ДискретнаяМатематика / Student Solutions Manual / chapter 8

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1. (f) (A \ B \ C) [ (B \ C \ A) [ (C \ A \ B) = f2; 3; 6g

1. (g) (A \ B) [ (A \ C) [ (B \ C) = f2; 3; 6g

1. (h) (A [ B [ C) =

1. (i) (A \ B [ C) [ (B \ A [ C) [ (C \ A [ B) [ A [ B [ C = f1; 4; 5g

1. (j) A \ B \ C = f1; 2; 3; 4; 5; 6g

1. (k) A \ B [ C = f1g

3.What is the probability that a randomly chosen integer is a member of the set of numbers divisible by 3? Not divisible by 5? Divisible by either 4 or 6?

3. Let A denote the event that the integer is divisible by 3; D the event that the integer is divisible by 5; B the event that it is divisible by 4; and C the event that it is divisible by 6. We are to determine P (A \D \(B [C)).

Note that ADc(B [ C) = ADcB [ ADcC ADcBC = ABC ABCD, where we use multiplication notation to denote set intersection, and we denote X by Xc.

The probability we seek is P (ABC) P (ABCD) = (1=12) (1=60) = (1=15).

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5.Assign a probability density function to the possible outcomes of adding the sum of the top faces after the roll of a pair of fair dice. What is the probability that both top faces have the same value?

P (f(1; 1); (2; 2); (3; 3); (4; 4); (5; 5); (6; 6)g = 1=6

7.A chain of home entertainment stores sells three di erent brands of DVD players. Fifty percent of its sales are brand 1, 30% are brand 2, and 20% are brand 3. Each manufacturer o ers a one-year warranty on parts and labor. It is known that 25% of brand 1's DVD players require warranty repair work, whereas the corresponding percentages for brands 2 and 3 are 20% and 10%, respectively.

(a)What is the probability that a randomly selected purchaser has bought a brand 1 DVD player that will need repair under warranty?

(b)What is the probability that a randomly selected purchaser has a DVD player that will need repair while under warranty?

(c)If a customer returns to the store with a DVD player that needs warranty repair work, what is the probability that it is a a brand 1 DVD

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player? A brand 2 DVD player? A brand 3 DVD player?

7. Let Ai denote the event that brand i was purchased, for i = 1; 2; 3. Then P (A1) = :5; P (A2) = :3; and P (A3) = :2. Let B denote the event that the DVD needs repair, and let B denote the event that the DVD does not need repair. Then P (BjA1) = :25; P (BjA2) = :2; and P (BjA3) = :1.

7. (a) P (A1 \ B) = P (BjA1)P (A1) = :125

7. (b) P(B) = P(brand 1 repair) + P(brand 2 repair) + P(brand 3 repair) = P (A1 \ B) + P (A2 \ B) + P (A3 \ B) = :125 + :060 + :020 = :205

7. (c) P (A1jB) = P (A1 \B)=P (B) = :125=:205 = :61; P (A2jB) = P (A2 \ B)=P (B) = :060=:205 = 29; P (A3jB) = 1 P (A1 \ B) P (A2jB) = :1

2

P(B|A ) = .25 1

P(B | A ) = .75 1

P(A ) = .5 1

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9.Show that the three following events based on the toss of two fair coins are independent: E1 is the event \even on the rst die." E2 is the event \even on the second die," E3 is the event \even sum.".

9. P (E1 \ E2) = 1=4 = P (E1)P (E2); P (E1 \ E3) = 1=4 = P (E1)P (E3); P (E2 \ E3) = 1=4 = P (E2)P (E3)

11.Let = f!1; !2; : : : ; !7g be a sample space that represents parcels of a large lot divided into sublots for sale. The percentage of the total area for each lot and the price for each lot is as follows:

De ne the random variable X to be the price of the lot for !1; !2; : : : ; !7: Find the expected value of X.

11. 890

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13.A computer store has purchased three computers of a certain type at $500 apiece. The computers then are sold for $1000 each. The manufacturer has agreed to repurchase any computers that remain unsold after one month for $200 each. Let X be the random variable that denotes the number

of computers sold. Suppose the probabilities for selling i computers for i = 0; 1; 2; 3 are p(0) = 0:1; p(1) = 0:2; p(2) = 0:3; and p(3) = 0:4. Let h(x) denote the pro t from selling X units. Find the expected value of h as well as the standard deviation and the variance.

13. E(h(x)) = h(0)p(0) + h(1)p(1) + h(2)p(2) + h(3)p(3) =

900(:1) 100(:2) + 700(:3) + 1500(:4) = 700

so = 2 and E(X2) = (0)2(:1) + (1)2(:2) + (2)2(:3) + (3)2(:4) = 5 so V (X) = 5 4 = 1, and h(x) = 800X 900 has variance (800)2 so = 800.

