MAT 540 Statistical Concepts for Research

81

Chapter 4

PROBABILITY 4.1 (a) (iv), (v) (e) (v) (b) (ii), (v) (f) (i) (c) (vi) (g) (iii), (v) (d) (vi) 4.2 (a) (i) (e) (ii) (b) (v) (f) (iv) (c) (iii) (d) (v) 4.3 (a) (ii) (b) (iii) (c) (i) 4.4 (a) {0, 1, 3}S =

Note: The number of correct identifications cannot be 2 because whenever any two are correctly identified, the third one cannot be wrong.

(b) {0, 1, 2,...}S =

(c) Denoting t = duration of satisfactory work, say in days, { : 0}S t t= ≥ .

(a) and (b) are discrete, (c) is continuous. 4.5 (a) {0, 1}

(b) {0, 1,..., 344}

(c) { : 90 425.4}t t< <

82 CHAPTER 4. PROBABILITY

4.6 The President’s assignment is admissible. The Vice President’s assignment is not

permissible. Although the total probability is 1, the listed outcomes are not all distinct. In particular, ‘win Project 1’ includes the outcome ‘win both’ so P(win Project 1) must not be less than P(win both).

4.7 (a) Let us identify Bob, John, Linda, and Sue by their initials B, J, L, and S,

respectively. We make a tree diagram: S {BJ, BL, BS, JB, JL, JS, LB, LJ, LS, SB, SJ, SL}=

(b) A {LB, LJ, LS}= , B {JL, LJ, JS, SJ, LS, SL}=

4.8 (a) The tree diagram is (b) A {HH, HTH, THH, TTHH}= , B {HTT, THT, TTHH}=

83

4.9 1 2 3( ) ( ) ( ) 0.3 0.4 0.2 0.9.P e P e P e+ + = + + = Since ( ) 1P S = , we must have

4( ) 1 0.9 0.1.P e = − =

4.10 Since ( ) 1P S = , and 1 2,e e , and 3e have the same probabilities, each has the

probability 1/3. 4.11 (a) yes (b) no, because the sum of probabilities is less than 1 (c) yes

4.12 (a) Denoting Monday by 1e and so on, we are given that the probabilities of

1 2 3 4 5, , , , and e e e e e are equally likely and that the probabilities of 6e and 7e

are equal but each is twice as likely as any of the first five outcomes. Since 1 1 1 1 1 2 2 9+ + + + + + = , we have

1 2 3 4 5

1( ) ( ) ( ) ( ) ( )

9P e P e P e P e P e= = = = = and 6 7

2( ) ( )

9P e P e= =

(b) 1 2 3 4 5

5( ) ( ) ( ) ( ) ( ) ( )

9P A P e P e P e P e P e= + + + + =

4.13 Denote May by 1e and so on. Because 1 3 6 10 20+ + + = , we have

1 2 3 4

1 3 6 10( ) , ( ) , ( ) , ( )

20 20 20 20P e P e P e P e= = = =

so that 1 2

4( ) ( ) ( ) 0.2.

20P A P e P e= + = =

4.14 (a) P(good weather) 3

0.754

= = (b) P(delay) 7

0.710

= =

4.15 The relative frequencies are based on a very large number of cases and will

therefore be very good approximations to the probabilities. Since

P(weekday) + P(weekend) 1= and P(weekend) is approximately 0.257, the probability of a weekday birth is

approximately 1 0.257 0.743− = . 4.16 There were a large number 41,131, of permits issued, so the relative frequency is a

good approximation to the probability. We estimate that the probability is

8,8450.215

41,131=

that a turkey will be harvested with a specific single permit.


4.17 (a) The tree diagram is

1 2 8{ , , ..., }S e e e=

(b) Assuming the coins are all fair, all the elementary outcomes are equally likely.

1 2 8

1( ) ( ) ( )

8P e P e P e= = ⋅⋅⋅ = =

(c) [Exactly one head] 4 6 7{ , , }e e e= , its probability is 3

8.

4.18 There are 4 letters of which 2 are vowels. Since all letters are equally likely to be

selected, the probability of selecting a vowel is 2

0.5.4

=

4.19 (a) Let 1 2,e e , and 3e denote the outcomes of getting a ticket numbered 1, 2, and 3,

respectively. Then 1 2 3{ , , }S e e e= . Since all 8 tickets are equally likely to be

drawn, and there are 2 tickets with number 1, we have 1

2( ) .

8P e = Likewise,

2

3( )

8P e = and 3

3( )

8P e = .

(b) [Odd-numbered ticket drawn] 1 3{ , }e e= , so the probability is

1 3

2 3 5( ) ( ) 0.625

8 8 8P e P e+ = + = = .

