Download Chapter 15

AP Statistics Objectives Ch15
Understand conditional
probability.
Understand concept of
independence.
Know how and when to apply
General Addition Rule.
Know how and when to apply
General Multiplication Rule.
AP Statistics Objectives Ch15
Know how to find probabilities
for compound events
Know how to make and use a tree
diagram to understand conditional
probabilities and reverse
conditioning.
Vocabulary
Sample Space
Conditional Probability
Independence
Tree Diagram
Chapter 15 Notes
Chp 15
Extra
Practice Practice
Chapter 15
Assignment
Chapter 14
Assignment Part 1
Answers
Chapter 14
Assignment Part 2
Answers
1. Addition Rule
The Addition Rule only works if the two events
are disjoint.
If two events A & B are disjoint,
P(A OR B) = P(A) + P(B)
A
B
1. Addition Rule
If two events A & B are disjoint,
P(A OR B) = P(A) + P(B)
A
B
2. General Addition Rule
For any two events A & B,
P(A OR B) = P(A) + P(B) – P(A & B)
A
B
Ex 1) A large auto center sells cars made by many
different manufacturers. Three of these are
Honda, Nissan, and Toyota.
Suppose that P(H) = .25, P(N) = .18, P(T) = .14.
a. Are these disjoint events?
Yes, the cars only have one manufacturer.
b. P(H or N or T) = .25 + .18+ .14 = .57
c. P(not (H or N or T)) = 1 - .57 = .43
Ex. 2) Musical styles other than rock and pop
are becoming more popular. A survey of
college students finds that the probability
they like country music is .40. The
probability that they liked jazz is .30 and
that they liked both is .10. What is the
probability that they like country or jazz?
Ex. 2) Musical styles other than rock and pop are becoming more
popular. A survey of college students finds that the probability they
like country music is .40. The probability that they liked jazz is
.30 and that they liked both is .10. What is the probability that they
like country or jazz?
a. Are these events disjoint?
They are not disjoint events. The students
can like both music types.
b. What is the probability that they like country or
jazz?
J
C
.3
.1
.2
P(C ∪
or J) = P(C) + P(J) – P(C ∩&J)
= .4 + .3 – .1 = .6
3. Independent
Two events are independent if knowing that one will
occur (or has occurred) does not change the
probability that the other occurs
– A randomly selected student is female.
What is the probability she plays soccer for SHS?
Independent,
P(student plays soccer) = P(stud. plays soccer given they are female).
– A randomly selected student is female.
What is the probability she plays football for SHS?
Not Independent,
P(student plays ftball) ≠ P(stud. plays ftball given they are female)
4. The Multiplication Rule only works
if the two events are independent of
each other.
P(A & B) = P(A) ∙ P(B)
4. The Multiplication Rule only works if the two
events are independent of each other.
P(A & B) = P(A) ∙ P(B)
5. General Multiplication Rule:
For any two events A & B,
P(A & B) = P(A) ∙ P(B|A)
P(B|A) is read…
“The probability of B given A”.
Ex. 3) A certain brand of light bulbs are
defective five percent of the time. You randomly
pick a package of two such bulbs off the shelf of
a store.
a. Can you assume they are independent?
Yes, since they are randomly selected.
b. What is the probability that both bulbs are
defective? P(Defect & Defect) =
= (0.05)(0.05)
= 0.0025
Ex. 3) A certain brand of light bulbs are defective five percent of the time. You
randomly pick a package of two such bulbs off the shelf of a store.
a. Can you assume they are independent?
Yes, since they are randomly selected.
b. What is the probability that both bulbs are defective?
0.95
2nd Bulb
good
0.05
2nd Bulb
defective
1st Bulb
0.95
0.05 defective
2nd Bulb
good
0.05
2nd Bulb
defective
0.95
1st Bulb
good
(0.05)(0.05)
= 0.0025
Ex. 3) A certain brand of light bulbs are defective five percent of the time. You
randomly pick a package of two such bulbs off the shelf of a store.
c. What is the probability that exactly one bulb is defective?
P(exactly one D) = P(D & DC) or P(DC & D)
= (.05)(.95) + (.95)(.05)
= 0.095
Ex. 3) A certain brand of light bulbs are defective five percent of the time. You
randomly pick a package of two such bulbs off the shelf of a store.
c. What is the probability that exactly one bulb is defective?
0.95
2nd Bulb
good
0.05
2nd Bulb
defective
(.95)(.05) =
0.475
1st Bulb
0.95
0.05 defective
2nd Bulb
good
(.05)(.95) =
+ 0.475
0.05
2nd Bulb
defective
0.95
1st Bulb
good
0.095
Ex. 3) A certain brand of light bulbs are defective five percent of the time. You
randomly pick a package of two such bulbs off the shelf of a store.
d) What is the probability that at least one bulb is defective?
P(at least one D) = P(D & DC) or P(DC & D) or (D & D)
= (.05)(.95) + (.95)(.05) + (.05)(.05)
= .0975
“At Least One”
The probability that at least one
outcome happens is 1 minus the
probability that no outcomes happen.
P(at least 1) = 1 – P(none)
Ex. 3 revisited) A certain brand of light bulbs are defective five percent of the time. You
randomly pick a package of two such bulbs off the shelf of a store.
d. What is the probability that at least one bulb is defective?
P(at least one D) = 1 – P(DC & DC)
= 1 – (.95*.95)
= .0975
Ex. 3) A certain brand of light bulbs are defective five percent of the time. You
randomly pick a package of two such bulbs off the shelf of a store.
d) What is the probability that at least one bulb is defective?
0.95
2nd Bulb
good
0.05
2nd Bulb
defective
(.95)(.05) =
0.0475
1st Bulb
0.95
0.05 defective
2nd Bulb
good
(.05)(.95) =
+ 0.0475
0.05
2nd Bulb
defective
(.05)(.05) =
+ 0.0025
0.95
1st Bulb
good
0.0975
Example 4
If P(A) = 0.45, P(B) = 0.35, and A & B are
independent. Find P(A or B).
a. Is A & B disjoint?
