1. Understand conditional probability.
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.

2. Know how to find probabilities for compound events
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.

3. Conditional Probability Independence Tree Diagram
Vocabulary
Sample Space
Conditional Probability
Independence
Tree Diagram

4. Chapter 15 Notes Chp 15 Practice Extra Practice
Chapter 15 Assignment
Chapter 14 Assignment Part 1
Answers
Chapter 14 Assignment Part 2
Answers

5. If two events A & B are disjoint, P(A OR B) = P(A) + P(B)
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

6. If two events A & B are disjoint, P(A OR B) = P(A) + P(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

7. Suppose that P(H) = .25, P(N) = .18, P(T) = .14.
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) =
= .57
c. P(not (H or N or T)) =
= .43

8. 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 What is the probability that they like country or jazz?

9. P(C or J) = P(C) + P(J) – P(C &J) = .4 + .3 – .1 = .6
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 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? C J .3 .1 .2
P(C or J) = P(C) + P(J) – P(C &J) ∪ ∩
= – .1 = .6

10. 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?
What is the probability she plays football for SHS?
Independent,
P(student plays soccer) = P(stud. plays soccer given they are female).
Not Independent,
P(student plays ftball) ≠ P(stud. plays ftball given they are female)

11. 4. The Multiplication Rule only works if the two events are independent of each other.
P(A & B) = P(A) ∙ P(B)

12. P(A & B) = P(A) ∙ P(B|A) 5. General Multiplication Rule:
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”.

13. 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?
b. What is the probability that both bulbs are defective?
Yes, since they are randomly selected.
P(Defect & Defect) =
= (0.05)(0.05) =

14. = 0.0025 (0.05)(0.05) 2nd Bulb good 0.95 1st Bulb good 0.95 defective
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?
b. What is the probability that both bulbs are defective?
Yes, since they are randomly selected.
0.05
0.95
2nd Bulb
good
defective
0.05
0.95
1st Bulb
good
defective
(0.05)(0.05) =

15. P(exactly one D) = P(D & DC) or P(DC & D) = (.05)(.95) + (.95)(.05)
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

16. 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?
0.05
0.95
2nd Bulb
good
defective
0.05
0.95
1st Bulb
good
defective
(.95)(.05) =
0.475 +
0.475
(.05)(.95) =
0.095

17. d) What is the probability that at least one bulb is defective?
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

18. “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)

19. P(at least one D) = 1 – P(DC & DC) = 1 – (.95*.95) = .0975
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) =

20. 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?
2nd Bulb
good
0.95
1st Bulb
good
2nd Bulb
defective
0.95
(.95)(.05) =
0.0475
0.05
1st Bulb
defective
2nd Bulb
good
0.95
(.05)(.95) = +
0.0475
0.05
2nd Bulb
defective
(.05)(.05) = +
0.0025
0.05
0.0975

21. 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) = (.35) =
0.6425

22. Example 4 : If P(A) = 0. 45, P(B) = 0. 35, and A & B are independent
Example 4 : If P(A) = 0.45, P(B) = 0.35, and A & B are independent. Find P(A or B).
P(A&B) =
P(A)*P(B) B A
= (0.45)(0.35)
0.2925
0.1925 =
0.1575
0.2925
0.45 – =
0.6425
0.35 – =

23. Example 6
If P(A) = 0.45, P(B) = 0.35, and A & B are independent. Find P(A or B). A
NOT A B
0.35
NOT B
0.45
0.1575
0.65
1 – 0.35 = 0.65
0.55
1.00
1 – 0.45 = 0.55
P(A&B) =
P(A)*P(B)
= (0.45)(0.35) =

24. 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

25. 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?
Dr. Pepper
P(at least one winning symbol) =
1 – P(no winning symbols)
= .4686

26. 6. Conditional Probability
A probability that takes into account a given condition
P(B|A) = 𝑷(𝑨∩𝑩) 𝑷(𝑨) = 𝑨𝑵𝑫 𝑮𝑰𝑽𝑬𝑵