85 4.20 Referring to the tree diagram of Exercise 4.17 (a), suppose the first toss corresponds

to yours and the second and third to your friends’. S has 8 elementary outcomes which are all equally likely.

(a) The event ‘you will have to pay’ consists of all those elementary outcomes whose

first letter is different from the other two. Since there are two such elementary outcomes HTT 4( )e and THH 5( )e the required probability is

4 5

2 1({ , })

8 4P e e = =

.

(b) [Three tosses yield the same result] 1 8{HHH, TTT} { , }e e= = , so that

P(all three will share) 2 1

8 4= = .

(Note: The answer to Part (a) remains the same no matter which specific toss is identified as yours.)

4.21 (a) Each elementary outcome is a pair of numbers, the first corresponds to the

white die and the second to the colored die. {(1, 5), (2, 4), (3, 3), (4, 2), (5, 1)}A =

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

{(2, 6), (4, 6), (6, 6), (1, 5), (3, 5), (5, 5),

(2, 4), (4, 4), (6, 4), (1, 3), (3, 3), (5, 3),

(2, 2), (4, 2), (6, 2), (1, 1), (3, 1), (5, 1)}

C =

{(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)}D = .

(b) Probability 1

36 for each elementary outcome.

(c) 5

( )36

P A = , 6 1

( )36 6

P B = = , 18 1

( )36 2

P C = = , 6 1

( )36 6

P D = =

4.22 All 34 slots are equally likely so each has probability 1/34. (a) There are 18 slots which are not black. The probability that a black slot will

not occur is then 18

34.

(b) The event ‘win’ corresponds to the occurrence of a black slot and its probability

is 16

34.


4.23 (a) The tree diagram is

So, { }S= TT, TF, TI, FT, FF, FI, IT, IF, II .

(b) Because the student selects the answers at random, the 9 elementary outcomes

in S are all equally likely, each has a probability 1

9. Let us suppose that the

correct answers are T for Q1 and T for Q2. Then, the event “one correct answer” has the composition {TF, TI, FT, IT} , so that

P(one correct answer) 4

9= .

Note: Whatever be the correct answers for Q1 and Q2, there will be four cases in which one marked answer will match and one will not match.

4.24 (a) The 50 states are equally likely to be selected. The number of states with a birth

rate under 16 (per thousand) is 7 23 16 46+ + = , so that

P(under 16) 46

0.9250

= = .

(b) The number of states with a birth rate under 18 but not under 14 is 16 3 19+ = , so the required probability is

P(under 18 but not under 14) 19

0.3850

= =

(c) The number of states with a birth rate 16 or over is 3 1 4+ = , so the required probability is

P(16 or over) 4

0.0850

= = .

4.25 (a) The 15 persons are equally likely to be selected. Among them there is only

one of blood group AB, so that 1

[AB]15

P = .

87 (b) The number of persons of blood group either A or B is 5 6 11+ = , so that the

required probability is 11

15.

(c) P[not O] 5 6 1 12

15 15

+ += = .

4.26 Let N be a no response. {A, NA, NNA, NNNA, NNNN}S = .

4.27 {N, YN, YYN, YYYN, YYYYN, YYYYY}S =

4.28 (a) For simplicity, we consider all 12 months to be of equal size (duration). Then

the assumption of even distribution of births makes the 12 months equally likely. The probability of a birth in November or December is 2 /12 0.167≈ . For a more precise calculation, consider 365 days of the year equally likely. November and December together consist of 61 days, so the required

probability is 61

0.16712365

= .

(b) The relative frequencies are given below and the probabilities, determined by the uniform probability model for days, are shown in parentheses.

Jan. 0.080 (0.085) July 0.086 (0.085) Feb. 0.075 (0.077) Aug. 0.089 (0.085) Mar. 0.084 (0.085) Sept. 0.088 (0.082) Apr. 0.080 (0.082) Oct. 0.083 (0.085) May 0.084 (0.085) Nov. 0.081 (0.082) June 0.085 (0.082) Dec. 0.084 (0.085)

1.000 August and September appear to be somewhat higher and January lower than

what the uniform probability model suggests. 4.29 (a) Letting c, b, and v denote “compliance”, “borderline case”, and “violation”,

respectively,

1 2 9 1 2 3 1 2{ , ,..., , , , , , }S c c c b b b v v= .

(b) The 14 elementary outcomes are equally likely, and two of them, namely 1v and

2v , constitute the event that a violator is detected. The probability is

2 /14 0.143= .


4.30 (a) The tree diagram is { , , , }S aa aA Aa AA= .

