Homework 3 Solution, due July 16


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1 Homework 3 Solution, due July 16 Problems from old actuarial exams are marked by a star. Problem 1*. Upon arrival at a hospital emergency room, patients are categorized according to their condition as critical, serious, or stable. In the past year: 10% of the emergency room patients were critical; 30% of the emergency room patients were serious; the rest of the emergency room patients were stable; 40% of the critical patients died; 10% of the serious patients died; 1% of the stable patients died. Given that a patient survived, what is the probability that the patient was categorized as serious upon arrival? Solution. Let F 1 = {patient is critical}, F 2 = {patient is serious}, F 3 = {patient is stable}, A = {patient survived}. Then Thus, by the second Bayes formula P(F 2 A) = P(F 1 ) = 0.1, P(F 2 ) = 0.3, P(F 3 ) = 0.6, P(A F 1 ) = 0.6, P(A F 2 ) = 0.9, P(A F 3 ) = P(F 2 )P(A F 2 ) P(F 1 )P(A F 1 ) + P(F 2 )P(A F 2 ) + P(F 3 )P(A F 3 ) = = 29% Problem 2*. Among a large group of patients recovering from shoulder injuries, it is found that 22% visit both a physical therapist and a chiropractor, whereas 12% visit neither of these. The probability that a patient visits a chiropractor exceeds by 0.14 the probability that a patient visits a physical therapist. Determine the probability that a randomly chosen member of this group visits a physical therapist. Solution. Suppose A = {patient visits physical therapist}, B = {patient visits chiropractor}. Then by conditions of the problem P(A B) = 0.22, P((A B) c ) = 0.12, and We know that P(B) = P(A) P(A B) = P(A) + P(B) P(A B), 1
2 and P(A B) = = P(A) + P(B) = 1.1, and 2P(A) = 1.1. So P(A) = 0.48 Problem 3*. An urn contains 10 balls: 4 red and 6 blue. A second urn contains 16 red balls and an unknown number of blue balls. A single ball is drawn from each urn. The probability that both balls are the same color is Calculate the number of blue balls in the second urn. Solution. Let the number of blue balls in the second urn be x. Then the probability that balls drawn from each urn are both red is The probability that both balls are blue is x. 6 x x. Their sum is the probability that the balls have the same color: Solving for x, we get: x + 6 x x = (16 + x) = x, 1.6x = 6.4, x = 4 Problem 4*. The probability that a visit to a primary care physicians (PCP) office results in neither lab work nor referral to a specialist is 35%. Of those coming to a PCPs office, 30% are referred to specialists and 40% require lab work. Determine the probability that a visit to a PCPs office results in both lab work and referral to a specialist. Solution. Let A = {lab work}, B = {specialist}. We know that P((A B) c ) = 0.35, P(A) = 0.4, P(B) = 0.3. Thus, P(A B) = = P(A B) = P(A) + P(B) P(A B) = 0.05 Problem 5*. An insurance company issues life insurance policies in three separate categories: standard, preferred, and ultrapreferred. Of the companys policyholders, 50% are standard, 40% are preferred, and 10% are ultrapreferred. Each standard policyholder has probability of dying in the next year, each preferred policyholder has probability of dying in the next year, and each ultrapreferred policyholder has probability of dying in the next year. A policyholder dies in the next year. What is the probability that the deceased policyholder was ultrapreferred? 2
3 Solution. Let F 1 = {standard}, F 2 = {preferred}, F 3 = {ultrapreferred}, A = {deceased next year}. Then P(F 1 ) = 0.5, P(F 2 ) = 0.4, P(F 3 ) = 0.1, By the second Bayes formula, P(F 3 A) = P(A F 1 ) = 0.01, P(A F 2 ) = 0.005, P(A F 3 ) = P(A F 3 )P(F 3 ) P(A F 1 )P(F 1 ) + P(A F 2 )P(F 2 ) + P(A F 3 )P(F 3 ) = 1 71 = 1.41% Problem 6*. For two events A and B, you are given Determine P(A). P(A B) = 0.7, P(A B c ) = 0.9. Solution. Let C = A B and D = A B c. Then C D = Ω ( everything ), and C D = A. Applying the inclusionexclusion formula to C and D, we have: P(A) = P(C D) = P(C) + P(D) P(C D) = = 0.6 Problem 7*. An actuary is studying the prevalence of three health risk factors, denoted by A, B, and C, within