27. P(Sport | Female) = 𝑷(𝑺𝒑𝒐𝒓𝒕 ∩ 𝑭𝒆𝒎𝒂𝒍𝒆) 𝑷(𝑭𝒆𝒎𝒂𝒍𝒆)
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 If the student is female, what is the probability that she plays sports?
P(Sport | Female) = 𝑷(𝑺𝒑𝒐𝒓𝒕 ∩ 𝑭𝒆𝒎𝒂𝒍𝒆) 𝑷(𝑭𝒆𝒎𝒂𝒍𝒆)
= .𝟏 .𝟓𝟏 = .1961

28. .𝟏 .𝟓𝟏 = .1961 Sport NO Sport Female Male 0.10 0.41 0.51 0.49 1.00
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 If the student is female, what is the probability that she plays sports?
0.51 – 0.10 = 0.41
Sport
NO Sport
Female
Male
0.10
0.41
0.51
0.49
1 – 0.51 = 0.49
1.00
.𝟏 .𝟓𝟏 = .1961

29. .31 = 𝒙 .𝟒𝟗 .1519 = x P(Sport | Male) = 𝑷(𝑺𝒑𝒐𝒓𝒕 ∩𝑴𝒂𝒍𝒆) 𝑷(𝑴𝒂𝒍𝒆)
Ex 7) b. The probability that a randomly selected student plays sports if they are male is 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
= x

30. x = .1519 𝒙 .𝟒𝟗 = .𝟑𝟏 Sport NO Sport Female Male 0.10 0.41 0.51 0.49
Ex 7) b. The probability that a randomly selected student plays sports if they are male is 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
Male
0.10
0.41
0.51
0.49
1.00
𝒙 .𝟒𝟗 = .𝟑𝟏
x = .1519

31. Sport NO Sport Female Male 0.41 0.51 0.49 1.00
OK. Let’s finish the table…
Sport
NO Sport
Female
Male
0.10
0.41
0.51
0.1519
0.3381
0.49
0.2519
0.7481
1.00
0.49 – = = =

32. P(Female | Sports) = 𝑷(𝑺𝒑𝒐𝒓𝒕 ∩ 𝑭𝒆𝒎𝒂𝒍𝒆) 𝑷(𝑺𝒑𝒐𝒓𝒕)
c. What’s the probability that someone who plays sports is female?
Sport
NO Sport
Female
Male
0.10
0.41
0.51
0.1519
0.3381
0.49
0.2519
0.7481
1.00
P(Female | Sports) = 𝑷(𝑺𝒑𝒐𝒓𝒕 ∩ 𝑭𝒆𝒎𝒂𝒍𝒆) 𝑷(𝑺𝒑𝒐𝒓𝒕)
= .𝟏 .𝟐𝟓𝟏𝟗 = .397

33. Probabilities from two way tables
Example 8
Probabilities from two way tables
Stu Staff Total
American
European
Asian
Total
a. What is the probability that the driver is a student?
≈𝟓𝟒.𝟑%

34. Probabilities from two way tables
Example 8
Probabilities from two way tables
Stu Staff Total
American
European
Asian
Total
b. What is the probability that the driver drives a European car?
≈𝟏𝟐.𝟓%

35. Probabilities from two way tables
Disjoint so: P(Am or Asian) = P(Am) + P(Asian)
Probabilities from two way tables
Stu Staff Total
American
European
Asian
Total
c. What is the probability that the driver drives an American or Asian car?
Disjoint
≈𝟖𝟕.𝟓%

36. Probabilities from two way tables
Not Disjoint so:
P(Staff or Asian) = P(Staff) + P(Asian) – P(Staff & Asian)
Probabilities from two way tables
Stu Staff Total
American
European
Asian
Total
d. What is the probability that the driver is staff or drives an Asian car?
Not Disjoint
𝑃 𝑆𝑡𝑎𝑓𝑓 𝑜𝑟 𝐴𝑠𝑖𝑎𝑛 = −47 359
= 𝟐𝟏𝟗 𝟑𝟓𝟗
≈𝟔𝟏%