(b) Under random selection, the four elementary outcomes of S are equally likely.

The probability is 1

4 for each.

(c) 1

[ ] ({ })4

P Short P aa= = .

4.31 (a) The successive days cannot be considered as independent trials. The rate on

one day is the same as, or is very close to, the rate on the next day. The results for successive days are not independent. There may also be a trend in rates over the year.

(b) Cars brought in with other problems are more likely to have an emission

problem. For instance, there would be too many old cars in this sample. 4.32 (a) We identify the persons by F, S and T. To list the sample space, we find all

possible ways of assigning the gift certificates. We take the correct assignment as $100 for F, $50 for S and $25for T.

$100 $50 $25 F S T (# of correct gifts) F S T 3 (e1) F T S 1 (e2) S F T 1 (e3) S T F 0 (e4) T F S 0 (e5) T S F 1 (e6)

1 2 3 4 5 6{ , , , , , }S e e e e e e=

(b) 2 3 6{ , , }A e e e= , 4 5{ , }B e e=

89 4.33 (a) Since the gift certificates are assigned at random, all 6 elementary outcomes are

equally likely, so that each has probability 1/6.

(b) 3 1

( )6 2

P A = = , 2 1

( )6 3

P B = = .

4.34 January, February, and April have the lowest estimated probability

0.080 0.075 0.080 0.235+ + = 4.35 (a) The Venn diagram is

(b) (i) 6 7{ , }AB e e=

(ii) 2 3 4 5{ , , , }B e e e e=

(iii) 4 5{ , }AB e e=

(iv) 1 4 5 6 7{ , , , , }A B e e e e e∪ =

4.36 (a) The Venn diagram is

(b) (i) 1 4 5 8{ , , , }B e e e e= , ( ) 0.08 0.12 0.14 0.14 0.48P B = + + + =

(ii) 6{ }BC e= , ( ) 0.14P BC =


(iii) 1 2 5 6 7 8{ , , , , , }A C e e e e e e∪ = ,

( ) 0.08 0.12 0.14 0.14 0.14 0.14 0.6P A C∪ = + + + + + =

(iv) 3 4 6 8{ , , , }A C e e e e∪ = ,

( ) 0.12 0.12 0.14 0.14 0.52P A C∪ = + + + =

4.37 (a) 1 2 3 4 5 7{ , , , , , }C e e e e e e= , ( ) 0.08 0.12 0.12 0.12 0.14 0.14 0.72P C = + + + + + =

(b) 2 6 7{ , , }AB e e e= , ( ) 0.12 0.14 0.14 0.40P AB = + + =

(c) 1 5{ , }AB e e= , ( ) 0.08 0.14 0.22P AB = + =

(d) 3 4{ , }AC e e= , ( ) 0.12 0.12 0.24P AC = + =

4.38 (a) DM

(b) DMG (Note: ‘Fail to get offer’ is the complement of ‘offer’)

(c) GM

4.39 (a) Denote by 1 2 3, , ,e e e and 4e the elementary outcomes that the person hired is

candidate number 1, 2, 3, and 4, respectively. The Venn diagram is given in Figure 4.2.

(b) 1 3 4{ , , }A B e e e∪ = and 3{ }AB e=

4.40 (a) A female candidate is hired, 2 3{ , }C e e= .

(b) The candidate hired is male and is not a social science major, 4{ }CA e= .

(c) The candidate hired is either a social science major or a female,

1 2 3{ , , }A C e e e∪ = .

91 4.41 (a) ( ) 0.04 0.2 0.06 0.30P A = + + = and ( ) 0.04 0.2 0.06 0.06 0.36P B = + + + =

Also, since 5 8{ , }AB e e= , ( ) 0.2 0.06 0.26P AB = + = .

(b) ( ) ( ) ( ) ( ) 0.30 0.36 0.26 0.40P A B P A P B P AB∪ = + − = + − =

(c) Since 1 5 8 2 9{ , , , , }A B e e e e e∪ = ,

( ) 0.04 0.2 0.06 0.04 0.06 0.40P A B∪ = + + + + =

(d) ( ) 1 ( ) 1 0.36 0.64P B P B= − = − = . Alternatively, since 1 3 4 6 7{ , , , , }B e e e e e= ,

( ) 0.04 0.2 0.2 0.1 0.1 0.64P B = + + + + = .

