CHAPTER 2
PROBABILITY
Introduction
This chapter introduces the concept of probability. It is a central part of statistics and one that gives many
students second thoughts about dropping the course. Some of the basic concepts you will find to be
straightforward (e.g., sample space, subjective and objective probabilities, complements of an event, etc.)
while other concepts (joint and conditional probabilities and Bayes’ theorem) may prove to be somewhat
confusing at first glance. Perseverance on your part will get you through the chapter and on to the more
application-oriented topics in the subsequent chapters.
Applicable Excel Templates used in this Chapter:
Bayes Revision.xls
Contingency Table.xls
Probability of at Least 1.xls
Permutation & Combination.xls
Applicable MegaStat commands:
MEGASTAT→ Probability → Counting Rules
Applicable MINITAB commands:
NONE
2-7.
Sample
Space
first toss
second toss > first
First Toss Second Toss > First
1
2
3
4
5
6
2
3
4
5
6
3
4
5
6
4
5
6
5
6
6
none
There are 36 possible outcomes tossing two dice.
There are 15 possible outcomes where the second
toss is greater than the first.
P(Second Toss > First) = 15/36 = 0.417
30
2-8.
Let R be “exposed to radio advertisement.” Let T be “exposed to television advertisement.”
a. Then RT represents the event that a randomly selected person will be exposed to either a
radio or a television advertisement, or both.
b. Then RT represents the event that a randomly selected person will be exposed to both a
radio and a television advertisement.
2-12.
We are given that 5 million Blackberry users were unable to use their devices. We also know
that there are 18 million users of handheld devices of this kind. If a user is chosen at random,
what is the probability that their device will not work?
P(device not working) = 5MK / 18M = 0.2778
2-13.
Continuing with the Blackberry problem in 2-12, 3 million of the 18 million users could not use
their devices as cellphones. An additional 1 million could not use their devices as either a
cellphone or a data device. Determine the probability of a randomly selected user not being able
to use their device as either a cellphone or a data device.
P(nonfunctioning cell phone) = 3/18 = 0.1667
P(nonfunctioning data device and nonfunctioning cell phone) = 1/18 = 0.0556
P(nonfunctioning data device or nonfunctioning cell phone) = 0.2778 + 0.1667 – 0.0556 =
0.3889
2-18.
Given that P(Detect 1st) = .98 and P(Detect 2nd) = .94 and P(Detect 1st and Detect 2nd) = .93,
compute the probability that at least one of the two will be detected (that is, the 1st or the 2nd or
both will be detected).
In general, P(A  B) = P(A) + P(B)  P(A  B).
Here, P(Detect 1st  Detect 2nd) =
P(Detect 1st) + P(Detect 2nd)  P(Detect 1st  Detect 2nd) = .98 + .94  .93 = 0.99.
2-20.
We are given age and sex data for 20 managers.
34F, 49M, 27M, 63F, 33F, 29F, 45M, 46M, 30F, 39M, 42M, 30F, 48M, 35F, 32F, 37F, 48F,
50M, 48F, 61F
A manager will be chosen at random.
a. Compute the probability the manager will be either a woman or over 50 years old, or both
 (P(F  50) = P(F) + P(50) - P(F  50)
=
2-24.
12 Females 2 people over fifty 2 Females over fifty 12
= 0.60.



