Solutions for Chapter 3 of Statistics for Business & Economics

The following are full solutions to chapter 3 in “Statistics for Business and Economics” By Newbold, Carlson & Thorne. I do not guarantee the correctness of any of the answers presented here. If you found a mistake, have a comment, or would like to ask me anything, I’m available by mail: me (at) shayacrich (dot) com.

3.1. $$\overline{A}=[E_2, E_4, E_5, E_7, E_8, E_{10}]$$
3.2.a. $$A \cap B = [E_3, E_9]$$
3.2.b. $$A \cup B = [E_1, E_2, E_3, E_7, E_8, E_9]$$
3.2.c. It’s not, because \(E_4\) to \(E_6\), as well as \(E_{10}\) are not covered.

3.3.a. $$A \cap B = [E_4, E_5, E_6, E_{10}]$$
3.3.b. $$A \cup B = [E_1, E_2, E_4, E_5, E_6, E_7, E_8, E_{10}]$$
3.3.c. It’s not, because \(E_3\) and \(E_{10}\) are missing.

3.4.a. $$A \cap B = [E_3, E_6]$$
3.4.b. $$A \cup B = [E_3, E_4, E_5, E_6, E_9, E_{10}]$$
3.4.c. It’s not.

3.5.a. \(\overline{A}\) is the event “it will be 4 or less days before the machinery becomes available”.
3.5.b. \(A \cap B\) is the event it will be 5 days before the machine becomes available.
3.5.c. \(A \cup B\) is collectively exhaustive (any number of days before the machinery becomes available.
3.5.d. \(A \cap B\) is not the empty set, but rather contains the outcome of 5 days, so A and B are not mutually exclusive.
3.5.e. A and B are collectively exhaustive because any outcome is either below 6 or above 4.
3.5.f. According to Table 3.2, \(B \cap \overline{A} = B - (A \cap B)\), so \((A \cap B) \cup (\overline{A} \cap B) = (A \cap B) \cup (B - (A \cap B)) = B\).
An alternative demonstration is thus: \((A \cap B) \cup (\overline{A} \cap B) = (A \cup \overline{A}) \cap B = S \cap B = B\).
As for our specific A and B, \(A \cap B\) is only day 5, and \(\overline{A} \cap B\) is 4 days or less, so the union of the two is anything that’s less than 6, or B.
3.5.g. I can’t think of a mathematical way to describe this, but it’s clear intuitively, that \(\overline{A} \cap B\) is all of B, except for the intersection with A, and the union of that with A is all of B and all of A, so \(A \cup B\).
As for our specific A and B, \(\overline{A} \cap B\) is 4 days or less, and A is more than 4 days, so the union is the entire sample space, and \(A \cup B\) is collectively exhaustive so it too equals the entire sample space.

3.6.a. \(A \cap B\) matches \(O_1\) and \(\overline{A} \cap B\) matches \(O_3\), their union is exactly \(b = [O_1, O_3]\)
3.6.b. \(\overline{A} \cap B = [O_3]\) and \(A = [O_1, O_2]\), so the union is \(A \cup B = [O_1, O_2, O_3]\).

3.7.a. $$[(M_1, M_2), (M_1, M_3), (M_1, T_1), (M_1, T_2), (M_2, M_3), (M_2, T_1), (M_2, T_2), (M_3, T_1), (M_3, T_2), (T_1, T_2)]$$
3.7.b. $$A = [(M_1, T_1), (M_1, T_2), (M_2, T_1), (M_2, T_2), (M_3, T_1), (M_3, T_2), (T_1, T_2)]$$
3.7.c. $$B=[(M_1, M_2), (M_1, M_3), (M_2, M_3), (T_1, T_2)]$$
3.7.d. $$\overline{A}=[(M_1, M_2), (M_1, M_3), (M_2, M_3)]$$
3.7.e. \(A \cap B = [(T_1, T_2)]\), whereas \(\overline{A} \cap B = [(M_1, M_2), (M_1, M_3), (M_2, M_3)]\). Therefore, the union is all four outcomes that make up B.
3.7.f. \(\overline{A} \cap B = [(M_1, M_2), (M_1, M_3), (M_2, M_3)]\), and \(A = [(M_1, T_1), (M_1, T_2), (M_2, T_1), (M_2, T_2), (M_3, T_1), (M_3, T_2), (T_1, T_2)]\), so the union is \(S = [(M_1, M_2), (M_1, M_3), (M_2, M_3), (M_1, T_1), (M_1, T_2), (M_2, T_1), (M_2, T_2), (M_3, T_1), (M_3, T_2), (T_1, T_2)]\), whereas \(S = A \cup B\), so the two subsets match.

3.8. 35/66 = 0.53030303

3.9. 36/120 = 0.3

3.10. 675/1820 = 0.370879121

3.11. 20000/120000 = 0.166666667

3.12. 1/9 * 1/9 = 1/81

3.13.a. 0.68
3.13.b. 0.73
3.13.c. 0.32
3.13.d. 0.41
3.13.e. 1

3.14.a. 0.54
3.14.b. 0.18
3.14.c. Rate of return will be less than 10%.
3.14.d. 0.46
3.14.e. The empty set.
3.14.f. 0
3.14.g. Rate of return will be at least 10% or negative.
3.14.h. 0.72
3.14.i. Yes, because \(A \cap B = \emptyset\)
3.14.j. No, \(A \cup B\) do not cover 0% – 10%.

3.15.a. 0.5
3.15.b. 0.25
3.15.c. 0.25

3.16. \(A \cup B = [E_1, E_2, E_3, E_7, E_8, E_9]\), so the \(P(A \cup B) = 6/10 = 0.6\), whereas \(P(A) = P(B) = 4/10 = 0.4\), so \(P(A) + P(B) = 0.8\).

3.17.a. 0.86
3.17.b. 0.91
3.17.c. 0.14
3.17.d. 1
3.17.e. 0.77
3.17.f. No, any number of complaints between 1 to 9 falls under both.
3.17.g. Yes, \(P(A \cup B) = 1\)

3.18.a. 0.87
3.18.b. 0.35
3.18.c. The five classes cover all possible outcomes, and P(S) = 1.

3.19. $$P(A \cap B) = 0.25$$

3.20. $$P(A \cap B) = 0$$

3.21. $$P(A \cap B) = 0.24$$

3.22. $$P(A \cup B) = 0.75$$

3.23. \(P(A|B) = 0.67\). This does not equal P(A), so the events are not statistically independent.

3.24. \(P(A|B) = 0.8 = P(A)\), therefore, the events are statistically independent.

