1. Conditional Probability
AS91585

2. The language of probability questions.
Section One
The language of probability questions.

3. The question, "Do you smoke?" was asked of 100 people.
Yes No
Total
Male 19 41 60
Female 12 28 40 31 69
100
The results are shown in the table.

4. (i) What is the probability of a randomly selected individual being a male who smokes?.
Yes No
Total
Male 19 41 60
Female 12 28 40 31 69
100

5. (i) What is the probability of a randomly selected individual being a male who smokes?.
Yes No
Total
Male 19 41 60
Female 12 28 40 31 69
100
This is just a joint probability question:
The number of "Male and Smoke" divided by the total = 19/100 = 0.19

6. (ii) What is the probability of a randomly selected individual being a male ?.
Yes No
Total
Male 19 41 60
Female 12 28 40 31 69
100

7. (ii) What is the probability of a randomly selected individual being a male ?.
Yes No
Total
Male 19 41 60
Female 12 28 40 31 69
100
This is the total for male divided by the total = 60/100 = 0.60.
Since no mention is made of smoking or not smoking, it includes all the cases.

8. (iii) What is the probability of a randomly selected individual smoking?.
Yes No
Total
Male 19 41 60
Female 12 28 40 31 69
100

9. (iii) What is the probability of a randomly selected individual smoking?.
Yes No
Total
Male 19 41 60
Female 12 28 40 31 69
100
Again, since no mention is made of gender, it is the total who smoke divided by the total = 31/100 = 0.31.

10. (iv) What is the probability of a randomly selected male smoking?.
Yes No
Total
Male 19 41 60
Female 12 28 40 31 69
100

11. (iv) What is the probability of a randomly selected male smoking?.
Yes No
Total
Male 19 41 60
Female 12 28 40 31 69
100
This time, you're told that you have a male, so we only consider the males. What is the probability that the male smokes? Well, 19 males smoke out of 60 males, so 19/60

12. (v) What is the probability that a randomly selected smoker is male?.
Yes No
Total
Male 19 41 60
Female 12 28 40 31 69
100

13. (v) What is the probability that a randomly selected smoker is male?.
Yes No
Total
Male 19 41 60
Female 12 28 40 31 69
100
This time, you're told that you have a smoker and asked to find the probability that the smoker is also male. There are 19 male smokers out of 31 total smokers, so 19/31

14. Section Two
Creating tables

15. Question 2
There are three major manufacturing companies that make a product: Aberations, Brochmailians, and Chompielians. Aberations has a 50% market share, and Brochmailians has a 30% market share. 5% of Aberations' product is defective, 7% of Brochmailians' product is defective, and 10% of Chompieliens' product is defective.

16. Choose 1000 for the total Company Good Defective Total Aberations
Brochmailians
Chompielians
1000
Use A, B and C to save writing time

17. Aberations has a 50% market share, and Brochmailians has a 30% market share.
Company
Good
Defective
Total A
500 B
300 C
200
1000
Fill in gaps as you go.

18. Company Good Defective Total A 25 500 B 21 300 C 20 200 66 1000
5% of Aberations' product is defective, 7% of Brochmailians' product is defective, and 10% of Chompieliens' product is defective.
Company
Good
Defective
Total A 25
500 B 21
300 C 20
200 66
1000

19. Company Good Defective Total A 475 25 500 B 279 21 300 C 180 20 200
5% of Aberations' product is defective, 7% of Brochmailians' product is defective, and 10% of Chompieliens' product is defective.
Company
Good
Defective
Total A
475 25
500 B
279 21
300 C
180 20
200
934 66
1000
Fill in the gaps

20. What is the probability a randomly selected product is defective?
Company
Good
Defective
Total A
475 25
500 B
279 21
300 C
180 20
200
934 66
1000

21. (i) What is the probability a randomly selected product is defective?
Company
Good
Defective
Total A
475 25
500 B
279 21
300 C
180 20
200
934 66
1000
66/1000=0.066

22. Company Good Defective Total A 475 25 500 B 279 21 300 C 180 20 200
(ii) What is the probability that a defective product came from Brochmailians?
Company
Good
Defective
Total A
475 25
500 B
279 21
300 C
180 20
200
934 66
1000

23. P(Brochmailian|Defective)
(ii) What is the probability that a defective product came from Brochmailians?
Company
Good
Defective
Total A
475 25
500 B
279 21
300 C
180 20
200
934 66
1000
P(Brochmailian|Defective)
= P(Brochmailian and Defective) / P(Defective)
= 21/66 = 7/22 = (approx).

