a homeowner) = 1000/2000 = 0.50 is not equal to P (a homeowner | drives to work) = 824/1505 = 0.5475, driving to work and whether the respondent is a homeowner or a renter are not statistically independent.

Exercise [4.20 in 9th] (Not in 8th edition):The table reads:

Book Airline Tickets on the
Internet?
Research Airline Tickets on the Internet? / Yes / No / Total
Yes / 88 / 124 / 212
No / 20 / 168 / 188
Total / 108 / 292 / 400

Find a) the probability that someone who researches ticket prices on the internet books on the internet, b) the probability that someond who books on the internet researches prices on the internet and c) the difference between a) and b).

Solution:Define the following events:

“Books airline tickets on the Internet.”

“Does notbook airline tickets on the Internet”

“Researches airline prices on the Internet.”

“Does not research airline prices on the Internet.”

If we use the events above, we get the table below. If we divide by 400, we get the second table.

According to the Instructor’s Solutions Manual(edited)

(a) P(book tickets on the internet | research ticket prices on the internet) = 88/212= 0.4151

(b) P(researches ticket prices on the internet | book tickets on the internet) = 88/108 = 0.8148

(d) Since P(book tickets on the internet) = 108/400 = 0.27 is not equal to P(book tickets on the internet | research ticket prices on the internet) = 88/212 = 0.4151, researching airline ticket prices on the internet and booking airline tickets on the internet are not statistically independent.

Exercise 4.24 [4.22 in 9th] (Not in 8th edition):Of 56 white workers terminated, 29 claimed bias. Of 407 black workers terminated, 126 claimed bias. Find a) the probability that a white worker claims bias,b) the probability that a worker who has claimed bias is white, c) the difference between the meaning of a) and b) and d) whether claiming bias and being white are independent.

Solution:Given: 56 + 407 = 463 workers were terminated. Of these 56 were white, so 56 out of 463 or 12.10% were white. 29 + 126 =155 claimed bias.

According to the Instructor’s Solutions Manual

(a)P(claimed bias | white ) = 29/56 = 0.5179

(b)P(white | claim bias) = 29/155 = 0.1871

(c)The conditional events are reversed.

(d)Since P(white | claim bias) = 0.1871 is not equal to P(white) = 0.1210, being white and claiming bias are not statistically independent.

As you may have guessed, I would be much happier with a formal solution using tables. Let us define the following events: - ‘white’ and ‘claims bias’ The problem says ‘Of 56 white workers terminated, 29 claimed bias’ so It also says “Of 407 black workers terminated, 126 claimed bias” or It asks for . If we do this as a table and use numbers rather than probabilities we find. You can read the probabilities from the table.

In terms of probabilities, we may recall that This means that Bayes rule says . We need to find So .

Exercise4.30 [4.28 in 9th] (4.26 in 8th edition):If , , find .

Solution:According to the Instructor’s Solutions Manual(using for )

Exercise 4.31 [4.29 in 9th] (4.27 in 8th edition):If , , find .

Solution:According to the Instructor’s Solutions Manual

Exercise 4.33 [4.31 in 9th] (4.29 in 8th edition):The problem says that husbands watch television in prime time 60% of the time. When the husband is watching TV, the wife is watching 40% of the time and when the husband is not watching television, the wife is watching TV 30% of the time. It asks for a) the probability that, if the wife is watching TV, the husband is also watching and b) the probability that the wife is watching.

Solution:According to the Instructor’s Solutions Manual, If we define the following, H = husband watching and W = wife watching, the facts that are given in the problem are and . We deduce that

(a)

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To do this by the box method, note that of 100 husbands 60%, or 60 husbands are watching in prime time. So 40 are not watching.

Of the 60 wives whose husbands who are watching television, 40% or 24 are also watching television. Of the 40 wives whose husbands are not watching television, 30% or 12 are watching television.

If we just add up the stuff in the box, and fill in the blanks we get a complete table. We find that 36 wives are watching TV. 24 of these have husbands that are also watching, so that our conditional probability is

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(b)P(W) = 0.24 + 0.12 = 0.36. Using the box, we find that 36 women out of 100 are watching.

