Detailed Solutions to Recommended Chapter 6 Problems
Statistics 213 (L03) Detailed Solutions to Recommended Chapter 6 Problems 1. a) An experiment is a special observation under controlled conditions in order to achieve a value for a random variable. b) The sample space is the set of all possible outcomes of an experiment. c) Outcomes are the results from an experiment. d) Events are outcomes or combinations of possible outcomes that the researcher judges can be classified as single occurrences of interest. 2. (1) Classical or a priori probability: This type of probability can be calculated through knowledge of the process involved, by making theoretical counts of the outcomes that are possible and those that constitute the event. (Examples: (a) A coin is tossed. What is the probability that it lands on ‘Heads’? (b) What is the probability of drawing a red ace from a standard deck of 52 cards?) (2) Empirical probability: This type of probability is derived from relative frequencies as observed in the past, e.g., based on survey or census data. (Examples: (a) Estimate the probability that a Canadian watching T.V. this Saturday will be watching Hockey Night in Canada. (b) What is the probability that a randomly selected Statistics student at your institution will pass the course?) (3) Subjective probability: This type of probability is a numerical judgment by an individual or a group regarding the outcome of a variable for which no a priori or empirical probability can be calculated (although relevant evidence, if available, may be considered in the process). (Examples: (a) What is the probability that the Ottawa Senators will win the Stanley Cup in the next season? 2. If a homeowner stops using pesticides, what is the probability that his or her lawn will be destroyed by grubs?) 3. a) Collection of all 36 possible outcomes when two dice were rolled. b) Possible answers are if the first die and second die display, respectively, (i) 1 and 2; (ii) 3 and 6; (iii) 6 and 4; (iv) 5 and 5. c) All possible outcomes where the two dice together add to 7 (e.g., a 1 and a 6). d) The probability that the event of interest occurs is 6/36 = 0.1667, because there are 6 possible combinations of two dice to get a sum of 7. 4. a) (Suppose there are exactly 200 participants, and your area is Whitby.) The sample space consists of the genders of all 200 participants in the Terry Fox run in Whitby.
b) Four (of many) possible outcomes are shown below: •
101 participants are female; 99 participants are male
•
50 participants are female; 150 participants are male
•
99 participants are female; 101 participants are male
•
150 participants are female; 50 participants are male
c) If data were available for the proportions of women who have registered for that run in past years, we could estimate an empirical probability for that proportion for this year—but, unfortunately, the relevant data are not given in the problem. (We can’t calculate an a priori probability, because we lack sufficient knowledge about the experiment. Suppose, for example, that there are only 110 males in the vicinity of Whitby; then the outcome “8 participants are female; 192 participants are male” would have zero probability. On the other hand, if there thousands of males and just hundreds of females in Whitby fit enough to run, then “8 participants are female; 192 participants are male” might be reasonably probable.) 5. a) The (stop or don’t-stop) actions taken by all 500 drivers. b) Possible answers are: •
200 drivers stop fully; 300 drivers do not stop fully.
•
400 drivers stop fully; 100 drivers do not stop fully.
•
125 drivers stop fully; 375 drivers do not stop fully.
•
430 drivers stop fully; 70 drivers do not stop fully.
6. a) Probability that a selected student clearly enjoyed the concert = 162/250 = 0.648. b) Probability that a selected student stated a clear opinion about whether they enjoyed the concert =
162 + 67 229 = = 0.916 . 250 250 7. Subjective probability.
8. a) P(Ace of Hearts) = 1/52 = 0.0192
Ranks of Cards
Suit
Ace
Other
TOTAL
Hearts
1
12
13
Other
3
36
39
TOTAL
4
48
52
b) In a standard deck of 52 playing cards, the “suit” Spade consists of 13 cards; thus P(any Spade) = 13/52 = 0.25. c) In a standard deck of 52 playing cards, there are four “suits”—each of which contains exactly one ‘10’, one ‘Jack’, one ‘Queen’, one ‘King’ and one ‘Ace’. So the number of possible ways to draw any of those cards, disregarding its suit = (5 cards of interest) × (4 “suits”) = 20. Therefore, P(any one of a 10, a Jack, a Queen, a King, or an Ace) = 20/52 = 0.3846. 9. a) P(red marble) = 4/8 = 0.5. b)
The first Marble Selected The second marble selected
\/
1st Red
2nd Red
3rd Red
4th Red
1st White
2nd White
1st Blue
2nd Blue
1st White
√
√
2nd White
√
√
1st Red 2nd Red 3rd Red 4th Red
1st Blue
√
√
2nd Blue
√
√
P(one blue and one white—in either order) = 8/64 = 0.125.
10. a) P(winning response is “Mainland China”) = 68/312 = 0.2179 b) The list of countries, as such, is a mutually exclusive list. Each region or country is listed only once; the options are clearly defined and separated from one another. However, if the interest is on “sets of countries chosen” by individuals, there is no mutually exclusive list – because one individual’s choice-set might include more than one entry in the list. We don’t have enough information to find the probability that half of the respondents chose Mainland China, because there is no information about the number of respondents. Each respondent was allowed to check off more than one destination. We have no information about how often respondents checked off more than one destination. 11. a) (All combinations without an X in the table below are possible. Combinations that add up to 58 show a check mark.)
First Element Chosen Second Element Chosen \/
4
4
X
20
38
20
X
√
38
√
X
12
56 88
P(sum = 58) = 2/30 = 0.0667.