NO, independent events cannot be disjoint
So…. P(A or B) = P(A) + P(B) – P(A & B)
b. How do we find P(A & B)?
Because they are independent, we multiply.
So…. P(A or B) = P(A) + P(B) – P(A)∗ 𝐏(B)
P(A or B) = .45 + .35 - .45(.35) = 0.6425
Example 4 : If P(A) = 0.45, P(B) = 0.35, and A & B
are independent. Find P(A or B).
B
A
0.2925
0.1575
0.1925
P(A&B) = P(A)*P(B)
= (0.45)(0.35)
= 0.1575
0.2925
+ 0.1575
+ 0.1925
0.45 – 0.1575 = 0.2925
0.35 – 0.1575 = 0.1925
0.6425
Ex 5) Suppose I will pick two cards from a standard deck
without replacement. What is the probability that I select two
spades?
a. Are the cards independent?
NO, once I select 1 card the probability changes.
P(1st Spade) = 13/52
P(2nd Spade | Spade 1st) = 12/51
“Probability of 2nd Spade given Spade 1st “
b. P(Spade & Spade) = P(1st Spade) · P(2nd Spade | Spade 1st )
= 13/52 · 12/51 = 0.059
Ex 6) For a sales promotion the manufacturer
places winning symbols under the caps of 10% of
all Dr. Pepper bottles. You buy a six-pack. What
is the probability that you win something?
P(at least one winning symbol) =
1 – P(no winning symbols)
1 - .96 = .4686
6. Conditional Probability
A probability that takes into account a
given condition
P(B|A) =
𝑷(𝑨∩𝑩)
𝑷(𝑨)
=
𝑨𝑵𝑫
𝑮𝑰𝑽𝑬𝑵
Ex 7) a. In a recent study it was found
that the probability that a randomly
selected student is a girl is .51 and is a
girl and plays sports is .10. If the
student is female, what is the probability
that she plays sports?
P(Sport | Female) =
𝑷(𝑺𝒑𝒐𝒓𝒕 ∩ 𝑭𝒆𝒎𝒂𝒍𝒆)
𝑷(𝑭𝒆𝒎𝒂𝒍𝒆)
=
.𝟏
.𝟓𝟏
= .1961
Ex 7) a. In a recent study it was found that the probability that a
randomly selected student is a girl is .51 and is a girl and plays
sports is .10. If the student is female, what is the probability that she
plays sports?
0.51 – 0.10 = 0.41
Sport NO Sport
Female 0.10
0.41
Male
.𝟏
= .1961
. 𝟓𝟏
0.51
0.49
1.00
1 – 0.51 = 0.49
Ex 7) b. The probability that a randomly
selected student plays sports if they are
male is .31. What is the probability that the
student is male and plays sports if the
probability that they are male is .49?
P(Sport | Male) =
𝑷(𝑺𝒑𝒐𝒓𝒕 ∩𝑴𝒂𝒍𝒆)
𝑷(𝑴𝒂𝒍𝒆)
.31 =
𝒙
.𝟒𝟗
.1519 = x
Ex 7) b. The probability that a randomly selected student
plays sports if they are male is .31. What is the probability
that the student is male and plays sports if the probability
that they are male is .49?
Sport NO Sport
Female 0.10
0.41
Male
𝒙
= . 𝟑𝟏
. 𝟒𝟗
x = .1519
0.51
0.49
1.00
OK. Let’s finish the table…
Sport NO Sport
Female 0.10
0.41
Male 0.1519 0.3381
0.2519 0.7481
0.49 – 0.1519 = 0.3381
0.51
0.49
1.00
0.10 + 0.1519 = 0.2519
0.41 + 0.3381 = 0.7481
c. What’s the probability that someone who plays sports is
female?
Sport NO Sport
Female 0.10
0.41
Male 0.1519 0.3381
0.2519 0.7481
P(Female | Sports) =
=
0.51
0.49
1.00
𝑷(𝑺𝒑𝒐𝒓𝒕 ∩ 𝑭𝒆𝒎𝒂𝒍𝒆)
𝑷(𝑺𝒑𝒐𝒓𝒕)
.𝟏
.𝟐𝟓𝟏𝟗
= .397
Example 8
Probabilities from two way tables
American
European
Asian
Total
Stu
107
33
55
195
Staff
105
12
47
164
Total
212
45
102
359
a. What is the probability that the driver is a
student?
195
P (Student ) 
359
≈ 𝟓𝟒. 𝟑%
Example 8
Probabilities from two way tables
American
European
Asian
Total
Stu
107
33
55
195
Staff
105
12
47
164
Total
212
45
102
359
b. What is the probability that the driver drives
a European car?
45
P (European ) 
359
≈ 𝟏𝟐. 𝟓%
Disjoint so: P(Am or Asian) = P(Am) + P(Asian)
Probabilities from two way tables
American
European
Asian
Total
Stu
107
33
55
195
Staff
105
12
47
164
Total
212
45
102
359
c. What is the probability that the driver drives an
American or Asian car?
Disjoint
212  102 314
P( American or Asian ) 

359
359
≈ 𝟖𝟕. 𝟓%
Not Disjoint so:
P(Staff or Asian) = P(Staff) + P(Asian) – P(Staff & Asian)
Probabilities from two way tables
American
European
Asian
Total
Stu
107
33
55
195
Staff
105
12
47
164
Total
212
45
102
359
d. What is the probability that the driver is staff or
drives an Asian car?
Not Disjoint
164 + 102 − 47
𝑃 𝑆𝑡𝑎𝑓𝑓 𝑜𝑟 𝐴𝑠𝑖𝑎𝑛 =
359
=
𝟐𝟏𝟗
𝟑𝟓𝟗
≈ 𝟔𝟏%
Joint Distribution:
P(Staff and Asian Car)
Probabilities from two way tables
American
European
Asian
Total
Stu
107
33
55
195
Staff
105
12
47
164
Total
212
45
102
359
e. What is the probability that the driver is staff
and drives an Asian car?