Exercise 4.68 (Not in 10th or 8th edition): In 1997, 24.0% of all highway fatalities involved a rollover. 15.8% of all accidents involving a rollover involved SUVs, vans and pickups. Given that a rollover was not involved, 5.6% of fatalities involved SUVs, vans and pickups. Define the following events:

{Fatality involved an SUV, van or pickup}

{Fatality involved a rollover}

a. Use Bayes’ theorem to find the probability that the fatality involved a rollover, given that the

fatality involved an SUV, van or pickup.

b. Compare the result in (a) to the probability that the fatality involved a rollover, and comment on

whether on whether SUVs, vans and pickups are more prone to rollover accidents.

Solution:Given:

Try this: 24% of all highway fatalities involved a rollover. 76% of fatalities did not involve a rollover. Since 15.8% of rollover accidents involved an SUV etc., out of 100 accidents 24 involved a rollover and (.158) 24 = 3.792 involved both a rollover and an SUV. Also, out of the 76 accidents out of 100 that did not involve a rollover, 5.6% or (.056) 76 = 4.256 involved an SUV. Thus 4.256 + 3.792 = 8.048 accidents involved an SUV and 3.792 also had a rollover, so the fraction of SUV accidents that involved a rollover was 3.792 out of 8.048 or 47.12%.

Bayes’ rule says . And

= .08048.

According to the Instructor’s Solutions Manual

4.65(a) P(B | A) = (0.158)(0.24)/((0.158)(0.24)+(0.056)(0.76) = 0.4712

(b) Since the probability of a fatality involved a rollover given that the fatality involved an SUV, van or pickup is 0.4712, which is almost twice the probability of a fatality involved a rollover with any vehicle type at 0.24, SUV's, vans or pickups are generally more prone to rollover accidents.

PROBLEM H8: Explain why mutually exclusive events are also dependent events.

Solution:If and are mutually exclusive . If they are independent . If and and and are independent, can be zero? Think about it!

PROBLEM H9:(McClave, Benson and Sincich): Flip a coin three times. Define the following events: ,, and . You have seen a tree diagram for this in class that showed that . We can do this in a more sophisticated way now. Let be a head on the first try, be a head on the second try and be a head on the third try. These three events are independent so, by the extended Multiplication Rule, . This applies to all the possible events, which are ,,,,,, and . All are .

a) Find ,,,,,, and .

b) Use your answers in a) to calculate , and .

c) Which pairs of events are independent?

Solution:a) ,, (Since heads and tails are equally likely),,, and .

c) Testing for independence.

Pair / Conditional Probability / Probability

Note that the definition of independence for and is . Because of the inequality, none of the above pairs are independent.

Downing and Clark, pg. 85, Application 4: The probability of at least one head on the first three of 5 coin flips.

Solution: You only need to consider the first three tosses. We have shown in class that . Now we want .

Downing and Clark, pg. 85, Application 5: If you roll a die 3 times, what is the chance of a 1 on at least one of the 3 tosses.

Solution: The probability of 1 on 1 toss is , so the probability of not getting 1 on 1 toss is . The probability of not getting 1 on all 3 tosses is . Since a 1 on at least one of the 3 tosses is the complement of not getting 1 on all 3 tosses, its probability is .

Downing and Clark, pg. 85, Application 8: Find the probability of at least one head on flips. Let . Then the probability of all tails on flips is

. So the probability of not getting all tails is .

Downing and Clark, pg. 85, Application 9: If your 2 favorite prime-time shows are on different networks and there are 21 prime-time slots, what is the chance that they will both be on at the same time?

Solution: Arbitrarily assume that the first show is on at 8PM on Monday. Since this is one of the 21 slots, the chance that the other is on at the same time is . This is actually the answer to the problem, since it would be true no matter what time the first show was on. Note: If you look at the next two problems, you will see that the probability that both shows appear at 8PM on Monday is , but you have to do this for all 21 possible slots, so the answer is .

Downing and Clark, pg. 85, Application 10: If your 2 favorite prime-time shows are on different networks, what is the chance that they will both be on at the same night?

Solution: By the same logic as in the previous problem, the answer is .
Downing and Clark, pg. 85, Application 11: If your 2 favorite prime-time shows are on different networks, what is the chance that they will both be shown on Monday?

Solution: The probability that the first show will be televised on Monday is . The probability that the second show will appear on the same day is also as we showed in the previous problem. Therefore the joint probability of the two events occurring is

Downing and Clark, pg. 85, Application 44: If you roll a die times, what is the probability that at least one 1 will turn up?

Solution: ‘At least one’ is a code word for ‘not none!’ On any one roll the probability of not rolling a 1 is so the probability of not rolling a 1 on tries is The probability of the complement of this event is

If you want to work this out, the probability rises surprisingly slowly. If , the probability is If , the probability is . If , the probability is . If , the probability is . Only when , does the probability rise to 99%.

