Chapter 3 Probability
An experiment/ random phenomenon is an act or process of observation where the outcome is
uncertain. eg Play a game with an opponent, spin a coin, roll a die, run a red light, measure a
person’s height
The sample space of an experiment is the collection of all possible outcomes. Sample space is
denoted S
eg expt = roll a six-sided die once. Then S = { 1, 2, 3, 4, 5, 6 }
An event is a collection of outcomes. Events are often denoted by letters near the beginning of
the alphabet
eg expt = roll a six-sided die once.
Let E be the event of rolling an even number.
Let F be the event of rolling a number that’s at least 5.
Define E and F.
Soln: E = { 2, 4, 6 }
F={
5, 6
}
The union of two events A and B, written as A or B, is the collection of all outcomes that are in
A or B or both A and B -- that is, outcomes that make A occur, or B occur, or both A and B
occur. Union is denoted ∪ (or v)
eg E ∪ F = an even number or a number that’s at least 5
= { 2, 4, 6, 5 }
The intersection of two events A and B, written as A and B, is the collection of all outcomes that
are in both A and B—that is, outcomes that make both A and B occur. Intersection is denoted ∩
1
E and F contains the outcomes that are in both E and F.
eg E ∩ F = an even number that’s at least 5 = { 6 }
A Venn diagram is a visual representation of the sample space and events of interest for an
experiment.
eg Draw a Venn diagram for the die-rolling experiment. Show E, F, E ∪ F and E∩F.
The complement of an event A consists of the outcomes of the experiment that are not in A.
Denoted 𝐴𝑐 . A’ or 𝐴̅
eg If E = even number. Then 𝐸 𝑐 =
odd number
= { 1, 3, 5
}
Note: The complement of the sample space S is the empty set, denoted ∅. ∅ = { } = 𝑆 𝑐
Two events are called mutually exclusive if they cannot happen at the same time –that is, if their
intersection is empty. Eg E = even number, O = odd number.
Then E∩O = ∅ so E and O are mutually exclusive.
2
The probability of an event is the likelihood that it will occur. The probability of an event E
occurring is denoted P(E)=P(even number) = .5
Eg Roll a fair six-sided die once. E = even number. Fair tells us that the 6 sides are equally
likely.
P(E) = .5 How do we know? P(E) = 3/6
Example: Brothers and sisters 28 respondents
18 people have one or more brothers (B)
17 people have one or more sisters (S)
3 people have no brothers and no sisters (Not B and Not S)
One or more brothers
No brothers
Total
One or more sisters
No sisters
Total
10
7
17
8
3
11
18
10
28
How is this table set up? First decide on the two basic questions (Brother status and sister status)
The individuals are people in the room now. They are classified according to two variables:


Brother status
Sister status
Suppose that one person is randomly selected from those in the “room” now.
randomly selected tells us that everyone has the same chance
Find:
18
P(B)= 28
17
P(S) = 28
10
𝑃(B ∩ S) = P(B and S) = 28
P(B ∪ S) = P(B or S) =
10+8+7
28
25
= 28
3
One or more brothers
No brothers
Total
One or more sisters
No sisters
Total
10
7
17
8
3
11
18
10
28
8
P(B ∩ S c ) = P(B and not S) = 28
P(B ∪ S c ) = P(B or Not S) =
10+8+3
28
21
= 28
3
P(Bc ∩ S c ) = P(Not B and Not S) = 28
P(Bc ∪ S c ) = P(Not B or Not S) =
7+3+8
28
18
= 28
Note: The Addition Rule is
𝑃(𝐴 ∪ 𝐵) = 𝑃(𝐴) + 𝑃(𝐵) − 𝑃(𝐴 ∩ 𝐵)
Check that the addition rule can be used to find P(B ∪ S) in the brothers and sisters example.
𝑃(𝑆 ∪ 𝐵) = 𝑃(𝑆) + 𝑃(𝐵) − 𝑃(𝑆 ∩ 𝐵) =
17 18 10 25
+
−
=
28 28 28 28
Note: Probabilities are calculated from two numbers.
Bottom number (denominator): What’s it out of?
Top number (numerator): How many (out of the bottom number) are there?
Exercise 3.64 p 143 Employee behaviour problems A survey of human resource officers
(HRO’s) at big companies was described in The Organisational Journal (Summer 2006). 55% of
the human resource officers mentioned problems with absenteeism amongst employees, 41% had
problems with turnover of employees and 22% had problems with both absenteeism and
turnover.