12
56
88
X
X X
b) (Checked-off combinations in the table add up to less than 60.)
First Element Chosen Second Element Chosen \/
4
12
20
38
4
X
√
√
√
12
√
X
√
√
20
√
√
X
√
38
√
√
√
X
56
56
88
X
88
X
P(sum < 60) = 12/30 = 0.4. 12. (The simplified compass assumed in this question has no printing on the background; the only label of interest is “N” (or “North”) printed on one side of the pointer—hopefully, on the north-seeking side..)
For each of the three compasses, the pointer label “North” may either be painted on the correct side of the pointer or on incorrect side. A possible table of outcomes for the labelling of the pointers is shown below:
The table shows 8 possible combinations of correct or incorrect labels for the three compass pointers. Only one of the 8 combinations is all are correct. Therefore, the probability = 1/8 = 0.125.
17. a)
b) P(late) = 0.13; (c) P(late | Health Sciences student) = 0.04/0.35 = 0.1143; (d) P(Health Sciences student | on time) = 0.31/0.87 = 0.3563. 18. a) P(ticket costs $25) = 70/250 = 0.28 b) P(paid $25 | paid more than $15) = 70/(60+70) = 0.5385 c) P(paid $15 | paid less than $20) = 75/(45+75) = 0.625 19. P(enjoyed concert | voiced an opinion) = 162/229 = 0.7074. 20. a) P(marble is red | red marble taken and not replaced) = 3/7 = 0.4286 b) P(marble is red | three red marbles taken and not replaced) = 1/5 = 0.2 21. a) P(low grade) = 18/85 = 0.2118. b) P(low grade | frequently absent) = 8/17 = 0.4706. c) P(frequently absent) = 17/85 = 0.20. d) P(frequently absent | high grade) = 2/19 = 0.1053. 22. The relevant data are summarized in this table: STATE
Mines_U04
Mines_S04
8
41
49
Other
578
730
1308
Totals
586
771
1357
Alabama
Totals
a) P(mine located in Alabama | active coal mine in 2004) = 49/1,357 = 0.0361. b) P(mine located in Alabama | active surface coal mine in 2004) = 41/771 = 0.0532. c) P(underground mine | active coal mine in 2004) = 586/1,357 = 0.4318. d) P(underground mine | active coal mine in Alabama in 2004) = 8/49 = 0.1633. 23. a) P(atomic number < 21 | alkaline earth element) = 3/6 = 0.5. b) P(atomic number of 12 | alkaline earth element whose atomic number is < 21) = 1/3 = 0.3333.
29. Note that according to the Multiplication Rule, two events A and B are independent, if P(A) x P(B) = P(A and B). a)
Independent, because P(A and B) = 0.14 = P(A) x P(B) = 0.20 x 0.70 = 0.14.
b) Dependent, because P(A and B) = 0.36 ≠ P(A) x P(B) = 0.80 x 0.50 = 0.40. 30. The empirical probability P(successful) = 0.81. The event “the student does not complete the course successfully” is the complement of the event “the student completes the course successfully”. Hence, P(not successful) = 1 – P(successful) = 1 – 0.81 = 0.19. 31. P(not win the scholarship) = 1 – P(win scholarship) = 1 – 0.05 = 0.95. 32. a) Assume for the calculation that the tests give their results independently from one another. P(A and B) = P(A) x P(B) = 0.85 x 0.90 = 0.765. b) Since both tests could show a correct result at the same time, the events A and B are not mutually exclusive. P(A or B) = P(A) + P(B) – P(A and B) = 0.85 + 0.90 – 0.765 = 0.985; c) P(not A and not B) = P(not A) x P(not B) = [1 – P(A)] x [1 – P(B)] = 0.15 x 0.1 = 0.015. d) The two events are independent, because the result of test 1 (event A) has no impact on the result of test 2 (event B), regardless of whether its result is correct or incorrect, and vice versa. (Note: It is possible to have two tests that interact with each other in some way, e.g. if one test introduces a drug that impacts the other test; but there is no information to suggest this is the case in the problem.) 33. a) P(both cards are aces) = P(ace) x P(ace | 1st card was an ace) =
4 3 × = 0.0045 . 52 51
b) P(both cards are face cards) = P(face card) x P(face card | 1st card was a face card) =
12 11 × = 0.0498 . 52 51 c) P(first card = ace or second card = ace or both) = P(1st ace and 2nd no ace) or P(1st no ace and 2nd ace) or P(1st and 2nd are aces) =
⎛ 4 48 ⎞ ⎛ 48 4 ⎞ ⎛ 4 3 ⎞ ⎜ × ⎟ + ⎜ × ⎟ + ⎜ × ⎟ = 0.0724 + 0.0724 + 0.0045 = 0.1493 . ⎝ 52 51 ⎠ ⎝ 52 51 ⎠ ⎝ 52 51 ⎠ ⎛ 48 47 ⎞ × ⎟ = 0.8507 . ⎝ 52 51 ⎠
d) P(neither card is an ace) = ⎜
⎛ 4 4⎞ × ⎟ = 0.0059 . ⎝ 52 52 ⎠
34. a) P(both cards are Aces) = P(Ace) x P(Ace) = ⎜
⎛ 12 12 ⎞ × ⎟ = 0.0533 . ⎝ 52 52 ⎠