𝟒𝟕
𝑃 𝑆𝑡𝑎𝑓𝑓 𝑎𝑛𝑑 𝐴𝑠𝑖𝑎𝑛 =
𝟑𝟓𝟗
≈ 𝟏𝟑. 𝟏%
Conditional Distribution:
P(Am Car|Student)
Probabilities from two way tables
American
European
Asian
Total
Stu
107
33
55
195
Staff
105
12
47
164
Total
212
45
102
359
f. If the driver is a student, what is the probability
that they drive an American car?
𝟏𝟎𝟕
𝑃 𝐴𝑚𝑒𝑟𝑖𝑐𝑎𝑛|𝑆𝑡𝑢𝑑𝑒𝑛𝑡 =
≈
𝟏𝟗𝟓
𝟓𝟒. 𝟗%
Conditional Distribution:
P(Student|Euro Car)
Probabilities from two way tables
American
European
Asian
Total
Stu
107
33
55
195
Staff
105
12
47
164
Total
212
45
102
359
g. What is the probability that the driver is a
student if the driver drives a European car?
𝟑𝟑
𝑃 𝑆𝑡𝑢𝑑𝑒𝑛𝑡|𝐸𝑢𝑟𝑜𝑝𝑒𝑎𝑛 =
𝟒𝟓
≈ 𝟕𝟑. 𝟑%
Time for Final Word on
INDEPENDENCE
7. We know that two events are
independent if the probability of
one doesn’t change just because
of the other event.
If Event A and Event B are
Independent, then…
𝑃 𝐴 ∩ 𝐵 = 𝑃 𝐴 ∗ 𝑃(𝐵)
𝑃 𝐵
𝑃(𝐴|𝐵)
So if
𝑃 𝐵
=
𝑃 𝐴
P(A) = P(A|B)
then Events A and B are independent
American
European
Asian
Total
Stu
107
33
55
195
Staff
105
12
47
164
Total
212
45
102
359
Probability
P(A)
8. Parts of a Tree Diagram
P(B|A)
Conditional
Probability
American & Student
107/359 29.81%
50.5%
49.5%
59.1%
European
45/359
P(A)*P(B|A)
Joint Probability
12.5%
28.4%
American & Staff
105/359 29.25%
European & Student
33/359 9.19%
73.3%
26.7%
Student
55/102 53.9%
46.1%
European & Staff
12/359 3.34%
Asian & Student
55/359 15.32%
Asian & Staff
47/359 13.09%
9. Draw with Replacement
Blue & Blue
25.0%
Blue & Green
25.0%
Green & Blue
25.0%
Green & Green
25.0%
10. Draw without Replacement
Blue & Blue
17%
Blue & Green
33%
Green & Blue
33%
Green & Green
17%
Example 9:
Management has determined that customers return
12% of the items assembled by inexperienced
employees, whereas only 3% of the items assembled
by experienced employees are returned. Due to
turnover and absenteeism at an assembly plant,
inexperienced employees assemble 20% of the items.
Construct a tree diagram or a chart for this data.
What is the probability that an item is returned?
If an item is returned, what is the probability that an
inexperienced employee assembled it?
Example 9: a. Construct a tree diagram.
Management has determined that customers return 12% of the items assembled by
inexperienced employees, whereas only 3% of the items assembled by experienced
employees are returned. Due to turnover and absenteeism at an assembly plant,
inexperienced employees assemble 20% of the items. a. Construct a tree diagram
for this data.
P(Inexp. AND Returned) =
= .20 * .12 = .024 = 2.4%
P(Inexp. AND Not Returned) =
= .20 * .88 = .176 = 17.6%
P(Exp. AND Returned) =
= .80 * .03 = .024 = 2.4%
P(Exp. AND Not Returned) =
= .80 * .97 = .776 = 77.6%
Example 9:
Management has determined that customers return 12% of the items
assembled by inexperienced employees, whereas only 3% of the items
assembled by experienced employees are returned. Due to turnover and
absenteeism at an assembly plant, inexperienced employees assemble
20% of the items. Construct a tree diagram or a chart for this data.
b. What is the probability that an item is returned?
P(Return) = P(Return & Inexp.) + P(Return & Exp.)
= 2.4% + 2.4% = 4.8%
Example 9:
Management has determined that customers return 12% of the
items assembled by inexperienced employees, whereas only 3% of
the items assembled by experienced employees are returned. Due to
turnover and absenteeism at an assembly plant, inexperienced
employees assemble 20% of the items. Construct a tree diagram or
a chart for this data.
c. If an item is returned, what is the probability that
an inexperienced employee assembled it?
P(Inexp. | Returned) =
Example 9:
Management has determined that customers return 12% of the items assembled by
inexperienced employees, whereas only 3% of the items assembled by experienced
employees are returned. Due to turnover and absenteeism at an assembly plant,
inexperienced employees assemble 20% of the items..
P(Returned | Inexp.) = .12
Not same as P(Inexp. | Returned)
P(Not Returned | Inexp.) = .88
P(Returned | Exp.) = .03
P(Not Returned | Exp.) = .97
Example 9:
Management has determined that customers return 12% of the
items assembled by inexperienced employees, whereas only 3% of
the items assembled by experienced employees are returned. Due to
turnover and absenteeism at an assembly plant, inexperienced
employees assemble 20% of the items. Construct a tree diagram or
a chart for this data.
c. If an item is returned, what is the probability that
an inexperienced employee assembled it?
P(Inexp. | Returned) = 𝑷(𝑰𝒏𝒆𝒙𝒑.∩ 𝑹𝒆𝒕𝒖𝒓𝒏𝒆𝒅)
𝑷(𝑹𝒆𝒕𝒖𝒓𝒏𝒆𝒅)
=
.𝟎𝟐𝟒
.𝟎𝟒𝟖
= .5 = 50%
Example 10
Dr. Carey has two bottles of sample pills on his desk for
the treatment of arthritic pain. He often grabs a bottle
without looking and takes the medicine. Since the first
bottle is closer to him, the chances of grabbing it are
0.60. He knows the medicine from this bottle relieves
the pain 70% of the time while the medicine in the
second bottle relieves the pain 90% of the time. What is
the probability that Dr. Carey grabbed the first bottle
given his pain was not relieved?