4.42 (a) 4 5 6 7( ) ( ) ( ) ( ) ( ) 0.06 0.06 0.2 0.23 0.55P A P e P e P e P e= + + + = + + + =

1 6 7( ) ( ) ( ) ( ) 0.15 0.2 0.23 0.58P B P e P e P e= + + = + + =

6 7( ) ( ) ( ) 0.2 0.23 0.43P AB P e P e= + = + =

(b) ( ) 1 ( ) 1 0.55 0.45P A P A= − = − =

( ) ( ) ( ) ( ) 0.55 0.58 0.43 0.7P A B P A P B P AB∪ = + − = + − =

(c) 1 2 3( ) ( ) ( ) ( ) 0.15 0.15 0.15 0.45P A P e P e P e= + + = + + =

1 4 5 6 7( ) ( ) ( ) ( ) ( ) ( )

0.15 0.06 0.06 0.2 0.23 0.7

P A B P e P e P e P e P e∪ = + + + +

= + + + + =

4.43 (a) The specified probabilities are entered in the table and underlined. The other

entries are obtained in part (b).

B B

A 0.12 0.13 0.25

A 0.38 0.37 0.75

0.50 0.50 1.00

(b) Since A AB AB= ∪ , a union of mutually exclusive events, we have

( ) ( ) ( )P A P AB P AB= + . Hence, for the given information,

0.25 0.12 ( )P AB= + . Thus, solving for ( )P AB yields:

( ) 0.25 0.12 0.13P AB = − = .

Similarly,

( ) 0.75 0.37 0.38P AB = − = , ( ) 1 (0.12 0.13 0.38) 0.37P AB = − + + =

4.44 (a) AB , ( ) 0.38P AB =

(b) AB , ( ) 0.37P AB =

(c) A B∪ ,


( ) ( ) ( ) ( ) 0.25 0.50 0.13 0.62P A B P A P B P AB∪ = + − = + − =

4.45 (a) The completed probability table is given below

B B

A 0.14 0.23 0.37

A 0.36 0.27 0.63

0.50 0.50 1.00

(b) ( ) 0.37 0.14 0.23P AB = − =

(c) ( ) ( ) ( ) ( ) 0.37 0.50 0.14 0.73P A B P A P B P AB∪ = + − = + − =

(d) ( ) ( ) ( )P AB AB P AB P AB∪ = + (union of incompatible events)

0.23 0.36 0.59= + = 4.46 No. If they were mutually exclusive we would have ( ) 0P AB = and

( )P A B∪ would then be ( ) ( ) 0.2 0.9 1.1P A P B+ = + = which cannot be a

probability.

4.47 (a) ( ) 0.4 0.25 0.65P A = + =

(b) ( ) 0.15P AB =

(c) ( ) ( ) 0.15 0.4 0.55P AB P AB+ = + =

4.48 The classification of the 32 students is shown in the following table:

Senior Graduate student Total

Male 16 4 20 Female 4 8 12

20 12 32 (a) Of the 32 students, 20 are seniors. Therefore, the probability that a randomly

selected student is a senior is

20(senior) 0.625

32P = =

(b) 4

(male graduate student) 0.12532

P = =

4.49 Denoting ‘violation’ by V and ‘compliance’ by C, the classification of the 18

restaurants is shown in the following table: Safety V C Total

Sanitary V 4 3 7 C 4 7 11

8 10 18

7

( ) 0.38918

P CC = =

93 4.50 The specified probabilities are

( ) 0.3 (i)P A =

( ) 0.6 (ii)P B =

( ) 0.5 (iii)P A B∪ =

(a) ( ) 1 ( ) 1 0.3 0.7P A P A= − = − =

(b) To find ( )P AB , we use (iii) and the addition law

0.5 ( ) ( ) ( ) ( )P A B P A P B P AB= ∪ = + −

0.3 (1 0.6) ( )P AB= + − − since ( ) 1 0.6P B = − by (ii)

Thus, we get ( ) 0.3 0.4 0.5 0.2P AB = + − = .

(c) To find ( )P AB , we use the fact that A AB AB= ∪ . So,

( ) ( ) ( )P A P AB P AB= + or 0.3 0.2 ( )P AB= + (from part (b))

Therefore, ( ) 0.3 0.2 0.1P AB = − = .

4.51 (a) ( ) 0.08 0.02 0.20 0.10 0.40P A = + + + =

( ) 0.15 0.10 0.08 0.02 0.35P B = + + + =

( ) 0.15 0.08 0.23P BC = + =

( ) 0.08P ABC =

(b) (i) Light case and above 40.

( ) 0.15 0.20 0.35P AB = + =

(ii) Either a light case or the parents are not diabetic or both.

( ) 0.15 0.10 0.15 0.20 0.02 0.10 0.72P A C∪ = + + + + + =

(iii) A light case, age is below 40 and parents are not diabetic.