20
20
20
20
Continuing with problem 2-12, determine the probability that a randomly chosen user could use
their Blackberry device.
From 2-12, we know the probability of a nonfunctioning device is 0.2778. To calculate the
probability of a functioning device, we need to use the complement rule:
31
P(functioning data device) = 1 – P(nonfunctioning data device) = 1 – 0.2778 = 0.7222
2-27.
If a large competitor buys a small firm, the firm’s stock will increase with probability 0.85. In
other words: P(stock will Rise | Bought by large firm) = 0.85.
The purchase of the company has a probability of 0.40 of taking place, or P(being Bought by a
large firm) = 0.40.
The probability that the purchase will take place and the firm’s stock will rise is determined by
the intersection of the two events, P(R  B), found by:
P(R  B) = P(R | B) P(B) = (.85)(.40) = 0.34
2-28.
If interest rates decrease, then the probability the market will go up is 0.80. Using the symbol ““
for “given,” this may be written P(market goes up  interest rates decrease) = 0.80. That is, the
conditional part is the interest rates; if they decrease, then the market goes up, so the given part is
interest rates decreasing. Also, we assume that the probability is 0.40 that interest rates will
decrease, which may be written P(interest rates decrease) = 0.40. Then, we wish to compute the
probability that the market will go up and the interest rates go down, so we seek P(market goes
up  interest rates go down). The conditional law is
P(A|B) =
P(A  B)
,
P(B)
so P(Market goes up | interest rates decrease) =
or P(M.up | Int.down) =
P(market goes up  interest rates decrease)
,
P(interest rates decrease)
P(M.up  Int.down )
P(M.up  Int.down )
,
 0.80 =
.40
P(Int.down )
and solving for the intersection, P(M.up  Int.down) = 0.80(0.40) = .32.
2-33.
Given the following table of counts:
Price
No Price
Increase
Increase
Paid
34
78
Not Paid
85
49
Total
119
127
Total
112
134
246
a. Compute the probability a randomly selected stock increased in price:
119 stocks increased
= .484.
246 total stocks
112 paid dividends
b. Compute P(paid dividends) =
= .455.
246 total stocks
c. Compute P(price increase  paid dividends)
P(price increase) =
=
34 had price increase and paid dividends
= .138.
246 total stocks
32
d. Compute P(not paid  no price increase)
=
49 not paid and no price increase
= .199.
246 total stocks
e. Given a price increase, compute the probability it also paid dividends:
P(paid dividendsprice increase) =
P(paid dividends  price increase)
P(price increase)
 34 paid dividends and price increase 


246 total stocks
 = .138 = .285.
= 
.484
 119 price increase 


 246 total stocks 
Another way to view this is from a reduced-space perspective.
34 paid dividends
= .286.
119 price increase
In the previous version we considered the 34 out of 246 compared to 119 out of 246, but in
the reduced space version, we recognize that “given price increase” restricts us to the 119
which had a price increase, and then out of this 119, what proportion also paid dividends?
It was 34 out of 119, or a proportion of .286. The viewpoints are equivalent, and the answers
differ due to rounding.
Compute P(paid dividendsprice increase) =
f.
P(increased in pricepaid no dividends)
=
P(increased  paid no dividends
P(paid no dividends)
 85 


.3455
246 
= 
=
= .6343.
.5447
 134 


 246 
g. Compute P(price increase  paid dividends) (i.e., either price increase or paid dividends or
both) =
P(price increase) + P(paid dividends)  P(price increase and paid dividend)
=
2-36.
119 112 34
= .801.