(a) What’s the probability that a randomly chosen HRO from this group had problems with
absenteeism or turnover?
(b) What’s the probability that a randomly chosen HRO from this group did not have
problems with absenteeism?
(c) What’s the probability that a randomly chosen HRO from this group had problems with
neither absenteeism nor turnover?
Solution: Classify the human resource officers according to 2 variables: (What are they?)
4
Problems with turnover (T)
Problems with absenteeism (A)
55% of the human resource officers mentioned problems with absenteeism amongst employees,
41% had problems with turnover of employees and 22% had problems with both absenteeism
and turnover.
T
Not T
Total
Get rid of the % signs
T
Not T
Total
A
22%
33%
55%
Not A
19%
26%
45%
Total
41%
59%
100%
A
22
33
55
Not A
19
26
45
Total
41
59
100
Now answer the questions
(a) What’s the probability that a randomly chosen HRO from this group had problems with
absenteeism or turnover?
P(A or T) =
22+19+33
100
74
= 100
(b) What’s the probability that a randomly chosen HRO from this group did not have problems
with absenteeism?
P(not A) =
45
100
(c) What’s the probability that a randomly chosen HRO from this group had problems with
neither absenteeism nor turnover?
26
P(not A and Not T) = 100
(d) What is the probability that a randomly selected human resource officer from this group has
problems with at least one of the two problems (absenteeism and turnover)?
P(A or T)
=
22+19+33
100
74
= 100
5
(e) What is the probability that a randomly selected human resource officer from this group has
problems with only one of the two problems (absenteeism and turnover)?
P(A and Not T) +P( T and Not A)=
19+33
100
52
= 100
Exercise 3.52 p141 Social networking. The Pew Research Internet Group Project (Dec 2013)
conducted a survey of adult internet users in the US. The survey found that the two most popular
social network sites in the US (amongst adult users of the Internet) were LinkedIn and Facebook:
71% had a Facebook account, 22% had a LinkedIn account and 18% used both LinkedIn and
Facebook.
a) Make an appropriate Venn Diagram.
b) What’s the chance that an adult internet user used at least one of LinkedIn and Facebook?
c) What’s the chance that an adult internet user uses exactly one of the two sites LinkedIn
and Facebook?
Solution:
(a) Draw a Venn Diagram using Paint.
(b) Use the diagram in (a).
(c) Use the diagram in (a).
Table method:
6
The Pew Research Internet Group Project (Dec 2013) conducted a survey of adult internet users
in the US. The survey found that the two most popular social network sites in the US (amongst
adult users of the Internet) were LinkedIn and Facebook: 71% had a Facebook account, 22% had
a LinkedIn account and 18% used both LinkedIn and Facebook.
What/who is being classified? Adult internet users in the US
What are they being classified according to? (ie what are the two variables?)
Facebook use (F)
LinkedIn use (L)
F
18
53
71
L
Not L
Total
Not F
4
25
29
Total
22
78
100
Now use the table to do (b) and (c).
(b) What’s the chance that an adult internet user used at least one of LinkedIn and Facebook?
P(L or F) = (18+4+53)/100= 75/100
(c) What’s the chance that an adult internet user uses exactly one of the two sites LinkedIn
and Facebook?
P(exactly one of F and L) = (4+53)/100= 57/100
Conditional Probability
(not out of all), conditional on some other event happening, incorporate other info.
Brothers and sisters
One or more sisters
No sisters
Total
One or more brothers 10
8
18
No brothers
7
3
10
Total
17
11
28
Suppose we select one person randomly from the 28 in the room now. We find that this person
has a brother. What’s the probability that (s)he has a sister?
What is this out of?
10
P(S|B) = 18
Answer *
7
Suppose we select one person randomly from the 28 in the room now. We find that this person
has a sister. What’s the probability that (s)he has a brother?
10
P(B|S)= 17
Note: The conditional probability of A occurring, given that B has occurred, is denoted P(A/B)
or P(A|B).
Useful formula: 𝑃(𝐴|𝐵) =
𝑃(𝐴 ∩ 𝐵)
𝑃(𝐵)
Find P(Sister|Brother) using this formula.
Suppose we select one person randomly from the 28 in the room now. We find that this person
has a brother. What’s the probability that (s)he has a sister?
