Chapter 4: Probability and Probability Distributions4.1 a. Subjective probabilityb. Relative frequencyc. Classicald. Relative frequencye. Subjective probabilityf. Subjective probabilityg. Classical4.2 Answers will vary depending on person’s experience with each situation.c. This is know as the “birthday problem.” Students will be surprised to learn that ina class of only 23 students the probability of 2 or more students having the samebirthday is greater than 50%.4.3 Using the binomial formula, the probability of guessing correctly 15 or more of the 20questions is 0.021.4.4 a. Positive Outcomes are 00 to 74; Negative Outcomes are 75 to 99.b. Let M be the number of sets of 20 2-digit numbers out of the 2000 sets in which wehave 15 or more “positive outcomes” (i.e., 15 or more two digit numbers in the range00 to 74). Then the probability would be approximated by M/2000.4.5 HHH, HHT, HTH, THH, TTH, THT, HTT, TTT4.6 a. A = {HTT, THT, TTH}; Thus, P(A) =38b. B = {HHH, HHT, HTH, THH, THT, HTT, TTH}; Thus, P(B) =78c. C = {TTT}; Thus, P(C) =184.7 a. P¡A¢= 1 −38=58P¡B¢= 1 −78=18P¡C¢= 1 −18=78b. Events A and B are not mutually exclusive because B contains A ⇒A ∩ B = A, which is not the empty set.4.8 a. P (A|B) = P (A ∩ B)/P (B) =38/78=37b. P (A|C) = P (A ∩ C)/P (C) = 0, since A ∩ C is empty.c. P (B|C) = P (B ∩ C)/P (C) = 0, since B ∩ C is empty.4.9 Because P (A|B) =376=38= P (A) ⇒ A and B are not independent.Because P (A|C) = 0 6=38= P (A) ⇒ A and C are not independent.Because P (B|C) = 0 6=78= P (B) ⇒ B and C are not independent.204.10 A = {1, 3, 5 } and B = {4, 5}a. P (A) = P (1) + P (3) + P (5) = 0.20 + 0.15 + 0.30 = 0.65P (B) = P (4) + P (5) = 0.10 + 0.30 = 0.40b. P (A ∩ B) = P (5) = 0.30c. P (A ∪ B) = P (A) + P (B) − P (A ∩ B) = 0.65 + 0.40 − 0.30 = 0.75Alternatively, P (A∪B) = P (1)+P(3)+P (4)+P (5) = 0.20+0.15 +0.10+0.30 = 0.754.11 No, since A and B are not mutually exclusive. Also, P (A ∩ B) = 0.30 6= 04.12 a. A: Generator 1 does not workb. B|A: Generator 2 does not work given that Generator 1 does not workc. A ∪ B: Generator 1 works or Generator 2 works or Both Generators work4.13 a. S = {F1F2, F1F3, F1F4, F1F5, F2F1, F2F3, F2F4, F2F5, F3F1, F3F2,F3F4, F3F5, F4F1, F4F2, F4F3, F4F5, F5F1, F5F2, F5F3, F5F4}b. Let T1be the event that the 1st firm chosen is stable and T2be the event that the2nd firm chosen is stable.P (T1) =35and P (T1) =25P (Both Stable) = P (T1∩ T2) = P (T2|T1)P (T1) = (24)(35) =620= 0.30Alternatively, if we designated F1and F2as the Shakey firms, then we could go tothe list of 20 possible outcomes and identify 6 pairs containing just Stable firms:F3F4, F3F5, F4F3, F4F5, F5F3, F5F4. Thus, the probability that both firms are Stableis620c. P (One of two firms is Shakey)= P (1st chosen is Shakey and 2nd chosen is Stable)+P (1st chosen is Stable and 2nd chosen is Shakey)= P (T1∩ T2) + P (T1∩ T2) = P (T2|T1)P (T1) + P (T2|T1)P (T1)= (34)(25) + (24)(35) =1220= 0.60Alternatively, in the list of 20 outcomes there are 12 pairs which consist of exactly1 Shakey firm (F1or F2) and exactly 1 Stable