37. Probabilities from two way tables
Joint Distribution:
P(Staff and Asian Car)
Probabilities from two way tables
Stu Staff Total
American
European
Asian
Total
e. What is the probability that the driver is staff and drives an Asian car?
𝑃 𝑆𝑡𝑎𝑓𝑓 𝑎𝑛𝑑 𝐴𝑠𝑖𝑎𝑛 = 𝟒𝟕 𝟑𝟓𝟗
≈𝟏𝟑.𝟏%

38. Probabilities from two way tables
Conditional Distribution:
P(Am Car|Student)
Probabilities from two way tables
Stu Staff Total
American
European
Asian
Total
f. If the driver is a student, what is the probability that they drive an American car?
𝑃 𝐴𝑚𝑒𝑟𝑖𝑐𝑎𝑛|𝑆𝑡𝑢𝑑𝑒𝑛𝑡 = 𝟏𝟎𝟕 𝟏𝟗𝟓
≈𝟓𝟒.𝟗%

39. Probabilities from two way tables
Conditional Distribution:
P(Student|Euro Car)
Probabilities from two way tables
Stu Staff Total
American
European
Asian
Total
g. What is the probability that the driver is a student if the driver drives a European car?
𝑃 𝑆𝑡𝑢𝑑𝑒𝑛𝑡|𝐸𝑢𝑟𝑜𝑝𝑒𝑎𝑛 = 𝟑𝟑 𝟒𝟓
≈𝟕𝟑.𝟑%

40. 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.

41. If Event A and Event B are Independent, then…
𝑃 𝐴∩𝐵 =𝑃 𝐴 ∗𝑃(𝐵)
𝑨𝒏𝒅 𝑮𝒊𝒗𝒆𝒏
𝑃 𝐵
𝑃 𝐵
𝑃(𝐴|𝐵) =
𝑃 𝐴
P(A) = P(A|B)
So if
then Events A and B are independent

42. Conditional Probability
Stu Staff Total
American
European
Asian
Total
8. Parts of a Tree Diagram
P(B|A)
P(A)*P(B|A)
Conditional Probability
Joint Probability
Probability
American & Student
107/359
P(A)
Student
107/212
50.5%
29.81%
Staff
105/212
American & Staff
105/359
American
212/359
49.5%
29.25%
59.1%
European & Student
33/359
Student
33/45
European
45/359
73.3%
9.19%
Staff
12/45
12.5%
European & Staff
12/359
26.7%
Asian
102/359
3.34%
Student
55/102
Asian & Student
55/359
28.4%
53.9%
15.32%
Staff
47/102
Asian & Staff
47/359
46.1%
13.09%

43. 9. Draw with Replacement Blue & Blue Blue 50.0% 25.0% Green Blue
Green & Green
25.0%

44. 10. Draw without Replacement
Blue & Blue
Blue
33.3%
17%
Green
Blue & Green
Blue
66.7%
50.0%
33%
Green & Blue
Blue
Green
66.7%
33%
50.0%
Green
33.3%
Green & Green
17%

45. 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?

46. 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%

47. 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%

48. 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) =

49. Example 9: P(Returned | Inexp.) = .12 Not same as P(Inexp. | Returned)
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

50. 𝑷(𝑰𝒏𝒆𝒙𝒑. ∩𝑹𝒆𝒕𝒖𝒓𝒏𝒆𝒅) 𝑷(𝑹𝒆𝒕𝒖𝒓𝒏𝒆𝒅) 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. 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(𝑹𝒆𝒕𝒖𝒓𝒏𝒆𝒅)
P(Inexp. | Returned) =
= .𝟎𝟐𝟒 .𝟎𝟒𝟖 = .5 = 50%

51. 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 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?