b) P(both cards are face cards) = P(face card) x P(face card) = ⎜
c) P(first card is an Ace or the second card is an Ace (or both)) = P(1st Ace and 2nd no Ace) or P(1st no Ace and 2nd Ace) or P(1st card and 2nd Ace) =
⎛ 4 48 ⎞ ⎛ 48 4 ⎞ ⎛ 4 4 ⎞ ⎜ × ⎟ + ⎜ × ⎟ + ⎜ × ⎟ = 0.0710 + 0.0710 + 0.0059 = 0.1479 . ⎝ 52 52 ⎠ ⎝ 52 52 ⎠ ⎝ 52 52 ⎠ d) P(neither card is an Ace) =
⎛ 48 48 ⎞ ⎜ × ⎟ = 0.8521 . ⎝ 52 52 ⎠
35. a) P(no | male) = 40/100 = 0.4. b) P(male and no) = 40/202 = 0.198. c) P(female and yes) = 86/202 = 0.4257. d) P(‘male and no or female and yes) = P(male and no) + P(female and yes) =
40 86 126 + = = 0.6238 . 202 202 202 36. a) Fruit
Yogurt
Totals
Grades 1-3
50
50
100
Grades 4-6
35
85
120
Totals
85
135
220
b) P(grades 4-6) = 120/220 = 0.5455. c) P(grades 1-3 and yogurt) = 50/220 = 0.2273. d) P(grades 1-3 or yogurt) = P(grades 1-3) + P(yogurt) – P(grades 1-3 and yogurt) =
100 135 50 185 + − = = 0.8409 . 220 220 220 220 e) P(grades 4-6 | yogurt) = 85/135 = 0.6296. 37. Since the probability of rolling a pair of sixes with two dice is independent of any former event that involves rolling the dice, find the probability P(roll a pair of sixes with two dice) = 1/36 = 0.0278. 41. a) P(at least one is struck by lightening) = P(1st is struck and 2nd is not) or P(1st is not and 2nd is struck) or P(1st is struck and 2nd is struck) = [0.00000032 x (1 – 0.00000032)] + [(1 – 0.00000032) x 0.00000032] + (0.00000032)2 = 0.00000064. b) P(both are struck by lightening) = 0.000000322 = 0.0000000000001. c) P(neither are struck by lightening) = (1 – 0.00000032)2 = 0.99999936. 45. P(A|B) =
P( B | A) × P( A) 0.05 × 0.81 = = 0.2893 . P( B) 0.14
46. Construct this contingency table:
10 24 × P(injured | infielder) × P(infielder) 24 68 a) Bayes: P(infielder | injured) = = = 0.3448 . 29 P(injured) 68 Contingency Table:
P(infielder | injured) = 10/29 = 0.3448.
b) Bayes: P(outfielder | injured) =
5 18 × P(injured | outfielder) × P(outfielder) 18 68 = = 0.1724 . 29 P(injured) 68 Contingency Table: P(outfielder | injured) = 5/29 = 0.1724.
12 20 × P(injured | pitcher) × P(pitcher) 20 68 c) Bayes: P(pitcher | injured) = = = 0.4138 . 29 P(injured) 68 Contingency Table: P(pitcher | injured) = 12/29 = 0.4138.
2 6 × P(injured | catcher) × P(catcher) 6 68 d) Bayes: P(catcher | injured) = = = 0.069 . 29 P(injured) 68 Contingency Table: P(catcher | injured) = 2/29 = 0.069. 47. Given: P(day) = 0.65; P(night) = 1 – 0.65 = 0.35; P(defective | day) = 0.1; P(defective | night) = 0.35 Preliminary calculations: P(day and defective) = P(day) x P(defective | day) = 0.65 x 0.1 = 0.065 P(night and defective) = P(night) x P(defective | night) = 0.35 x 0.35 = 0.1225 P(defective) = P(day and defective) + P(night and defective) = 0.065 + 0.1225 = 0.1875
Solution: P(night | defective) =
P(defective | night ) × P(night ) 0.35 × 0.35 = =0.1225/0.1875 = 0.6533. P(defective) 0.1875 48. Given: P(H) = 0.4; P(G) = 0.4; P(F) = 0.2; P(S|H) = 0.6; P(S|G) = 0.4; P(S|F) = 0.2 Preliminary Calculations: P(H and S) = P(H) x P(S | H) = 0.4 x 0.6 = 0.24 P(G and S) = P(G) x P(S | G) = 0.4 x 0.4 = 0.16 P(F and S) = P(F) x P(S | F) = 0.2 x 0.2 = 0.04 Assuming that exactly one of H, G, or F must occur: P(S) = P(H and S) + P(G and S) + P(F and S) = 0.24 + 0.16 + 0.04 = 0.44 a) P(H|S) =
P( S | H ) × P( H ) 0.6 × 0.4 = = 0.5455 . P( S ) 0.44
b) P(G|S) =
P( S | G ) × P(G ) 0.4 × 0.4 = = 0.3636 . P( S ) 0.44
c) P(F|S) =
P( S | F ) × P( F ) 0.2 × 0.2 = = 0.0909 . P( S ) 0.44
49. Preliminary Calculations: P(Siamese and curly tail kitten) = P(Siamese) x P(curly hair kitten | Siamese) = 0.35 x 0.03 = 0.0105 P(Ocicat and curly tail kitten) = P(Ocicat) x P(curly hair kitten | Ocicat) = 0.60 x 0.07 = 0.042 P(Devon Rex and curly tail kitten) = P(Devon Rex) x P(curly hair kitten | Devon Rex) = 0.05 x 0.04 = 0.002 Assuming that the only possibilities for who is the father are the Siamese cat, the Ocicat, and the Devon Rex: P(curly tail kitten) = P(Siamese and curly tail kitten) + P(Ocicat and curly tail kitten) + P(Devon Rex and curly tail kitten) = 0.0105 + 0.042 + 0.002 = 0.0545 a) P(the Siamese cat is the father of the litter | a kitten with a curly tail is born ) =
P(curly _ tail _ kitten | Siamese _ Father ) × P( Siamese _ Father ) 0.03 × 0.35 = = P(curly _ tail _ kitten) 0.0545 0.0105/0.0545 = 0.1927.
