Exam (elaborations) TEST BANK FOR Probability and Random Processes for

Exam (elaborations) TEST BANK FOR Probability and Random Processes for We first let C ⊂ A and show that for all B, (A∩B)∪C = A∩(B∪C). Write A∩(B∪C) = (A∩B)∪(A∩C), by the distributive law, = (A∩B)∪C, since C ⊂ A ⇒ A∩C = C. For the second part of the problem, suppose (A∩B)∪C = A∩(B∪C). We must show that C ⊂ A. Let ω ∈ C. Then ω ∈ (A ∩ B) ∪C. But then ω ∈ A ∩ (B ∪C), which implies ω ∈ A. 9. Let I := {ω ∈ Ω : ω ∈ A ⇒ ω ∈ B}. We must show that A∩I = A∩B. ⊂: Let ω ∈ A∩I. Then ω ∈ A and ω ∈ I. Therefore, ω ∈ B, and then ω ∈ A∩B. ⊃: Let ω ∈ A∩B. Then ω ∈ A and ω ∈ B. We must show that ω ∈ I too. In other words, we must show that ω ∈ A ⇒ ω ∈ B. But we already have ω ∈ B. 10. The function f :(−∞,∞) → [0,∞) with f(x) = x 3 is not well defined because not all values of f(x) lie in the claimed co-domain [0,∞). 11. (a) The function will be invertible if Y = [−1,1]. (b) {x : f(x) ≤ 1/2} = [−π/2,π/6]. (c) {x : f(x) 0} = [−π/2,0). 12. (a) Since f is not one-to-one, no choice of co-domain Y can make f : [0,π] → Y invertible. (b) {x : f(x) ≤ 1/2} = [0,π/6]∪[5π/6,π]. (c) {x : f(x) 0} = ∅. 13. For B ⊂ IR, f −1 (B) =    X, 0 ∈ B and 1 ∈ B, A, 1 ∈ B but 0 ∈/ B, A c , 0 ∈ B but 1 ∈/ B, ∅, 0 ∈/ B and 1 ∈/ B. 14. Let f :X → Y be a function such that f takes only n distinct values, say y1,...,yn. Let B ⊂ Y be such that f −1 (B) is nonempty. By definition, each x ∈ f −1 (B) has the property that f(x) ∈ B. But f(x) must be one of the values y1,...,yn, say yi . Now f(x) = yi if and only if x ∈ Ai := f −1 ({yi}). Hence, f −1 (B) = [ i:yi∈B Ai . 15. (a) f(x) ∈ B c ⇔ f(x) ∈/ B ⇔ x ∈/ f −1 (B) ⇔ x ∈ f −1 (B) c . (b) f(x) ∈ [∞ n=1 Bn if and only if f(x) ∈ Bn for some n; i.e., if and only if x ∈ f −1 (Bn) for some n. But this says that x ∈ [∞ n=1 f −1 (Bn). 4 Chapter 1 Problem Solutions (c) f(x) ∈ ∞ n=1 Bn if and only if f(x) ∈ Bn for all n; i.e., if and only if x ∈ f −1 (Bn) for all n. But this says that x ∈ ∞ n=1 f −1 (Bn). 16. If B = S i{bi} and C = S i{ci}, put a2i := bi and a2i−1 := ci . Then A = S i ai = B∪C is countable. 17. Since each Ci is countable, we can write Ci = S j ci j. It then follows that B := [∞ i=1 Ci = [∞ i=1 [∞ j=1 {ci j} is a doubly indexed sequence and is therefore countable as shown in the text. 18. Let A = S m{am} be a countable set, and let B ⊂ A. We must show that B is countable. If B = ∅, we’re done by definition. Otherwise, there is at least one element of B in A, say ak . Then put bn := an if an ∈ B, and put bn := ak if an ∈/ B. Then S n{bn} = B and we see that B is countable. 19. Let A ⊂ B where A is uncountable. We must show that B is uncountable. We prove this by contradiction. Suppose that B is countable. Then by the previous problem, A is countable, contradicting the assumption that A is uncountable. 20. Suppose A is countable and B is uncountable. We must show that A∪B is uncountable. We prove this by contradiction. Suppose that A ∪ B is countable. Then since B ⊂ A ∪ B, we would have B countable as well, contradicting the assumption that B is uncountable. 