246 246 246
According to a report, 65% of Americans are overweight or obese. The problem asks you to
determine the probability that in a group of five randomly selected Americans at least one is
overweight or obese. How do we start? Let’s determine what we do know. First, a random
sample of five people is selected, which implies independence. Second, it is given that 65% of
all Americans are overweight or obese. If this is so, then we also know that 35% are not
overweight or obese.
So how do we proceed? We could set up the entire sample space of all the possible outcomes for
five people being overweight or obese, starting at none are overweight up to all are overweight,
and then calculating the probability of each outcome and adding up the relevant probabilities.
33
This is the more time consuming way to do it. Let’s first determine the probability that no one in
the group of five is overweight or obese, which would be one of the outcomes in the sample
space. The probability that none of the five selected people are overweight is:
P(not overweight) = (0.35)(0.35)(0.35)(0.35)(0.35) = (0.35)5 = 0.0053 (approximately).
Since all the possible outcomes in our sample space must add up to 1.00, we can use the
complement rule to determine the sum of the remaining probabilities of the other outcomes. The
remaining outcomes include one person being overweight, two being overweight, three…,etc.
Since the question ask us to determine the probability that at least one is overweight, we simply
subtract our probability of none being overweight from 1.00:
P(at least one is overweight) = 1.00 – 0.00525 = 0.9947 (approximately).
Using the template (Probability of at least 1.xls), enter the probability of success (overweight) for
the sample of size 5 in column C. The result is shown cell H4.
Probability of at least one success from many independent trials.
Success Probs
1
2
3
4
5
2-38.
0.65
0.65
0.65
0.65
0.65
Prob. of at least one success
0.9947
We want to be sure that a package is delivered within one day so we decide to send the same
package by three different delivery services. The three delivery firms have different success
rates for on-time delivery: Firm A has a 90% success rate [P(A) = 0.90], Firm B has an 88%
success rate [P(B) = 0.88], and Firm C has a 91% success rate [P(C) = 0.91]. We want to
determine the probability that at least one of the packages arrives on time.
First, we assume independence in the events; i.e., the delivery by one service has no impact on
the delivery of the other two services.
Second, let’s determine the probability that none of the three packages are delivered on time. To
do this we need to calculate the failure rate for each firm. The failure rate for each firm is found
by subtracting their success rate, expressed in decimal format, from 1.00, or:
Firm A failure rate: 1.00  0.90 = 0.10
Firm B failure rate: 1.00  0.88 = 0.12
Firm C failure rate: 1.00  0.91 = 0.09
34
The probability that none of the packages will be delivered on time is found by the multiplication
of the three failure rates, since we assumed the events were independent of each other.
P(none are delivered on time) = (0.10)(0.12)(0.09) = 0.00108
Finally, the probability that at least one of the packages is delivered on time is same as asking for
the probability that one or two or all three packages were delivered on time. The easiest way to
do this is to use the “complement rule.” To determine the probability that at least one of the
packages is delivered on time we subtract the probability that none of the packages were
delivered on time from 1.00.
P(at least one arrives on time) = 1  P(all three fail to arrive)
= 1.00  (1  .90)(1  .88)(1  .91) = 1.00  0.00108 = 0.99892
Using the template (Probability of at least 1.xls), enter the probability of success for each
package delivery company in column C. The result is in cell H4.
Probability of at least one success from many independent trials.
Success Probs
1
2
3
0.9
0.88
0.91
Prob. of at least one success
0.9989
There is a 99.89% chance that at least one of the packages will be delivered on time.
2-42.
We are given the probabilities of three credit derivatives for making a profit, and we want to
know the probability of at least on will make a profit.
Using template: Probability of at least 1.xls
Enter the three probabilities in cells: C4:C6
Probability of at least one success from many independent trials.
Success Probs
1
2
3
0.9
0.75
0.6
Prob. of at least one success
0.9900
The probability of at least one of the three investments makes a profit is 0.9900
35
2-44.
This problem pertains to the data of problem 2-31, which provided the following table of the
number of claims at an insurance company.
East
South
Midwest
West
Totals
75
128
29
52
284
Hospitalization
233
514
104
251
1,102
Physician’s
visit
100
326
65
99
590
Outpatient
treatment
408
968
198
402
1,976
Totals
The question is whether the event “hospitalization” is independent of the event “Midwest.” We
know that in general (whether the events are independent or not) P(AB) = P(AB)P(B), but if
A and B are independent, P(AB) = P(A), since the fact that B is given or came first doesn’t
make any difference to factor A. Therefore, if P(AB) = P(A), then events A and B are
independent, and consequently, P(AB) = P(A)P(B).
a. One test for independence: The events are independent if
P(hospitalizationMidwest) = P(hospitalization).
1. P(hospitalizationMidwest) =
P(hospitalization  Midwest)
P(Midwest)
 29 hospitaliz ation and Midwest 