firm (F3or F4or F5). Thus, theprobability of exactly 1 Shakey firm is1220.d. P (Both Shakey) = P (T1∩ T2) = P (T2|T1)P (T1) = (14)(25) =220= 0.10Alternatively, in the list of 20 outcomes there are 2 pairs in which both firms areShakey (F1or F2). Thus, the probability that both firms are Shakey is220.4.14 P (A) = 192/(192 + 248) = 0.436; P (B) = 128/440 = 0.291;P (A ∩ B) = 48/440 = 0.1094.15 a. P (A) = P (none ∩ high) + P (little ∩ high) + P (some ∩ high) + P (extensive ∩ high)= 0.10 + 0.15 + 0.16 + 0.22 = 0.63P (B) = P (low ∩extensive) + P (medium ∩ extensive) + P (high ∩ extensive)= 0 .04 + 0.10 + 0.22 = 0.36P (C) = P (low ∩none) + P(low ∩ little) + P (medium ∩ none) + P (medium ∩ little)= 0.01 + 0.02 + 0.05 + 0.06 = 0.1421b. P (A|B) = P (A ∩ B)/P (B) = 0.22/0.36 = 0.611P (A|B) = P (A ∩ B)/P (B) = 0.41/(1 − .36) = 0.64P (B|C) = P (B ∩ C)/P (C) = 0.14/0.14 = 1c. P (A ∪ B) = P (A) + P (B) − P (A ∩ B) = 0.63 + 0.36 − 0.22 = 0.77P (A ∩ C) = 0; P (B ∩ C) = 04.16 a. P(Both customers pay in full) = (0.70)(0.70) = 0.49b. P(At least one of 2 customers pay in full) = 1 - P(neither customer pays in full) = 1- (1-0.70)(1-0.70) = 1 − (0.30)2= 0.914.17 Let A = event customer pays 1st month’s bill in full and B = event customer pays 2ndmonth’s bill in full. We are given thatP (A) = 0.70, P (B|A) = 0.95, P (B|A) = 0.90, P (B|A) = 1 − P (B|A) = 1 − 0.90 = 0.10a. P (A ∩ B) = P (B|A)P (A) = (0.95)(0.70) = 0.665b. P (B ∩ A) = P (B|A)P (A) = (0.90)(1 − 0.70) = 0.270c. P (pay exactly one month in full)= 1 − P (pays neither month or pays both months)= 1 − P (A ∩ B) − P (A ∩ B) = 1 - 0.665 - 0.27 = 0.0654.18 Let D be the event loan is defaulted, R1applicant is poor risk, R2fair risk, and R3goodrisk.P (D) = 0.01, P (R1|D) = 0.30, P (R2|D) = 0.40, P (R3|D) = 0.30P (D) = 0.01, P (R1|D) = 0.10, P (R2|D) = 0.40, P (R3|D) = 0.50P (D|R1) =P (R1|D)P (D)P (R1|D)P (D)+P (R1|D)P (D)=(0.30)(0.01)(0.30)(0.01)+(0.10)(0.99)= 0.02944.19 P (D|R2) =P (R2|D)P (D)P (R2|D)P (D)+P (R2|D)P (D)=(0.40)(0.01)(0.40)(0.01)+(0.40)(0.99)= 0.01The two probabilities are equal since the proportion of fair risk applicants is the same forboth the defaulted and nondefaulted loans.4.20 Let F be the event fire occurs and Tibe the event a type i furnace is in the home fori = 1, 2, 3, 4, where T4represent other types.P (T1|F ) =P (F |T1)P (T1)P (F |T1)P (T1)+P (F |T2)P (T2)+P (F |T3)P (T3)+P (F |T4)P (T4)=(0.05)(0.30)(0.05)(0.30)+(0.03)(0.25)+(0.02)(0.15)+(0.04)(0.30)= 0.404.21 a. Sensitivity: P (DA|C) = 50/53 = 0.943Specificity: P (DNA|RO) = 44/47 = 0.936b. P (C|DA) =(0.943)(0.00108)(0.943)(0.00108)+(0.021)(1−0.00108)= 0.046c. P (RO|DA) = 1 − P (C|DA) = 1 − 0.046 = .954d. P (RO|DNA) =(0.936)(1−0.00108)(0.936)(1−0.00108)+(0.019)(0.00108)= 0.999984.22 a. P (y = 3) =¡103¢(.2)3(.8)7= 0.20122b. P (y = 2) =¡42¢(.4)2(.6)2= 0.3456c. P (y = 12) =¡1612¢(.7)12(.3)4= 0.2044.23 a. a. P (y ≤ 4) = P (0) + P (1) + P (2) + P (3) + P (4)= 0.0168 + 0.0896 + 0.2090 + 0.2787 + 0.2322 = 0.8263b. P (y > 4) = 1 − P (y ≤ 4) = 1 − …
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