21. MATLAB. OMITTED. 22. MATLAB. Intuitive explanation: Using only the numbers 1,2,3,4,5,6, consider how many ways there are to write the following numbers: 2 = 1+1 1 way, 1/36 = 0.0278 3 = 1+2 = 2+1 2 ways, 2/36 = 0.0556 4 = 1+3 = 2+2 = 3+1 3 ways, 3/36 = 0.0833 5 = 1+4 = 2+3 = 3+2 = 4+1 4 ways, 4/36 = 0.1111 6 = 1+5 = 2+4 = 3+3 = 4+2 = 5+1 5 ways, 5/36 = 0.1389 7 = 1+6 = 2+5 = 3+4 = 4+3 = 5+2 = 6+1 6 ways, 6/36 = 0.1667 8 = 2+6 = 3+5 = 4+4 = 5+3 = 6+2 5 ways, 5/36 = 0.1389 9 = 3+6 = 4+5 = 5+4 = 6+3 4 ways, 4/36 = 0.1111 10 = 4+6 = 5+5 = 6+4 3 ways, 3/36 = 0.0833 11 = 5+6 = 6+5 2 ways, 2/36 = 0.0556 12 = 6+6 1 way, 1/36 = 0.0278 36 ways, 36/36 = 1 23. Take Ω := {1,...,26} and put P(A) := |A| |Ω| = |A| 26 . Chapter 1 Problem Solutions 5 The event that a vowel is chosen is V = {1,5,9,15,21}, and P(V) = |V|/26 = 5/26. 24. Let Ω := {(i, j) : 1 ≤ i, j ≤ 26 and i 6= j}. For A ⊂ Ω, put P(A) := |A|/|Ω|. The event that a vowel is chosen followed by a consonant is Bvc =  (i, j) ∈ Ω : i = 1,5,9,15, or 21 and j ∈ {1,...,26} {1,5,9,15,21} . Similarly, the event that a consonant is followed by a vowel is Bcv =  (i, j) ∈ Ω : i ∈ {1,...,26} {1,5,9,15,21} and j = 1,5,9,15, or 21 . We need to compute P(Bvc ∪Bcv) = |Bvc|+|Bcv| |Ω| = 5 ·(26−5)+(26−5)· 5 650 = 21 65 ≈ 0.323. The event that two vowels are chosen is Bvv =  (i, j) ∈ Ω : i, j ∈ {1,5,9,15,21} with i 6= j , and P(Bvv) = |Bvv|/|Ω| = 20/650 = 2/65 ≈ .031. 25. MATLAB. The code for simulating the drawing of a face card is % Simulation of Drawing a Face Card % n = 10000; % Number of draws. X = ceil(52*rand(1,n)); faces = (41 = X & X = 52); nfaces = sum(faces); fprintf(’There were %g face cards in %g draws.n’,nfaces,n) 26. Since 9 pm to 7 am is 10 hours, take Ω := [0,10]. The probability that the baby wakes up during a time interval 0 ≤ t1 t2 ≤ 10 is P([t1,t2]) := Z t2 t1 1 10 dω. Hence, P([2,10] c ) = P([0,2]) = R 2 0 1/10dω = 1/5. 27. Starting with the equations SN = 1+z+z 2 +···+z N−2 +z N−1 zSN = z+z 2 +···+z N−2 +z N−1 +z N , subtract the second line from the first. Canceling common terms leaves SN −zSN = 1−z N , or SN(1−z) = 1−z N . If z 6= 1, we can divide both sides by 1−z to get SN = (1−z N)/(1−z). 6 Chapter 1 Problem Solutions 28. Let x = p(1). Then p(2) = 2p(1) = 2x, p(3) = 2p(2) = 2 2 x, p(4) = 2p(3) = 2 3 x, p(5) = 2 4 x, and p(6) = 2 5 x. In general, p(ω) = 2 ω−1 x and we can write 1 = 6 ∑ ω=1 p(ω) = 6 ∑ ω=1 2 ω−1 x = x 5 ∑ ω=0 2 ω = 1−2 6 1−2 x = 63x. Hence, x = 1/63, and p(ω) = 2 ω−1/63 for ω = 1,...,6. 29. (a) By inclusion–exclusion, P(A∪B) = P(A)+P(B)−P(A∩B), which can be rearranged as P(A∩B) = P(A)+P(B)−P(A∪B). (b) Since P(A) = P(A∩B)+P(A∩B c ), P(A∩B c ) = P(A)−P(A∩B) = P(A∪B)−P(B), by part (a). (c) Since B and A∩B c are disjoint, P(B∪(A∩B c )) = P(B)+P(A∩B c ) = P(A∪B), by part (b). (d) By De Morgan’s law, P(A c ∩B c ) = P([A∪B] c ) = 1−P(A∪B). 30. We must check the four axioms of a probability measure. First, P(∅) = λP1(∅)+(1−λ)P2(∅) = λ · 0+(1−λ)· 0 = 0. Second, P(A) = λP1(A)+(1−λ)P2(A) ≥ λ · 0+(1−λ)· 0 = 0. Third, P  [∞ n=1 An  = λP1  [∞ n=1 An  +(1−λ)P2  [∞ n=1 An  = λ ∞ ∑ n=1 P1(An)+(1−λ) ∞ ∑ n=1 P2(An) = ∞ ∑ n=1 [λP1(An)+(1−λ)P2(An)] = ∞ ∑ n=1 P(An). Fourth, P(Ω) = λP1(Ω)+(1−λ)P2(Ω) = λ +(1−λ) = 1. 