Example 10
st
P(1
bottle  not relieved)
st
P(1 bottle|pain not relieved) 
P(pain not relieved)
.7
relieved
.3
not
1st
.6
.9
.4
relieved
2nd
.1
not
Example 10
st
P(1 bottle|pain not relieved)
=
=
.7
𝑷 𝟏𝒔𝒕 𝒃𝒐𝒕𝒕𝒍𝒆 ∩𝑵𝒐𝒕 𝒓𝒆𝒍𝒊𝒆𝒗𝒆𝒅
𝑃 𝑝𝑎𝑖𝑛 𝑁𝑜𝑡 𝑟𝑒𝑙𝑖𝑒𝑣𝑒𝑑
𝟎.𝟏𝟖
relieved
1st
.6
.3
.9
.4
not
= (.6)(.3) = 0.18
relieved
2nd
.1
𝑷 𝟏𝒔𝒕 𝒃𝒐𝒕𝒕𝒍𝒆 ∩ 𝑵𝒐𝒕 𝒓𝒆𝒍𝒊𝒆𝒗𝒆𝒅
not
Example 10
st
P(1 bottle|pain not relieved)
=
=
.7
𝑷 𝟏𝒔𝒕 𝒃𝒐𝒕𝒕𝒍𝒆 ∩𝑵𝒐𝒕 𝒓𝒆𝒍𝒊𝒆𝒗𝒆𝒅
𝑷 𝒑𝒂𝒊𝒏 𝑵𝒐𝒕 𝒓𝒆𝒍𝒊𝒆𝒗𝒆𝒅
𝟎.𝟏𝟖
≈ 𝟎. 𝟖𝟏𝟖2
𝟎.𝟐𝟐
relieved
1st
.6
.3
.9
.4
not
𝑷 𝟏𝒔𝒕 𝒃𝒐𝒕𝒕𝒍𝒆 ∩ 𝑵𝒐𝒕 𝒓𝒆𝒍𝒊𝒆𝒗𝒆𝒅
= (.6)(.3) = 0.18
relieved 𝑷 𝑵𝒐𝒕 𝒓𝒆𝒍𝒊𝒆𝒗𝒆𝒅 =.18 +.04 = 0.22
2nd
𝑷 𝟐𝒏𝒅 𝒃𝒐𝒕𝒕𝒍𝒆 ∩ 𝑵𝒐𝒕 𝒓𝒆𝒍𝒊𝒆𝒗𝒆𝒅
.1
not
= (.4)(.1) = 0.04
Example 10
st
P(1 bottle|pain not relieved)
=
=
.7
𝑷 𝟏𝒔𝒕 𝒃𝒐𝒕𝒕𝒍𝒆 ∩𝑵𝒐𝒕 𝒓𝒆𝒍𝒊𝒆𝒗𝒆𝒅
𝑷 𝒑𝒂𝒊𝒏 𝑵𝒐𝒕 𝒓𝒆𝒍𝒊𝒆𝒗𝒆𝒅
𝟎.𝟏𝟖
≈ 𝟎. 𝟖𝟏𝟖2
𝟎.𝟐𝟐
relieved
1st
.6
.3
.9
.4
not
𝑷 𝟏𝒔𝒕 𝒃𝒐𝒕𝒕𝒍𝒆 ∩ 𝑵𝒐𝒕 𝒓𝒆𝒍𝒊𝒆𝒗𝒆𝒅
= (.6)(.3) = 0.18
relieved 𝑷 𝑵𝒐𝒕 𝒓𝒆𝒍𝒊𝒆𝒗𝒆𝒅 =.18 +.04 = 0.22
2nd
𝑷 𝟐𝒏𝒅 𝒃𝒐𝒕𝒕𝒍𝒆 ∩ 𝑵𝒐𝒕 𝒓𝒆𝒍𝒊𝒆𝒗𝒆𝒅
.1
not
= (.4)(.1) = 0.04
Chp 15 p.362 #3
Real estate ads suggest that 64% of homes for sale have
garages, 21% have swimming pools, and 17% have both
features. What is the probability that a home for sale has:
a) a pool or a garage?
NOT DISJOINT
P(Garage U Pool) = P(G) + P(Pool) – P(G
∩ Pool)
= .64 + .21 - .17 = 68%
Pool
Garage
47%
17%
4%
64% - 17% = 47%
21% - 17% = 4%
P(Garage U Pool) = .47 + .17 + .04 = 68%
Chp 15 p.362 #3
Real estate ads suggest that 64% of homes for sale have garages, 21% have
swimming pools, and 17% have both features. What is the probability that a home
for sale has:
b) Neither a pool nor a garage?
P(NOT(Garage U Pool)) = P((Garage U Pool)C)
= 1 – P(Garage U Pool)
= 1 – .68 = 32%
Chp 15 p.362 #3
Real estate ads suggest that 64% of homes for sale have garages, 21% have
swimming pools, and 17% have both features. What is the probability that a home
for sale has:
c) a pool but no garage?
P(Pool ∩ Garagec) = 4%
Chp 15 p.363 #5
A check of dorm rooms on a large college campus revealed
that 38% had refrigerators, 52% had TVs, and 21% had
both a TV and a refrigerator. What’s the probability that a
randomly selected dorm room has :
a) A TV but no refrigerator?
Frig
TV
.31
.21
.17
52% - 21% = 31%
38% - 21% = 17%
P(TV ∩ Frigc) = 31%
Chp 15 p.363 #5
A check of dorm rooms on a large college campus revealed that 38% had
refrigerators, 52% had TVs, and 21% had both a TV and a refrigerator. What’s
the probability that a randomly selected dorm room has :
b) A TV or a refrigerator, but not both?