18
From above: P(B) = 28
10
𝑃(B ∩ S) = P(B and S) = 28
We want P(sister | brother)
𝑃(𝑆|𝐵) =
𝑃(𝑆 ∩ 𝐵)
𝑃(𝐵)
=
10/28
18/28
= 10/18 Compare with answer * above.
Note that P(S|B) is different from P(B|S)!!!
Exercise 3.82 p 155 Social robots An article in The International Conference on Social Robots
(Vol 6414, 2010) summarised aspects of the design of social robots as follows: a random sample
of 106 social robots had 63 that were built with legs only, 20 that were built with only wheels,
and 15 that were built without legs or wheels. If a social robot has wheels, what’s the probability
that it also has legs?
Solution:
Table method (use a table):
What is being classified? Social robots
What questions about them are of interest?
Do it have legs?
Does it have wheels?
8
a random sample of 106 social robots had 63 that were built with legs only, 20 that were built
with only wheels, and 15 that were built without legs or wheels.
W
Not W
L
8
63 legs and not wheels
Not L
20 wheels and not legs
15
Total
28
78
If a social robot has wheels, what’s the probability that it also has legs?
Total
71
35
106
What is this out of? (denominator) It’s out of the robots that have wheels. There are 28 of
them!
P( L | W
)=
8
28
Formula (use formulae):
a random sample of 106 social robots had 63 that were built with legs only, 20 that were built
with only wheels, and 15 that were built without legs or wheels. If a social robot has wheels,
what’s the probability that it also has legs?
Notation:
L= randomly chosen robot has legs
W = randomly chosen robot has wheels
What probabilities have been given?
P(L and not W ) =63/106
P(W and Not L) =20/106
P(Not L and Not W) 15/106
What probability is wanted? P(L|W)
9
Exercise 3.160 p176 Ancient pottery. Archeologists described (in Chance (Fall 2000)) 837
pieces of pottery found at the ancient Greek settlement of Phylakopi. 183 pieces of pottery were
painted. Of these 183 painted pieces of pottery, 14 were painted in a curvi-linear pattern, 165
were painted in a geometric design and 4 were painted in a naturalistic way.
a) Find the probability that a randomly selected piece of pottery is painted.
b) Given that a randomly selected piece of pottery is painted, what is the probability that it
is painted in a curvi-linear pattern?
Solution: What is being classified?
Not painted
837-183=654
(a) P(painted)
Painted 183
Curvilinear
14
=
Total
Geometric
165
Naturalistic
4
183
837
(b) P( curvi-linear pattern
| painted
)
=
10
14
183
837
Note: The so-called Multiplication Rule is
𝑃(𝐴 ∩ 𝐵) = 𝑃(𝐴/𝐵)𝑃(𝐵)
This is the conditional prob formula with the quantities re-arranged.
Exercise 3.166 (b) p176 Library card. A Harris poll found that 68% of all American adults
have a public library card. Also, 62% of men have library cards while 73% of women have
library cards. Assuming that there are equal numbers of men and women in the US, what is the
probability that a randomly chosen American adult is a woman who owns a library card?
Solution: Table method:
What is being classified? American adults
What variables are we using to classify Am adults?
Gender ( M and F)
Library card status (L = yes, Not L = no library card)
Note that 62% and 73% are conditional!! (not out of all) Conditional probs do NOT go into the
table!
M
F
Total
L
62% of 50=31
73% of 50=36.5
67.5
Not L
50-31=19
50-36.5=13.5
32.5
Total
50
50
100
Not L
190
135
325
Total
500
500
1000
In case we don’t like half people, multiply by 10:
M
F
Total
L
310
365
675
68% is redundant and mis-leading. They rounded 67.5% to 68%. ☹
What is the probability that a randomly chosen American adult is a woman who owns a library
card?
P(woman who owns a library card) =P(W and L) =
11
365
1000
Formulae:
A Harris poll found that 68% of all American adults have a public library card. Also, 62% of
men have library cards while 73% of women have library cards. Assuming that there are equal
numbers of men and women in the US
What has been given (must use correct notation)? We’ll ignore the 68% (redundant and
misleading)
L = randomly chosen Am adult has a library card
M= randomly chosen Am adult is male
F= randomly chosen Am adult is female
M|L or L|M?? P(L|M) =.62
P(L|F) =.73
P(M) =.5
P(F) =.5
what is the probability that a randomly chosen American adult is a woman who owns a library
card?