31. First, since ω0 ∈/ ∅, µ(∅) = 0. Second, by definition, µ(A) ≥ 0. Third, for disjoint An, suppose ω0 ∈ S n An. Then ω0 ∈ Am for some m, and ω0 ∈/ An for n 6= m. Then µ(Am) = 1 and µ(An) = 0 for n 6= m. Hence, µ S n An  = 1 and ∑n µ(An) = µ(Am) = 1. A similar analysis shows that if ω0 ∈/ S n An then µ S n An  and ∑n µ(An) are both zero. Finally, since ω0 ∈ Ω, µ(Ω) = 1. Chapter 1 Problem Solutions 7 32. Starting with the assumption that for any two disjoint events A and B, P(A ∪ B) = P(A)+P(B), we have that for N = 2, P  [ N n=1 An  = N ∑ n=1 P(An). (∗) Now we must show that if (∗) holds for any N ≥ 2, then (∗) holds for N +1. Write P N [ +1 n=1 An  = P  [ N n=1 An  ∪AN+1  = P  [ N n=1 An  +P(AN+1), additivity for two events, = N ∑ n=1 P(An)+P(AN+1), by (∗), = N+1 ∑ n=1 P(An). 33. Since An := Fn ∩F c n−1 ∩ ··· ∩F c 1 ⊂ Fn, it is easy to see that [ N n=1 An ⊂ [ N n=1 Fn. The hard part is to show the reverse inclusion ⊃. Suppose ω ∈ SN n=1 Fn. Then ω ∈ Fn for some n in the range 1,...,N. However, ω may belong to Fn for several values of n since the Fn may not be disjoint. Let k := min{n : ω ∈ Fn and 1 ≤ n ≤ N}. In other words, 1 ≤ k ≤ N and ω ∈ Fk , but ω ∈/ Fn for n k; in symbols, ω ∈ Fk ∩F c k−1 ∩ ··· ∩F c 1 =: Ak . Hence, ω ∈ Ak ⊂ SN n=1 An. The proof that S∞ n=1 An ⊂ S∞ n=1 Fn is similar except that k := min{n : ω ∈ Fn and n ≥ 1}. 34. For arbitrary events Fn, let An be as in the preceding problem. We can then write P  [∞ n=1 Fn  = P  [∞ n=1 An  = ∞ ∑ n=1 P(An), since the An are disjoint, = lim N→∞ N ∑ n=1 P(An), by def. of infinite sum, = lim N→∞ P  [ N n=1 An  = lim N→∞ P  [ N n=1 Fn  . 8 Chapter 1 Problem Solutions 35. For arbitrary events Gn, put Fn := G c n . Then P  ∞ n=1 Gn  = 1−P  [∞ n=1 Fn  , by De Morgan’s law, = 1− lim N→∞ P  [ N n=1 Fn  , by the preceding problem, = 1− lim N→∞  1−P  N n=1 Gn , by De Morgan’s law, = lim N→∞ P  N n=1 Gn  . 36. By the inclusion–exclusion formula, P(A∪B) = P(A)+P(B)−P(A∩B) ≤ P(A)+P(B). This establishes the union bound for N = 2. Now suppose the union bound holds for some N ≥ 2. We must show it holds for N +1. Write P N [ +1 n=1 Fn  = P  [ N n=1 Fn  ∪FN+1  ≤ P  [ N n=1 Fn  +P(FN+1), by the union bound for two events, ≤ N ∑ n=1 P(Fn)+P(FN+1), by the union bound for N events, = N+1 ∑ n=1 P(Fn). 37. To establish the union bound for a countable sequence of events, we proceed as follows. Let An := Fn ∩F c n−1 ∩ ··· ∩F c 1 ⊂ Fn be disjoint with S∞ n=1 An = S∞ n=1 Fn. Then P  [∞ n=1 Fn  = P  [∞ n=1 An  = ∞ ∑ n=1 P(An), since the An are disjoint, ≤ ∞ ∑ n=1 P(Fn), since An ⊂ Fn. 38. Following the hint, we put Gn := S∞ k=n Bk so that we can write P  ∞ n=1 [∞ k=n Bk  = P  ∞ n=1 Gn  = lim N→∞ P  N n=1 Gn  , limit property of P, Chapter 1 Problem Solutions 9 = lim N→∞ P(GN), since Gn ⊃ Gn+1, = lim N→∞ P  [∞ k=N Bk  , definition of GN, ≤ lim N→∞ ∞ ∑ k=N P(Bk), union bound. This last limit must be zero since ∑ ∞ k=1 P(Bk) ∞. 