b) P(the Ocicat is the father of the litter | a kitten with a curly tail is born ) =
P(curly _ tail _ kitten | Ocicat _ Father ) × P (Ocicat _ Father ) 0.07 × 0.60 = = P (curly _ tail _ kitten) 0.0545 0.042/0.0545 = 0.7706. c) P(the Devon Rex cat is the father of the litter | a kitten with a curly tail is born ) =
P(curly _ tail _ kitten | Devon _ Re x _ Father ) × P( Devon _ Re x _ Father ) 0.04 × 0.05 = P(curly _ tail _ kitten) 0.0545 = 0.002/0.0545 = 0.0367. 50. Given: P(‘Globe and Mail’) = 0.6; P(‘Toronto Sun’) = 0.3; P(‘Toronto Star’) = 0.1; P(article on environment | ‘Globe and Mail’) = 0.4; P(article on environment | ‘Toronto Sun’) = 0.3; P(article on environment | ‘Toronto Star’) = 0.7 Preliminary Calculations: P(‘Globe and Mail’ and article on environment) = P(‘Globe and Mail’) x P(article on environment | ‘Globe and Mail’) = 0.6 x 0.4 = 0.24 P(‘Toronto Sun’ and article on environment) = P(‘Toronto Sun’) x P(article on environment | ‘Toronto Sun’) = 0.3 x 0.3 = 0.09 P(‘Toronto Star’ and article on environment) = P(‘Toronto Star’) x P(article on environment | ‘Toronto Star’) = 0.1 x 0.7 = 0.07 Assuming that the only papers she reads are the ones listed in the problem: P(article on environment) = P(‘Globe and Mail’ and article on environment) + P(‘Toronto Sun’ and article on environment) + P(‘Toronto Star’ and article on environment) = 0.24 + 0.09 + 0.07 = 0.4 a) P(‘Globe and Mail’ | article on the environment) =
P(article on the environment |' Globe and Mail ' ) × P(' Globe and Mail ' ) 0.4 × 0.6 = P(article on the environment ) 0.4 = 0.24/0.4 = 0.600.
b) P(‘Toronto Sun’ | article on the environment) =
P(article on the environment |'Toronto Sun' ) × P('Toronto Sun' ) 0.3 × 0.3 = = 0.09/0.4 P(article on the environment ) 0.4 = 0.225. c) P(‘Toronto Star’ | article on the environment) =
P(article on the environment |'Toronto Star ' ) × P('Toronto Star ' ) 0.7 × 0.1 = = P(article on the environment ) 0.4 0.07/0.4 = 0.175. 51. Given: P(telephone) = 0.5; P(letter) = 0.3; P(personal visit) = 0.2; P(payment | telephone) = 0.60; P(payment | letter) = 0.30; P(payment | ‘personal visit’) = 0.80 Preliminary calculations: P(telephone and payment) = P(telephone) x P(payment | telephone) = 0.5 x 0.6 = 0.3 P(letter and payment) = P(letter) x P(payment | letter) = 0.3 x 0.3 = 0.09 P(personal visit and payment) = P(personal visit) x P(payment | personal visit) = 0.2 x 0.8 = 0.16 Assuming that exactly one of the contact methods will always be chosen: P(payment) = 0.3 + 0.09 + 0.16 = 0.55 a) P(telephone | payment) =
P( payment | telephone) × P(telephone) 0.6 × 0.5 = = 0.3/0.55 = P( payment ) 0.55
0.5455. b) P(letter | payment) =
P( payment | letter ) × P(letter ) 0.3 × 0.3 = = 0.09/0.55 = 0.1636. P( payment ) 0.55
c) P(personal visit | payment) = = 0.16/0.55 = 0.2909.
P( payment | personal visit ) × P( personal visit ) 0.8 × 0.2 = P( payment ) 0.55
52. a) P(absent infrequently | low grade) =
1 15 × P(low grade | absent infreqeuntly) × P(absent infrequently) 15 85 = = 0.0556 . 18 P(low grade) 85 b) P(medium absenteeism | low grade) =
9 53 × P(low grade | medium absenteeism) × P(medium absenteeism) 53 85 = = 0.5 . 18 P(low grade) 85 c) P(absent frequently | low grade) =
8 17 × P(low grade | absent frequently) × P(absent frequently) 17 85 = = 0.4444 . 18 P(low grade) 85 53. Given: P(A) = 0.3; P(B) = 0.4; P(C) = 0.2; P(D) = 0.1; P(X|A) = 0.4; P(X|B) = 0.2; P(X|C) = 0.1; P(X|D) = 0.5 Preliminary calculation:
(Assuming exactly one of A, B, C, or D occurs.)