What is wanted? P(F and L). We see that we can use the multiplication formula because we
already know conditional and unconditional probabilities.
P(F and L) = P(L|F)P(F)= .73*.5= .365 same as before.
Exercise 3.81 p155 Blood diamonds Global Research News (Mar 4, 2014) reported that one
quarter of all rough diamonds produced in the world are blood diamonds (ie produced in a war
zone in order to finance war-lords’ activities, or insurgencies or invading army). 90% of all
rough diamonds are processed in Surat, India. One-third of the diamonds processed in Surat are
blood diamonds.
(a) What’s the probability that a rough diamond is not a blood diamond?
(b) What’s the probability that a rough diamond is processed in Surat and is a blood
diamond?
Answer the questions given using probability rules. Then make a table. How do you know that at
least one of the numbers provided is incorrect?
Solution using probability rules:
What has been given?
12
one quarter of all rough diamonds produced in the world are blood diamonds (ie produced in a
war zone in order to finance war-lords’ activities, or insurgencies or invading army). 90% of all
rough diamonds are processed in Surat, India. One-third of the diamonds processed in Surat are
blood diamonds.
B = blood diamond P(B) = .25
S = diamond comes from Surat P(S)=.9
P(B/S) = 1/3
What is requested?
(a) What’s the probability that a rough diamond is not a blood diamond?
P(Not B) = 1-.25=.75
(b) What’s the probability that a rough diamond is processed in Surat and is a blood
diamond?
Multiplication Rule is
𝑃(𝐴 ∩ 𝐵) = 𝑃(𝐴/𝐵)𝑃(𝐵)
P(S and B) = P(B|S)P(S)= (1/3)* (.9) = .3
Solution using table:
one quarter of all rough diamonds produced in the world are blood diamonds (ie produced in a
war zone in order to finance war-lords’ activities, or insurgencies or invading army). 90% of all
rough diamonds are processed in Surat, India. One-third of the diamonds processed in Surat are
blood diamonds.
What is being classified? diamonds
How are they being classified?
Is it a blood diamond?
Does it come from Surat?
B
Not B
Total
S
1/3 of 90%=30%
60%
90%
Not S
Box 2
Box 5
10%
Total
25%
75%
100%
Problem: Line 1 does not add up properly. So at least one of the numbers is wrong!
13
Next: Independence
Events A and B are independent if 𝑃(𝐴|𝐵) = 𝑃(𝐴)
P(A|B)= the part of B occupied by A
P(A) = the part of S occupied by A
We can’t see independence in a Venn Diagram. If A and B are independent, then they must
overlap.
A and B are independent if the chance of A happening is not affected by B happening—that is,
the chance of A happening stays the same even when B has happened.
Eg Roll a fair six-sided die several times. P(six on the first roll) = 1/6
P(six on 2nd roll/ six on first roll) = 1/6
Successive rolls are independent.
Independence vs mutually exclusive: Are they the same? NO. Mutually exclusive means no
overlap. Independence means the probability of one event doesn’t change when you know that
the other has happened.
Example: Brothers and sisters. Are B and S independent?
B= One or more
brothers
No brothers
Total
S=One or more
sisters
10
No sisters
8
18
7
17
3
11
10
28
Events S and B are independent if 𝑃(𝑆|𝐵) = 𝑃(𝑆)
Or Events S and B are independent if 𝑃(𝐵|𝑆) = 𝑃(𝐵)
Let’s do the first one:
P(S|B) =10/18=.555555….
14
Total
P(S) = 17/28=.607…..
We see that these two numbers are different, so S and B are not independent (ie they are
dependent)
Example of independence:
Sister(s)
Brother(s)
9
No brothers
3
Total
12
Are B and S independent? P(B) = P(B|S)
No Sisters
6
2
8
Total
15
5
20
P(B)=15/20=.75
P( B | S )= 9 / 12 =.75
The part of all 20 taken up by B is the same as the part of S taken up by B
so B and S are independent.
OR P(S) =12/20 =.6
P( S | B ) = 9 /15 = .6
Exercise 3.86 p 155 Guilty decision-making (Note: Choose stated option = repair the car.)
The Journal of Behavioural Decision Making (Jan 2007) describes an experiment to investigate
how feeling guilty affects a person who has to make decisions. 171 students who volunteered for
the experiment were randomly divided into three groups. Each group had to complete a reading
or writing task which induced a particular emotional state: Guilt, Anger or Neutral. As soon as
the task had been completed, the subjects were presented with a decision-making problem in
which the both options have predominantly negative features-- for example, either spend money
on an old car or do nothing.