52. Example 10
relieved
not .7
1st .6 .3
relieved
not .9 .4
2nd .1

53. = 𝑷 𝟏 𝒔𝒕 𝒃𝒐𝒕𝒕𝒍𝒆 ∩𝑵𝒐𝒕 𝒓𝒆𝒍𝒊𝒆𝒗𝒆𝒅 𝑃 𝑝𝑎𝑖𝑛 𝑁𝑜𝑡 𝑟𝑒𝑙𝑖𝑒𝑣𝑒𝑑
Example 10
= 𝑷 𝟏 𝒔𝒕 𝒃𝒐𝒕𝒕𝒍𝒆 ∩𝑵𝒐𝒕 𝒓𝒆𝒍𝒊𝒆𝒗𝒆𝒅 𝑃 𝑝𝑎𝑖𝑛 𝑁𝑜𝑡 𝑟𝑒𝑙𝑖𝑒𝑣𝑒𝑑
= 𝟎.𝟏𝟖
relieved .7
1st .6 .3
𝑷 𝟏 𝒔𝒕 𝒃𝒐𝒕𝒕𝒍𝒆 ∩𝑵𝒐𝒕 𝒓𝒆𝒍𝒊𝒆𝒗𝒆𝒅
= (.6)(.3) = 0.18
not
relieved
not .9 .4
2nd .1

54. = 𝑷 𝟏 𝒔𝒕 𝒃𝒐𝒕𝒕𝒍𝒆 ∩𝑵𝒐𝒕 𝒓𝒆𝒍𝒊𝒆𝒗𝒆𝒅 𝑷 𝒑𝒂𝒊𝒏 𝑵𝒐𝒕 𝒓𝒆𝒍𝒊𝒆𝒗𝒆𝒅
Example 10
= 𝑷 𝟏 𝒔𝒕 𝒃𝒐𝒕𝒕𝒍𝒆 ∩𝑵𝒐𝒕 𝒓𝒆𝒍𝒊𝒆𝒗𝒆𝒅 𝑷 𝒑𝒂𝒊𝒏 𝑵𝒐𝒕 𝒓𝒆𝒍𝒊𝒆𝒗𝒆𝒅
= 𝟎.𝟏𝟖
= 𝟎.𝟏𝟖 𝟎.𝟐𝟐
≈𝟎.𝟖𝟏𝟖2
relieved .7
1st .6 .3
𝑷 𝟏 𝒔𝒕 𝒃𝒐𝒕𝒕𝒍𝒆 ∩𝑵𝒐𝒕 𝒓𝒆𝒍𝒊𝒆𝒗𝒆𝒅
= (.6)(.3) = 0.18
not
relieved
𝑷 𝑵𝒐𝒕 𝒓𝒆𝒍𝒊𝒆𝒗𝒆𝒅 = = 0.22 .9 .4
2nd
𝑷 𝟐 𝒏𝒅 𝒃𝒐𝒕𝒕𝒍𝒆 ∩𝑵𝒐𝒕 𝒓𝒆𝒍𝒊𝒆𝒗𝒆𝒅
= (.4)(.1) = 0.04 .1
not

55. = 𝑷 𝟏 𝒔𝒕 𝒃𝒐𝒕𝒕𝒍𝒆 ∩𝑵𝒐𝒕 𝒓𝒆𝒍𝒊𝒆𝒗𝒆𝒅 𝑷 𝒑𝒂𝒊𝒏 𝑵𝒐𝒕 𝒓𝒆𝒍𝒊𝒆𝒗𝒆𝒅
Example 10
= 𝑷 𝟏 𝒔𝒕 𝒃𝒐𝒕𝒕𝒍𝒆 ∩𝑵𝒐𝒕 𝒓𝒆𝒍𝒊𝒆𝒗𝒆𝒅 𝑷 𝒑𝒂𝒊𝒏 𝑵𝒐𝒕 𝒓𝒆𝒍𝒊𝒆𝒗𝒆𝒅
= 𝟎.𝟏𝟖
= 𝟎.𝟏𝟖 𝟎.𝟐𝟐
≈𝟎.𝟖𝟏𝟖2
relieved .7
1st .6 .3
𝑷 𝟏 𝒔𝒕 𝒃𝒐𝒕𝒕𝒍𝒆 ∩𝑵𝒐𝒕 𝒓𝒆𝒍𝒊𝒆𝒗𝒆𝒅
= (.6)(.3) = 0.18
not
relieved
𝑷 𝑵𝒐𝒕 𝒓𝒆𝒍𝒊𝒆𝒗𝒆𝒅 = = 0.22 .9 .4
2nd
𝑷 𝟐 𝒏𝒅 𝒃𝒐𝒕𝒕𝒍𝒆 ∩𝑵𝒐𝒕 𝒓𝒆𝒍𝒊𝒆𝒗𝒆𝒅
= (.4)(.1) = 0.04 .1
not