1,976 total

 = 0.0147
=
 198 Midwest 
0.1002


 1,97 6total 
P(hospitalizationMidwest) = 0.1465.
(This can also be computed more directly by
2. Also, P(hospitalization) =
29
= 0.1465.)
198
284
= 0.1437
1,976
3. Then: is P(hospitalizationMidwest) = P(hospitalization)? Here 0.1465  0.1437;
therefore the two events are not statistically independent. Where the hospitalization
occurs does make a difference.
b. Another test for independence: The two events are independent if
P(hospitalizationMidwest) = P(hospitalization)P(Midwest).
1. P(hospitalizationMidwest) =
29
= 0.0147.
1,976
2. P(hospitalization)P(Midwest)?
284
198
*
= 0.0144.
1,976 1,976
3. Since P(hospitalizationMidwest)  P(hospitalization)P(Midwest), because
0.0147  0.0144, the two events are not statistically independent.
36
Permutations and Combinations Introduction (Reference 2-52)
To determine how many ways there are to order potential participants, we use either the formula
for combinations or the formula for permutations. To decide which, we ask whether the order in
which they are selected matters. For instance, if selecting two people from among five (call them
A, B, C, D, and E), there are several approaches, depending on whether order makes a
difference.
Permutations.
If the order in which they are selected matters, we have a permutation. That is, a permutation of
ways to order the outcome occurs if the order of selection is important, such as “the first person
chosen is the chairman and the second person is treasurer.” Here the outcome AB (selecting A
first and then B) is different from the outcome BA (selecting B first) since in one case A is the
chairman and in the other A is the treasurer.
The formula for permutations is nPr =
n!
.
(n  r )!
For the example of selecting two from five,
5  4  3! 5  4
5!
= 20 permutations, which may be listed as:
3!
(5  2)!
1
AB
BA
CA
DA
EA
AC
BC
CB
DB
EB
AD
BD
CD
DC
EC
AE
BE
CE
DE
ED
and note that both AB and BA are counted as separate outcomes, because we are assuming order
matters; a permutation.
5P2
=
Combinations.
If, however, the order of selection does not matter, as in “the two people selected will be equal
members of the committee,” then a combination formula is used instead of a permutation
formula. Then, the outcome AB (A selected first) is the exact same outcome as BA (B selected
first), since in both cases A is a member and B is a member. So we cannot count AB and BA as
separate outcomes; AB and BA are the same outcome (with slightly different names). The
number of outcomes for the combination case is less than for the permutation case. The
n!
combination formula is nCr =
. For the example of selecting two items from five we
r!(n  r )!
(5)(4)(3)(2)(1) 120
5!
5!
have 5C2 =
=
=
=
= 10 combinations, which
2!(5  2  3)! (2)(1)(3)(2)(1) (2)(1)(3)(2)(1)
12
may be listed as AB AC AD AE BC BD BE CD CE DE.
(Note that BA is not listed, since it would be equivalent to AB, which is listed.)
2-50.
Use Template: Probability of at least 1.xls
To determine the probability that at least one driver makes it home safely is the same as
determining the probability that one or two or all three drivers make it home safely. The sample
space consists of four possible outcomes: none make it home safely, one makes it home safely,
two make it home safely, and all three make it home safely. Since we are interested in all the
possible outcomes except the one where none make it home safely, we can approach the problem
using the complement rule. First, we convert each probability of a successful trip from the
37
problem into a probability of an unsuccessful trip. Since driver one has a probability of 0.50 of
making a successful trip, then he has a (1 – 0.50 = 0.50) probability of not making it home. The
corresponding probabilities for driver #2 and #3 are: (1 – 0.25 = 0.75) and (1 – 0.20 = 0.80)
respectively. Using these probabilities of failures gives us:
P(at least one drives home safely) = 1 – P(none drive home safely)
= 1 – [(0.50)(0.75)(0.80)]
= 1 – 0.30 = 0.70
The template approach is much easier. Insert the given probabilities of success in cells C4:C6,
and the result is displayed in cell H4.
Success
Probs
1
2
3
2-52.
0.5
0.25
0.2
Prob. of at least one success
0.7000
To select the four representatives (one from each department), we use Rule 8 to give the number
of possible sequences. There are 55 to select from in manufacturing, 30 in distribution, 21 in
marketing, and 13 in management. The total number of possible outcomes is then
(55)(30)(21)(13) = 450,450 combinations.
(Another approach is to treat each department as a separate task of selecting one representative,
and use the combination formula, since the order doesn’t matter:)
Manufacturing: Select 1 from 55;
55C1
=
55!
(55)54!
= 55.

[(1!)(55  1  54)!]
54!
Distribution: Select 1 from 30;
30C1
=
30!
(30)29!
= 30.

[(1!)(30  1  29)!]
29!
Marketing: Select 1 from 21;
21C1
=
21!
(21)20!
= 21.

[(1!)(21  1  20)!]
20!
Management: Select 1 from 13;
13C1
=
13!
(13)12!
= 13.

[(1!)(13  1  12)!]
12!
Use Template: Permutation & Combination.xls
For this problem, we enter the respective department sizes under “n” and the desired number of
representatives from each department under “r” (in this case only one representative from each
department) in the cells pertaining to combinations (order is not important): F5:G8. The
combination values are displayed in cells H5:H8.
Combination
n
55
30
21
13
r
1
1
1
1
nCr
55
30
21
13
38
Now, there are 55 combinations from manufacturing, and for each one there are 30 from
distribution, and for each of these there are 21 from marketing, and for each of these there are 13
from management. Thus, there are
(55)(30)(21)(13) = 450,450 combinations.
Of course, using Rule 8 is much simpler than using Rule 11 four times, but we present this
approach just for practice in using combinations.
2-53.
9! = (9)(8)(7)(6)(5)(4)(3)(2)(1) = 362,880 different orders.
2-54.
In this problem, sequence or order is important, so we use the permutation formula. We wish to
select 8 out of 15, so:
15P8
=
(15)(14)(13)(12)(11)(10)(9)(8)7!
15!
15!
=