( ) 0.10P ABC = .

4.52 Let us denote the chemicals Arsenic, Barium, and Mercury by the letters A, B, and

M respectively, and indicate the concentrations by the subscripts ‘H’ for high and ‘L’ for low. For instance, a high concentration of Barium will be denoted by BH.


(a) Of the 58 landfills, the number with BH is 1 4 3 8 16+ + + = . Therefore,

16( ) 0.276

58HP B = = .

(b) The number of MHALBL landfills is 10, so 10

( ) 0.172.58

H L LP M A B = =

(c) There are three possibilities for landfills with two H’s one L. The number of

AHBLMH landfills is 5, the number of ALBHMH is 4, and the number of AHBHML is

3, so the total is 12. Therefore, P(two H’s and one L)12

0.20758

= = .

(d) There are three possibilities for landfills with one H and two L’s. The number of AHBLML landfills is 9, the number of ALBHML is 8, and the number of ALBLMH

is 10, so the total is 27. Therefore, P(one H and two L’s)27

0.46658

= =

4.53 (a) With the stated numbers identifying the gift boxes, the list is:

(1,1), (1,2), (1,3), (1,4), (1,5)

(2,1), (2,2), (2,3), (2,4), (2,5)

(3,1), (3,2), (3,3), (3,4), (3,5)

(4,1), (4,2), (4,3), (4,4), (4,5)

(5,1), (5,2), (5,3), (5,4), (5,5)

The 25 elementary outcomes are equally likely, so each has the probability 1

25.

(b) {(2,1), (2,2), (2,3), (2,4), (2,5), (3,1), (3,2), (3,3), (3, 4), (3,5),

(1, 2), (4, 2), (5, 2), (1,3), (4,3), (5,3)},

16So, ( ) .

25

A

P A

=

=

{(3,1), (3, 2), (3,3), (3, 4), (3,5), (4,1), (4, 2), (4,3), (4,4), (4,5),

(5,1), (5, 2), (5,3), (5,4), (5,5), (1,3), (1, 4), (1,5), (2,3), (2,4), (2,5)}

21So, ( ) .

25

B

P B

=

=

{(2,3), (2,4), (2,5), (1,3), (3,1), (3,2), (3,3), (3, 4), (3,5),

(4,2), (4,3), (5,2), (5,3)},

13So, ( ) .

25

AB

P AB

=

=

95

4.54 (a) C A B= ∪ 16 21 13 24

( ) ( ) ( ) ( )25 25 25 25

P C P A P B P AB= + − = + − =

(b) C has all the elementary outcomes of S except the single outcome (1,1). Hence,

24( )

25P C = .

4.55 It is reasonable to assume that a person with a college degree is more likely to

command a higher salary than a person without a college degree, so we would expect ( | ) ( )P A B P A> . If A and B are independent, then ( | ) ( )P A B P A= so by

the previous reasoning, A and B are not independent. 4.56 It is reasonable to expect ( | ) ( )P A B P A> , since the set of luxury car owners would

include a large percentage of the set of lawyers. By this reasoning, A and B are not independent for otherwise we would have ( | ) ( )P A B P A= .

4.57 (a) ( ) 0.12

( | ) 0.48( ) 0.25

P ABP B A

P A= = =

(b) From ( ) 0.25P A = and ( ) 0.12P AB = and the fact that A AB AB= ∪ (union of

mutually exclusive events), we calculate ( ) 0.25 0.12 0.13P AB = − = .

Therefore, since ( )

( | )( )

P ABP B A

P A= , we have

0.13( | ) 0.52

0.25P B A = = .

Alternatively, note that B and B are complementary events so that

( | ) 1 ( | ) 1 0.48 0.52P B A P B A= − = − =

(c) Observe that ( ) 1 ( ) 1 0.25 0.75P A P A= − = − = and

( ) ( ) ( ) 0.50 0.12 0.38P AB P B P AB= − = − = (since B AB AB= ∪ as in (b)).

Therefore, since ( )

( | )( )

P ABP B A

P A= , we have

0.38( | ) 0.51

0.75P B A = = .

4.58 (a) ( ) 0.14

( | ) 0.28( ) 0.50

P ABP A B

P B= = =

(b) ( ) 0.36

( | ) 0.57( ) 0.63

P ABP B A

P A= = =

4.59 Observe that ( ) 0.001

( | ) 0.0099( ) 0.101

P ABP B A

P A= = = and ( ) 0.05 0.001 0.051P B = + = .

Since ( ) ( | )P B P B A≠ , the events A and B are not independent.


