Hw 3 Solution

Math 394 B&C. Probability I. Summer 2014. Homework 3, due July 16

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 ac-cording 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 categorizedas serious upon arrival?

Solution. Let F1 = patient is critical, F2 = patient is serious, F3 = patient isstable, A = patient survived. Then

P(F1) = 0.1, P(F2) = 0.3, P(F3) = 0.6,

P(A | F1) = 0.6, P(A | F2) = 0.9, P(A | F3) = 0.99.

Thus, by the second Bayes formula

P(F2 | A) =P(F2)P(A | F2)

P(F1)P(A | F1) + P(F2)P(A | F2) + P(F3)P(A | F3)=

0.9 · 0.30.99 · 0.6 + 0.9 · 0.3 + 0.6 · 0.1

= 29%

Problem 2*. Among a large group of patients recovering from shoulder injuries, it isfound that 22% visit both a physical therapist and a chiropractor, whereas 12% visit neitherof these. The probability that a patient visits a chiropractor exceeds by 0.14 the probabilitythat a patient visits a physical therapist. Determine the probability that a randomly chosenmember of this group visits a physical therapist.

Solution. Suppose A = patient visits physical therapist, B = patient visits chiro-practor. Then by conditions of the problem

P(A ∩B) = 0.22, P((A ∪B)c) = 0.12,

andP(B) = P(A) + 0.14.

We know thatP(A ∪B) = P(A) + P(B)−P(A ∩B),

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and P(A ∪ B) = 1− 0.12 = 0.88. Therefore, P(A) + P(B) = 1.1, and 2P(A) + 0.14 = 1.1.

So P(A) = 0.48

Problem 3*. An urn contains 10 balls: 4 red and 6 blue. A second urn contains 16red balls and an unknown number of blue balls. A single ball is drawn from each urn. Theprobability that both balls are the same color is 0.44. Calculate the number of blue balls inthe second urn.

Solution. Let the number of blue balls in the second urn be x. Then the probabilitythat balls drawn from each urn are both red is

4

10

16

16 + x.

The probability that both balls are blue is

6

10

x

16 + x.

Their sum is the probability that the balls have the same color:

4

10

16

16 + x+

6

10

x

16 + x= 0.44.

Solving for x, we get:

4.4(16 + x) = 64 + 6x, 1.6x = 6.4, x = 4

Problem 4*. The probability that a visit to a primary care physicians (PCP) officeresults in neither lab work nor referral to a specialist is 35%. Of those coming to a PCPsoffice, 30% are referred to specialists and 40% require lab work. Determine the probabilitythat 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.

Therefore,P(A ∪B) = 1− 0.35 = 0.65.

Thus,P(A ∩B) = P(A) + P(B)−P(A ∪B) = 0.05

Problem 5*. An insurance company issues life insurance policies in three separatecategories: standard, preferred, and ultra-preferred. Of the companys policyholders, 50%are standard, 40% are preferred, and 10% are ultra-preferred. Each standard policyholderhas probability 0.010 of dying in the next year, each preferred policyholder has probability0.005 of dying in the next year, and each ultra-preferred policyholder has probability 0.001of dying in the next year. A policyholder dies in the next year. What is the probability thatthe deceased policyholder was ultra-preferred?

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Solution. Let F1 = standard, F2 = preferred, F3 = ultra-preferred, A = deceasednext year. Then

P(F1) = 0.5, P(F2) = 0.4, P(F3) = 0.1,

P(A | F1) = 0.01, P(A | F2) = 0.005, P(A | F3) = 0.001.

By the second Bayes’ formula,

P(F3 | A) =P(A | F3)P(F3)

P(A | F1)P(F1) + P(A | F2)P(F2) + P(A | F3)P(F3)=

1

71= 1.41%

Problem 6*. For two events A and B, you are given

P(A ∪B) = 0.7, P(A ∪Bc) = 0.9.

Determine P(A).