3.25. \(P(A|B) = 0.75\). The events are not statistically independent.

3.26. \(P(A|B) = 0.625\). The events are not statistically independent.

3.27. 1/9

3.28.a. 7! = 5040
3.28.b. 1/5040

3.29. 49*50 = 2450

3.30. 1/120

3.31. 0.2

3.32. N = 60, P = 1/60

3.33. 28

3.34.a. 42
3.34.b. 6
3.34.c. 6
3.34.d. 6/42, if considering only the lead role, the chance is 1/7, which is the same probability.
3.34.e. (6 + 6)/42 = 2/7. \(P(A \cup B) = P(A) + P(B) - P(A \cap B)\), and \(P(A \cap B) = \emptyset\), so we can add up 1/7 + 1/7 to reach the same result.

3.35.a. 150
3.35.b. 40/150 = 0.27
3.35.c. If A is the event that the craftsman brother gets selected, and B is the event that the labourer brother gets selected, then \(P(A) = 4/10\), \(P(B) = 10/15\) and \(P(A \cap B) = 40/150\), so \(P(A \cup B) = 120/150\), and its complement is the answer \(\overline{P(A \cup B)} = 30/150\)

3.36.a. 90
3.36.b. We look at the complement. There are 10 combinations of U.S funds that won’t under-perform, and 3 combinations of such international funds. So the probability of the complement is 30/90, and the probability of the original event in question is 60/90, or 0.67.

3.37. $$P(A \cup B) = 0.3 + 0.25 - 0.2 = 0.35$$

3.38. $$P(A \cup B) = 0.3 + 0.2 - 0.15 = 0.35$$

3.39.a. We take ‘unsuccessful’ to mean ‘no immediate action’, and if we call that A, then \(P(A) = 0.95\), and the probability of 4 consecutive outcomes that belong to A is 0.81.
3.39.b. We’d like the probability of “at least 4 unsuccessful calls” where here “unsuccessful” means anything not leading to a donation at all. Anything after those 4 calls work, because the question states that we’re looking for the first successful call after “at least 4 unsuccessful” ones. So it’s the same answer whether there were 4 unsuccessful calls and then a successful one, and if there were 10 unsuccessful calls and only then a successful one. The probability of any type of donation happening is 0.1, so the answer is \(0.9^4 = 0.65\)

3.40. Because B and C are mutually exclusive, \(P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(A \cap B) - P(A \cap C)\), and because A is independent of both, \(P(A \cap B) = P(A)P(B)\), and \(P(A \cap C) = P(A)P(C)\), so \(P(A \cup B \cup C) = 0.069\)

3.41. For independent events, \(P(A \cap B) = P(A)P(B)\), so 0.833.

3.42.a. $$P(B|A) = \frac{P(A \cap B)}{P(B)}=\frac{0.1}{0.18}=0.555$$
3.42.b. $$P(A|B) = \frac{P(A \cap B)}{P(A)}=\frac{0.1}{0.12}=0.833$$

3.43. If A is “item is defective”, and B is “inspector accepted item”, then \(P(B|A) = 0.8, P(A \cap B) = 0.01\), so \(P(A) = 0.125\)

3.44. Let A be “analyst is successful at stocks” and let B be “analyst is successful at bonds”. \(P(A) = 1/12, P(B) = 1/20\). Because they’re independent, \(P(B \cap A) = P(A)P(B)=1/240\). So \(P(A \cup B) = 1/12 + 1/20 - 1/240 = 31/240\)

3.45. Let A be the event “loan is for a high risk client”, and B will be “loan in default”. So \(P(A) = 0.15, P(B) = 0.05, P(A \cap B) = 0.02\), and the answer is \(P(B|A) = 0.02/0.15 = 0.133\)

3.46.a. $$P(A \cup B) = 0.4 + 0.5 - 0 = 0.9$$
3.46.b. $$P(A \cup C) = 0.8 + 0.5 - 0.4*0.8 = 0.88$$
3.46.c. \(P(C|B) = \frac{P(B \cap C)}{P(B)} = 0.75\), so \(P(B \cap C) = 0.375\) and \(P(A \cup B) = 0.5 + 0.8 - 0.375 = 0.925\)

3.47. The number of combinations for events A and B are 10 each, so the probabilities are 0.1. The number of combinations for getting both A and B is 210, so the overall probability is: \(P(A \cup B) = P(A) + P(B) - P(A \cap B) = 0.1 + 0.1 - \frac{1}{210} = 0.95\), and the analyst’s claim is a bit presumptuous.

3.48.a. Let A be the event “occurred on Monday” and B be the event “occurred on the last hour of the shift”, so \(P(A) = 0.3, P(B) = 0.2, P(A \cap B) = 0.04, P(A) = P(A \cap B) \cup P(A \cap \overline{B}) \Rightarrow 0.3 = 0.04 + P(A \cap \overline{B}) \Rightarrow P(A \cap \overline{B}) = 0.26. P(\overline{B}|A) = \frac{P(A \cap \overline{B})}{P(A)} = \frac{0.26}{0.3} = 0.8667\)
3.48.b. \(P(A)P(B) = (0.3)(0.2) = 0.06 \neq 0.04 = P(A \cap B)\), so no.

3.49.a. Let A be the event “signed up for reading class” and B be the event “signed up for math class”. \(P(A) = 0.4, P(B) = 0.5, P(B|A) = 0.3, P(A \cap B) = P(B|A)P(A) = (0.3)(0.4) = 0.12\)
3.49.b. $$P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{0.12}{0.5} = 0.24$$
3.49.c. $$P(A \cup B) = P(A) + P(B) - P(A \cap B) = 0.4 + 0.5 - 0.12 = 0.78$$
3.49.d. \(P(A)P(B) = (0.4)(0.5) = 0.2 \neq 0.12 = P(A \cap B)\), so no.