39. In this problem, the probability of an interval is its length. (a) P(A0) = 1, P(A1) = 2/3, P(A2) = 4/9 = (2/3) 2 , and P(A3) = 8/27 = (2/3) 3 . (b) P(An) = (2/3) n . (c) Write P(A) = P  ∞ n=1 An  = lim N→∞ P  N n=1 An  , limit property of P, = lim N→∞ P(AN), since An ⊃ An+1, = lim N→∞ (2/3) N = 0. 40. Consider the collection consisting of the empty set along with all unions of the form S i Aki for some finite subsequence of distinct elements from {1,...,n}. We first show that this collection is a σ-field. First, it contains ∅ by definition. Second, since A1,...,An is a partition, [ i Aki c = [ i Ami , where mi is the subsequence {1,...,n} {ki}. Hence, the collection is closed under complementation. Third, [∞ n=1 [ i Akn,i  = [ j Amj , where an integer l ∈ {1,...,n} is in {mj} if and only if kn,i = l for some n and some i. This shows that the collection is a σ-field. Finally, since every element in our collection must be contained in every σ-field that contains A1,...,An, our collection must be σ(A1,...,An). 41. We claim that A is not a σ-field. Our proof is by contradiction: We assume A is a σ-field and derive a contradiction. Consider the set ∞ n=1 [0,1/2 n ) = {0}. Since [0,1/2 n ) ∈ Cn ⊂ An ⊂ A , the intersection must be in A since we are assuming A is a σ-field. Hence, {0} ∈ A . Now, any set in A must belong to some An. By the 10 Chapter 1 Problem Solutions preceding problem, every set in An must be a finite union of sets from Cn. However, the singleton set {0} cannot be expressed as a finite union of sets from any Cn. Hence, {0} ∈/ A . 42. Let Ai := X −1 ({xi}) for i = 1,...,n. By the problems mentioned in the hint, for any subset B, if X −1 (B) 6= ∅, then X −1 (B) = [ i:xi∈B Ai ∈ σ(A1,...,An). It follows that the smallest σ-field containing all the X −1 (B) is σ(A1,...,An). 43. (a) F =  ∅,A,B,{3},{1,2},{4,5},{1,2,4,5},Ω (b) The corresponding probabilities are 0,5/8,7/8,1/2,1/8,3/8,1/2,1. (c) Since {1} ∈/ F, P({1}) is not defined. 44. Suppose that a σ-field A contains an infinite sequence Fn of sets. If the sequence is not disjoint, we can construct a new sequence An that is disjoint with each An ∈ A . Let a = a1,a2,... be an infinite sequence of zeros and ones. Then A contains each union of the form [ i:ai=1 Ai . Furthermore, since the Ai are disjoint, each sequence a gives a different union, and we know from the text that the number of infinite sequences a is uncountably infinite. 45. (a) First, since ∅ is in each Aα, ∅ ∈ T α Aα. Second, if A ∈ T α Aα, then A ∈ Aα for each α, and so A c ∈ Aα for each α. Hence, A c ∈ T α Aα. Third, if An ∈ A for all n, then for each n and each α, An ∈ Aα. Then S n An ∈ Aα for each α, and so S n An ∈ T α Aα. (b) We first note that A1 = {∅,{1},{2},{3,4},{2,3,4},{1,3,4},{1,2},Ω} and A2 = {∅,{1},{3},{2,4},{2,3,4},{1,2,4},{1,3},Ω}. It is then easy to see that A1 ∩ A2 = {∅,{1},{2,3,4},Ω}. (c) First note that by part (a), T A :C ⊂A A is a σ-field, and since C ⊂ A for each A , the σ-field T A :C ⊂A A contains C . Finally, if D is any σ-field that contains C , then D is one of the A s in the intersection. Hence, C ⊂ A :C ⊂A A ⊂ D. Thus T A :C ⊂A A is the smallest σ-field that contains C . Chapter 1 Problem Solutions 11 46. The union of two σ-fields is not always a σ-field. Here is an example. Let Ω := {1,2,3,4}, and put F :=  ∅,{1,2},{3,4},Ω and G :=  ∅,{1,3},{2,4},Ω . Then F ∪G =  ∅,{1,2},{3,4},{1,3},{2,4},Ω is not a σ-field since it does not contain {1,2} ∩ {1,3} = {1}. 