Notice that this is an experiment because they were randomly divided into three groups.
15
(Note: Choose stated option = repair the car.)
a) Assuming that the respondent has been assigned to the guilty state, find the probability
that he/she chooses to repair the car.
P( Repair |Guilty )
=
45
57
b) Assuming that the respondent does not choose to repair the car, find the probability that
he/she is in the anger state.
P( Anger | not repair the car
)=
50
111
c) Are the events repair the car and guilty state independent?
Events A and B are independent if 𝑃(𝑅𝑒𝑝𝑎𝑖𝑟|𝐺𝑢𝑖𝑙𝑡𝑦) = 𝑃(𝑅𝑒𝑝𝑎𝑖𝑟)
P( Repair |Guilty )
=
45
57
= .82456….
P(Repair)=60/171 = .35087….
So the events repair the car and guilty state are dependent (ie Not independent)
Exercise 3.93 p157 Red snapper (Using independence, trees) Restaurants sometimes serve less
expensive fish in place of expensive fish like red snapper. According to Nature (July 15, 2004),
researchers at the University of North Carolina used DNA analysis to decide whether fish sold as
red snapper by vendors across the US was actually red snapper. They found that 77% of the fish
was not actually red snapper but some other cheaper fish which looked like red snapper.
a) Find the probability of being served red snapper if you order red snapper at a restaurant.
b) Find the probability that at least one out of five randomly selected restaurant customers
who ordered red snapper will be served red snapper.
(a) P(red snapper) = 1- 0.77 =.23
Also: Find the probability that at least one out of two customers is served snapper.
Make a tree using Paint.
16
Bayes’ Theorem. Suppose that 𝐵1 , 𝐵2 , … 𝐵𝑘 partition the sample space. A is any event in the
sample space. Then
𝑃(𝐵𝑖 /𝐴) =
𝑃(𝐴/𝐵𝑖 )𝑃(𝐵𝑖 )
𝑃(𝐴/𝐵1 )𝑃(𝐵1 ) + 𝑃(𝐴/𝐵2 )𝑃(𝐵2 ) + … 𝑃(𝐴/𝐵𝑘 )𝑃(𝐵𝑘 )
17
Example T-shirt factory: Betty makes 30% of the T-shirts; Tom makes 20% and Jane makes the
rest (50%). 2% of Betty’s T-shirts are defective; 3% of Tom’s T-shirts are defective and 5% of
Jane’s T-shirts are defective. Choose one T-shirt at random from the day’s production
(a)
(b)
(c)
(d)
Find the probability that it’s a defective T-shirt made by Betty.
Find the probability that it’s a defective T-shirt made by Jane.
Find the probability that it’s a defective T-shirt.
Suppose you find that the chosen T-shirt is defective. What’s the chance that Betty
made it?
Method I: Table
Betty makes 30% of the T-shirts; Tom makes 20% and Jane makes the rest (50%). 2% of Betty’s
T-shirts are defective; 3% of Tom’s T-shirts are defective and 5% of Jane’s T-shirts are
defective. Choose one T-shirt at random from the day’s production.
What is being classified? T-shirts
What questions do we need to ask ourselves in order to classify the T-shirts?
Who made it?
Is it defective?
Betty makes 30% of the T-shirts; Tom makes 20% and Jane makes the rest (50%). 2% of Betty’s
T-shirts are defective; 3% of Tom’s T-shirts are defective and 5% of Jane’s T-shirts are
defective. Choose one T-shirt at random from the day’s production.
Suppose we have 1000 T-shirts
Def
Not Def
Total
Betty
2% of 300=6
294
300
Tom
3% of 200=6
194
200
18
Jane
5% of 500 =25
475
500
Total
37
963
1000
(a)
Find the probability that it’s a defective t-shirt made by Betty.
Ans: P(Def made by Betty) =
(b)
Find the probability that it’s a defective t-shirt made by Jane.
Ans: P(Def made by Jane) =
(c)
25
1000
Find the probability that it’s a defective t-shirt.
Ans: P(Def) =
(d)
6
1000
37
1000
Suppose you find that the chosen t-shirt is defective. What’s the chance that Betty
made it?