24. (iii) Are these events independent?
Company
Good
Defective
Total A
475 25
500 B
279 21
300 C
180 20
200
934 66
1000

25. (iii) Are these events independent?
Company
Good
Defective
Total A
475 25
500 B
279 21
300 C
180 20
200
934 66
1000
No. If they were, then P(Brochmailians|Defective)=0.318 would have to equal the P(Brochmailians)=0.30, but it doesn’t i.e. if independent:

26. SECTION THREE
Practicing Methods

27. Using Tree diagrams
There are three major manufacturing companies that make a product: Aberations, Brochmailians, and Chompielians. Aberations has a 50% market share, and Brochmailians has a 30% market share. 5% of Aberations' product is defective, 7% of Brochmailians' product is defective, and 10% of Chompieliens' product is defective.

28. (i) What is the probability a randomly selected product is defective?
There are three major manufacturing companies that make a product: Aberations, Brochmailians, and Chompielians. Aberations has a 50% market share, and Brochmailians has a 30% market share. 5% of Aberations' product is defective, 7% of Brochmailians' product is defective, and 10% of Chompieliens' product is defective.

29. 0.05 D A
Not D
0.5 D
0.07
0.3 B
Not D
0.2
0.1 C D
Not D

30. (ii) What is the probability that a defective product came from Brochmailians?
There are three major manufacturing companies that make a product: Aberations, Brochmailians, and Chompielians. Aberations has a 50% market share, and Brochmailians has a 30% market share. 5% of Aberations' product is defective, 7% of Brochmailians' product is defective, and 10% of Chompieliens' product is defective.

31. 0.05 D A
Not D
0.5 D
0.07
0.3 B
Not D
0.2
0.1 C D
Not D

32. Question Three
Players are strongly advised to warm up before playing sports games to reduce their risk of injury from playing the game.
For a particular sports team of 20 players:
• 14 of the players warmed up before the last game
• 5 of the players were injured during the last game
• 2 of the players did not warm up and were not injured during the last game.
Using this information, calculate the probability that a randomly chosen player from the team was injured, given that the player did not warm up before the last game.

33. Venn Diagram
Injured
Warmed up 20 13 1 4 2

34. Table – initial information
Warmed Up
Did not warm up
Totals
Injured 5
Not Injured 2 14 20

35. Table fill the gaps Warmed Up Did not warm up Totals Injured 1 4 5
Not Injured 13 2 15 14 6 20

36. Question 4
On a certain type of aircraft the warning lights (showing green for normal and red for trouble) for the engines are accurate 90% of the time. If there are problems with the engines on 2% of all flights, find the probability that there is a fault with an engine, given that the warning light shows red.

37. We can also create a table: Assume we look at 1000 flights
Red
Green
Trouble 20 OK
980
Totals
1000
“problems with the engines on 2% of flights”

38. We can also create a table: Assume we look at 1000 flights
Red
Green
Trouble 18 20 OK
882
980
Totals
1000
“warning lights are accurate 90% of the time”

39. We can also create a table: Assume we look at 1000 flights
Red
Green
Totals
Trouble 18 2 20 OK 98
882
980
116
884
1000
Finish the table

40. find the probability that there is a fault with an engine, given that the warning light shows red.
Green
Totals
Trouble 18 2 20 OK 98
882
980
116
884
1000

41. R
0.9 T
0.02
0.1 G R
0.1
0.98 T’
0.9 G

42. SECTION FOUR
READING TABLES

43. Question 5
One of the biggest problems with conducting a mail survey is the poor response rate. In an effort to reduce nonresponse, several different techniques for formatting questionnaires have been proposed. An experiment was conducted to study the effect of the questionnaire layout and page size on response in a mail survey. The results are given below.