P(X) = P(A and X) + P(B and X) + P(C and X) + P(D and X) = (0.3 x 0.4) + (0.4 x 0.2) + (0.2 x 0.1) + (0.1 x 0.5) = 0.12 + 0.08 + 0.02 + 0.05 = 0.27 a) P(A|X) =
P( X | A) × P( A) 0.4 × 0.3 = = 0.4444 . P( X ) 0.27
b) P(B|X) =
P( X | B) × P( B) 0.2 × 0.4 = = 0.2963 . P( X ) 0.27
c) P(C|X) =
P( X | C ) × P(C ) 0.1 × 0.2 = = 0.0741 . P( X ) 0.27
d) P(D|X) =
P( X | D) × P( D) 0.5 × 0.1 = = 0.1852 . P( X ) 0.27
57. Given: P(banished) = 0.667; P(surviving | banished) = 0.2; P(surviving) = 0.43; P(banished | surviving) =
P(surviving | banished) × P(banished) 0.2 × 0.667 = = 0.3102 . P(surviving) 0.43
b) Four (of many) possible outcomes are shown below: •
101 participants are female; 99 participants are male
•
50 participants are female; 150 participants are male
•
99 participants are female; 101 participants are male
•
150 participants are female; 50 participants are male
c) If data were available for the proportions of women who have registered for that run in past years, we could estimate an empirical probability for that proportion for this year—but, unfortunately, the relevant data are not given in the problem. (We can’t calculate an a priori probability, because we lack sufficient knowledge about the experiment. Suppose, for example, that there are only 110 males in the vicinity of Whitby; then the outcome “8 participants are female; 192 participants are male” would have zero probability. On the other hand, if there thousands of males and just hundreds of females in Whitby fit enough to run, then “8 participants are female; 192 participants are male” might be reasonably probable.) 5. a) The (stop or don’t-stop) actions taken by all 500 drivers. b) Possible answers are: •
200 drivers stop fully; 300 drivers do not stop fully.
•
400 drivers stop fully; 100 drivers do not stop fully.
•
125 drivers stop fully; 375 drivers do not stop fully.
•
430 drivers stop fully; 70 drivers do not stop fully.
6. a) Probability that a selected student clearly enjoyed the concert = 162/250 = 0.648. b) Probability that a selected student stated a clear opinion about whether they enjoyed the concert =
162 + 67 229 = = 0.916 . 250 250 7. Subjective probability.
8. a) P(Ace of Hearts) = 1/52 = 0.0192
Ranks of Cards
Suit
Ace
Other
TOTAL
Hearts
1
12
13
Other
3
36
39
TOTAL
4
48
52
b) In a standard deck of 52 playing cards, the “suit” Spade consists of 13 cards; thus P(any Spade) = 13/52 = 0.25. c) In a standard deck of 52 playing cards, there are four “suits”—each of which contains exactly one ‘10’, one ‘Jack’, one ‘Queen’, one ‘King’ and one ‘Ace’. So the number of possible ways to draw any of those cards, disregarding its suit = (5 cards of interest) × (4 “suits”) = 20. Therefore, P(any one of a 10, a Jack, a Queen, a King, or an Ace) = 20/52 = 0.3846. 9. a) P(red marble) = 4/8 = 0.5. b)
The first Marble Selected The second marble selected
\/
1st Red
2nd Red
3rd Red
4th Red
1st White
2nd White
1st Blue
2nd Blue
1st White
√
√
2nd White
√
√
1st Red 2nd Red 3rd Red 4th Red
1st Blue
√
√
2nd Blue
√
√
P(one blue and one white—in either order) = 8/64 = 0.125.
10. a) P(winning response is “Mainland China”) = 68/312 = 0.2179 b) The list of countries, as such, is a mutually exclusive list. Each region or country is listed only once; the options are clearly defined and separated from one another. However, if the interest is on “sets of countries chosen” by individuals, there is no mutually exclusive list – because one individual’s choice-set might include more than one entry in the list. We don’t have enough information to find the probability that half of the respondents chose Mainland China, because there is no information about the number of respondents. Each respondent was allowed to check off more than one destination. We have no information about how often respondents checked off more than one destination. 11. a) (All combinations without an X in the table below are possible. Combinations that add up to 58 show a check mark.)
First Element Chosen Second Element Chosen \/
4
4
X
20
38
20
X
√
38
√
X
12
56 88
P(sum = 58) = 2/30 = 0.0667.
12
56
88
X
X X
b) (Checked-off combinations in the table add up to less than 60.)
First Element Chosen Second Element Chosen \/
4
12
20
38
4
X
√
√
√
12
√
X
√
√
20
√
√
X
√
38
√
√
√
X
56
56
88
X
88
X
P(sum < 60) = 12/30 = 0.4. 12. (The simplified compass assumed in this question has no printing on the background; the only label of interest is “N” (or “North”) printed on one side of the pointer—hopefully, on the north-seeking side..)
For each of the three compasses, the pointer label “North” may either be painted on the correct side of the pointer or on incorrect side. A possible table of outcomes for the labelling of the pointers is shown below:
The table shows 8 possible combinations of correct or incorrect labels for the three compass pointers. Only one of the 8 combinations is all are correct. Therefore, the probability = 1/8 = 0.125.