PROBLEM H7:A machine has two components with the following probabilities of failure:

Day / Component 1 / Component 2
1 / .25 / 0
2 / .25 / .50
3 / .25 / .50
4 / .25 / 0

Notice that, since neither component has a life of more than four days, the machine cannot last beyond day 4 if it requires both or either component to function. Thus the probabilities of it failing on the four days must add to 1.

a. If the machine requires both components to operate, it cannot fail on day 4, sincecomponent 2 cannot last beyond day 3. What is the probability of the machine failing onday 1? day 2? day 3?

b. Instead of assuming that the machine needs both components to work, assume that itwill operate as long as either component 1 or component 2 is working. Notice that this means that it cannot fail on day 1. What about the probability of failure on day 2? day 3?day 4?

Solution: Let the table below define events .

Day / Component 1
Fails / Component 2
Fails
1 /
2 / /
3 / /
4 /

For Example, Event is 'Component 1 fails on Day 2.' Though Events are mutually exclusive, as are events , events involving component 1 are independent of events involving component 2. This means for example that . In Part a of the problem, where the machine needs both components to work, if event occurs, the machine fails on day 2. If we check each possible joint event, we find that the event 'Machine fails on Day 2,' is the union of Events ,,and . Because these joint events are mutually exclusive, we can find the probability of their union by adding their probabilities. The table below names the joint events, gives their probabilities and tells when the machine fails under the assumptions of Part a and Part b.

Joint Event / Probability / Under Assumptions of Part a, Machine fails on day / Under Assumptions of Part b, Machine fails on day
/ .125 / 1 / 2
/ .125 / 2 / 2
/ .125 / 2 / 3
/ .125 / 2 / 4
/ .125 / 1 / 3
/ .125 / 2 / 3
/ .125 / 3 / 3
/ .125 / 3 / 4

Note that in Part a the component that fails first determines when the machine fails, while in Part b, the component that fails last determines when the machine fails.

The final step is to add together all the probabilities associated with each day, so that, for example, there are two joint events that lead to failure on day 2 in Part b. This means that the probability of machine failure on day 2 is . The last table summarizes the results.

Day / Probability of Failure under
Assumptions of Part a / Probability of Failure under
Assumptions of Part b.
1 / .250 / 0
2 / .500 / .250
3 / .250 / .500
4 / 0 / .250

If we handle this with a joint probability table we can write it as:

Events / / / Total
/ .125 / .125 / .25
/ .125 / .125 / .25
/ .125 / .125 / .25
/ .125 / .125 / .25
Total / .50 / .50 / 1.00

In Part a, where both components are needed for the machine to operate the machine will fail on day 1 only if event occurs. It will fail on the third day only if both components last until the third day, that is only if event occurs at the same time as or . The probability of event is .25 and the probability of either event or is .125+.125=.25. So the probability of failure on day 2 is .50. Similar reasoning can be used in part b.

Appendix: Solutions to problems that were in the 8th edition.

Exercise 4.18: Again, we have a contingency table.

According to the Instructor’s Solutions Manual(edited)

(a)P(has travel/entertainment credit card | has bank credit card) = 60/120 = 1/2 = 0.5

(b)P(has bank credit card | does not have travel/entertainment credit card) = 60/125

= 12/25 = 0.48

(c)Since P(has travel/entertainment credit card | has bank credit card) = 60/120 or 0.5 and

P(has travel/entertainment credit card) = 75/200 or 0.375, the two events are not statistically independent

Exercise 4.19: Again, we have a contingency table.

According to the Instructor’s Solutions Manual(edited)

(a)P(did not received product in time | satisfied) = 33/1230 = 0.0269

(b) P(satisfied | received product in time) = 1197/1324 = 0.904

(a) while it is known that the customer did receive the product in time for the holiday.

(d)Since P(satisfied) = 1230/1500 = 0.82 which is not equal to P(satisfied | received product in time) = 0.904, being satisfied with their experience and receiving their product in time for the holidays are not statistically independent.

Exercise 4.21:According to the Instructor’s Solutions Manual

(a) P(does not trade online | bullish) = = 240/585 = 0.410

(b) P(does not trade online | not bullish) = 260/415 = 0.627

= 240/585 = 0.410, being bullish on the market and the type of investor are not statistically independent.