4.60 (a) Because ( ) ( ) ( )P AB P AB P B+ = , we have ( ) 0.32 0.20 0.12P AB = − =

Likewise, ( ) ( ) ( ) 0.55 0.12 0.43P AB P A P AB= − = − =

Also, ( ) 1 ( ) 1 0.55 0.45P A P A= − = − =

( ) ( ) ( ) 0.45 0.20 0.25P AB P A P AB= − = − =

B B

A 0.12 0.43 0.55

A 0.20 0.25 0.45

0.32 0.68 1.00

(b) ( ) 0.43

( | ) 0.632( ) 0.68

P ABP A B

P B= = =

4.61 (a) ( ) 1 ( ) 1 0.4 0.6P A P A= − = − =

(b) ( ) ( ) ( | ) 0.25 0.7 0.175P AB P B P A B= = × =

(c) ( ) ( ) ( ) ( ) 0.4 0.25 0.175 0.475P A B P A P B P AB∪ = + − = + − =

4.62 (a) ( ) 0.12 0.08 0.05 0.25P A = + + =

( ) 0.08 0.25 0.33P M = + =

( ) 0.08P AM =

( ) 0.25 0.33 0.08 0.50P A M∪ = + − =

(b) ( ) ( ) 0.25 0.33 0.0825 ( )P A P M P AM= × = ≠ . Not independent.

4.63

2 Green 3 Red → 2, without replacement

5 (a) We denote G for green, R for red and attach subscripts to identify the order of

the draws. Since the event A, a green ball appears in the first draw, has nothing

to do with the second draw, we identify 1A G= so

1

2( ) ( ) 0.4

5P A P G= = =

The event 2B G= is the union of 1 2G G and 1 2R G .

1 2 1 2 1

2 1 2( ) ( ) ( | )

5 4 20P G G P G P G G= = × =

1 2 1 2 1

3 2 6( ) ( ) ( | )

5 4 20P R G P R P G R= = × =

Hence, 2

2 6 8( ) ( ) 0.4

20 20 20P B P G= = + = =

97

(b) 1 2

2( ) ( ) 0.1

20P AB P G G= = = . On the other hand, ( ) ( ) 0.4 0.4 0.16P A P B = × = ,

and this is different from P(AB). Therefore, A and B are not independent. 4.64 Under the scheme of sampling with replacement, the contents of the urn remain

unchanged at each draw.

(a) 1

2( ) ( ) 0.4

5P A P G= = =

1 2 1 2 1

2 2 4( ) ( ) ( | )

5 5 25P G G P G P G G= = × =

1 2 1 2 1

3 2 6( ) ( ) ( | )

5 5 25P R G P R P G R= = × =

Adding, we get 2

4 6 10( ) ( ) 0.4

25 25 25P B P G= = + = =

(b) 1 2

4( ) ( ) 0.16

25P AB P G G= = =

( ) ( ) 0.4 0.4 0.16 ( )P A P B P AB= × = = .

Since these probabilities are equal, the events are independent. 4.65 We use the symbols M for male, F for female, U for unemployed and E for

employed. (a) ( ) 0.6P M =

( | ) 0.051P U M =

( | ) 0.043P U F =

(b) ( ) 0.051 0.6 0.0306P UM = × = and ( ) 0.043 (1 0.6) 0.0172P UF = × − =

Adding these we obtain ( ) 0.0306 0.0172 0.0478P U = + = , so the overall rate of

unemployment is 4.8%. (c) To find ( | )P F U , we use the results ( ) 0.0478P U = and ( ) 0.0172P UF = and

obtain ( ) 0.0172

( | ) 0.360( ) 0.0478

P FUP F U

P U= = = .

4.66 Let A = running out of gas and B = electronic starting system doesn’t work. We are given that ( ) 0.03, ( ) 0.01P A P B= = .

(a) By independence, ( ) ( ) ( ) (0.97)(0.99) 0.9603P A B P A P B= = = .

(b) If the two problems are related in the sense that running out of gas would increase the probability of the starting system failing, then independence would be a poor assumption.


4.67 (a) If independent, ( ) ( ) ( ) 0.6 0.22 0.132P AB P A P B= = × = so

( ) ( ) ( ) ( ) 0.6 0.22 0.132 0.688P A B P A P B P AB∪ = + − = + − =

(b) If mutually exclusive, ( ) 0P AB = so

( ) ( ) ( ) 0.6 0.22 0.82P A B P A P B∪ = + = + =

(c) If A and B are mutually exclusive, we have AB A= so

( ) 0.6( | ) 0.769

( ) 0.78

P ABP A B

P B= = = .

