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Introductory Statistics

David Diez, Mine Çetinkaya-Rundel, Christopher D Barr

Section A.3 Probability

Solution 1. Exercise 1
  1. False. These are independent trials.
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  2. False. There are red face cards.
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  3. True. A card cannot be both a face card and an ace.
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Solution 2. Exercise 3
  1. 10 tosses. Fewer tosses mean more variability in the sample fraction of heads, meaning there’s a better chance of getting at least 60\% heads.
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  2. 100 tosses. More flips means the observed proportion of heads would often be closer to the average, 0.50, and therefore also above 0.40.
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  3. 100 tosses. With more flips, the observed proportion of heads would often be closer to the average, 0.50.
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  4. 10 tosses. Fewer flips would increase variability in the fraction of tosses that are heads.
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Solution 3. Exercise 5
  1. \(0.5^{10}\) = 0.00098.
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  2. \(0.5^{10}\) = 0.00098.
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  3. \(P\)(at least one tails) = \(1 - P\)(no tails) = \(1 - (0.5^{10}) \approx 1 - 0.001 = 0.999\text{.}\)
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Solution 4. Exercise 7
  1. No, there are voters who are both independent and swing voters.
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  2. [Figure: A Venn diagram is shown for variables "Independent" and "Swing", where the two circles representing the variable are partially overlapping. The region of the "Independent" circle not overlapping the other circle is labeled with "24". The region of the "Swing" circle not overlapping the other circle is labeled with "12". The region where the two circles overlap is labeled with "11".].
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  3. Each Independent voter is either a swing voter or not. Since 35\% of voters are Independents and 11\% are both Independent and swing voters, the other 24\% must not be swing voters.
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  4. 0.47.
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  5. 0.53.
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  6. P(Independent) \(\times\) P(swing) = \(0.35\times0.23 = 0.08\text{,}\) which does not equal P(Independent and swing) = 0.11, so the events are dependent.
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Solution 5. Exercise 9
  1. If the class is not graded on a curve, they are independent. If graded on a curve, then neither independent nor disjoint -- unless the instructor will only give one A, which is a situation we will ignore in parts.
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  2. and.
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  3. .
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  4. They are probably not independent: if you study together, your study habits would be related, which suggests your course performances are also related.
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  5. No. See the answer to part.
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  6. when the course is not graded on a curve. More generally: if two things are unrelated (independent), then one occurring does not preclude the other from occurring.
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Solution 6. Exercise 11
  1. \(0.16 + 0.09 = 0.25\text{.}\)
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  2. \(0.17 + 0.09 = 0.26\text{.}\)
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  3. Assuming that the education level of the husband and wife are independent: \(0.25 \times 0.26 = 0.065\text{.}\) You might also notice we actually made a second assumption: that the decision to get married is unrelated to education level.
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  4. The husband/wife independence assumption is probably not reasonable, because people often marry another person with a comparable level of education. We will leave it to you to think about whether the second assumption noted in part.
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  5. is reasonable.
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Solution 7. Exercise 13
  1. No, but we could if A and B are independent. (b-i) 0.21. (b-ii) 0.79. (b-iii) 0.3.
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  2. No, because 0.1 \(\ne\) 0.21, where 0.21 was the value computed under independence from part.
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  3. .
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  4. 0.143.
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Solution 8. Exercise 15
  1. No, 0.18 of respondents fall into this combination.
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  2. \(0.60 + 0.20 - 0.18 = 0.62\text{.}\)
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  3. \(0.18 / 0.20 = 0.9\text{.}\)
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  4. \(0.11 / 0.33 \approx 0.33\text{.}\)
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  5. No, otherwise the answers to.
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  6. and.
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  7. would be the same.
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  8. \(0.06 / 0.34 \approx 0.18\text{.}\)
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Solution 9. Exercise 17
  1. No. There are 6 females who like Five Guys Burgers.
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  2. \(162 / 248 = 0.65\text{.}\)
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  3. \(181 / 252 = 0.72\text{.}\)
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  4. Under the assumption of a dating choices being independent of hamburger preference, which on the surface seems reasonable: \(0.65 \times 0.72 = 0.468\text{.}\)
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  5. \((252 + 6 - 1)/500 = 0.514\text{.}\)
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Solution 10. Exercise 19