47. Let Ω denote the positive integers, and let A denote the collection of subsets A such that either A or A c is finite. (a) Let E denote the subset of even integers. Then E does not belong to A since neither E nor E c (the odd integers) is a finite set. (b) To show that A is closed under finite unions, we consider two cases. First suppose that A1,...,An are all finite. Then     [n i=1 Ai     ≤ n ∑ i=1 |Ai | ∞, and so Sn i=1 Ai ∈ A . In the second case, suppose that some A c j is finite. Then  [n i=1 Ai c = n i=1 A c i ⊂ A c j . Hence, the complement of Sn i=1 Ai is finite, and so the union belongs to A . (c) A is not a σ-field. To see this, put Ai := {2i} for i = 1,2,.... Then S∞ i=1 Ai = E ∈/ A by part (a). 48. Let Ω be an uncountable set. Let A denote the collection of all subsets A such that either A is countable or A c is countable. We show that A is a σ-field. First, the empty set is countable. Second, if A ∈ A , we must show that A c ∈ A . There are two cases. If A is countable, then the complement of A c is A, and so A c ∈ A . If A c is countable, then A c ∈ A . Third, let A1,A2,... belong to A . There are two cases to consider. If all An are countable, then S n An is also countable by an earlier problem. Otherwise, if some A c m is countable, then write  [∞ n=1 An c = ∞ n=1 A c n ⊂ A c m. Since the subset of a countable set is countable, we see that the complement of S∞ n=1 An is countable, and thus the union belongs to A . 49. (a) Since (a,b] = T∞ n=1 (a,b+ 1 n ), and since each (a,b+ 1 n ) ∈ B, (a,b] ∈ B. (b) Since {a} = T∞ n=1 (a− 1 n ,a+ 1 n ), and since each (a− 1 n ,a+ 1 n ) ∈ B, the singleton {a} ∈ B. 12 Chapter 1 Problem Solutions (c) Since by part (b), singleton sets are Borel sets, and since A is a countable union of Borel sets, A ∈ B; i.e., A is a Borel set. (d) Using part (a), write λ (a,b]
= λ  ∞ n=1 (a,b+ 1 n )  = lim N→∞ λ  N n=1 (a,b+ 1 n )  , limit property of probability, = lim N→∞ λ (a,b+ 1 N )
, decreasing sets, = lim N→∞ (b+ 1 N )−a, characterization of λ, = b−a. Similarly, using part (b), we can write λ {a}
= λ  ∞ n=1 (a− 1 n ,a+ 1 n )  = lim N→∞ λ  N n=1 (a− 1 n ,a+ 1 n )  , limit property of probability, = lim N→∞ λ (a− 1 N ,a+ 1 N )
, decreasing sets, = lim N→∞ 2/N, characterization of λ, = 0. 50. Let I denote the collection of open intervals, and let O denote the collection of open sets. We need to show that σ(I ) = σ(O). Since I ⊂ O, every σ-field containing O also contains I . Hence, the smallest σ-field containing O contains I ; i.e., I ⊂ σ(O). By the definition of the smallest σ-field containing I , it follows that σ(I ) ⊂ σ(O). Now, if we can show that O ⊂ σ(I ), then it will similarly follow that σ(O) ⊂ σ(I ). Recall that in the problem statement, it was shown that every open set U can be written as a countable union of open intervals. This means U ∈ σ(I ). This proves that O ⊂ σ(I ) as required. 