17. a)
b) P(late) = 0.13; (c) P(late | Health Sciences student) = 0.04/0.35 = 0.1143; (d) P(Health Sciences student | on time) = 0.31/0.87 = 0.3563. 18. a) P(ticket costs $25) = 70/250 = 0.28 b) P(paid $25 | paid more than $15) = 70/(60+70) = 0.5385 c) P(paid $15 | paid less than $20) = 75/(45+75) = 0.625 19. P(enjoyed concert | voiced an opinion) = 162/229 = 0.7074. 20. a) P(marble is red | red marble taken and not replaced) = 3/7 = 0.4286 b) P(marble is red | three red marbles taken and not replaced) = 1/5 = 0.2 21. a) P(low grade) = 18/85 = 0.2118. b) P(low grade | frequently absent) = 8/17 = 0.4706. c) P(frequently absent) = 17/85 = 0.20. d) P(frequently absent | high grade) = 2/19 = 0.1053. 22. The relevant data are summarized in this table: STATE
Mines_U04
Mines_S04
8
41
49
Other
578
730
1308
Totals
586
771
1357
Alabama
Totals
a) P(mine located in Alabama | active coal mine in 2004) = 49/1,357 = 0.0361. b) P(mine located in Alabama | active surface coal mine in 2004) = 41/771 = 0.0532. c) P(underground mine | active coal mine in 2004) = 586/1,357 = 0.4318. d) P(underground mine | active coal mine in Alabama in 2004) = 8/49 = 0.1633. 23. a) P(atomic number < 21 | alkaline earth element) = 3/6 = 0.5. b) P(atomic number of 12 | alkaline earth element whose atomic number is < 21) = 1/3 = 0.3333.
29. Note that according to the Multiplication Rule, two events A and B are independent, if P(A) x P(B) = P(A and B). a)
Independent, because P(A and B) = 0.14 = P(A) x P(B) = 0.20 x 0.70 = 0.14.
b) Dependent, because P(A and B) = 0.36 ≠ P(A) x P(B) = 0.80 x 0.50 = 0.40. 30. The empirical probability P(successful) = 0.81. The event “the student does not complete the course successfully” is the complement of the event “the student completes the course successfully”. Hence, P(not successful) = 1 – P(successful) = 1 – 0.81 = 0.19. 31. P(not win the scholarship) = 1 – P(win scholarship) = 1 – 0.05 = 0.95. 32. a) Assume for the calculation that the tests give their results independently from one another. P(A and B) = P(A) x P(B) = 0.85 x 0.90 = 0.765. b) Since both tests could show a correct result at the same time, the events A and B are not mutually exclusive. P(A or B) = P(A) + P(B) – P(A and B) = 0.85 + 0.90 – 0.765 = 0.985; c) P(not A and not B) = P(not A) x P(not B) = [1 – P(A)] x [1 – P(B)] = 0.15 x 0.1 = 0.015. d) The two events are independent, because the result of test 1 (event A) has no impact on the result of test 2 (event B), regardless of whether its result is correct or incorrect, and vice versa. (Note: It is possible to have two tests that interact with each other in some way, e.g. if one test introduces a drug that impacts the other test; but there is no information to suggest this is the case in the problem.) 33. a) P(both cards are aces) = P(ace) x P(ace | 1st card was an ace) =
4 3 × = 0.0045 . 52 51
b) P(both cards are face cards) = P(face card) x P(face card | 1st card was a face card) =
12 11 × = 0.0498 . 52 51 c) P(first card = ace or second card = ace or both) = P(1st ace and 2nd no ace) or P(1st no ace and 2nd ace) or P(1st and 2nd are aces) =
⎛ 4 48 ⎞ ⎛ 48 4 ⎞ ⎛ 4 3 ⎞ ⎜ × ⎟ + ⎜ × ⎟ + ⎜ × ⎟ = 0.0724 + 0.0724 + 0.0045 = 0.1493 . ⎝ 52 51 ⎠ ⎝ 52 51 ⎠ ⎝ 52 51 ⎠ ⎛ 48 47 ⎞ × ⎟ = 0.8507 . ⎝ 52 51 ⎠
d) P(neither card is an ace) = ⎜
⎛ 4 4⎞ × ⎟ = 0.0059 . ⎝ 52 52 ⎠
34. a) P(both cards are Aces) = P(Ace) x P(Ace) = ⎜
⎛ 12 12 ⎞ × ⎟ = 0.0533 . ⎝ 52 52 ⎠
b) P(both cards are face cards) = P(face card) x P(face card) = ⎜
c) P(first card is an Ace or the second card is an Ace (or both)) = P(1st Ace and 2nd no Ace) or P(1st no Ace and 2nd Ace) or P(1st card and 2nd Ace) =
⎛ 4 48 ⎞ ⎛ 48 4 ⎞ ⎛ 4 4 ⎞ ⎜ × ⎟ + ⎜ × ⎟ + ⎜ × ⎟ = 0.0710 + 0.0710 + 0.0059 = 0.1479 . ⎝ 52 52 ⎠ ⎝ 52 52 ⎠ ⎝ 52 52 ⎠ d) P(neither card is an Ace) =
⎛ 48 48 ⎞ ⎜ × ⎟ = 0.8521 . ⎝ 52 52 ⎠
35. a) P(no | male) = 40/100 = 0.4. b) P(male and no) = 40/202 = 0.198. c) P(female and yes) = 86/202 = 0.4257. d) P(‘male and no or female and yes) = P(male and no) + P(female and yes) =
40 86 126 + = = 0.6238 . 202 202 202 36. a) Fruit
Yogurt
Totals
Grades 1-3
50
50
100
Grades 4-6
35
85
120
Totals
85
135
220
b) P(grades 4-6) = 120/220 = 0.5455. c) P(grades 1-3 and yogurt) = 50/220 = 0.2273. d) P(grades 1-3 or yogurt) = P(grades 1-3) + P(yogurt) – P(grades 1-3 and yogurt) =
100 135 50 185 + − = = 0.8409 . 220 220 220 220 e) P(grades 4-6 | yogurt) = 85/135 = 0.6296. 37. Since the probability of rolling a pair of sixes with two dice is independent of any former event that involves rolling the dice, find the probability P(roll a pair of sixes with two dice) = 1/36 = 0.0278. 41. a) P(at least one is struck by lightening) = P(1st is struck and 2nd is not) or P(1st is not and 2nd is struck) or P(1st is struck and 2nd is struck) = [0.00000032 x (1 – 0.00000032)] + [(1 – 0.00000032) x 0.00000032] + (0.00000032)2 = 0.00000064. b) P(both are struck by lightening) = 0.000000322 = 0.0000000000001. c) P(neither are struck by lightening) = (1 – 0.00000032)2 = 0.99999936. 45. P(A|B) =
P( B | A) × P( A) 0.05 × 0.81 = = 0.2893 . P( B) 0.14
46. Construct this contingency table:
10 24 × P(injured | infielder) × P(infielder) 24 68 a) Bayes: P(infielder | injured) = = = 0.3448 . 29 P(injured) 68 Contingency Table:
P(infielder | injured) = 10/29 = 0.3448.