P(TV XOR Refrig) =
17% + 31% = 48%
XOR:
“Exclusive OR” used in Electrical Engineering. Not needed for AP Stats
Chp 15 p.363 #5
A check of dorm rooms on a large college campus revealed that 38% had
refrigerators, 52% had TVs, and 21% had both a TV and a refrigerator. What’s
the probability that a randomly selected dorm room has :
c) Neither a TV nor a refrigerator?
P(NOT(TV U Refrig)) =
= 1 – P(TV U Refrig)
= 1 – (.17 + .21 + .31)
= 1 - .69
= .31
Chp 15 p.363 #11
The probabilities that an adult American man has high
blood pressure and /or cholesterol are shown in the table.
Cholesterol
Blood Pressure
High
OK
Total
High
0.11
0.21
0.32
OK
0.16
0.52
0.68
0.27
0.73
1.00
Total
Chp 15 p.363 #11
Cholesterol
The probabilities that an adult American man has high blood pressure
and /or cholesterol are shown in the table.
Blood Pressure
High
OK
High
0.11
0.21
0.32
OK
0.16
0.52
0.68
0.27
0.73
1.00
Total
Total
a) What’s the
probability that a
man has both
conditions?
P(High BP ∩ High Chol) = 11%
Chp 15 p.363 #11
Cholesterol
The probabilities that an adult American man has high blood pressure
and /or cholesterol are shown in the table.
Blood Pressure
High
OK
High
0.11
0.21
0.32
OK
0.16
0.52
0.68
0.27
0.73
1.00
Total
Total
P(High BP) = 27%
b) What’s the
probability that he
has high blood
pressure?
Chp 15 p.363 #11
Cholesterol
The probabilities that an adult American man has high blood pressure
and /or cholesterol are shown in the table.
Blood Pressure
High
OK
High
0.11
0.21
0.32
OK
0.16
0.52
0.68
0.27
0.73
1.00
Total
Total
c) What’s the
probability that a
man with high
blood pressure has
high cholesterol?
Chp 15 p.363 #11
Cholesterol
The probabilities that an adult American man has high blood pressure
and /or cholesterol are shown in the table.
Blood Pressure
High
OK
High
0.11
0.21
0.32
OK
0.16
0.52
0.68
0.27
0.73
1.00
Total
Total
P(High Chol | High BP) =
c) What’s the
probability that a
man with high
blood pressure has
high cholesterol?
𝑨𝑵𝑫
𝑮𝑰𝑽𝑬𝑵
=
.𝟏𝟏
= 40.7%
.𝟐𝟕
Chp 15 p.363 #11
Cholesterol
The probabilities that an adult American man has high blood pressure
and /or cholesterol are shown in the table.
Blood Pressure
High
OK
High
0.11
0.21
0.32
OK
0.16
0.52
0.68
0.27
0.73
1.00
Total
d) What’s the
probability that a man
has high blood
pressure if it’s known
that he has high
cholesterol?
Chp
15
p.363
#11
The probabilities that an adult American man has high blood pressure
Cholesterol
and /or cholesterol are shown in the table.
Blood Pressure
High
OK
High
0.11
0.21
0.32
OK
0.16
0.52
0.68
0.27
0.73
1.00
Total
P(High BP | High Chol) =
=
d) What’s the
probability that a man
has high blood
pressure if it’s known
that he has high
cholesterol?
𝑨𝑵𝑫
𝑮𝑰𝑽𝑬𝑵
.𝟏𝟏
= 34.37%
.𝟑𝟐
Chp
15
p.363
#11
The probabilities that an adult American man has high blood pressure
Cholesterol
and /or cholesterol are shown in the table.
Blood Pressure
High
OK
High
0.11
0.21
0.32
OK
0.16
0.52
0.68
0.27
0.73
1.00
Total
P(High BP | High Chol) =
=
d) What’s the
probability that a man
has high blood
pressure if it’s known
that he has high
cholesterol?
𝑨𝑵𝑫
𝑮𝑰𝑽𝑬𝑵
.𝟏𝟏
= 34.37%
.𝟑𝟐
Chapter 15 Assignment
pp. 363-364 #8&14, 22
pp. 365-366 #28, 36&38, 43
6. Commercial airplanes have an excellent safety record. Nevertheless,
there are crashes occasionally, with the loss of many lives. In the weeks
following a crash, airlines often report a drop in the number of
passengers, probably because people are afraid to risk flying.
a) A travel agent suggests that, since the law of averages makes it
highly unlikely to have two plane crashes within a few weeks of each
other, flying soon after a crash is the safest time. What do you think?
6. Commercial airplanes have an excellent safety record. Nevertheless,
there are crashes occasionally, with the loss of many lives. In the weeks
following a crash, airlines often report a drop in the number of
passengers, probably because people are afraid to risk flying.
b) If the airline industry proudly announces that it has set a new
record for the longest period of safe flights, would you be reluctant
to fly? Are the airlines due to have a crash?
8. On January 20, 2000, the International Gaming Technology company
issued a press release:
(LAS VEGAS, Nev.)—Cynthia Jay was smiling ear to ear as she walked
into the news conference at The Desert Inn Resort in Las Vegas today,
and well she should. Last night, the 37-year-old cocktail waitress won
the world’s largest slot jackpot—$34,959,458—on a Megabucks
machine. She said she had played $27 in the machine when the jackpot
hit. Nevada Megabucks has produced 49 major winners in its 14-year
history. The top jackpot builds from a base amount of $7 million and can
be won with a 3-coin ($3) bet.
a) How can the Desert Inn afford to give away millions of dollars on
a $3 bet?
8. On January 20, 2000, the International Gaming Technology company
issued a press release:
(LAS VEGAS, Nev.)—Cynthia Jay was smiling ear to ear as she walked
into the news conference at The Desert Inn Resort in Las Vegas today,
and well she should. Last night, the 37-year-old cocktail waitress won
the world’s largest slot jackpot—$34,959,458—on a Megabucks
machine. She said she had played $27 in the machine when the jackpot
hit. Nevada Megabucks has produced 49 major winners in its 14-year
history. The top jackpot builds from a base amount of $7 million and can
be won with a 3-coin ($3) bet.
b) Why did the company issue a press release? Wouldn’t most
businesses want to keep such a huge loss quiet?