3.50. Let A be “new customer”, and be be “used rival”, so \(P(A) = 0.15, P(B|A) = 0.8, P(B) = 0.6, P(A \cap B) = P(B|A)P(A) = (0.8)(0.15) = 0.12, P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{0.12}{0.6} = 0.2\)
3.51. $$P(B) = 0.2, P(A|B) = 0.8, P(A \cap B) = P(D) = 0.16, P(D \cap C) = P(A \cap B \cap C) = 0.02, P(C|D) = \frac{P(C \cap D)}{P(C)} = \frac{0.02}{0.16} = 0.125$$

3.52. 0.05

3.53. 0.05

3.54. 0.05

3.55. 0.20

3.56. $$\frac{0.05}{0.3} = 0.1667$$

3.57. $$\frac{0.1}{0.4} = 0.25$$

3.58. $$\frac{0.1}{0.25} = 0.4$$

3.59. 4

3.60. 1

3.61. $$\frac{P(A|B_1)}{P(A|B_2)} = \frac{0.8}{0.4} = 2$$

3.62. $$\frac{P(A|B_1)}{P(A|B_2)} = \frac{0.4}{0.2} = 2$$

3.63. $$\frac{P(A|B_1)}{P(A|B_2)} = \frac{0.2}{0.4} = 0.5$$

3.64.a. 0.12
3.64.b. $$\frac{P(A|B)}{P(B)} = \frac{0.19}{0.27} = 0.703$$
3.64.c. No, \(P(A)P(B) = (0.27)(0.79) = 0.2133 \neq 0.19 = P(A \cap B)\)
3.64.d. \(\frac{P(A|B)}{P(B)} = \frac{0.07}{0.21} = 0.33\)
3.64.e. No, \(P(A)P(B) = (0.19)(0.21) = 0.0399 \neq 0.07 = P(A \cap B)\)
3.64.f. 0.79
3.64.g. 0.27
3.64.h. $$P(A \cup B) = P(A) + P(B) - P(A \cap B) = 0.79 + 0.27 - 0.19 = 0.87$$

3.65.a. 0.3
3.65.b. 0.38
3.65.c. Let A be a high prediction and B be a high outcome, $$ P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{0.23}{0.38} = 0.605 $$
3.65.d. $$P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.23}{0.3} = 0.766$$
3.65.e. Let b be a low outcome, \(P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.01}{0.3} = 0.033\)

3.66.a. 0.25
3.66.b. 0.32
3.66.c. Let A be “traded” and B will be “never reads”, so $$P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{0.04}{0.25} = 0.16$$
3.66.d. $$P(A|B) = \frac{P(A \cap B)}{P(A)} = \frac{0.04}{0.32} = 0.125$$
3.66.e. Let B be “regularly reads the paper”, so \(P(B) = 0.34, P(\overline{B}) = 0.66, P(A \cap \overline{B}) = P(A) - P(A \cap B) = 0.32 - 0.18 = 0.14, P(A|\overline{B}) = \frac{P(A \cap \overline{B})}{P(\overline{B})} = \frac{0.14}{0.66} = 0.212\)

3.67.a. Let D be “defective”. Because A, B & C are mutually exclusive and collectively exhaustive, \(P(D) = P(D \cap A) \cup P(D \cap B) \cup P(D \cap C) = P(D \cap A) + P(D \cap B) + P(D \cap C) - P((D \cap A) \cap (D \cap B)) - P((D \cap A) \cap (D \cap C)) - P((D \cap B) \cap (D \cap C)) + P((D \cap A) \cap (D \cap B) \cap (D \cap C)) = 0.02 + 0.05 + 0.03 - 0 - 0 - 0 + 0 = 0.1\)
3.67.b. Let G be “good. Because G and D are mutually exclusive and collectively exhaustive, \(P(B) = P(G \cap B) \cup P(D \cap B) = P(G \cap B) + P(D \cap B) - P((G \cap B) \cap (D \cap B)) = 0.3 + 0.05 - 0 = 0.35\)
3.67.c. $$P(D|B) = \frac{P(D \cap B)}{P(B)} = \frac{0.05}{0.35} = 0.142$$
3.67.d. $$P(B|D) = \frac{P(D \cap B)}{P(D)} = \frac{0.05}{0.1} = 0.5$$
3.67.e. No, for example: $$P(A|D) = \frac{P(A \cap D)}{P(D)} = \frac{0.02}{0.1} = 0.2 \neq P(A) = 0.29$$ And the same goes for other intersections.
3.67.f. \(P(G|A) = 0.931, P(G|B) = 0.857, P(G|C) = 0.916\), so A.

3.68.a. 0.32
3.68.b. 0.25
3.68.c. Let W be “worked on additional problems”, so \(P(A|W) = \frac{P(A \cap W)}{P(W)} = \frac{0.12}{0.32} = 0.375\)
3.68.d. $$P(W|A) = \frac{P(A \cap W)}{P(A)} = \frac{0.12}{0.25} = 0.48$$
3.68.e. Let D be “expects a grade below C”. Because C and D are mutually exclusive, \(P(C \cup D) = P(C) + P(D) - P(C \cap D) = 0.38 + 0.1 - 0 = 0.48\), so \(P((C \cup D) \cap W) = P(C \cap W) \cup P(D \cap W) = P(C \cap W) + P(D \cap W) - P((C \cap W) \cap (D \cap W)) = 0.12 + 0.02 - 0 = 0.14, P((C \cup D)|W) = \frac{P((C \cup D) \cap W)}{P(W)} = \frac{0.14}{0.32} = 0.4375\)
3.68.f. No, for example: \(P(A|W) = \frac{P(A \cap W)}{P(W)} = \frac{0.12}{0.32} = 0.375 \neq P(A) = 0.25\), and the same goes for other intersections.

3.69.a. 0.77
3.69.b. 0.19
3.69.c. Let S be “single”, and L be “left the job within the year, so \(P(L|S) = \frac{P(L \cap S)}{P(S)} = \frac{0.06}{0.23} = 0.26\)
3.69.d. \(P(\overline{S}|\overline{L}) = \frac{P(\overline{S} \cap \overline{L})}{P(\overline{L})} = \frac{0.64}{0.81} = 0.79\)

3.70.a. 0.76 3.70.b. 0.77 3.70.c. 0.1

3.71.

Men Women
Joined0.0280.0540.082
Not Joined0.3720.5460.918
Total0.40.61

a. 0.082
b. Let W be “women” and J be “joined the club”, so \(P(W|J) = \frac{P(W \cap J)}{P(J)} = \frac{0.054}{0.082} = 0.658\)

3.72. Let G be “significant growth”, and I1, I2 & I3 will greater, similar and lower interest events.

I1 I2 I3
\(G\)0.0250.30.120.445
\(\overline{G}\)0.2250.30.030.555
Total0.250.60.151

a. 0.025
b. 0.445
c. \(P(I3|G) = \frac{P(I3 \cap G)}{P(G)} = \frac{0.12}{0.445} = 0.269\)

3.73. \(P(H) = 0.42, P(S) = 0.22, P(S|H) = 0.34\)
a. $$P(H \cap S) = P(S|H)P(H) = (0.34)(0.22) = 0.0506$$
b. $$P(H \cup S) = P(S) + P(H) - P(S \cap H) = 0.22 + 0.42 - 0.0506 = 0.5894$$
c. $$P(H|S) = \frac{P(H \cap S)}{P(S)} = \frac{0.0506}{0.22} = 0.23$$