56. 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)
= = 68%
Garage
Pool
64% - 17% = 47%
47%
17% 4%
21% - 17% = 4%
P(Garage U Pool) = = 68%

57. Chp 15 p.362 #3 P(NOT(Garage U Pool)) = P((Garage U Pool)C)
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%

58. Chp 15 p.362 #3 P(Pool ∩ Garagec) = 4% c) a pool but no garage?
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%

59. 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? TV
Frig
52% - 21% = 31%
.31
.21
.17
38% - 21% = 17%
P(TV ∩ Frigc) =
31%

60. Chp 15 p.363 #5 P(TV XOR Refrig) = 17% + 31% = 48%
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

61. Chp 15 p.363 #5 P(NOT(TV U Refrig)) = = 1 – P(TV U Refrig)
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 – ( ) =
= .31

62. Chp 15 p.363 #11
The probabilities that an adult American man has high blood pressure and /or cholesterol are shown in the table.
Blood Pressure
High OK
0.11
0.21
0.16
0.52
Total
Cholesterol
0.32
0.68
Total
0.27
0.73
1.00

63. Chp 15 p.363 #11
The probabilities that an adult American man has high blood pressure and /or cholesterol are shown in the table.
Blood Pressure
High OK
0.11
0.21
0.16
0.52
a) What’s the probability that a man has both conditions?
Total
Cholesterol
0.32
0.68
Total
0.27
0.73
1.00
P(High BP ∩ High Chol) =
11%

64. Chp 15 p.363 #11 P(High BP) = 27% High OK 0.11 0.21 0.16 0.52
The probabilities that an adult American man has high blood pressure and /or cholesterol are shown in the table.
Blood Pressure
High OK
0.11
0.21
0.16
0.52
b) What’s the probability that he has high blood pressure?
Total
Cholesterol
0.32
0.68
Total
0.27
0.73
1.00
P(High BP) =
27%

65. Chp 15 p.363 #11
The probabilities that an adult American man has high blood pressure and /or cholesterol are shown in the table.
Blood Pressure
High OK
0.11
0.21
0.16
0.52
c) What’s the probability that a man with high blood pressure has high cholesterol?
Total
Cholesterol
0.32
0.68
Total
0.27
0.73
1.00

66. = .𝟏𝟏 .𝟐𝟕 = 40.7% Chp 15 p.363 #11 𝑨𝑵𝑫 𝑮𝑰𝑽𝑬𝑵 High OK 0.11 0.21 0.16
The probabilities that an adult American man has high blood pressure and /or cholesterol are shown in the table.
Blood Pressure
High OK
0.11
0.21
0.16
0.52
c) What’s the probability that a man with high blood pressure has high cholesterol?
Total
Cholesterol
0.32
0.68
Total
0.27
0.73
1.00
𝑨𝑵𝑫 𝑮𝑰𝑽𝑬𝑵
P(High Chol | High BP) =
= .𝟏𝟏 .𝟐𝟕 = 40.7%

67. Chp 15 p.363 #11
The probabilities that an adult American man has high blood pressure and /or cholesterol are shown in the table.
Blood Pressure
High OK
0.11
0.21
0.16
0.52
d) What’s the probability that a man has high blood pressure if it’s known that he has high cholesterol?
Cholesterol
0.32
0.68
Total
0.27
0.73
1.00

68. = .𝟏𝟏 .𝟑𝟐 = 34.37% Chp 15 p.363 #11 𝑨𝑵𝑫 𝑮𝑰𝑽𝑬𝑵 High OK 0.11 0.21 0.16
The probabilities that an adult American man has high blood pressure and /or cholesterol are shown in the table.
Blood Pressure
High OK
0.11
0.21
0.16
0.52
d) What’s the probability that a man has high blood pressure if it’s known that he has high cholesterol?
Cholesterol
0.32
0.68
Total
0.27
0.73
1.00
𝑨𝑵𝑫 𝑮𝑰𝑽𝑬𝑵
P(High BP | High Chol) =
= .𝟏𝟏 .𝟑𝟐 = %