a population of women. For each of the three factors, the probability is 0.1 that a woman in the population has only this risk factor (and no others). For any two of the three factors, the probability is 0.12 that she has exactly these two risk factors (but not the other). The probability that a woman has all three risk factors, given that she has A and B, is 1/3. What is the probability that a woman has none of the three risk factors, given that she does not have risk factor A? Solution. We know that P(A B c C c ) = P(A c B C c ) = P(A c B c C) = 0.1, and Also, P(A B C c ) = P(A B c C) = P(A c B C) = P(A B C A B) = 1 3. P(A B C) = 1 P(A B), 3 P(A B C c ) = P(A B) P(A B C) = 2 P(A B), 3 so P(A B) = (3/2)0.12 = 0.18, and P(A B C) = (1/3)0.18 = P(A) = P(A B C)+P(A B c C)+P(A B C c )+P(A B c C c ) = = 0.4. And P(A c ) = 0.6. Now, P(A B C) = =
4 Finally, P((A B C) c ) = = P((A B C) c A c ) = = 7 15 Problem 8*. A large pool of adults earning their first drivers license includes 50% lowrisk drivers, 30% moderaterisk drivers, and 20% highrisk drivers. Because these drivers have no prior driving record, an insurance company considers each driver to be randomly selected from the pool. This month, the insurance company writes 4 new policies for adults earning their first drivers license. What is the probability that these 4 will contain at least two more highrisk drivers than lowrisk drivers? Solution. There are the following combinations which satisfy the condition that there are at least two more highrisk drivers than lowrisk drivers: HHHH, HHHL, HHHM, HHMM (arrangements of letters can vary, we just stated the quantity of each letter). The HHHH case has probability The HHHL case has probability , because L can be in four different places. Similarly, the HHHM case has probability Finally, the HHMM case has probability , because the number of ways to choose places for two M letters in ( 4 2) = 6. The answer is the sum of all these four probabilities, which is Problem 9*. An insurer offers a health plan to the employees of a large company. As part of this plan, the individual employees may choose exactly two of the supplementary coverages A, B, and C, or they may choose no supplementary coverage. The proportions of the companys employees that choose coverages A, B, and C are 1/4, 1/3, and 5/12, respectively. Determine the probability that a randomly chosen employee will choose no supplementary coverage. Solution. Let C 0 be the event that people choose A and B: C 0 = A B. Similarly for A 0 = B C and B 0 = A C. Finally, let N = (A B C) c be the event that the individual chooses nothing. These four are nonintersecting events, by conditions of the problem. Then we know that A 0 B 0 C 0 N = Ω is everything. P(A) = P(B 0 ) + P(C 0 ), P(B) = P(A 0 ) + P(C 0 ), P(C) = P(A 0 ) + P(B 0 ). So Thus, P(A 0 ) + P(B 0 ) + P(C 0 ) = 1 2 (P(A) + P(B) + P(C)) = 1 2 ( ) = P(N) = 1 P(A 0 ) P(B 0 ) P(C 0 ) = 1 2 Problem 10*. A random person has a heart disease with probability 25%. A person with a heart disease is a smoker with probability twice as a person without a heart disease. What is the conditional probability that a smoker has a heart disease? 4
5 Solution. Let A = {smoker}, B = {heart disease}. Then P(B) = 0.25, and P(A B) = 2P(A B c ). P(A B) P(B) Since P(B) = 0.25 and P(B c ) = 0.75, we get: = 2 P(A Bc ). P(B c ) 4P(A B) = 8 3 P(A Bc ) P(A B c ) = 3 2 P(A Bc ). P(A) = P(A B) + P(A B c ) = 5 P(A B). 2 So P(B A) = P(B A)/P(A) = 2 5 = 40% Problem 11. Toss three fair coins. Are the following events A and B independent? (i) A = {first and second tosses resulted in exactly one H}, B = {second and third tosses resulted in exactly one H}; (ii) A = {first and second tosses resulted in exactly one H}, B = {second and third tosses resulted in at least one H}; (iii) A = {first and second tosses resulted in at least one H}, B = {second and third tosses resulted in at least one H}; (iv) A = {first toss in H}, B = {second and third tosses resulted in at least one H}. Solution. There are eight elementary outcomes: Ω = {HHH, HHT, HT H, T HH, T T H, T HT, HT T, T T T }, each with probability 1/8. (i) A = {HT T, T HT, HT H, T HH}, B = {HT H, T T H, HHT, T HT }, A B = {T HT, HT H}. So P(A) = 4/8 = 1/2, and P(B) = 4/8 = 1/2, while P(A B) = 2/8 = 1/4. Yes, they are independent. (ii) A = {HT T, T HT, HT H, T HH}, B = {HT H, T T H, HHT, T HT, HHH, T HH}, A B = {T HT, HT H, T HH}. So P(A) = 4/8 = 1/2, P(B) = 6/8 = 3/4, and P(A B) = 3/8. Yes, they are independent. (iii) A = {HHH, HHT, HT H, HT T, T HH, T HT }, B = {HT H, T T H, HHT, T HT, HHH, T HH}, A B = {T HT, HT H, T HH}. So P(A) = 6/8 = 3/4, and P(B) = 3/4, while P(A B) = 3/8. Dependent. (iv) Independent, because A depends on the first toss, B depends on the second and third tosses, but the coin does not have a memory. Formally: A = {HT H, HT T, HHH, HHT }, B = {HT H, T T H, HHT, T HT, HHH, T HH}, A B = {HT H, HT T, HHH}. So P(A) = 4 8 = 1 2, P(B) = 6 8 = 3 4, P(A B) = 3 8 = P(A)P(B). Problem 12. Genes related to albinism are denoted by A (nonalbino) and a (albino). Each person has two genes. People with AA, Aa or aa are nonalbino, while people with aa are albino. The nonalbino gene A is called dominant. Any parent passes one of two genes (selected randomly with equal probability) to his offspring. If a person has either Aa 5
6 or aa, then he is called a carrier: he is a nonalbino, but he can carry the albino gene to his offsprings. (i) If a person is nonalbine, what is the conditional probability that he is a carrier? (ii) If a couple consists of a nonalbino and an albino, what is the probability that their offsping will be albino? (iii) Suppose that a couple consists of two nonalbinos. They have two offsprings. What is the probability that both offsprings are albinos? Solution. (i) Nonalbino = {AA, Aa, aa}, carrier = {Aa, aa}. P(Nonalbino) = 3/4, P(Carrier) = 2/4. So the conditional probability is (2/4)/(3/4) = 2/3 (ii) Let F 1 = {the nonalbino parent is a carrier}, F 2 = F c 1, A = {offspring albino}. Then P(A F 2 ) = 0, because the nonalbino parent who is not a carrier has AA, so he cannot pass the gene a to his offspring. Also, P(A F 1 ) = 1 2, because the albino parent will always pass the gene a to his offspring, while a nonalbino carrier will do this with probability 1/2. Finally, P(F 1 ) = 2/3, P(F 2 ) = 1/3. by the first Bayes formula P(A) = P(A F 1 )P(F 1 ) + P(A F 2 )P(F 2 ) = 1 3 (iii) F 1 = {both parents carriers}, A = {both offsprings albinos}, F 2 = F c 1. Then P(F 1 ) = (2/3)(2/3) = 4/9. If both parents are carriers, then each will pass the a gene with probability 1/2 to his child. So the probability of one chiled being albino is 1/4, and P(A F 1 ) = (1/4) 2 = 1/16; also, P(A F 2 ) = 0, for the same reason as in (ii). by the first Bayes formula P(A) = P(A F 1 )P(F 1 ) + P(A F 2 )P(F 2 ) = = 1 36 Problem 13. Is it right that the following statement are true for all possible events A, B, C? (Venn diagrams might be helpful.) (i) A A B; (ii) A A B; (iii) A c A = ; (iv) A c A = Ω; (v) A c = Ω \ A; (vi) (A \ B) (B \ A) = A B; (vii) A (B C) = (A B) (A C); (viii) A \ (B C) = (A \ B) (A \ C). Solution. (i) Yes. (ii) No. (iii) Yes. (iv) Yes. (v) Yes. (vi) No. (vii) Yes. (viii) No. Problem 14. Consider two coins: a fair one and a magic one, which always falls H. Suppose you pick a coin at random (each coin with equal probability 1/2) and toss it n 6
7 times. All of the tosses are heads. You would like to convince me that the coin is magic, but I will believe you only if the probability that it is magic (given the result that all n tosses are H) exceeds 99%. How large the number n of tosses should be? Solution. F 1 = {magic}, F 2 = {fair}, A = {n heads}. Then P(F 1 ) = P(F 2 ) = 1/2, and P(A F 1 ) = 1, P(A F 2 ) = (1/2) n. By the second Bayes formula, P(F 2 A) = This is less than 0.01 when n 7 P(F 2 )P(A F 2 ) P(F 1 )P(A F 1 ) + P(F 2 )P(A F 2 ) = = n. (1/2)n 1 + (1/2) n Problem 15. Let Ω = {1, 2,..., 8}, A = {2, 4, 6, 8}, B = {1, 2, 3, 4}, C = {1, 3, 5, 7}. Find the following events: (i) A B; (ii) (A B C) c ; (iii) (A \ B) C; (iv) A \ C; (v) C \ A c. Solution. (i) A B = Ω; (ii) (A B C) c = c = Ω; (iii) (A \ B) C = ; (iv) A \ C = {2, 4, 6, 8}; (v) C \ A c =. 7
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