4.68 The classification of the 18 fast food restaurants is shown in the table

Safety V C Total

Sanitary V 4 3 7 C 4 7 11 8 10 18

(a) Denote the events A = violation of sanitary, and B = violation of safety

standards.

To find ( | )P A B we can use the definition of conditional probability

( )( | )

( )

P ABP A B

P B=

Calculating

( )P B = P[compliance of safety]10

18=

( )P AB = P[violation of sanitary and compliance of safety]3

18=

we get 3 /18 3

( | ) 0.310 /18 10

P A B = = = .

Alternatively, we note that the selected restaurant must be one of the 10 that are

in compliance of safety standards. Of these, there are 3 that violate sanitary

standards so the probability is 3

0.310

= .

(b) The number of restaurants that have at least one violation is 4 4 3 11+ + = . Of

these 11 restaurants, 3 are in compliance of safety standards so the required

probability is 3

0.27311

= .

99 4.69 We use the symbol D for defective and G for good, and attach subscripts to identify

the order of selection. (a) Here the second selection is irrelevant. If one air conditioner is selected at

random, the probability of its being defective is 1

3( ) 0.25

12P D = = .

(b) 1 2 1 2 1

3 9 27( ) ( ) ( | ) 0.205

12 11 132P D G P D P G D= = × = = .

(c) 1 2 1 2 1

3 2 6( ) ( ) ( | ) 0.045

12 11 132P D D P D P D D= = × = =

(d) The event 2D is the union of two incompatible events:

2 1 2 1 2D D D G D= ∪ so 2 1 2 1 2( ) ( ) ( )P D P D D P G D= + .

We have already calculated 1 2( )P D D in part (c). In the same way

1 2 1 2 1

9 3 27( ) ( ) ( | )

12 11 132P G D P G P D G= = × = . Thus,

2

6 27 33( ) 0.25

132 132 132P D = + = = .

(e) [exactly one defective] 1 2 1 2D G G D= ∪ .

We have already found that 1 2 1 2

27( ) ( )

132P D G P G D= = .

Hence, P(exactly one defective)27 27 54

0.409132 132 132

= + = = .

4.70 (a) 1 2 3 1 2 1 3 1 2( ) ( ) ( | ) ( | )

9 8 7 5040.382

12 11 10 1320

P G G G P G P G G P G G G=

= × × = =

(b) 1 2 3

9 8 3 216( ) 0.164

12 11 10 1320P G G D = × × = =

(c) The event [2 good and 1 defective] is the union of the three incompatible events

1 2 3G G D 1 2 3G D G and 1 2 3D G G .

In part (b) we have calculated 1 2 3

216( )

1320P G G D = . This is also the probability

of each of the other two events. For instance,

1 2 3 1 2 1 3 1 2

9 3 8 216( ) ( ) ( | ) ( | )

12 11 10 1320P G D G P G P D D P G G D= = × × = .

Hence, the required probability is 216

3 0.4911320

× = .


4.71 The classification of the 20 rats is shown in the following table:

Infected Not infected (I) (N) Total Male (M) 7 5 12 Female (F) 2 6 8

9 11 20

(a) There are 9 infected rats of which 2 are females. Therefore, 2

( | )9

P F I = .

Alternatively, we can use the definition of conditional probability

( ) 2 / 20 2( | )

( ) 9 / 20 9

P FIP F I

P I= = = .

(b) There are 12 males of which 7 are infected, so 7

( | )12

P I M = .

(c) 9 12 7

( ) , ( ) , ( ) 0.3520 20 20

P I P M P IM= = = =

9 12

( ) ( ) 0.2720 20

P I P M = × =

Since ( ) ( ) ( )P IM P I P M≠ , the events are not independent.

4.72 Let S = satisfactory visit and N = unsatisfactory visit.

We are given that ( ) 0.8P S = and ( ) 0.2P N = .

The possible outcomes are SS, SN, NS, NNS, and NNN. (a) Assuming independence, we have:

( ) ( ) ( ) (0.8)(0.8) 0.64

( ) ( ) ( ) (0.8)(0.2) 0.16

( ) ( ) ( ) (0.2)(0.8) 0.16

( ) ( ) ( ) ( ) (0.2)(0.2)(0.8) 0.032

( ) ( ) ( ) ( ) (0.2)(0.2)(0.2) 0.008

P SS P S P S

P SN P S P N

P NS P N P S

P NNS P N P N P S

P NNN P N P N P N

= = =

= = =

= = =

= = =

= = =

(b)

( )(at least two unsatisfactory visits) or

( ) ( )

0.032 0.008 0.040

P P NNS NNN

P NNS P NNN

=

= +

= + =

101

(c)

( )

( )

( ) ( )

1 satisfactory visit 1 unsatisfactory visit

1 no satisfactory visits 1 unsatisfactory visit

( and ) ( )1 ( ) 1 1

0.0081 0.978

1 0.64

P

P

P NNN SS P NNNP NNN SS

P SS P SS

≥ ≥

= − ≥

= − = − = −

= − =−

4.73 (a) BC, ( ) 0P BC = since B and C are incompatible.