12&14. In a large Introductory Statistics lecture hall, the professor
reports that 55% of the students enrolled have never taken a Calculus
course, 32% have taken only one semester of Calculus, and the rest have
taken two or more semesters of Calculus. The professor randomly
assigns students to groups of three to work on a project for the course.
i. What is the probability that the first groupmate you meet has studied
a) two or more semesters of Calculus?
12&14. In a large Introductory Statistics lecture hall, the professor reports that 55% of the
students enrolled have never taken a Calculus course, 32% have taken only one semester of
Calculus, and the rest have taken two or more semesters of Calculus. The professor randomly
assigns students to groups of three to work on a project for the course.
i. What is the probability that the first groupmate you meet has studied
a) two or more semesters of Calculus?
2+ Sem
1 Sem
55 + 32 = 87
13%
32%
100 – 87 = 13%
55%
12&14. In a large Introductory Statistics lecture hall, the professor
reports that 55% of the students enrolled have never taken a Calculus
course, 32% have taken only one semester of Calculus, and the rest have
taken two or more semesters of Calculus. The professor randomly
assigns students to groups of three to work on a project for the course.
i. What is the probability that the first groupmate you meet has studied
b) some Calculus?
12&14. In a large Introductory Statistics lecture hall, the professor reports that 55% of the
students enrolled have never taken a Calculus course, 32% have taken only one semester of
Calculus, and the rest have taken two or more semesters of Calculus. The professor randomly
assigns students to groups of three to work on a project for the course.
i. What is the probability that the first groupmate you meet has studied
b) some Calculus?
2+ Sem
1 Sem
13%
32%
55%
55 + 13 = 68%
12&14. In a large Introductory Statistics lecture hall, the professor
reports that 55% of the students enrolled have never taken a Calculus
course, 32% have taken only one semester of Calculus, and the rest have
taken two or more semesters of Calculus. The professor randomly
assigns students to groups of three to work on a project for the course.
i. What is the probability that the first groupmate you meet has studied
c) no more than one semester of Calculus?
12&14. In a large Introductory Statistics lecture hall, the professor reports that 55% of the
students enrolled have never taken a Calculus course, 32% have taken only one semester of
Calculus, and the rest have taken two or more semesters of Calculus. The professor randomly
assigns students to groups of three to work on a project for the course.
i. What is the probability that the first groupmate you meet has studied
c) no more than one semester of Calculus?
1 Sem
2+ Sem
13%
32%
55%
55 + 32 = 87%
12&14. In a large Introductory Statistics lecture hall, the professor
reports that 55% of the students enrolled have never taken a Calculus
course, 32% have taken only one semester of Calculus, and the rest have
taken two or more semesters of Calculus. The professor randomly
assigns students to groups of three to work on a project for the course.
ii. What is the probability that of your other two groupmates,
a) neither has studied Calculus?
12&14. In a large Introductory Statistics lecture hall, the professor
reports that 55% of the students enrolled have never taken a Calculus
course, 32% have taken only one semester of Calculus, and the rest have
taken two or more semesters of Calculus. The professor randomly
assigns students to groups of three to work on a project for the course.
ii. What is the probability that of your other two groupmates,
b) both have studied at least one semester of Calculus?
12&14. In a large Introductory Statistics lecture hall, the professor
reports that 55% of the students enrolled have never taken a Calculus
course, 32% have taken only one semester of Calculus, and the rest have
taken two or more semesters of Calculus. The professor randomly
assigns students to groups of three to work on a project for the course.
ii. What is the probability that of your other two groupmates,
c) at least one has had more than one semester of Calculus?
18&20. A Gallup Poll in June 2004 asked 1005 U.S. adults how likely
they were to read Bill Clinton’s autobiography My Life. Here’s how they
responded:
Response
Number
Will definitely read it
90
Will probably read it
211
Will probably not read it
322
Will definitely not read it
382
i. If we select a person at random from this sample of 1005 adults,
a) what is the probability that the person responded “Will definitely
not read it”?
18&20. A Gallup Poll in June 2004 asked 1005 U.S. adults how likely
they were to read Bill Clinton’s autobiography My Life. Here’s how they
responded:
Response
Number
Will definitely read it
90
301
Will probably read it
211
Will probably not read it
322
Will definitely not read it
382
i. If we select a person at random from this sample of 1005 adults,
b) what is the probability that the person will probably or definitely
read it?
18&20. A Gallup Poll in June 2004 asked 1005 U.S. adults how likely
they were to read Bill Clinton’s autobiography My Life. Here’s how they
responded:
Response
Number
Will definitely read it
90
301
Will probably read it
211
Will probably not read it
322
Will definitely not read it
382
ii. Let’s call someone who responded that they would definitely or
probably read it a “likely reader” and the other two categories, “unlikely
readers”. If we select two people at random from this sample,
a) what is the probability that both are likely readers?
18&20. A Gallup Poll in June 2004 asked 1005 U.S. adults how likely
they were to read Bill Clinton’s autobiography My Life. Here’s how they
responded:
Response
Number
Will definitely read it
90
301
Will probably read it
211
Will probably not read it
322
704
Will definitely not read it
382
ii. Let’s call someone who responded that they would definitely or
probably read it a “likely reader” and the other two categories, “unlikely
readers”. If we select two people at random from this sample,
b) what is the probability that neither is a likely reader?
18&20. A Gallup Poll in June 2004 asked 1005 U.S. adults how likely
they were to read Bill Clinton’s autobiography My Life. Here’s how they
responded:
Response
Number
Will definitely read it
90
Will probably read it
211
Will probably not read it
322
Will definitely not read it
382
ii. Let’s call someone who responded that they would definitely or
probably read it a “likely reader” and the other two categories, “unlikely
readers”. If we select two people at random from this sample,
c) what is the probability that one is a likely reader and one isn’t?