3.74. Let U be the event of graduating at the top 10% of the class.

$$>Q_1$$ $$Q_2 \cup Q_3$$ $$Q_4$$
\(U\)0.1750.250.050.475
\(\overline{U}\)0.0750.250.20.525
Total0.250.50.251

a. 0.475 b. $$P(Q_1|U) = \frac{P(Q_1 \cap U)}{P(U)} = \frac{0.175}{0.475} = 0.368$$
c. We can reach this result in one of two ways: c.1. $$P(\overline{Q_1}|\overline{U}) = 1 - P(Q_1|\overline{U}) = 1 - \frac{P(Q_1 \cap \overline{U})}{P(\overline{U})} = 1 - \frac{0.075}{0.525} = 0.857$$
c.2. Because \(Q4\) and \((Q2 \cup Q3)\) are mutually exclusive, then \(P((Q_2 \cup Q_3) \cap \overline{U}) \cup P(Q_4 \cap \overline{U}) = P((Q_2 \cup Q_3) \cap \overline{U}) + P(Q_4 \cap \overline{U}) - P(((Q_2 \cup Q_3) \cap \overline{U}) \cap (Q_4 \cap \overline{U})) = P((Q_2 \cup Q_3) \cap \overline{U}) + P(Q_4 \cap \overline{U}) - 0 = P((Q_2 \cup Q_3) \cap \overline{U}) + P(Q_4 \cap \overline{U})\), so we can do: \(P(\overline{Q_1}|\overline{U}) = P(((Q_2 \cup Q_3) \cup Q_4)|\overline{U}) = \frac{P(((Q_2 \cup Q_3) \cup Q_4) \cap \overline{U})}{P(\overline{U})} = \frac{P((Q_2 \cup Q_3) \cap \overline{U}) \cup P(Q_4 \cap \overline{U})}{P(\overline{U})} = \frac{0.2 + 0.25}{0.525} = 0.857\)

3.75.a. Let H be “high sales” and F will be “favorable reaction”. \(P(H|F) = \frac{P(H \cap F}{P(F)} = \frac{0.173}{0.303} = 0.570957096\)
3.75.b. Let L be “low sales” and U will be “unfavorable reaction”. \(P(L|U) = \frac{P(L \cap U}{P(U)} = \frac{0.141}{0.272} = 0.518382353\)
3.75.c. Let N be ‘neutral”, because N and F are mutually exclusive: \(P(L \cap \overline{U}) = P(L \cap (N \cup F)) = P((L \cap N) \cup (L \cap F)) = P(L \cap N) + P(L \cap F) - P((L \cap N) \cap (L \cap F)) = P(L \cap N) + P(L \cap F) - 0 = P(L \cap N) + P(L \cap F)\) $$ P(L|\overline{U}) = \frac{P(L \cap \overline{U})}{P(\overline{U})} = \frac{P(L) - P(L \cap U)}{P(\overline{U})} = \frac{0.296 - 0.141}{0.272} = 0.569852941 $$
3.75.d. $$P(\overline{U}|L) = \frac{P(L \cap \overline{U})}{P(L)} = \frac{0.155}{0.296} = 0.523648649$$

3.76. Let M1 and M2 be the machines, M1 being the faulty one, and let F be the event of a faulty piece.

$$M_1$$ $$M_2$$
\(F\)0.0400.04
\(\overline{F}\)0.360.60.96
Total0.40.61

$$P(M_1|F) = \frac{P(M_1 \cap \overline{F})}{P(\overline{F})} = \frac{0.36}{0.96} = 0.375$$

3.77.a. Let E be the event of finding a course enjoyable, and let V be the event of receiving strong positive evaluations.
$$P(E|V)^3 = (\frac{P(E \cap V)}{P(V)})^3 = (\frac{P(E \cap V)}{P(V \cap E) \cup P(V \cap \overline{E})})^3 = (\frac{0.42}{0.42 + 0.075})^3 = (0.848)^3 = 0.6108$$
3.77.b. $$1 - P(\overline{E}|V)^3 = 1 - (\frac{P(\overline{E} \cap V)}{P(V)})^3 = 1 - (\frac{P(\overline{E} \cap V)}{P(V \cap E) \cup P(V \cap \overline{E})})^3 = 1 - (\frac{0.075}{0.42 + 0.075})^3 = 1 - (0.151)^3 = 0.996521691$$

For all the following exercises, it should be assumed that A1 and A2 are complements, and that B1 and B2 are complements.

3.78. $$P(A_2) = 0.6, P(A_1|B_1) = \frac{P(B_1|A_1)P(A_1)}{P(B_1|A_1)P(A_1) + P(B_1|A_2)P(A_2)} = \frac{0.24}{0.24 + 0.42} = 0.36$$

3.79. $$P(A_2) = 0.2, P(A_1|B_1) = \frac{P(B_1|A_1)P(A_1)}{P(B_1|A_1)P(A_1) + P(B_1|A_2)P(A_2)} = \frac{0.48}{0.48 + 0.04} = 0.923$$

3.80. $$P(A_2) = 0.5 \\ P(B_1) = P(B_1|A_1)P(A_1) + P(B_1|A_2)P(A_2) = (0.4)(0.5) + (0.7)(0.5) = 0.55 \\ P(B_2) = 1 - P(B_1) = 0.45\\ P(A_1 \cap B_1) = P(B_1|A_1)P(A_1) = (0.4)(0.5) = 0.2 \\ P(A_1) = P(A_1 \cap B_1) + P(A_1 \cap B_2) \Rightarrow P(A_1 \cap B_2) = P(A_1) - P(A_1 \cap B_1) = 0.5 - 0.2 = 0.3 \\ P(B_2|A_1) = \frac{P(A_1 \cap B_2)}{P(A_1)} = \frac{0.3}{0.5} = 0.6 \\ P(A_1|B_2) = \frac{P(B_2|A_1)P(A_1)}{P(B_2)} = \frac{(0.6)(0.5)}{0.45} = 0.67$$

3.81. $$P(A_2) = 0.6 \\ P(B_1) = P(B_1|A_1)P(A_1) + P(B_1|A_2)P(A_2) = (0.6)(0.4) + (0.7)(0.6) = 0.66 \\ P(B_2) = 1 - P(B_1) = 0.34 \\ P(B_1 \cap A_2) = P(B_1|A_2)P(A_2) = (0.7)(0.6) = 0.42 \\ P(A_2) = P(A_2 \cap B_1) + P(A_2 \cap B_2) \Rightarrow P(A_2 \cap B_2) = P(A_2) - P(A_2 \cap B_1) = 0.6 - 0.42 = 0.18 \\ P(A_2|B_2) = \frac{P(A_2 \cap B_2)}{P(B_2)} = \frac{0.18}{0.34} = 0.529$$