Solution. Let C = A ∪ B and D = A ∪ Bc. Then C ∪ D = Ω (”everything”), andC ∩D = A. Applying the inclusion-exclusion formula to C and D, we have:

P(A) = P(C ∩D) = P(C) + P(D)−P(C ∪D) = 0.7 + 0.9− 1 = 0.6

Problem 7*. An actuary is studying the prevalence of three health risk factors, denotedby A, B, and C, within a population of women. For each of the three factors, the probabilityis 0.1 that a woman in the population has only this risk factor (and no others). For any twoof the three factors, the probability is 0.12 that she has exactly these two risk factors (butnot the other). The probability that a woman has all three risk factors, given that she hasA 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 ∩Bc ∩ Cc) = P(Ac ∩B ∩ Cc) = P(Ac ∩Bc ∩ C) = 0.1,

andP(A ∩B ∩ Cc) = P(A ∩Bc ∩ C) = P(Ac ∩B ∩ C) = 0.12.

Also,

P(A ∩B ∩ C | A ∩B) =1

3.

Therefore,

P(A ∩B ∩ C) =1

3P(A ∩B),

P(A ∩B ∩ Cc) = P(A ∩B)−P(A ∩B ∩ C) =2

3P(A ∩B),

so P(A ∩B) = (3/2)0.12 = 0.18, and P(A ∩B ∩ C) = (1/3)0.18 = 0.06. Therefore,

P(A) = P(A∩B∩C)+P(A∩Bc∩C)+P(A∩B∩Cc)+P(A∩Bc∩Cc) = 0.06+0.12+0.12+0.1 = 0.4.

And P(Ac) = 0.6. Now,

P(A ∪B ∪ C) = 3 · 0.1 + 3 · 0.12 + 0.06 = 0.72.

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Therefore,P((A ∪B ∪ C)c) = 1− 0.72 = 0.28.

Finally,

P((A ∪B ∪ C)c | Ac) =0.28

0.6=

7

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Problem 8*. A large pool of adults earning their first drivers license includes 50% low-risk drivers, 30% moderate-risk drivers, and 20% high-risk drivers. Because these drivershave no prior driving record, an insurance company considers each driver to be randomlyselected from the pool. This month, the insurance company writes 4 new policies for adultsearning their first drivers license. What is the probability that these 4 will contain at leasttwo more high-risk drivers than low-risk drivers?

Solution. There are the following combinations which satisfy the condition that thereare at least two more high-risk drivers than low-risk drivers: HHHH, HHHL, HHHM, HHMM(arrangements of letters can vary, we just stated the quantity of each letter).

The HHHH case has probability 0.24.The HHHL case has probability 4 · 0.230.3, because L can be in four different places.Similarly, the HHHM case has probability 4 · 0.23 · 0.5.Finally, the HHMM case has probability 6 · 0.22 · 0.32, because the number of ways to

choose places for two M letters in(42

)= 6.

The answer is the sum of all these four probabilities, which is 0.0488

Problem 9*. An insurer offers a health plan to the employees of a large company. Aspart of this plan, the individual employees may choose exactly two of the supplementarycoverages A, B, and C, or they may choose no supplementary coverage. The proportionsof 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 nosupplementary coverage.

Solution. Let C0 be the event that people choose A and B: C0 = A ∩ B. Similarly forA0 = B ∩C and B0 = A∩C. Finally, let N = (A∪B ∪C)c be the event that the individualchooses nothing. These four are non-intersecting events, by conditions of the problem. Thenwe know that A0 ∪B0 ∪ C0 ∪N = Ω is ”everything”. Therefore,

P(A) = P(B0) + P(C0), P(B) = P(A0) + P(C0), P(C) = P(A0) + P(B0).

So

P(A0) + P(B0) + P(C0) =1

2(P(A) + P(B) + P(C)) =

1

2

(1

4+

1

3+

5

12

)=

1

2.

Thus,

P(N) = 1−P(A0)−P(B0)−P(C0) =1

2

Problem 10*. A random person has a heart disease with probability 25%. A personwith 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?

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Solution. Let A = smoker, B = heart disease. Then P(B) = 0.25, and P(A | B) =2P(A | Bc). Therefore,

P(A ∩B)

P(B)= 2

P(A ∩Bc)

P(Bc).