7!
(15  8)! 7!
= 259,459,200 different sequences.
Use Template: Permutation & Combination.xls
In this problem, order is important. Therefore, we need to use the permutations section of the
template. Data is entered in cells B5:C5 and the permutation result is displayed in cell D5. (You
will need to increase the width of cell D5 to see the value by first unprotecting the sheet. Under
Tools|Protection select Unprotect Sheet. Then move your cursor to the column heading “D E”
until it changes to a vertical line with a dual arrow line horizontally through it. Holding down
the left-button on the mouse, slide the cursor to the right until the value changes from scientific
notation to its basic numerical display.)
Permutation
n
15
2-56.
r
8
nPr
259459200
In selecting two from seven, where the order doesn’t matter, we use the formula for combinations
7C2
=
n!
7!
7!
(7)(6)5! (7)(6)
= 21 combinations,




(r!)(n  r)! 2!(7  2  5)! 2! 5! (2)(1)(5!)
2
or 21 ways to get pairs out of seven populations.
2-61.
In setting up this problem we must convert the word statements to probabilities. Something
equals 0.95, and it is a conditional probability because it says that if tests show no side effects,
then the probability is 0.95 that the FDA will approve. So P(approveno side effects) = 0.95.
Also, if the tests indicate side effects, then the probability is 0.50 that the FDA will approve, so
*9
39
2-62.
P(approveside effects) = 0.50. Finally, there is 0.20 probability that the tests will show side
effects so P(side effects) = 0.20. Of course, either a test will indicate side effects or it won’t, so
P(no side effects) = 1  P(side effects) = 1  0.20 = 0.80.
Now, using the law of total probability (Rule 13), we want to find the probability the drug will be
approved, which is not a conditional probability.
P(approve) = [P(approve given side effects)] times [ the likelihood of tests showing side effects]
+ [P(approve give no side effects)] times [the likelihood of showing no side effects].
P(approve) = P(approveside effects)P(side effects)
+ P(approveno side effects)P(no side effects)
= 0.50(0.20) + 0.95(0.80) = .86.
Bayes’ Theorem
Bayes’ Theorem allows us to compute certain probabilities by reversing the order of events.
2-65.
This problem pertains to an electronic door lock system. Note that an over-bar such as
should open means a logical “not,” which here would be “should not open.”
The first step is to write down what is known and the compliments; for some problems these
compliments will be needed, and it is useful to write them down each time. It is given that 90%
of the time the door should open, so
P(should open) = 0.90 and the compliment is P( should open ) = 0.10.
Then, given that the door should open, 98% of the time a green light will appear, so
P(green lightshould open) = P(GOSO) = .98 and its compliment is
P( green light should open) = P( GO SO) = .02.
(How do we know that the compliment of P(GLSO) is P( GL SO) rather than P(GL SO )? We
can think of a tree of events, with the given event coming first.
(GIVEN
SO)
SO
P(GLSO) = .98
GL
The sum of probabilities off of one node equals
one, so .98 + .02 = 1.00
GL
P( GL SO) = .02
So the compliment of P(GLSO) = .98 is P( GL SO) = .02.
40
Similarly, 5% of the time when (or given) the door should not open, a green light will appear (an
error) so:
P(green light should open ) = P(GL SO ) = .05 and its compliment is P( GL  SO ) = .95.
The six probabilities we have are:
P(SO) = 0.90
Given:
P( SO ) = 0.10
Compliments:
P(GLSO) = .98
P( GL SO) = .02
P(GL SO ) = .05
P( GL  SO ) = .95
Now we want P(door will open given the light is green), or P(SOGL), which is the reverse
order of the probabilities we have. Bayes’ gives us a way to compute the reverse order
probabilities. Using the formula for Bayes’ Theorem,
P(SO|GL) =
=
P(GL | SO)P(SO)
P(GL | SO)P(SO)  P(GL | SO)P(SO)
(.98)(.90)
.882
=
= .9944.
(.98)(.90)  (.05)(.10)
.887
Another approach to using Bayes’ Theorem provides practice using the laws of probability and
indicates how Bayes’ Theorem is derived.
We seek P(SOSL), which we can write (using Rule 5) as:
P(SO|GL) =
P(SO  GL)
.
P(GL)
a. Rewrite the denominator as P(GL) = P(GLSO) + P(GL SO ) using the law of total
probability. Now rewrite the three intersections or joint probabilities:
P(SOGL) = P(SOGL)P(GL) but it also = P(GLSO)P(SO).
Which form should we choose? The one for which we know the probabilities:
P(SOGL) = P(GLSO)P(SO).
b. Use the same approach to rewrite each of the joint probabilities:
P(SOGL) =
rewritten as =
P(SO  GL )
P(SO  GL)
rewritten as =
P(GL)