3.82. $$P(A_2) = 0.4 \\ P(A_1|B_1) = \frac{P(B_1|A_1)P(A_1)}{P(B_1|A_1)P(A_1) + P(B_1|A_2)P(A_2)} = \frac{(0.6)(0.4)}{(0.6)(0.6) + (0.4)(0.4)} = 0.461$$

3.83. Let A be “received the material” and let B be “adopted the book”. \(P(\overline{A}) = 0.2 \\ P(A|B) = \frac{P(B|A)P(A)}{P(B|A)P(A) + P(B|\overline{A})P(\overline{A})} = \frac{(0.3)(0.8)}{(0.3)(0.8) + (0.1)(0.2)} = 0.923\)

3.84. Let P1, P2 and P3 be stocks that performed better, same or worse than last year, and let R be stocks that were rated as good by the analyst. $$P(P_1|R) = \frac{P(R|P_1)P(P_1)}{P(R|P_1)P(P_1) + P(R|P_2)P(P_2) + P(R|P_3)P(P_3)} = \frac{(0.4)(0.25)}{(0.4)(0.25) + (0.2)(0.5) + (0.1)(0.25)} = 0.44$$

3.85. $$P(B) = P(A \cap B) \cup P(\overline{A} \cap B) \\ \Rightarrow P(\overline{A} \cap B) = P(B) - P(A \cap B) \\ \Rightarrow P(\overline{A}|B) = \frac{P(\overline{A} \cap B)}{P(B)} = \frac{P(B)}{P(B)} - \frac{P(A \cap B)}{P(B)} = 1 - P(A|B)$$
Let F be the event of the process functioning correctly, and let D be the event of a defective bulb. $$P(D) = P(D|F)P(F) + P(D|\overline{F})P(\overline{F}) = (0.1)(0.9) + (0.1)(0.5) = 0.14 \\ P(F|D) = \frac{P(D|F)P(F)}{P(D)} = \frac{(0.1)(0.9)}{0.14} = 0.642$$ $$P(\overline{D}) = 1 - P(D) = 0.86 \\ P(\overline{D}|F) = 1 - P(D|F) = 0.9 \text{ (proof at beginning of question)} \\ P(F|\overline{D}) = \frac{P(\overline{D}|F)P(F)}{P(\overline{D})} = \frac{(0.9)(0.9)}{0.86} = 0.94$$

3.86. Don’t purchase corpses of dead animals. It’s not rational behaviour for a self-interested economic agent.
As for the question, let A1 be the event of a chicken coming from Free Range Farms, and let A2 be the event of a chicken coming from Big Foods. Let B be the event of a chicken weighing less than 3 pounds. $$ P(\overline{B}|A_1) = 1 - P(B|A_1) = 0.9 \text{ (according to proof at 3.85)} \\ P(A_2) = 1 - P(A_1) = 0.6 \\ P(B) = P(B|A_1)P(A_1) + P(B|A_2)P(A_2) = (0.1)(0.4) + (0.2)(0.6) = 0.16 \\ P(\overline{B}) = 1 - P(B) = 0.84 \\ P(A_1|\overline{B}) = \frac{P(\overline{B}|A_1)P(A_1)}{P(\overline{B})} = \frac{(0.9)(0.4)}{0.84} = 0.428$$
Calculating the chances of 3 chickens out of 5 matching the previous event requires the use of the binomial distribution formula, which as far as I’ve seen, has not yet been introduced in this book. $$P = \binom{5}{3}0.428^3(1 - 0.428)^2 = 0.256$$

3.87. This question is so poorly phrased that I can’t even approach it.

3.88. Mutually exclusive events are events that can’t both happen at the same time. Independence is when the occurrence of one event has no affect on the probability of another.

3.89.a. True, \(\overline{P(A) \cup P(B)} = \overline{P(A) + P(B) - P(A \cap B)} = \overline{P(A) + P(B) - (P(A) - P(A \cap \overline{B})} = \overline{P(B) + P(A \cap \overline{B})} = \overline{P(B) + P(\overline{B}) - P(\overline{A} \cap \overline{B})} = \overline{1 - P(\overline{A} \cap \overline{B})} = \overline{\overline{P(\overline{A} \cap \overline{B})}} = P(\overline{A} \cap \overline{B})\)>
3.89.b. False, because collectively exhaustive events might not be mutually exclusive, in which case their sum is greater than their union. The union of collectively exhaustive events is equal to 1, and if their sum is greater than that, it cannot equal 1.
3.89.c. True, \(\frac{n!}{x!(n-x!)} = \frac{n!}{(n-x)!(n-(n-x))!}\)
3.89.d. True, according to Bayes Theorem, \(P(A|B) = \frac{P(B|A)P(A)}{P(B)} = P(B|A)\frac{P(A)}{P(B)} = P(B|A)\)
3.89.e. True, \(P(A) = P(\overline{A}) = 1-P(A) \Rightarrow 2P(A) = 1 \Rightarrow P(A) = 0.5 \text{ and } P(\overline{A}) = 1-P(A) = 0.5\)
3.89.f. True, we’ll prove for A, \(P(A|B) = P(A)=1-P(\overline{A}) \Rightarrow P(\overline{A}) = 1-P(A|B)=P(\overline{A}|B)\)
3.89.g. False, suppose A and B are not collectively exhaustive, we’ll get a probability greater than 1, while P(S) = 1, \(P(A \cup B) = P(A) + P(B) < 1 \text{ and } P(\overline{A}) + P(\overline{B}) = 1-P(A) + 1-P(B) = 2-(P(A)+P(B)) > 1\)

3.90. Conditional probability is the ratio of occurrences of some event A, out of all the times that some other event B has occurred. Sometimes in real world data all we have is conditional probabilities and we’d like to calculate the individual probabilities, and sometimes it’s the other way around.

3.91. Given a subjective idea of the chances of some event A happening, if we know that some other event B has happened, we’d like to update out prediction as to the likelihood of event A. Bayes theorem provides us with a tool in these scenarios.