b) Bayes: P(outfielder | injured) =
5 18 × P(injured | outfielder) × P(outfielder) 18 68 = = 0.1724 . 29 P(injured) 68 Contingency Table: P(outfielder | injured) = 5/29 = 0.1724.
12 20 × P(injured | pitcher) × P(pitcher) 20 68 c) Bayes: P(pitcher | injured) = = = 0.4138 . 29 P(injured) 68 Contingency Table: P(pitcher | injured) = 12/29 = 0.4138.
2 6 × P(injured | catcher) × P(catcher) 6 68 d) Bayes: P(catcher | injured) = = = 0.069 . 29 P(injured) 68 Contingency Table: P(catcher | injured) = 2/29 = 0.069. 47. Given: P(day) = 0.65; P(night) = 1 – 0.65 = 0.35; P(defective | day) = 0.1; P(defective | night) = 0.35 Preliminary calculations: P(day and defective) = P(day) x P(defective | day) = 0.65 x 0.1 = 0.065 P(night and defective) = P(night) x P(defective | night) = 0.35 x 0.35 = 0.1225 P(defective) = P(day and defective) + P(night and defective) = 0.065 + 0.1225 = 0.1875
Solution: P(night | defective) =
P(defective | night ) × P(night ) 0.35 × 0.35 = =0.1225/0.1875 = 0.6533. P(defective) 0.1875 48. Given: P(H) = 0.4; P(G) = 0.4; P(F) = 0.2; P(S|H) = 0.6; P(S|G) = 0.4; P(S|F) = 0.2 Preliminary Calculations: P(H and S) = P(H) x P(S | H) = 0.4 x 0.6 = 0.24 P(G and S) = P(G) x P(S | G) = 0.4 x 0.4 = 0.16 P(F and S) = P(F) x P(S | F) = 0.2 x 0.2 = 0.04 Assuming that exactly one of H, G, or F must occur: P(S) = P(H and S) + P(G and S) + P(F and S) = 0.24 + 0.16 + 0.04 = 0.44 a) P(H|S) =
P( S | H ) × P( H ) 0.6 × 0.4 = = 0.5455 . P( S ) 0.44
b) P(G|S) =
P( S | G ) × P(G ) 0.4 × 0.4 = = 0.3636 . P( S ) 0.44
c) P(F|S) =
P( S | F ) × P( F ) 0.2 × 0.2 = = 0.0909 . P( S ) 0.44
49. Preliminary Calculations: P(Siamese and curly tail kitten) = P(Siamese) x P(curly hair kitten | Siamese) = 0.35 x 0.03 = 0.0105 P(Ocicat and curly tail kitten) = P(Ocicat) x P(curly hair kitten | Ocicat) = 0.60 x 0.07 = 0.042 P(Devon Rex and curly tail kitten) = P(Devon Rex) x P(curly hair kitten | Devon Rex) = 0.05 x 0.04 = 0.002 Assuming that the only possibilities for who is the father are the Siamese cat, the Ocicat, and the Devon Rex: P(curly tail kitten) = P(Siamese and curly tail kitten) + P(Ocicat and curly tail kitten) + P(Devon Rex and curly tail kitten) = 0.0105 + 0.042 + 0.002 = 0.0545 a) P(the Siamese cat is the father of the litter | a kitten with a curly tail is born ) =
P(curly _ tail _ kitten | Siamese _ Father ) × P( Siamese _ Father ) 0.03 × 0.35 = = P(curly _ tail _ kitten) 0.0545 0.0105/0.0545 = 0.1927.