44. Format
Responses
Nonresponses
Total
Typewritten (small page)
Typewritten (large page)
Typeset
(small page)
(large page) 86
191 72
192 57 97 69 92
143
288
141
284
541
315
856

45. (a) What proportion of the sample responded to the questionnaire?
Format
Responses
Nonresponses
Total
Typewritten (small page)
Typewritten (large page)
Typeset
(small page)
(large page) 86
191 72
192 57 97 69 92
143
288
141
284
541
315
856

46. (a) What proportion of the sample responded to the questionnaire?
Format
Responses
Nonresponses
Total
Typewritten (small page)
Typewritten (large page)
Typeset
(small page)
(large page) 86
191 72
192 57 97 69 92
143
288
141
284
541
315
856

47. b. What proportion of the sample received the typeset small-page version?
Format
Responses
Nonresponses
Total
Typewritten (small page)
Typewritten (large page)
Typeset
(small page)
(large page) 86
191 72
192 57 97 69 92
143
288
141
284
541
315
856

48. What proportion of those who received a typeset large-page version actually responded to the questionnaire?
Format
Responses
Nonresponses
Total
Typewritten (small page)
Typewritten (large page)
Typeset
(small page)
(large page) 86
191 72
192 57 97 69 92
143
288
141
284
541
315
856

49. What proportion of the sample received a typeset large-page questionnaire and responded?
Format
Responses
Nonresponses
Total
Typewritten (small page)
Typewritten (large page)
Typeset
(small page)
(large page) 86
191 72
192 57 97 69 92
143
288
141
284
541
315
856

50. (e) What proportion of those who responded to the questionnaire actually received a type-written large page questionnaire?
Format
Responses
Nonresponses
Total
Typewritten (small page)
Typewritten (large page)
Typeset
(small page)
(large page) 86
191 72
192 57 97 69 92
143
288
141
284
541
315
856

51. f. By looking at the response rates for each of the four formats, what do you conclude from the study?
Format
Responses
Nonresponses
Total
Typewritten (small page)
Typewritten (large page)
Typeset
(small page)
(large page) 86
191 72
192 57 97 69 92
143
288
141
284
541
315
856
Type set (Large page) gave the best response rate at 68% with typewritten (large page) almost the same at 66% and Type set (Small page) was the worst at 51%. As a margin of error is likely I would conclude that the response rates seem to be better for large page.

52. SECTION FIVE
Harder PROBLEMS

53. Question 6
A cab was involved in a hit-and-run accident at night. There are two cab companies that operate in the city, a Blue Cab company and a Green Cab company. It is known that 85% of the cabs in the city are Green and 15% are Blue. A witness at the scene identified the cab involved in the accident as a Blue Cab. The witness was tested under similar visibility conditions, and made correct colour identifications in 80% of the trial instances. What is the probability that the cab involved in the accident was a Blue cab as stated by the witness?

54. If your answer was 80%, you are in the majority
If your answer was 80%, you are in the majority. The 80% answer shows how we have a tendency to primarily consider only the last evidence given to us, ignoring earlier evidence.

55. If we are simply told that a cab was involved in a hit-and-run accident, and are not given the information about the witness, then the majority of us will correctly estimate the probability of it being a Blue cab as 15%.

56. Given new evidence (the 80% reliable witness) we throw away the first calculation and base our answer solely on the reliability of the witness. We do this to simplify the calculation, but in this case it leads to the wrong answer.

57. There are four possible scenarios.
Green (85%) and correctly identified as Green (80%). Chance is 68%
Green (85%) and misidentified as Blue (20%). Chance is 17%
Blue (15%) and correctly identified as Blue (80%). Chance is 12%
Blue (15%) and misidentified as Green (20%). Chance is 3%

58. Conditional probability
In this case, we know that the witness said it was a Blue cab, so we only need to consider those cases where the cab was identified as Blue.

59. That means it was either a misidentified Green (17%) or a correctly identified Blue (12%). So the chance that it was actually Blue is the chance of it being correctly identified as Blue (12%) over the chance that it was identified as Blue, whichever colour it actually was (12% + 17%, or 29%). That means that the chance of it being Blue, after being identified as Blue, is 12/29, or about 41%.