Ans: P( Betty | Def ) =
6
37
Method II: Tree. Use Paint
19
Method III: formulae
Betty makes 30% of the T-shirts; Tom makes 20% and Jane makes the rest (50%). 2% of Betty’s
T-shirts are defective; 3% of Tom’s T-shirts are defective and 5% of Jane’s T-shirts are
defective. Choose one T-shirt at random from the day’s production.
Given: P(B) =.3 P(T) = .2 P(J) = .5
P(D|B) = .02 P(D|T)= .03 P(D|J) = .05
(a)
Find the probability that it’s a defective T-shirt made by Betty.
P(D and B) = P(D|B)P(B)=.02*.3=.006 Mult Rule
(b)
Find the probability that it’s a defective T-shirt made by Jane.
P(D and J) = P(D|J)P(B)=.05*.5=.025 Mult Rule
(c)
Find the probability that it’s a defective T-shirt.
Also P(D and T) =.2*.03=.006
P(D) = P(D and B)+P(D and T)+P(D and J) = .006+.006+.025= .037 Law of Total Prob
(d)
Suppose you find that the chosen T-shirt is defective. What’s the chance that Betty
made it?
P(B|D) = P(B and D)/ P(D) = .006/.037
Or Bayes’ Rule:
20
P(B) =.3 P(T) = .2 P(J) = .5
P(D|B) = .02 P(D|T)= .03 P(D|J) = .05
𝑃(𝐵/𝐷) =
𝑃(𝐷/𝐵)𝑃(𝐵)
=
𝑃(𝐷/𝐵)𝑃(𝐵) + 𝑃(𝐷/𝑇)𝑃(𝑇) + 𝑃(𝐷/𝐽)𝑃(𝐽)
.02∗.3
= .02∗.3+.03∗.2+.05∗.5 = .006/.037 as before.
Exercise 3.191 p181 Pregnancy tests Suppose that 75% of all women who take a pregnancy test
really is pregnant. There’s a 2% chance that a particular pregnancy test gives a false positive
result and a 99% chance that it gives a true positive result. Janet has received a positive result.
What’s the chance that she really is pregnant?
Solution: Table:
Tree:
21
Exercise 3.139 p172 Athlete doping Biostatisticians at the University of Texas demonstrated
the use of Bayes’ Rule in the context of testosterone abuse amongst Olympic athletes, reported in
Chance (Spring 2004): Suppose that 100 out of 1000 athletes are using testosterone illegally.
Suppose that 50 of the users would test positive for testosterone (ie 50 are true positives) and 9 of
the non-users would also test positive (ie 9 are false positives).
a) Find the sensitivity of this drug test—that is, Given that the athlete is a user, find the
probability of a positive result.
b) Find the specificity of this drug test – that is, Given that the athlete is a non-user, find the
probability of a negative result.
c) Find the positive predictive value of this drug test – that is, find the probability that an
athlete really is using testosterone illegally, given that he/she has received a positive test
result.
Solution:
Which method? (Table or tree or formulae?) Table is simplest
What are we classifying? Athletes
What questions will we ask about each athlete?
Does the athlete use testosterone? Use not use
What is the test result? Pos or neg
Suppose that 100 out of 1000 athletes are using testosterone illegally. Suppose that 50 of the
users would test positive for testosterone (ie 50 are true positives) and 9 of the non-users would
also test positive (ie 9 are false positives).
Use testosterone
Don’t use
Total
Pos result
50
9
59
Neg result
50
891
941
Total
100
900
1000
a) Find the sensitivity of this drug test—that is, Given that the athlete is a user, find the
probability of a positive result.
P( Pos| user) = 50/ 100 =.5 Not very sensitive because it picks up only half of them!
b) Find the specificity of this drug test – that is, Given that the athlete is a non-user, find the
probability of a negative result.
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P(neg result| non-user) = 891 /900
Use testosterone
Don’t use
Total
Pos result
50
9
59
Neg result
50
891
941
Total
100
900
1000
c) Find the positive predictive value of this drug test – that is, find the probability that an
athlete really is using testosterone illegally, given that he/she has received a positive test
result.
P( Using |Pos result
)= 50/59
Review Exercises Chapter 3 3.9, 3.11, 3.23, 3.47, 3.53, 3.55, 3.57, 3.59, 3.61, 3.71, 3.73, 3.79,
3.83, 3.85, 3.87, 3.91, 3.95, 3.97, 3.101, 3.141, 3.145, 3.163, 3.179, 3.181, 3.183, 3.189
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