b) P(the Ocicat is the father of the litter | a kitten with a curly tail is born ) =
P(curly _ tail _ kitten | Ocicat _ Father ) × P (Ocicat _ Father ) 0.07 × 0.60 = = P (curly _ tail _ kitten) 0.0545 0.042/0.0545 = 0.7706. c) P(the Devon Rex cat is the father of the litter | a kitten with a curly tail is born ) =
P(curly _ tail _ kitten | Devon _ Re x _ Father ) × P( Devon _ Re x _ Father ) 0.04 × 0.05 = P(curly _ tail _ kitten) 0.0545 = 0.002/0.0545 = 0.0367. 50. Given: P(‘Globe and Mail’) = 0.6; P(‘Toronto Sun’) = 0.3; P(‘Toronto Star’) = 0.1; P(article on environment | ‘Globe and Mail’) = 0.4; P(article on environment | ‘Toronto Sun’) = 0.3; P(article on environment | ‘Toronto Star’) = 0.7 Preliminary Calculations: P(‘Globe and Mail’ and article on environment) = P(‘Globe and Mail’) x P(article on environment | ‘Globe and Mail’) = 0.6 x 0.4 = 0.24 P(‘Toronto Sun’ and article on environment) = P(‘Toronto Sun’) x P(article on environment | ‘Toronto Sun’) = 0.3 x 0.3 = 0.09 P(‘Toronto Star’ and article on environment) = P(‘Toronto Star’) x P(article on environment | ‘Toronto Star’) = 0.1 x 0.7 = 0.07 Assuming that the only papers she reads are the ones listed in the problem: P(article on environment) = P(‘Globe and Mail’ and article on environment) + P(‘Toronto Sun’ and article on environment) + P(‘Toronto Star’ and article on environment) = 0.24 + 0.09 + 0.07 = 0.4 a) P(‘Globe and Mail’ | article on the environment) =
P(article on the environment |' Globe and Mail ' ) × P(' Globe and Mail ' ) 0.4 × 0.6 = P(article on the environment ) 0.4 = 0.24/0.4 = 0.600.
b) P(‘Toronto Sun’ | article on the environment) =
P(article on the environment |'Toronto Sun' ) × P('Toronto Sun' ) 0.3 × 0.3 = = 0.09/0.4 P(article on the environment ) 0.4 = 0.225. c) P(‘Toronto Star’ | article on the environment) =
P(article on the environment |'Toronto Star ' ) × P('Toronto Star ' ) 0.7 × 0.1 = = P(article on the environment ) 0.4 0.07/0.4 = 0.175. 51. Given: P(telephone) = 0.5; P(letter) = 0.3; P(personal visit) = 0.2; P(payment | telephone) = 0.60; P(payment | letter) = 0.30; P(payment | ‘personal visit’) = 0.80 Preliminary calculations: P(telephone and payment) = P(telephone) x P(payment | telephone) = 0.5 x 0.6 = 0.3 P(letter and payment) = P(letter) x P(payment | letter) = 0.3 x 0.3 = 0.09 P(personal visit and payment) = P(personal visit) x P(payment | personal visit) = 0.2 x 0.8 = 0.16 Assuming that exactly one of the contact methods will always be chosen: P(payment) = 0.3 + 0.09 + 0.16 = 0.55 a) P(telephone | payment) =
P( payment | telephone) × P(telephone) 0.6 × 0.5 = = 0.3/0.55 = P( payment ) 0.55
0.5455. b) P(letter | payment) =
P( payment | letter ) × P(letter ) 0.3 × 0.3 = = 0.09/0.55 = 0.1636. P( payment ) 0.55
c) P(personal visit | payment) = = 0.16/0.55 = 0.2909.
P( payment | personal visit ) × P( personal visit ) 0.8 × 0.2 = P( payment ) 0.55
52. a) P(absent infrequently | low grade) =
1 15 × P(low grade | absent infreqeuntly) × P(absent infrequently) 15 85 = = 0.0556 . 18 P(low grade) 85 b) P(medium absenteeism | low grade) =
9 53 × P(low grade | medium absenteeism) × P(medium absenteeism) 53 85 = = 0.5 . 18 P(low grade) 85 c) P(absent frequently | low grade) =
8 17 × P(low grade | absent frequently) × P(absent frequently) 17 85 = = 0.4444 . 18 P(low grade) 85 53. Given: P(A) = 0.3; P(B) = 0.4; P(C) = 0.2; P(D) = 0.1; P(X|A) = 0.4; P(X|B) = 0.2; P(X|C) = 0.1; P(X|D) = 0.5 Preliminary calculation:
(Assuming exactly one of A, B, C, or D occurs.)
P(X) = P(A and X) + P(B and X) + P(C and X) + P(D and X) = (0.3 x 0.4) + (0.4 x 0.2) + (0.2 x 0.1) + (0.1 x 0.5) = 0.12 + 0.08 + 0.02 + 0.05 = 0.27 a) P(A|X) =
P( X | A) × P( A) 0.4 × 0.3 = = 0.4444 . P( X ) 0.27
b) P(B|X) =
P( X | B) × P( B) 0.2 × 0.4 = = 0.2963 . P( X ) 0.27
c) P(C|X) =
P( X | C ) × P(C ) 0.1 × 0.2 = = 0.0741 . P( X ) 0.27
d) P(D|X) =
P( X | D) × P( D) 0.5 × 0.1 = = 0.1852 . P( X ) 0.27
57. Given: P(banished) = 0.667; P(surviving | banished) = 0.2; P(surviving) = 0.43; P(banished | surviving) =
P(surviving | banished) × P(banished) 0.2 × 0.667 = = 0.3102 . P(surviving) 0.43