51. MATLAB. Chips from S1 are 80% reliable; chips from S2 are 70% reliable. 52. Observe that N(Ow,S1) = N(OS1)−N(Od,S1) = N(OS1)  1− N(Od,S1) N(OS1)  and N(Ow,S2) = N(OS2)−N(Od,S2) = N(OS2)  1− N(Od,S2) N(OS2)  . Chapter 1 Problem Solutions 13 53. First write P(A|B∩C)P(B|C) = P(A∩[B∩C]) P(B∩C) · P(B∩C) P(C) = P([A∩B]∩C) P(C) = P(A∩B|C). From this formula, we can isolate the equation P(A|B∩C)P(B|C) = P([A∩B]∩C) P(C) . Multiplying through by P(C) yields P(A|B∩C)P(B|C)P(C) = P(A∩B∩C). 54. (a) P(MM) = 140/(140+60) = 140/200 = 14/20 = 7/10 = 0.7. Then P(HT) = 1−P(MM) = 0.3. (b) Let D denote the event that a workstation is defective. Then P(D) = P(D|MM)P(MM)+P(D|HT)P(HT) = (.1)(.7)+(.2)(.3) = .07+.06 = 0.13. (c) Write P(MM|D) = P(D|MM)P(MM) P(D) = .07 .13 = 7 13 . 55. Let O denote the event that a cell is overloaded, and let B denote the event that a call is blocked. The problem statement tells us that P(O) = 1/3, P(B|O) = 3/10, and P(B|O c ) = 1/10. To find P(O|B), first write P(O|B) = P(B|O)P(O) P(B) = 3/10 · 1/3 P(B) . Next compute P(B) = P(B|O)P(O)+P(B|O c )P(O c ) = (3/10)(1/3)+(1/10)(2/3) = 5/30 = 1/6. We conclude that P(O|B) = 1/10 1/6 = 6 10 = 3 5 = 0.6. 56. The problem statement tells us that P(R1|T0) = ε and P(R0|T1) = δ. We also know that 1 = P(Ω) = P(T0 ∪T1) = P(T0)+P(T1). The problem statement tells us that these last two probabilities are the same; hence they are both equal to 1/2. To find P(T1|R1), we begin by writing P(T1|R1) = P(R1|T1)P(T1) P(R1) . 14 Chapter 1 Problem Solutions Next, we note that P(R1|T1) = 1−P(R0|T1) = 1−δ. By the law of total probability, P(R1) = P(R1|T1)P(T1)+P(R1|T0)P(T0) = (1−δ)(1/2)+ε(1/2) = (1−δ +ε)/2. So, P(T1|R1) = (1−δ)(1/2) (1−δ +ε)/2 = 1−δ 1−δ +ε . 57. Let H denote the event that a student does the homework, and let E denote the event that a student passes the exam. Then the problem statement tells us that P(E|H) = .8, P(E|H c ) = .1, and P(H) = .6. We need to compute P(E) and P(H|E). To begin, write P(E) = P(E|H)P(H)+P(E|H c )P(H c ) = (.8)(.6)+(.1)(1−.6) = .48+.04 = .52. Next, P(H|E) = P(E|H)P(H) P(E) = .48 .52 = 12 13 . 58. The problem statement tells us that P(AF|CF) = 1/3, P(AF|C c F ) = 1/10, and P(CF) = 1/4. We must compute P(CF|AF) = P(AF|CF)P(CF) P(AF) = (1/3)(1/4) P(AF) = 1/12 P(AF) . To compute the denominator, write P(AF) = P(AF|CF)P(CF)+P(AF|C c F )P(C c F ) = (1/3)(1/4)+(1/10)(1−1/4) = 1/12+3/40 = 19/120. It then follows that P(CF|AF) = 1 12 · 120 19 = 10 19 . 59. Let F denote the event that a patient receives a flu shot. Let S, M, and R denote the events that Sue, Minnie, or Robin sees the patient. The problem tells us that P(S) = .2, P(M) = .4, P(R) = .4, P(F|S) = .6, P(F|M) = .3, and P(F|R) = .1. We must compute P(S|F) = P(F|S)P(S) P(F) = (.6)(.2) P(F) = .12 P(F) . Chapter 1 Problem Solutions 15 Next, P(F) = P(F|S)P(S)+P(F|M)P(M)+P(F|R)P(R) = (.6)(.2)+(.3)(.4)+(.1)(.4) = .12+.12+.04 = 0.28. Thus, P(S|F) = 12 100 · 100 28 = 3 7 . 