Likely Reader
Likely Reader
Not Likely Reader
(301/1005)*(301/1005) (704/1005)*(301/1005)
Not Likely Reader
(301/1005)*(704/1005) (704/1005)*(704/10050
18&20. A Gallup Poll in June 2004 asked 1005 U.S. adults how likely
they were to read Bill Clinton’s autobiography My Life. Here’s how they
responded:
Response
Number
Will definitely read it
90
Will probably read it
211
Will probably not read it
322
Will definitely not read it
382
ii. Let’s call someone who responded that they would definitely or
probably read it a “likely reader” and the other two categories, “unlikely
readers”. If we select two people at random from this sample,
d) What assumption did you make in computing the probabilities?
18&20. A Gallup Poll in June 2004 asked 1005 U.S. adults how likely
they were to read Bill Clinton’s autobiography My Life. Here’s how they
responded:
Response
Number
Will definitely read it
90
Will probably read it
211
Will probably not read it
322
Will definitely not read it
382
ii. Let’s call someone who responded that they would definitely or
probably read it a “likely reader” and the other two categories, “unlikely
readers”. If we select two people at random from this sample,
e) Explain why you think that assumption is reasonable.
18&20. A Gallup Poll in June 2004 asked 1005 U.S. adults how likely
they were to read Bill Clinton’s autobiography My Life. Here’s how they
responded:
Response
Number
Will definitely read it
90
Will probably read it
211
Will probably not read it
322
Will definitely not read it
382
ii. Let’s call someone who responded that they would definitely or
probably read it a “likely reader” and the other two categories, “unlikely
readers”. If we select two people at random from this sample,
e) Explain why you think that assumption is reasonable.
24. The American Red Cross says that aout 45% of the U.S.
population has Type O blood, 40% Type A, 11% Type B, and
the rest Type AB.
a) Someone volunteers to give blood. What is the probability
that this donor
i. has Type AB blood?
24. The American Red Cross says that aout 45% of the U.S.
population has Type O blood, 40% Type A, 11% Type B, and
the rest Type AB.
a) Someone volunteers to give blood. What is the probability
that this donor
ii. has Type A or Type B?
24. The American Red Cross says that aout 45% of the U.S.
population has Type O blood, 40% Type A, 11% Type B, and
the rest Type AB.
a) Someone volunteers to give blood. What is the probability
that this donor
iii. is not Type O?
24. The American Red Cross says that aout 45% of the U.S.
population has Type O blood, 40% Type A, 11% Type B, and
the rest Type AB.
b) Among four potential donors what is the probability that
i. all are Type O?
24. The American Red Cross says that aout 45% of the U.S.
population has Type O blood, 40% Type A, 11% Type B, and
the rest Type AB.
b) Among four potential donors what is the probability that
ii. no one is Type AB?
24. The American Red Cross says that aout 45% of the U.S.
population has Type O blood, 40% Type A, 11% Type B, and
the rest Type AB.
b) Among four potential donors what is the probability that
iii. they are not all Type A?
24. The American Red Cross says that aout 45% of the U.S.
population has Type O blood, 40% Type A, 11% Type B, and
the rest Type AB.
b) Among four potential donors what is the probability that
iv. least one person is Type B?
26. Some of your answers above depended on the assumption
that the outcomes described were disjoint; that is, they could
not both happen at the same time. Other answers depended on
the assumption that the events were independent; that is, the
occurrence of one of them doesn’t affect the probability of the
other. Do you understand the difference between disjoint and
independent?
a) If you examine one person, are the events that the
person is Type A and that the person is Type B disjoint or
independent or neither.
26. Some of your answers above depended on the assumption
that the outcomes described were disjoint; that is, they could
not both happen at the same time. Other answers depended on
the assumption that the events were independent; that is, the
occurrence of one of them doesn’t affect the probability of the
other. Do you understand the difference between disjoint and
independent?
b) If you examine two people, are the events that the first is
Type A and the second Type B disjoint or independent or
neither?
26. Some of your answers above depended on the assumption
that the outcomes described were disjoint; that is, they could
not both happen at the same time. Other answers depended on
the assumption that the events were independent; that is, the
occurrence of one of them doesn’t affect the probability of the
other. Do you understand the difference between disjoint and
independent?
c) Can disjoint events ever be independent? Explain.
30. To get to work, a commuter must cross train tracks. The
time the train arrives varies slight from day to day, but the
commuter estimates he’ll get stopped on about 15% of work
days. During a certain 5-day work week, what is the
probability that he
a) gets stopped on Monday and again on Tuesday?
30. To get to work, a commuter must cross train tracks. The
time the train arrives varies slight from day to day, but the
commuter estimates he’ll get stopped on about 15% of work
days. During a certain 5-day work week, what is the
probability that he
b) gets stopped for the first time on Thursday?
30. To get to work, a commuter must cross train tracks. The
time the train arrives varies slight from day to day, but the
commuter estimates he’ll get stopped on about 15% of work
days. During a certain 5-day work week, what is the
probability that he
c) get stopped every day?
30. To get to work, a commuter must cross train tracks. The
time the train arrives varies slight from day to day, but the
commuter estimates he’ll get stopped on about 15% of work
days. During a certain 5-day work week, what is the
probability that he
d) gets stopped at least once during the week?
32. Census reports for a city indicate that 62% of residents
classify themselves as Christian, 12% as Jewish, and 16% as
members of other religions (Muslims, Buddhists, etc.). The
remaining residents classify themselves as non-religious. A
polling organization seeking information about public
opinions wants to be sure to talk with people holding a variety
of religious views, and makes random phone calls. Among the
first four people they call, what is the probability they reach
a) all Christians?
32. Census reports for a city indicate that 62% of residents
classify themselves as Christian, 12% as Jewish, and 16% as
members of other religions (Muslims, Buddhists, etc.). The
remaining residents classify themselves as non-religious. A
polling organization seeking information about public
opinions wants to be sure to talk with people holding a variety
of religious views, and makes random phone calls. Among the
first four people they call, what is the probability they reach
b) no Jews?
