Answer Key for Homework 2 (Due Date: February 6, 2004 in Class)

Answer Key for Homework 2 (Due date: February 6, 2004 in class)

Use the description of example 2.3 and the given sample space to answer the following questions.

(a) Let A be the event that at least 5 pumps in use at each station. Write the outcomes in A. Is A an simple event? What is the probability of A?

A={(5,5),(5,6),(6,5),(6,6)} compound event

P(A)=4/49=0.0816

(b) Let B be the event that the difference of the pumps in use at two stations is 1 in absolute value. Write the outcomes in B. Is B a simple event? What is the probability of B?

B={(0,1),(1,0),(2,1),(1,2),(2,3)(3,2),(3,4),(4,3),(4,5),(5,4),(5,6),(6,5)} compound event

P(B)=12/49=0.2449

AÇB={(5,6),(6,5)} A and B are not mutually exclusive

P(AÇB)=2/49=0.0408

Exercise 2.6

(a) S={3,4,5,13,14,15,23,24,25,123,124,125,213,214,215}

(b) A={3,4,5}

(d) C={3,4,5,23,24,25}

Exercise 2.14

A: randomly selected household in a certain community does not exceed the lifeline usage in January

B: randomly selected household in a certain community does not exceed the lifeline usage in July

P(A)=0.8 P(B)=0.7 P(AÈB)=0.9

(a) P(AÇB)=P(A)+P(B)-P(AÈB)=0.8+0.7-0.9=0.6

(b) P(AÇB’)+P(A’ÇB)= (P(A)-P(AÇB))+ (P(B)-P(AÇB))= 0.8+0.7-2(0.6)=0.3

Exercise 2.19

A: Inspector A founds the defectives

B: Inspector B founds the defectives

AÈB: at least one inspector discovers the defectives

A' ÇB' =(AÈB)': neither inspector discovers the defectives

A' ÇB =B- A ÇB: Inspector B finds the defectives but not inspector A

P(A)=724/10000 P(B)=751/10000 P(AÈB)=1159/10000

then P(A ÇB)=(724+751-1159)/10000=316/10000=0.0316

(a) P(A' ÇB')=1-P(A ÈB)=8841/10000=0.8841

(b) P(A' ÇB )=P(B)-P(A ÇB)=435/10000=0.0435

Exercise 2.25

P(A)=0.7 P(AÈB)=0.85 P(AÈC)=0.9

P(B)=0.8 P(BÈC)=0.95

P(C)=0.75 P(AÈBÈC)=0.98

(a) P(AÈBÈC)=0.98

(b) P(A’ÇB’ÇC’)=1-0.98=0.02

(d) P(AÇB’ÇC’) +P(A’ÇBÇC’)+ P(A’ÇB’ÇC)=3P(AÈBÈC)- P(AÈB)-P(AÈC)- P(BÈC)=0.24

Exercise 2.30 (this question is not in the 5th edition. The library has the 6th edition on the reserve)

8 zinfandel, 10 merlot, 12 cabernet

(a) =8(7)(6)=336

(b) =593775

(d) P(two bottles of each variety)=(answer in (c))/(answer in (b))=83160/593775=0.14

(e) P(all 6 are the same variety)=P(all are zinfandel)+P(all are merlot)+P(all are cabernet)= =0.002

Exercise 2.32

5 types of receiver, 4 types of compact disc players, 3 types of speakers, 4 types of cassette deck

(a) 5(4)(3)(4)=240

(b) 1(1)(3)(4)=12

(d) at least one sony=all possible selections – none are sony =240-108=132

(e) P(at least one sony)=132/240=0.55

P(exactly one sony)=P(either the receiver is sony or the compact disc player or the cassette deck)=[1(3)(3)(3)+4(1)(3)(3)+4(3)(3)1]/240=99/240=0.4125

Exercise 2.50

(a) 0.05

(b) 0.12

(d) P(medium)=0.49 P(print)=0.25

(e) P(medium |short-sleeved plaid)=0.08/0.15=0.5333

(f) P(short-sleeved | medium plaid)=0.08/0.18=0.4444

P(long-sleeved | medium plaid)=0.10/0.18=0.5556

Exercise 2.59

P(A1:regular unleaded)=0.40 P(B:fill the thank |A1)=0.30

P(A2:extra unleaded)=0.35 P(B:fill the thank |A2)=0.60

P(A3:premium unleaded)=0.25 P(B:fill the thank |A3)=0.50

(a) P(A2ÇB)=P(B|A2)P(A2)=0.6(0.35)=0.21

(b) P(B)=P(A1ÇB)+ P(A2ÇB) +P(A3ÇB)=0.3(0.4)+0.6(0.35)+0.5(0.25)=0.455

P(A2 | B)= P(A2ÇB)/P(B)= 0.6(0.35)/0.455=0.4615

P(A3 | B)= P(A3ÇB)/P(B)= 0.5(0.25)/0.455=0.2747

Exercise 2.72

P(both are O)=0.44(0.44)=0.1936

P(both match)=P(both A)+P(both B)+P(both AB)+P(both O)

=0.42(0.42)+0.10(0.10)+0.04(0.04)+0.44(0.44)=0.3816

Exercise 2.78

P(system works)=1-(0.1)2(1-0.92)=0.9981

Exercise 2.80 (this question is not in the 5th edition. The library has the 6th edition on the reserve)

Let’s identified the outcomes as (face number on red , face number on green)

A={(3,1),(3,2),(3,3),(3,4),(3,5),(3,6)} P(A)=6/36=1/6

B={(1,4),(2,4),(3,4),(4,4),(5,4),(6,4)} P(B)=6/36=1/6

C={(1,6)(6,1),(2,5),(5,2),(3,4),(4,3)} P(C)=6/36=1/6

A∩B={(3,4)} P(A∩B)=1/36=P(A)P(B) then A and B are pairwise independent

A∩C={(3,4)} P(A∩C)=1/36=P(A)P(C) then A and C are pairwise independent

B∩C={(3,4)} P(B∩C)=1/36=P(B)P(C) then B and C are pairwise independent

A∩B∩C={(3,4)} P(A∩ B∩C)=1/36 ¹P(A)P(B)P(C)=1/216 then A, B and C are not mutually independent

Exercise 2.82 (Exercise 2.80 in the 5th edition. In the 6th edition, they have used 70% passing the inspection instead of the 60% passing the inspection in the 5th edition. You should use 70% to answer)

(a) P(A1∩A2∩A3)=P(A1)P(A2)P(A3)=0.73=0.343

(b) 1-P(all three pass)=1-0.73=0.657

(d) P(none of those passes)+P(exactly 1 passes)= 0.33+0.189=0.216

(e) P(A1∩A2∩A3 | A1UA2UA3)=P(A1∩A2∩A3)/P(A1UA2UA3)

= P(A1∩A2∩A3)/[1-P(none passes)] =0.343/[1-0.33] =0.3525