P(GL  SO)  P(GL  SO)
P(GL | SO)P(SO)
P(GL | SO)P(SO)  P(GL | SO)P(SO)
and inserting probabilities,
=
.98(.90)
.882
=
= .9944.
.882  .005
.98(.90)  .05(.10)
41
2-67.
Here we extend the Bayes’ Theorem to three sets. First, write down the probabilities and their
complements. Given “high” economic situation, indicator rises with probability 0.60, so
P(indicator riseshigh) = 0.60 and its complement is P( indicator rises high) = 0.40.
Similarly, P( indicator rises medium) = 0.30
and its complement is P( indicator rises medium) = 0.70.
The listing of the probabilities is:
P(IRhigh) = .60
P(IRmed) = .30
P( IR high) = .40
and P(high) = .15
P( IR med) = .70
P(med) = .70
P(IRlow) = .10
P( IR low) = .90
P(low) = .15
Compute P(high economic given indicator has risen) or P(highIR), which is the reverse order of
events. Using the formula,
P(high|IR) =
=
2-76.
P(IR | high )P(high )
P(IR | high )P(high )  P(IR | med)P(med)  P(IR | low )P(low )
.09
.09
(.60)(.15)
=
=
= .2857.
(.60)(.15)  (.30)(.70)  (.10)(.15)
.09  .21  .015 .315
For CPA exams, it is known that P(pass exam) = 0.60, and P(pass examget job offer) = .40.
Compute P (given pass exam, then get job offer).
Find (using Rule 5),
P(get job offerpass exam) =
2-79.
P(get job offer  pass exam) .40
=
= .67.
.60
P(pass exam)
To test for independence: we know that in general (whether the two events are independent or
not) P(AB) = P(AB)P(B), but if A and B are independent, P(AB) = P(A), since the fact that
B is given or came first doesn’t make any difference to A. Therefore, if P(AB) = P(A), then
events A and B are independent, and consequently, P(AB) = P(A)P(B).
For this example, P(engineering most desirable) = .35 and P(design most desirable) = .50.
Also, P(engineering most desirabledesign most desirable) = .25.
Now, engineering and design are independent if
P(engineering most desirabledesign most desirable) =
P(engineering most desirable)P(design most desirable).
Then, 0.25  [(.35)(.50) = .175].
So since they are not equal, then
engineering and design priorities are not statistically independent.
42
2-81.
Use Template: Probability of at least 1.xls
This is similar to problem 2-50 in that we want to know what the probability that a randomly
chosen individual will be exposed to at least one method of advertising during a given week.
Manually, we would determine the probability of not being exposed to each type of
advertisement and then utilize the complement rule. The probability of not being exposed to a
billboard ad is 1 – 0.1 = 0.90, of not being exposed to a magazine ad is 1 – 0.15 = 0.85, and of
not being exposed to a radio ad is 1 – 0.20 = 0.80.
Therefore, the probability of being exposed to at least one type of ad is:
P(being exposed to at least one ad) = 1 – P(not being exposed to any ad)
= 1 – [(0.90)(0.85)(0.80)]
= 1 – 0.612
= 0.388
The template approach: insert the given probabilities of success in cells C4:C6, and the result is
displayed in cell H4.
Success
Probs
1
2
3
2-92.
0.1
0.15
0.2
Prob. of at least one success
0.3880
This is another Bayes’ Theorem problem, using sample data from a university. Let “fr” =
freshman, “so” = sophomore, and so on. We have
P(fr) = .30
P(so) = .35
P(jr) = .20
P(sr) = .15
Also, P(given the person is a freshman, the grade is an A) = .20, so
P(Afr) = .20, P(Aso) = .30, P(Ajr) = .35, P(Asr) = .40.
Now, compute P(given a student got an A, the student is a sr), or P(srA).
P(sr | A)
=
P(A | sr)P(sr)
P(A | sr)P(sr)  P(A | fr)P(fr)  P(A | so)P(so)  P(A | jr )P( jr )
=
(.40)(.15)
(.40)(.15)  (.20)(.30)  (.30)(.35)  (.35)(.20)
=
.06
.06
=
= .2034 .
.06  .06  .105  .07
.295
43
2-101. We can display the following information given in the problem in table format as follows:
Kwik Save
893
107
1000
open
close
total
Somerfield
0
424
424
total
893
531
1424
a. Given that a store is closing, what is the probability that it is a Kwik Save?
There are 531 stores that are closing; 107 of which are Kwik Save stores.
The probability that it will be a Kwik Save store is found by:
107
 0.202
531
b. What is the probability that a randomly chosen store is closing or a Kwik Save? We know
there are 531 stores closing out of the 1424 total stores. And we know there are 1000 Kwik
Save stores of the 1424 stores. Furthermore, we know that 107 of the Kwik Save stores will
be closing so we do not want to count them twice. Either we count them in the total number
of stores closing or in the total Kwik Save stores, but not in both categories. The probability
is found by:
P(KS  Closing) = P(KS) + P(Closing)  P(KS  Closing)
531 1000 107 1424