32. Census reports for a city indicate that 62% of residents
classify themselves as Christian, 12% as Jewish, and 16% as
members of other religions (Muslims, Buddhists, etc.). The
remaining residents classify themselves as non-religious. A
polling organization seeking information about public
opinions wants to be sure to talk with people holding a variety
of religious views, and makes random phone calls. Among the
first four people they call, what is the probability they reach
c) at least one person who is non-religious?
36. You shuffle a deck of cards, and then start turning them
over one at a time. The first one is red. So is the second. And
the third. In fact, you are surprised to get 10 red cards in a row.
You start thinking, “The next one is due to be black!”
a) Are you correct in thinking, that there’s a higher
probability that the next card will be black than red?
Explain.
36. You shuffle a deck of cards, and then start turning them
over one at a time. The first one is red. So is the second. And
the third. In fact, you are surprised to get 10 red cards in a row.
You start thinking, “The next one is due to be black!”
b) Is this an example of the Law of Large Numbers?
Explain.
36. You shuffle a deck of cards, and then start turning them
over one at a time. The first one is red. So is the second. And
the third. In fact, you are surprised to get 10 red cards in a row.
You start thinking, “The next one is due to be black!”
b) Is this an example of the Law of Large Numbers?
Explain.
Extra Practice
Sixty seniors at a local high school are taking AP
Statistics. Forty seniors at the same high school
are taking AP Economics. If ten seniors at the
high school take both classes and there are 500
seniors, answer the following questions:
A. P(Stat) =
Extra Practice
Sixty seniors at a local high school are taking AP
Statistics. Forty seniors at the same high school
are taking AP Economics. If ten seniors at the
high school take both classes and there are 500
seniors, answer the following questions:
A. P(Stat) =
60
500
=
6
50
= 12%
Extra Practice
Sixty seniors at a local high school are taking AP
Statistics. Forty seniors at the same high school
are taking AP Economics. If ten seniors at the
high school take both classes and there are 500
seniors, answer the following questions:
B. P(Econ) =
Extra Practice
Sixty seniors at a local high school are taking AP
Statistics. Forty seniors at the same high school
are taking AP Economics. If ten seniors at the
high school take both classes and there are 500
seniors, answer the following questions:
B. P(Econ) =
40
500
=
4
50
= 8%
Extra Practice
Sixty seniors at a local high school are taking AP
Statistics. Forty seniors at the same high school
are taking AP Economics. If ten seniors at the
high school take both classes and there are 500
seniors, answer the following questions:
C. P( Stat ∩ Econ) =
Extra Practice
Sixty seniors at a local high school are taking AP
Statistics. Forty seniors at the same high school
are taking AP Economics. If ten seniors at the
high school take both classes and there are 500
seniors, answer the following questions:
C. P( Stat ∩ Econ) =
10
500
=
1
50
= 2%
Extra Practice
Sixty seniors at a local high school are taking AP
Statistics. Forty seniors at the same high school
are taking AP Economics. If ten seniors at the
high school take both classes and there are 500
seniors, answer the following questions:
D. P(Not Stat) =
E. P(Not Econ) =
Extra Practice
Sixty seniors at a local high school are taking AP
Statistics. Forty seniors at the same high school
are taking AP Economics. If ten seniors at the
high school take both classes and there are 500
seniors, answer the following questions:
D. P(Not Stat) = 100% – 12% = 88%
E. P(Not Econ) = 100% – 8% = 92%
Extra Practice
Sixty seniors at a local high school are taking AP
Statistics. Forty seniors at the same high school
are taking AP Economics. If ten seniors at the
high school take both classes and there are 500
seniors, answer the following questions:
F.
P( Stat U Econ)
= P(Stat) + P(Econ) – P(Stat ∩ Econ)
Extra Practice
Sixty seniors at a local high school are taking AP
Statistics. Forty seniors at the same high school
are taking AP Economics. If ten seniors at the
high school take both classes and there are 500
seniors, answer the following questions:
F.
P( Stat U Econ)
= P(Stat) + P(Econ) – P(Stat ∩ Econ)
= 12% + 8% –
2%
= 18%
Extra Practice
Sixty seniors at a local high school are taking AP
Statistics. Forty seniors at the same high school
are taking AP Economics. If ten seniors at the
high school take both classes and there are 500
seniors, answer the following questions:
G. P(Stat | Econ) =
𝑃(𝑆𝑡𝑎𝑡 ∩ 𝐸𝑐𝑜𝑛)
𝑃(𝐸𝑐𝑜𝑛)
Extra Practice
Sixty seniors at a local high school are taking AP
Statistics. Forty seniors at the same high school
are taking AP Economics. If ten seniors at the
high school take both classes and there are 500
seniors, answer the following questions:
G. P(Stat | Econ) =
𝑃(𝑆𝑡𝑎𝑡 ∩ 𝐸𝑐𝑜𝑛)
𝑃(𝐸𝑐𝑜𝑛)
.02 1
=
.08 4
=
= .25 = 25%
Extra Practice
Sixty seniors at a local high school are taking AP
Statistics. Forty seniors at the same high school
are taking AP Economics. If ten seniors at the
high school take both classes and there are 500
seniors, answer the following questions:
H.
P(Econ | Stat)
=
𝑃(𝐸𝑐𝑜𝑛 ∩ 𝑆𝑡𝑎𝑡)
𝑃(𝑆𝑡𝑎𝑡)
Extra Practice
Sixty seniors at a local high school are taking AP
Statistics. Forty seniors at the same high school
are taking AP Economics. If ten seniors at the
high school take both classes and there are 500
seniors, answer the following questions:
H.
P(Econ | Stat)
𝑃(𝐸𝑐𝑜𝑛 ∩ 𝑆𝑡𝑎𝑡)
=
𝑃(𝑆𝑡𝑎𝑡)
.02
1
=
.12
6
=
= .1667 = 16.67%
Note: the P(A|B) is not always equal to the P(B|A).