3.92.a. True, \(P(A \cup B) \geq P(A)=P(A \cap B) + P(A \cap \overline{B}) \geq P(A \cap B)\)
3.92.b. True, \(P(A \cup B) = P(A) + P(A) + P(A \cap B) \geq P(A) + P(B)\)
3.92.c. True, \(P(A) = P(A \cap B) + P(A \cap \overline{B}) \geq P(A \cap B)\)
3.92.d. True, if we can say that the intersection of an event with itself equals itself, then: \(P(A) = P(A \cap A) + P(A \cap \overline{A}) = P(A \cap A) = P(A) \Rightarrow P(A \cap \overline{A})= 0\)
3.92.e. False. For any event A, \(0 \leq P(A) \leq 1\), so two events A and B can both have a probability of 1, and their sum will be greater than 1, particularly, 2.
3.92.f. False. \(P(A \cap B) = 0 \Rightarrow P(A \cup B) = P(A) + P(B) + P(A \cap B) = P(A) + P(B)\), but \(P(A) + P(B) = 1\) only if A and B are complements, not in every case where they are mutually exclusive.
3.92.g. False. \(P(A \cup B) = P(A)+P(B)-P(A \cap B) = 1 \Rightarrow P(A \cap B) = P(A)+P(B)-1\). Therefore the intersection equals zero only if \(P(A) + P(B) = 1\), which is only the case if A and B are complements.

3.93. A joint probability is the probability that the outcome will be in the intersection between two events. Marginal probabilities are the probabilities of the distinct events. Conditional probabilities are the probabilities that some event A will happen out of all occurrences where event B does. So for example, given the following events: “participant is a vegetarian” and “participant is a vegan”, the probability that someone is a vegetarian is a marginal probability. The probability that someone who is a vegetarian is a vegan is a conditional probability and the probability that someone is both a vegetarian and a vegan is a joint probability.

3.94.a. False. See example 3.23. above. \(P(T_1|D_1)=0.1 \gt 0.18 = P(T_1)\)
3.94.b. False. Suppose event A with probability of 0.5. \(P(A\cap\overline{A})=0 \Rightarrow P(A|\overline{A})=\frac{P(A\cap\overline{A})}{P(\overline{A})}=0 \neq P(A)\)
3.94.c. \0 \leq (P(B) \leq 1\), Therefore for any \(P(B) \geq 0, P(A|B) = \frac{P(A \cap B)}{P(B)} \geq P(A \cap B)\)
3.94.d. False. \( P(A \cap B) = P(A|B)P(B) \leq P(A)P(B) \Rightarrow P(A|B) \leq P(A) \), and in example 3.23, we see a counter-example where \(P(T_1|D_1) = 0.9 \gt 0.18 = P(T_1)\)
3.94.e. In subjective probability, if an event B is assumed to have some probability P(B), we say this is a prior probability, and given knowledge that event A has happened, we update that probability to P(B|A). The assumption that the posterior probability must be at least as large as the prior probability means \(P(B|A) \geq P(B)\), which we’ve already disproved in section a of this question.

3.95. $$P(A \cup B) = P(A) + P(B) – P(A \cap B) = P(A) + P(B) – P(A|B)P(B) = P(A) + P(B)[1-P(A|B)]$$

3.96.a. $$P(A \cap B) = P(A|B)P(B) = 0.08$$
3.96.b. The events are not independent because \(P(A|B) = 0.4 \neq 0.3 = P(A)\)
3.96.c. $$P(B|A) = \frac{P(A|B)P(B)}{P(A)} = \frac{(0.4)(0.2)}{0.3} = 0.26667$$
3.96.d. $$P(\overline{A}) = P(\overline{A} \cap B) + P(\overline{A} \cap \overline{B}) \Rightarrow P(\overline{A} \cap \overline{B}) = P(\overline{A}) – P(\overline{A} \cap B)$$ $$ P(B) = P(A \cap B) + P(\overline{A} \cap B) \Rightarrow P(\overline{A} \cap B) = P(B) – P(A \cap B)$$ $$ P(\overline{A} \cap \overline{B}) = P(\overline{A}) – P(\overline{A} \cap B) = P(\overline{A}) – P(B) + P(A \cap B) = 0.7 – 0.2 + 0.08 = 0.58$$

3.97.a. Given event A that the thinner wire will arrive within a week, and event B that the thicker wire will arrive within a week, we have: $$P(A \cap B) = (0.6)P(B)$$ $$(0.6)P(B)=(0.4)P(A) \Rightarrow P(A) = 1.5P(B)$$ $$P(A \cup B) = P(A) + P(B) – P(A \cap B) = 1.5P(B) + P(B) – 0.6P(B) = 1.9P(B) \Rightarrow P(B) = 0.42$$
3.97.b. $$P(A) = \frac{P(A|B)P(B)}{P(B|A)} = \frac{(0.6)(0.42)}{0.4} = 0.63$$
3.97.c. $$P(A \cap B) = P(A|B)P(B) = (0.6)(0.42) = 0.252$$

3.98.a. Given event A “employee has an MBA” and event B “employee is over 35”: $$P(A \cap B) = P(B|A)P(A) = 0.105$$
3.98.b. $$P(A|B) = \frac{P(B|A)P(A)}{P(B)} = 0.2625$$
3.98.c. $$P(A \cup B) = P(A) + P(B) – P(A \cap B) = 0.35 + 0.4 – 0.105 = 0.645$$ 3.98.d. $$P(\overline{A} \cap B) = P(B) – P(A \cap B) = 0.4 – 0.105 = 0.295$$ $$P(\overline{A}|B) = \frac{P(\overline{A} \cap B)}{P(B)} = \frac{0.295}{0.4} = 0.7375$$ 3.98.e. No, \[P(B|A) = 0.3 \neq 0.4 = P(B)\]
3.98.f. No, \[P(A \cap B) = 0.105 \neq 0\]
3.98.g. No, \[P(A \cup B) = P(A) + P(B) – P(A \cap B) = 0.645 \neq 1\]

3.99.a. Let A be the event of someone ordering a vegetarian meal, and B will be the event of a customer being a student. $$P(A \cap B) = P(A|B)P(B) = (0.25)(0.5)=0.125$$ 3.99.b. $$P(B|A) = \frac{P(A|B)P(B)}{P(A)} = \frac{(0.25)(0.5)}{0.35} = 0.357$$ 3.99.c. $$P(\overline{A} \cap \overline{B}) = 1-P(A \cup B) = 1-(P(A) + P(B) – P(A \cap B) = 0.275$$ 3.99.d. No, \[P(A|B) = 0.25 \neq 0.35 = P(A)\]
3.99.e. No, \[P(A \cap B) = 0.125 \neq 0\]
3.99.f. No, \[P(A \cup B) = P(A) + P(B) – P(A \cap B) = 0.725 \neq 1\]