 1.00
1424 1424 1424 1424
We are 100% certain that a randomly chosen store will either be a Kwik Save or will be
closing. This makes sense since all of the Somerfield stores will be closing.
c. What is the probability that a randomly selected store is not closing given that it is a
Somerfield store? The probability is 0, since all the Somerfield stores will be closing.

2-105. There are three machines that produce the same part but with different levels of precision (non
defects). Furthermore, one part is selected at random from the bin containing all parts produced,
and it is identified as being either defective or good. You need to determine which machine
produced the part.
To determine the answers to parts 1) and 2), we first start with the template: Bayesian
Revisions.xls. Given the components of the problem, we will use the fourth worksheet,
“Empirical”, for the empirical conditional probability approach. In the green cells, fill in the
information contained in the problem. The template is designed to handle up to eight different
sources of probabilities, but we will be using only three. (Be sure that the green cells in columns
F-J are blank.)
Use template: Bayes Revision.xls
Step 1: enter the names of the machines in cells C3 through E3.
Step 2: enter the initial percentages for the output produced by each machine in cells C5-E5.
(These are the prior probabilities.)
Step 3: in cells A8-A13 enter all the possible outcomes of the production process. In this
problem there are only two outcomes: defective or good.
Step 4: in cells C8-E8 enter the information given in the problem on the percentage of defective
parts produced by each machine.
44
Step 5: At this point you will see an error message appear below each column in row 15 stating
that “the Total must be 1”. The conditional probabilities for each machine must add to 1.00. If
we know the percentage of defective parts produced by each machine, then we also know the
percentage of good parts produced. Simply subtract the percentage of defective parts from 1.00
and enter this value in cells C9-E9. (The error message will disappear.)
This is all the information that needs to be entered in the template. All results are presented in
red in cells C26-E27. Row 26 contains the probabilities for whichever condition was entered in
cell A8, in this case it is defective parts. Row 27 contains the probabilities for the condition
specified in cell A9: good parts. To answer question (1), what is the probability that the
defective part was produced by machine A, simply read the value in cell C26. There is a 38.38%
probability that the part was produced by machine A. For part (2), the probability that the good
part was produced by machine B is 38.17%.
Bayesian Revision based on Empirical Conditional Probabilities
Machine A
s1
Prior Probability
0.26
Machine B
s2
0.38
Machine C
s3
0.36
s4 s5 s6 s7 s8
s2
s3
s4 s5 s6 s7 s8
Total
1
Conditional Probabilities
s1
defective
good
P(I1 | .)
P(I2 | .)
P(I3 | .)
P(I4 | .)
P(I5 | .)
Total
0.08
0.92
0.05
0.95
0.04
0.96
1
1
1
0
0
0 0 0
s1
0.0208
0.2392
s2
0.0190
0.3610
s3
0.0144
0.3456
s4 s5 s6 s7 s8
s1
0.3838
0.2529
s2
0.3506
0.3817
s3
0.2657
0.3654
s4 s5 s6 s7 s8
Joint Probabilities
I1
I2
I3
I4
I5
Posterior Probabilities
P(. | I1)
P(. | I2)
P(. | I3)
P(. | I4)
P(. | I5)
45
Marginal
0.0542
0.9458