60. (a) Let Ω = {1,2,3,4,5} with P(A) := |A|/|Ω|. Without loss of generality, let 1 and 2 correspond to the two defective chips. Then D := {1,2} is the event that a defective chip is tested. Hence, P(D) = |D|/5 = 2/5. (b) Your friend’s information tells you that of the three chips you may test, one is defective and two are not. Hence, the conditional probability that the chip you test is defective is 1/3. (c) Yes, your intuition is correct. To prove this, we construct a sample space and probability measure and compute the desired conditional probability. Let Ω := {(i, j,k) : i j and k 6= i,k 6= j}, where i, j,k ∈ {1,2,3,4,5}. Here i and j are the chips taken by the friend, and k is the chip that you test. We again take 1 and 2 to be the defective chips. The 10 possibilities for i and j are 23 24 25 34 35 45 For each pair in the above table, there are three possible values of k: 125 124 123 Hence, there are 30 triples in Ω. For the probability measure we take P(A) := |A|/|Ω|. Now let Fi j denote the event that the friend takes chips i and j with i j. For example, if the friend takes chips 1 and 2, then from the second table, k has to be 3 or 4 or 5; i.e., F12 = {(1,2,3),(1,2,4),(1,2,5)}. The event that the friend takes two chips is then T := F12 ∪F13 ∪F14 ∪F15 ∪F23 ∪F24 ∪F25 ∪F34 ∪F35 ∪F45. Now the event that you test a defective chip is D := {(i, j,k) : k = 1 or 2 and i j with i, j 6= k}. 16 Chapter 1 Problem Solutions We can now compute P(D|T) = P(D∩T) P(T) . Since the Fi j that make up T are disjoint, |T| = 10·3 = 30 and P(T) = |T|/|Ω| = 1. We next observe that D∩T = ∅∪[D∩F13]∪[D∩F14]∪[D∩F15] ∪[D∩F23]∪[D∩F24]∪[D∩F25] ∪[D∩F34]∪[D∩F35]∪[D∩F45]. Of the above intersections, the first six intersections are singleton sets, and the last three are pairs. Hence, |D∩T| = 6·1+3·2 and so P(D∩T) = 12/30 = 2/5. We conclude that P(D|T) = P(D ∩ T)/P(T) = (2/5)/1 = 2/5, which is the answer in part (a). Remark. The model in part (c) can be used to solve part (b) by observing that the probability in part (b) is P(D|F12 ∪F13 ∪F14 ∪F15 ∪F23 ∪F24 ∪F25), which can be similarly evaluated. 61. (a) If two sets A and B are disjoint, then by definition, A∩B = ∅. (b) If two events A and B are independent, then by definition, P(A∩B) = P(A)P(B). (c) If two events A and B are disjoint, then P(A ∩ B) = P(∅) = 0. In order for them to be independent, we must have P(A)P(B) = 0; i.e., at least one of the two events must have zero probability. If two disjoint events both have positive probability, then they cannot be independent. 62. Let W denote the event that the decoder outputs the wrong message. Of course, W c is the event that the decoder outputs the correct message. We must find P(W) = 1 − P(W c ). Now, W c occurs if only the first bit is flipped, or only the second bit is flipped, or only the third bit is flipped, or if no bits are flipped. Denote these disjoint events by F100, F010, F001, and F000, respectively. Then P(W c ) = P(F100 ∪F010 ∪F001 ∪F000)