Since P(B) = 0.25 and P(Bc) = 0.75, we get:

4P(A ∩B) =8

3P(A ∩Bc) ⇒ P(A ∩Bc) =

3

2P(A ∩Bc).

Therefore,

P(A) = P(A ∩B) + P(A ∩Bc) =5

2P(A ∩B).

So P(B | A) = P(B ∩ A)/P(A) = 25

= 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,HTH, THH, TTH, THT,HTT, TTT,

each with probability 1/8.(i) A = HTT, THT,HTH, THH, B = HTH, TTH,HHT, THT, A∩B = THT,HTH.

So P(A) = 4/8 = 1/2, and P(B) = 4/8 = 1/2, while P(A ∩ B) = 2/8 = 1/4. Yes, they areindependent.

(ii) A = HTT, THT,HTH, THH, B = HTH, TTH,HHT, THT,HHH, THH, A∩B = THT,HTH, THH. 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,HTH,HTT, THH, THT, B = HTH, TTH,HHT, THT,HHH, THH,A ∩ B = THT,HTH, THH. 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 thirdtosses, but the coin does not have a memory. Formally: A = HTH,HTT,HHH,HHT,B = HTH, TTH,HHT, THT,HHH, THH, A ∩B = HTH,HTT,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 (non-albino) and a (albino).Each person has two genes. People with AA, Aa or aA are non-albino, while people withaa are albino. The non-albino gene A is called dominant. Any parent passes one of twogenes (selected randomly with equal probability) to his offspring. If a person has either Aa

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or aA, then he is called a carrier: he is a non-albino, but he can carry the albino gene to hisoffsprings.

(i) If a person is non-albine, what is the conditional probability that he is a carrier?(ii) If a couple consists of a non-albino and an albino, what is the probability that their

offsping will be albino?(iii) Suppose that a couple consists of two non-albinos. They have two offsprings. What

is the probability that both offsprings are albinos?

Solution. (i) Non-albino = AA, Aa, aA, carrier = Aa, aA. P(Non-albino) = 3/4,

P(Carrier) = 2/4. So the conditional probability is (2/4)/(3/4) = 2/3

(ii) Let F1 = the non-albino parent is a carrier, F2 = F c1 , A = offspring albino. Then

P(A | F2) = 0,

because the non-albino parent who is not a carrier has AA, so he cannot pass the gene a tohis offspring. Also,

P(A | F1) =1

2,

because the albino parent will always pass the gene a to his offspring, while a non-albinocarrier will do this with probability 1/2. Finally, P(F1) = 2/3, P(F2) = 1/3. Therefore, bythe first Bayes’ formula

P(A) = P(A | F1)P(F1) + P(A | F2)P(F2) =1

3

(iii) F1 = both parents carriers, A = both offsprings albinos, F2 = F c1 . Then P(F1) =

(2/3)(2/3) = 4/9. If both parents are carriers, then each will pass the a gene with probability1/2 to his child. So the probability of one chiled being albino is 1/4, and P(A | F1) =(1/4)2 = 1/16; also, P(A | F2) = 0, for the same reason as in (ii). Therefore, by the firstBayes’ formula

P(A) = P(A | F1)P(F1) + P(A | F2)P(F2) =1

16

4

9=

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) Ac ∩ A = ∅;(iv) Ac ∪ A = Ω;(v) Ac = 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

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times. All of the tosses are heads. You would like to convince me that the coin is magic, butI will believe you only if the probability that it is magic (given the result that all n tossesare H) exceeds 99%. How large the number n of tosses should be?

Solution. F1 = magic, F2 = fair, A = n heads. Then P(F1) = P(F2) = 1/2, andP(A | F1) = 1, P(A | F2) = (1/2)n. By the second Bayes’ formula,

P(F2 | A) =P(F2)P(A | F2)

P(F1)P(A | F1) + P(F2)P(A | F2)=

(1/2)n

1 + (1/2)n

=1

1 + 2n.

This is less than 0.01 when n ≥ 7

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 \ Ac.

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 \ Ac = ∅.

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Hw 3 Solution