60. The chance that it was actually Green is the remaining 59%.

61. But with a witness who is 80% reliable, how can he be so likely to get it wrong? The catch is that the small chance of his incorrect identification is swamped by the huge number of Green cabs, which just make it so much more likely that any cab in the city is Green.

62. Basically, with a compound probability like this you have to be careful to check out the contribution of both the correct (correctly identified Blue) and the incorrect (misidentified Green) terms. Otherwise, you may miss a large contribution which works against your intuition.

63. If only 5% of the cabs in the city are Blue, the chances drop to 4/23, or 17%. In other words, if only 5% of the cabs are Blue and our 80% reliable witness identifies a Blue cab in an accident, there is only a 17% chance that he's actually right. Our 80% reliable witness is 5 times more likely to be wrong than right!

64. Question 7a
A survey of newspaper purchasing patterns in a particular region found that 3⁄5 of the households surveyed purchased a daily newspaper during the previous week. The survey also found that in 7⁄10 of the households where a daily newspaper was purchased during the previous week, a Sunday paper was also purchased. A Sunday paper was purchased in 1⁄4 of households where no daily newspaper was purchased.

65. What is the probability that, in a household chosen at random from those surveyed:
a daily newspaper was purchased or a Sunday newspaper was purchased (but not both)?
(ii) a daily newspaper was purchased, given that a Sunday newspaper was purchased?

66. Table Daily Paper No Daily Paper Totals Sunday paper 42 10 52
No Sunday Paper 18 30 48 60 40
100

67. Daily Paper No Daily Paper Totals Sunday paper 42 10 52
(i) a daily newspaper was purchased or a Sunday newspaper was purchased (but not both)?
Daily Paper
No Daily Paper
Totals
Sunday paper 42 10 52
No Sunday Paper 18 30 48 60 40
100

68. Daily Paper No Daily Paper Totals Sunday paper 42 10 52
(ii) a daily newspaper was purchased, given that a Sunday newspaper was purchased?
Daily Paper
No Daily Paper
Totals
Sunday paper 42 10 52
No Sunday Paper 18 30 48 60 40
100

69. 7b In a different survey of 120 households, where 70 of them purchased a daily newspaper, the following information was obtained.
Households may contain both primary and secondary students.

70. (i) Find the probability that a household chosen at random from this sample contains both primary and secondary school students.

71. Find the probability that a household chosen at random from this sample contains both primary and secondary school students.
There are 120 households but 146 listed which means 26 are counted twice. Answer 26/120

72. (ii) Demonstrate mathematically that the event ‘the household contains both primary and secondary school students’ and the event ‘the household purchases a daily newspaper’ are not independent.

73. 70 OUT OF 120 purchased a daily paper 14 were counted twice
Demonstrate mathematically that the event ‘the household contains both primary and secondary school students’ and the event ‘the household purchases a daily newspaper’ are not independent.
70 OUT OF 120 purchased a daily paper
14 were counted twice

74. Events are not independent

75. (iii)
Find the probability that a household contains at least one secondary school student, given that the household purchases at least one daily newspaper.

76. Conditional probability P(S/D)
Find the probability that a household contains at least one secondary school student, given that the household purchases at least one daily newspaper.
Conditional probability P(S/D)

77. Conditional probability P(S/D)
Find the probability that a household contains at least one secondary school student, given that the household purchases at least one daily newspaper.
Conditional probability P(S/D)

78. Scholarship 2010 Q3a
Statsmobiles may be classified as “two door” or “at least three door” models. Fifty-five percent of all two-door Statsmobiles are non air-conditioned and 30% of all Statsmobiles are both air-conditioned and have at least three doors.
The proportion of air-conditioned Statsmobiles that have at least three doors is the same as the overall proportion of Statsmobiles that have at least three doors.
Find the value of this proportion.

79. Using a tree diagram NAC 0.55 2 door 1-p 0.45 AC 1-q NAC p 3+ doors q
0.3

80. There are some 2 door cars so p = 2/3
NAC
0.55
2 door
1-p
0.45 AC
1-q
NAC p 3+
doors q AC
0.3