3.100.a. Let A be ‘exceeds 160 acres’ and let B be ‘owned by persons over 50 years old’. \[P(A \cap B) = P(B|A)P(A) = (0.55)(0.2) = 0.11\]
3.100.b. $$P(A \cup B) = P(A) + P(B) – P(A \cap B) = 0.2 + 0.6 – 0.11 = 0.69$$ 3.100.c. $$P(A|B) = \frac{P(B|A)P(A)}{P(B)} = \frac{(0.55)(0.2)}{0.6} = 0.18333$$ 3.100.d. No, \[ P(B|A) = 0.55 \neq 0.6 P(B)\]

3.101.a. Let H be the event of an employee only having highschool training, and let M be the event of an employee being male. \[P(A \cap M) = P(H|M)P(M) = 0.48\]
3.101.b. Let G be the event of an employee having graduate training and let W be the event of an employee being a woman. W and M are mutually exclusive and collectively exhaustive, so \[P(G) = P(G \cap M) + P(G \cap W) = P(G|M)P(M) + P(G|W)P(W) = (0.1)(0.8) + (0.15)(0.2) = 0.11\] 3.101.c. $$P(M|G) = \frac{P(G|M)P(M)}{P(G)} = \frac{(0.1)(0.8)}{0.11} = 0.727$$ 3.101.d. No, for example, \[P(G|M) = 0.1 \neq 0.11 = P(G)\] 3.101.e. $$P(W|\overline{G})=\frac{P(W \cap \overline{G})}{P(\overline{G})} = \frac{P(W) – P(W \cap G)}{P(\overline{G})} = \frac{P(W) – P(G|W)P(W)}{P(\overline{G})} = 0.191$$

3.102.a. Let F be the event of an employee favoring the plan, W be the event of an employee being a woman, and N be the event of an employee being a night-shift worker. \[P(W \cap F) = P(F|W)P(W) = (0.4)(0.3) = 0.12\]
3.102.b. $$P(N \cup W) = P(N) + P(W) – P(N \cap W) = 0.5 + 0.3 – 0.12 = 0.68$$ 3.102.c. No, \[P(W|N) = 0.2 \neq 0.3 = P(W)\]
3.102.d. $$P(N|W) = \frac{P(W|N)P(N)}{P(W)} = \frac{(0.2)(0.5)}{0.3} = 0.333$$ 3.102.e. $$P(\overline{N} \cap \overline{F}) = 1 – P(N \cup F) = 1 – (P(N) + P(F) – P(N \cap F)) = 1 – P(N) – (P(F|W)P(W) + P(F|\overline{W})P(\overline{W})) + P(F|N)P(N) = 1 – 0.5 – (0.4)(0.3) – (0.5)(0.7) + (0.65)(0.5) = 0.355$$

3.103.a. \[\binom{16}{12} = \frac{16!}{12!(16-12)!} = 1820\]
3.103.b. The number of possibilities of 8 men and 4 women is \[\binom{8}{4} = \frac{8!}{4!(8-4)!} = 70\], and the number of possibilities of 7 men and 5 women is \[\binom{8}{7}\binom{8}{5} = \frac{8!}{7!(8-7)!\frac{8!}{5!(8-5)!} = 448\], so in total there are 518 possibilities and the probability is \[\frac{518}{1820} = 0.28\]

3.104.a. \[\binom{12}{2} = \frac{12!}{2!(12-2)!} = 66\]
3.104.b. \[\frac{1}{12} + \frac{1}{12} = \frac{1}{6}\]

3.105. Let A be the event of a stock being up two years after it has been purchased, and let B be the event of Mr. Roberts receiving the first year bonus for a stock. Then we have the following: $$P(A) = 0.4 \Rightarrow P(\overline{A}) = 0.6$$ $$P(B|A) = 0.6$$ $$P(B|\overline{A}) = 0.4$$ $$P(B) = P(A \cap B) + P(\overline{A} \cap B) = P(B|A)P(A) + P(B|\overline{A})P(\overline{A}) = 0.48$$

3.106.a. Let C be the event of a patient being cured, and let T be the event of a patient receiving the treatment. $$P(C \cap T) = P(C|T)P(T) = (0.75)(0.1) = 0.075$$ 3.106.b. $$P(T|C) = \frac{P(C|T)P(T)}{P(C)} = \frac{P(C|T)P(T)}{P(C|T)P(T) + P(C|\overline{T})P(\overline{T})} = \frac{(0.75)(0.1)}{(0.75)(0.1) + (0.5)(0.9)} = 0.1428$$ 3.106.c. $$\frac{1}{\frac{100!}{10!(100-10)!}} = \frac{10!(100-10)!}{100!}$$

3.107.a. The probability of renewals (R) in January (J), is \[P(R|J) = (0.08)(0.81) + (0.41)(0.79)+ (0.06)(0.6) + (0.45)(0.21) = 0.5192\]
3.107.b. $$P(R|J) = (0.1)(0.8) + (0.57)(0.76)+ (0.24)(0.51) + (0.09)(0.14) = 0.6482$$ 3.107.c. The probability of renewal has risen, but in most categories the percentage of renewals dropped, only to be leveled a the decrease in the share of subscriptions under subscription service, which have the lowest renewal rate. So in the long term this is not necessarily good news.

3.108. Let A be the event of a passenger carrying illegal amounts of liquor, and let B be the event of a passenger identified by the system. $$P(A|B) = \frac{P(B|A)P(A)}{P(B)} = \frac{P(B|A)P(A)}{P(B|A)P(A) + P(B|\overline{A})P(\overline{A})} = \frac{(0.8)(0.2)}{(0.8)(0.2) + (0.2)(0.8)} = 0.5$$ It appears that the system produces the same results as picking passengers at random.