  1. [Figure: A tree diagram with a primary branch "Can construct box plots?" and a secondary branch "Passed?". The primary "Can construct box plots" branching has two possibilities of "Yes" with probability 0.8 and "No" with probability 0.2. Each of these branches has two secondary branches. The "Yes" primary branch breaks into branches for "Yes" (for Passed) that has a conditional probability of 0.86 with a Yes-and-Yes final probability of 0.688, and a "No" secondary branch with a conditional probability of 0.14 with a Yes-and-No final probability of 0.112. The "No" primary branch from "Can construct box plots" has a branch of "Yes" that has a conditional probability of 0.65 with a No-and-Yes final probability of 0.13, and a "No" secondary branch with a conditional probability of 0.35 with a No-and-No final probability of 0.07.].
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  2. 0.84.
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Solution 11. Exercise 21
0.0714. Even when a patient tests positive for lupus, there is only a 7.14\% chance that he actually has lupus. House may be right. [Figure: A tree diagram with a primary branch "Lupus" and a secondary branch "Result" for the test of Lupus. The primary "Lupus" branching has two possibilities of "Yes" with probability 0.02 and "No" with probability 0.98. Each of these branches has two secondary branches. The "Yes" primary branch breaks into branches for "Yes" (for Result) that has a conditional probability of 0.98 with a Yes-and-Yes final probability of 0.0196, and a "No" secondary branch with a conditional probability of 0.02 with a Yes-and-No final probability of 0.0004. The "No" primary branch from "Lupus" has a secondary branch of "Yes" that has a conditional probability of 0.26 with a No-and-Yes final probability of 0.2548, and a "No" secondary branch with a conditional probability of 0.74 with a No-and-No final probability of 0.7252.]
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Solution 12. Exercise 23
  1. 0.3.
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  2. 0.3.
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  3. 0.3.
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  4. \(0.3\times0.3=0.09\text{.}\)
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  5. Yes, the population that is being sampled from is identical in each draw.
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Solution 13. Exercise 25
  1. \(2 / 9 \approx 0.22\text{.}\)
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  2. \(3 / 9 \approx 0.33\text{.}\)
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  3. \(\frac{3}{10} \times \frac{2}{9} \approx 0.067\text{.}\)
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  4. No, e.g. in this exercise, removing one chip meaningfully changes the probability of what might be drawn next.
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Solution 14. Exercise 27
\(P(^1\)leggings, \(^2\)jeans, \(^3\)jeans\() = \frac{5}{24} \times \frac{7}{23} \times \frac{6}{22} = 0.0173\text{.}\) However, the person with leggings could have come 2nd or 3rd, and these each have this same probability, so \(3 \times 0.0173 = 0.0519\text{.}\)
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Solution 15. Exercise 29
  1. 13.
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  2. No, these 27 students are not a random sample from the university’s student population. For example, it might be argued that the proportion of smokers among students who go to the gym at 9 am on a Saturday morning would be lower than the proportion of smokers in the university as a whole.
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Solution 16. Exercise 31
  1. E(X) = 3.59. SD(X) = 9.64.
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  2. E(X) = -1.41. SD(X) = 9.64.
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  3. No, the expected net profit is negative, so on average you expect to lose money.
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Solution 17. Exercise 33
5\% increase in value.
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Solution 18. Exercise 35
E = -0.0526. SD = 0.9986.
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Solution 19. Exercise 37
Approximate answers are OK. (a) \((29+32)/144 = 0.42\text{.}\) (b) \(21/144 = 0.15\text{.}\) (c) \((26+12+15)/144 = 0.37\text{.}\)
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Solution 20. Exercise 39
  1. Invalid. Sum is greater than 1.
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  2. Valid. Probabilities are between 0 and 1, and they sum to 1. In this class, every student gets a C.
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  3. Invalid. Sum is less than 1.
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  4. Invalid. There is a negative probability.
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  5. Valid. Probabilities are between 0 and 1, and they sum to 1.
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  6. Invalid. There is a negative probability.
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Solution 21. Exercise 41
0.8247. [Figure: A tree diagram with a primary branch "HIV" and a secondary branch "Result" for the test of HIV. The primary "HIV" branching has two possibilities of "Yes" with probability 0.259 and "No" with probability 0.741. Each of these branches has two secondary branches. The "Yes" primary branch breaks into secondary branches for "Yes" (for Result) that has a conditional probability of 0.997 with a Yes-and-Yes final probability of 0.2582, and a "No" secondary branch with a conditional probability of 0.003 with a Yes-and-No final probability of 0.0008. The "No" primary branch from "HIV" has a secondary branch of "Yes" for "Result" that has a conditional probability of 0.074 with a No-and-Yes final probability of 0.0548, and a "No" secondary branch with a conditional probability of 0.926 with a No-and-No final probability of 0.6862.]
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Solution 22. Exercise 43
  1. E = \\(3.90. SD = \\)0.34.
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  2. E = \\(27.30. SD = \\)0.89.
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Solution 23. Exercise 45
\(Var\left(\frac{X_1 + X_2}{2}\right)\) \(= Var\left(\frac{X_1}{2} + \frac{X_2}{2}\right)\) \(= \frac{Var(X_1)}{2^2} + \frac{Var(X_2)}{2^2}\) \(= \frac{\sigma^2}{4} + \frac{\sigma^2}{4}\) \(= \sigma^2 / 2\)
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Solution 24. Exercise 47
\(Var\left(\frac{X_1 + X_2 + \dots + X_n}{n}\right)\) \(= Var\left(\frac{X_1}{n} + \frac{X_2}{n} + \dots + \frac{X_n}{n}\right)\) \(= \frac{Var(X_1)}{n^2} + \frac{Var(X_2)}{n^2} + \dots + \frac{Var(X_n)}{n^2}\) \(= \frac{\sigma^2}{n^2} + \frac{\sigma^2}{n^2} + \dots + \frac{\sigma^2}{n^2}\) (there are \(n\) of these terms) \(= n \frac{\sigma^2}{n^2}\) \(= \sigma^2 / n\)
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