Using Discrete Mathematics in Computer Science

1.Consider sending a job through the series-parallel system of processors as shown:

p

4

p 2

The system consists of four stages connected in sequence. Each stage i consists of identical processors connected in parallel. Each processor at stage i has probability pi of being up. A job makes it through the system if and only if at least one processor at each stage is up. We want to determine the probability that a job makes it through the system.

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(a)Model this situation with a cross product sample space = 12 3 4: Describe in detail the component sample spaces i; their probability density functions fi; and the cross product probability density f: (Hint: It may be useful to assign labels to the individual processors.)

(b)Let Ai i be the event \at least one of the processors at stage i is up." List the elements of i belonging to Ai: What is P (Ai)?

(c)Let Bi i be the event \none of the processors at stage i is up." List the elements of Bi: What is P (Bi)?

(d)Describe the event Ai i that corresponds to Ai i: What is P (Ai )?

(e)Describe the event Bi that corresponds to Bi : What is P (Bi )?

(f)Describe in detail the event E that at least one processor at each stage is up, and write an expression for its probability. (You may use complements of events.)

1. (a) Label the processors s1; s2; ; s7 by beginning at the far left of the diagram and moving top to bottom. So for example, processors s2 and s3 both have probability p2 of being up, while processor s4 has probability p3 of being up. De ne the sample spaces i as follows:

1 = fs1; s1g

2 = fs2s3; s2s3; s2s3; s2s3g

3 = fs4; s4g

4 = fs5s6s7; s5s6s7; s5s6s7; s5s6s7; s5s6s7; s5s6s7; s5s6s7; s5s6s7g

where si denotes that processor i is up while si denotes that processor i is

1. (b) The event Ai i that at least one processor at stage i is up is given by:

A1 = fs1g

A2 = fs2s3; s2s3; s2s3g

A3 = fs4g;

A4 = fs5s6s7; s5s6s7; s5s6s7; s5s6s7; s5s6s7; s5s6s7; s5s6s7g

The probability of each event is just the sum of the probabilities of the individual outcomes in the events. Thus, P (A1) = p1, P (A2) = 2p2 p22, P (A3) = p3, and P (A4) = 3p4 3p24 + p34.

1. (c) The event Bi i that none of the processors at stage i are up is given by:

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1 = fs1; s1g

2 = fs2s3; s2s3; s2s3; s2s3g3 = fs4s5; s4s5; s4s5; s4s5g4 = fs5s6; s5s6; s5s6; s5s6g

where si denotes that processor i is up while si denotes that processor i is down. We now de ne the probability densities fi for each i. f1(s1) =

p1, f1(s1) = 1 p1. f2(s2s3) = p22, f2(s2s3) = f2(s2s3) = p2(1 p2), f2(s2s3) = (1 p2)2. f3(s4s5) = p3p4, f3(s4s5) = p3(1 p4), f3(s4s5) =

(1 p3)p4, f3(s4s5) = (1 p3)(1 p4), f4(s6s7) = p24, f4(s6s7) = f4(s6s7) = p4(1 p4), f4(s6s7) = (1 p4)2.

3. (b) The event Ai i that at least one processor at stage i is up is given by:

A1 = fs1g

A2 = fs2s3; s2s3; s2s3g

A3 = fs4s5; s4s5; s4s5g

A4 = fs6s7; s6s7; s6s7g

The probability of each event is just the sum of the probabilities of the individual outcomes in the events. Thus, P (A1) = p1, P (A2) = 2p2 p22, P (A3) = p3p4 + p3(1 p4) + (1 p3)p4, and P (A4) = 2p4 p24.

3. (c) The event Bi i that none of the processors at stage i are up is given by:

B1 = fs1g

B2 = fs2s3g

B3 = fs4s5g

B4 = fs6s7g

P (B1) = 1 p1, P (B2) = (1 p2)2, P (B3) = (1 p3)(1 p4), P (B4) = (1 p4)2

3. (d) The event Ai is the cross product of j for all j 6= i and Ai.

A1 = A1 2 3 4

A2 = 1 A2 3 4

A3 = 1 2 A3 4

A4 = 1 2 3 A4

P (Ai ) = P (Ai ) for 1 i 4

3. (e) The event Bi is the cross product of j for all j 6= i and Bi.

B1 = B1 2 3 4

B2 = 1 B2 3 4

B3 = 1 2 B3 4

B4 = 1 2 3 B4

P (Bi ) = P (Bi) for 1 i 4

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3. (f) Note that E = A1 A2 A3 A4 = B1 B2 B3 B4. Thus,

P(E) = P (A1)P (A2)P (A3)P (A4)

=1 P (B1)P (B2)P (B3)P (B4)

=(1 p1)(1 p2)2(1 p3)(1 p4)(1 p4)2

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