69. = .𝟏𝟏 .𝟑𝟐 = 34.37% Chp 15 p.363 #11 𝑨𝑵𝑫 𝑮𝑰𝑽𝑬𝑵 High OK 0.11 0.21 0.16
The probabilities that an adult American man has high blood pressure and /or cholesterol are shown in the table.
Blood Pressure
High OK
0.11
0.21
0.16
0.52
d) What’s the probability that a man has high blood pressure if it’s known that he has high cholesterol?
Cholesterol
0.32
0.68
Total
0.27
0.73
1.00
𝑨𝑵𝑫 𝑮𝑰𝑽𝑬𝑵
P(High BP | High Chol) =
= .𝟏𝟏 .𝟑𝟐 = %

70. Chapter 15 Assignment
pp #8&14, 22
pp #28, 36&38, 43

71. 6. Commercial airplanes have an excellent safety record
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?

72. 6. Commercial airplanes have an excellent safety record
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?

73. 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?

74. 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?

75. 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?

76. 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?
1 Sem
2+ Sem
= 87
13%
32%
100 – 87 = 13%
55%

77. 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?

78. 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?
= 68%
1 Sem
2+ Sem
13%
32%
55%

79. 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?

80. 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?
= 87%
1 Sem
2+ Sem
13%
32%
55%

81. 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?

82. 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?

83. 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?

84. 18&20. A Gallup Poll in June 2004 asked 1005 U. S
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”?

85. 18&20. A Gallup Poll in June 2004 asked 1005 U. S
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
301
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?

86. 18&20. A Gallup Poll in June 2004 asked 1005 U. S
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
301
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?

87. 18&20. A Gallup Poll in June 2004 asked 1005 U. S
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
301
704
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?

88. 18&20. A Gallup Poll in June 2004 asked 1005 U. S
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
Not Likely Reader
(301/1005)*(301/1005)
(704/1005)*(301/1005)
(301/1005)*(704/1005)
(704/1005)*(704/10050

89. 18&20. A Gallup Poll in June 2004 asked 1005 U. S
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?

90. 18&20. A Gallup Poll in June 2004 asked 1005 U. S
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.

91. 18&20. A Gallup Poll in June 2004 asked 1005 U. S
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.

92. 24. The American Red Cross says that aout 45% of the U. S
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.
Someone volunteers to give blood. What is the probability that this donor
i. has Type AB blood?

93. 24. The American Red Cross says that aout 45% of the U. S
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.
Someone volunteers to give blood. What is the probability that this donor
ii. has Type A or Type B?

94. 24. The American Red Cross says that aout 45% of the U. S
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.
Someone volunteers to give blood. What is the probability that this donor
iii. is not Type O?

95. 24. The American Red Cross says that aout 45% of the U. S
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?

96. 24. The American Red Cross says that aout 45% of the U. S
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?

97. 24. The American Red Cross says that aout 45% of the U. S
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?

98. 24. The American Red Cross says that aout 45% of the U. S
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?

99. 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.

100. 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?

101. 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.

102. 30. To get to work, a commuter must cross train tracks
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?

103. 30. To get to work, a commuter must cross train tracks
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?

104. 30. To get to work, a commuter must cross train tracks
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?

105. 30. To get to work, a commuter must cross train tracks
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?

106. 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?

107. 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?

108. 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?

109. 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.

110. 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.

111. 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.

112. 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) =

113. 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) = = = 12%

114. 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) =

115. 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) = = = 8%

116. 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) =

117. 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) = = = 2%

118. 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:
P(Not Stat) =
E. P(Not Econ) =

119. 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%

120. Not Disjoint 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)
Not Disjoint

121. Not Disjoint 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% % – %
= 18%
Not Disjoint

122. 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)

123. 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)
= = 1 4
= .25 = 25%

124. 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)

125. 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)
= = 1 6
= = 16.67%
Note: the P(A|B) is not always equal to the P(B|A).