3.109. Let A be the event of a person having contracted the disease, and let B be the event of a positive test result. $$P(B|\overline{A}) = 1 – P(\overline{B}|\overline{A}) = 0.2$$ $$P(A|B) = \frac{P(B|A)P(A)}{P(B)} = \frac{P(B|A)P(A)}{P(B|A)P(A) + P(B|\overline{A})P(\overline{A})} = \frac{(0.8)(0.08)}{(0.8)(0.08) + (0.2)(0.92)} = 0.248$$

3.110. Let A be the event of a sale, and let B be the event of an existing customer. $$P(A|B) = \frac{P(B|A)P(A)}{P(B)} = \frac{P(B|A)P(A)}{P(B|A)P(A) + P(B|\overline{A})P(\overline{A})} = \frac{(0.7)(0.4)}{(0.7)(0.4) + (0.5)(0.6)} = 0.482$$

3.111. Let A be the event of a final A grade, and let B be the event of an A on the midterm. $$P(A|B) = \frac{P(B|A)P(A)}{P(B)} = \frac{P(B|A)P(A)}{P(B|A)P(A) + P(B|\overline{A})P(\overline{A})} = \frac{(0.7)(0.2)}{(0.7)(0.2) + (0.1)(0.8)} = 0.63636$$

3.112.a. $$P(O_w|F_w) = \frac{P(O_w \cap F_w)}{P(F_w)} = \frac{0.149}{0.29} = 0.513$$ 3.112.b. $$1 – P(O_i|F_i) = 1 – \frac{P(O_i \cap F_i)}{P(F_i)} = 1 – \frac{0.21}{0.391} = 0.462$$

3.113.a. Let G be the event of a student who will graduate, and let A be the event of an entering freshmen. $$P(A \cap G) = P(G|A)P(A) = (0.62)(0.73) = 0.4526$$ 3.113.b. $$P(G) = P(G|A)P(A) + P(G|\overline{A})P(\overline{A}) = (0.62)(0.73) + (0.78)(0.27) = 0.6632$$ 3.113.c. $$P(A \cup G) = P(A) + P(G) – P(A \cap G) = 0.73 + 0.6632 – 0.4526 = 0.9406$$ 3.113.d. No, \[P(G|\overline{A}) = 0.78 \neq 0.6632 = P(G)\]

3.114.a. Let S be the event of a store being successful, and let G be the event of a good assessment. $$P(G) = P(G|S)P(S) + P(G|\overline{S})P(\overline{S}) = (0.7)(0.6) + (0.2)(0.4) = 0.5$$ 3.114.b. $$P(S|G) = \frac{P(G|S)P(S)}{P(G)} = \frac{(0.7)(0.6)}{0.5} = 0.84$$ 3.114.c. No, \[P(G|S) = 0.7 \neq 0.5 = P(G)\]\]
3.114.d. We’ll calculate the probability of no store being successful, as \[0.4^5 = 0.01024\], therefore, the probability of at least one being successful is \[1 – 0.01024 = 0.98976\]

3.115.a. Let A be the event of a customer ordering wine, and let \[C_r, C_o, C_n\] be the events that each type of customer visits. $$P(A) = P(A|C_r)P(C_r) + P(A|C_o)P(C_o) + P(A|C_n)P(C_n) = (0.7)(0.5) + (0.5)(0.4) + (0.3)(0.1) = 0.58$$ 3.115.b. $$P(C_r|A) = \frac{P(A|C_r)P(C_r)}{P(A)} = \frac{(0.7)(0.5)}{0.58} = 0.603$$ 3.115.c. $$P(C_o|A) = \frac{P(A|C_o)P(C_o)}{P(A)} = \frac{(0.5)(0.4)}{0.58} = 0.344$$

3.116.a. Let A be the event of a purchase and let the events of customers falling into the categories be \[C_h, C_c, C_o\]. $$P(A) = P(A|C_h)P(C_h) + P(A|C_c)P(C_c) + P(A|C_o)P(C_o) = (0.2)(0.3) + (0.6)(0.5) + (0.8)(0.2) = 0.52$$ 3.116.b. $$P(C_h|A) = \frac{P(A|C_h)P(C_h)}{P(A)} = \frac{(0.2)(0.3)}{0.52} = 0.115$$

3.117. $$\frac{\binom{8}{5}}{\binom{16}{5}} = 0.012$$

3.118.a. Let C be the event of someone being guilty of a crime, and let G be the event of someone wearing gloves. $$P(C|G) = \frac{P(G|C)P(C)}{P(G)} = \frac{P(G|C)P(C)}{P(G|C)P(C) + P(G|\overline{C})P(\overline{C})} = \frac{(0.6)(0.5)}{(0.6)(0.5) + (0.8)(0.5)} = \frac{3}{7}$$ 3.118.b. A jury should not convict based on evidence that has such low accuracy.

3.119. Let the types of error events be \[E_d, E_m, E_o\] and let F be the event of a failure. $$P(E_d|F) = \frac{P(F|E_d)P(E_d)}{P(F)} = \frac{P(F|E_d)P(E_d)}{P(F|E_d)P(E_d) + P(F|E_m)P(E_m) + P(F|E_o)P(E_o)} = \frac{(0.6)(0.5)}{(0.6)(0.5) + (0.7)(0.3) + (0.3)(0.2)} = 0.526$$

3.120. Let N be the event of a new operating system introduced, and let G be the event of growth. $$P(G|N) = \frac{P(N|G)P(G)}{P(N)} = \frac{P(N|G)P(G)}{P(N|G)P(G) + P(N|\overline{G})P(\overline{G})} = \frac{(0.3)(0.7)}{(0.3)(0.7) + (0.1)(0.3)} = 0.875$$

3.121. Let D be the event of lumber having defects and let the event of lumber coming from each supplier be the events \[S_n, S_m, S_s\]. Then: $$P(S_n) = P(S_n|D)P(D) + P(S_n|\overline{D})P(\overline{D}) = (0.3)(0.2) + (0.4)(0.8) = 0.38$$ $$P(S_m) = P(S_m|D)P(D) + P(S_m|\overline{D})P(\overline{D}) = (0.5)(0.2) + (0.2)(0.8) = 0.26$$ $$P(S_s) = P(S_s|D)P(D) + P(S_s|\overline{D})P(\overline{D}) = (0.2)(0.2) + (0.4)(0.8) = 0.36$$ $$P(\overline{D}|S_n) = 1 – P(D|S_n) = 1 – \frac{P(S_n|D)P(D)}{P(S_n)} = 1 – \frac{(0.3)(0.2)}{0.38} = 0.1578$$ $$P(\overline{D}|S_m) = 1 – P(D|S_m) = 1 – \frac{P(S_m|D)P(D)}{P(S_m)} = 1 – \frac{(0.5)(0.2)}{0.26} = 0.3846$$ $$P(\overline{D}|S_s) = 1 – P(D|S_s) = 1 – \frac{P(S_s|D)P(D)}{P(S_s)} = 1 – \frac{(0.2)(0.2)}{0.36} = 0.1111$$

3.122. Let R be the event of an acre with regular plowing and let H be the event of high yields. $$P(H|R) = 1 – P(\overline{H}|R) = 0.6$$ $$P(R|H) = \frac{P(H|R)P(R)}{P(H)} = \frac{P(H|R)P(R)}{P(H|R)P(R) + P(H|\overline{R})P(\overline{R})} = \frac{(0.6)(0.4)}{(0.6)(0.4) + (0.5)(0.6)} = 0.4444$$