(b) , ( ) ( ) ( ) ( )A B P A B P A P B P AB∪ ∪ = + −

Now, by independence, ( ) ( ) ( )P AB P A P B= 0.7 0.2 0.14= × =

Hence, ( ) 0.7 0.2 0.14 0.76P A B∪ = + − =

(c) , ( ) 1 ( ) 1 0.2 0.8B P B P B= − = − =

(d) , ( ) 0ABC P ABC = since B and C are incompatible and so are AB and C.

4.74 Let N denote the event that the blood type of a person is not O, and let A = at least

one type O. While A contains many elementary outcomes, A has only one:

{ }A NNNN= .

We have ( ) 1 0.4 0.6P N = − = . Since the population is large and only four persons

are selected, the results of the different draws can be taken to be independent. We therefore assign the probability

4( ) ( ) ( ) ( ) ( ) (0.6) 0.1296P NNNN P N P N P N P N= × × × = = .

Therefore, ( ) 1 0.1296 0.8704P A = − = .

4.75 Denote the events by: S = the cooling system functions S1 = the primary unit functions F1 = the primary unit fails S2 = the back-up unit functions F2 = the back up unit fails

Then S can be expressed as the union of two incompatible events: 1 1 2( )S S F S= ∪

As such, 1 1 2( ) ( ) ( )P S P S P F S= + . Next, observe that 1( ) .999P S = and

1 2 1 2 1( ) ( ) ( | ) 0.001 0.910 0.00091P F S P F P S F= = × = .

Therefore, ( ) 0.999 0.00091 0.99991P S = + = .


4.76 Let I denote the event that a bill is irregular and N denote the event that it is not. (a) We have ( ) 0.05, ( ) 0.95P I P N= =

P(a batch is passed) ( )P NNNN=

( ) ( ) ( ) ( )P N P N P N P N= (by independence)

4(0.95) 0.8145= = .

(b) Here, ( ) 0.2P I = and so, ( ) 0.8P N = .

P(a batch is passed) 4(0.8) 0.4096= = .

4.77 (a) Denoting the success and failure in each test by S and F, respectively, the

sample space is conveniently listed with the tree diagram below. To calculate the probabilities we note that ( ) 0.8P S = and ( ) 1 0.8 0.2P F = − = .

Since the tests are independent, we calculate ( ) ( ) ( ) ( ) 0.8 0.8 0.8 0.512P SSS P S P S P S= = × × =

( ) ( ) ( ) ( ) 0.8 0.8 0.2 0.128P SSF P S P S P F= = × × = , etc.

The results are shown in the column of probability in the above figure. (b) P(at least two successes) = P(2 successes) + P(3 successes) (0.128 0.128 0.128) 0.512 0.896= + + + = .

103 4.78 The 20 landfills in the MH group (that is, high in Mercury) have the following

classification: Barium High Low Total Arsenic High 1 5 6 Low 4 10 14

Total 5 15 20 (= the number of MH) (a) Of the 20 landfills that are MH, there are 5 that are also BH.

Therefore, 5

( | ) 0.2520

H HP B M = = .

(b) 10

( | ) 0.520

L L HP A B M = =

(c) P(Either AH or BH | MH) 10

0.520

= = .

4.79 Denote the events

S = strep-throat and A = allergy We are given that ( ) 0.25, ( ) 0.4P S P A= = , and ( ) 0.1P SA = .

(a) ( ) ( ) ( ) ( ) 0.25 0.4 0.1 0.55P S A P S P A P SA∪ = + − = + − =

(b) ( ) ( ) 0.25 0.4 0.10 ( )P S P A P SA= × = = , so the events are independent.

4.80 Here the sample space

{ , , , }S HH HT TH TT=

consists of four elementary outcomes, each having the probability 1

4.

The composition of the three events are:

{ , }, { , }, { , }A HH HT B HH TH C HH TT= = =

(a) We have 2 1

( ) ( ) ( )4 2

P A P B P C= = = =

1

{ }, ( ) ( ) ( )4

AB HH P AB P A P B= = = so A and B are independent.

1

{ }, ( ) ( ) ( )4

AC HH P AC P A P C= = = so A and C are independent.

1

{ }, ( ) ( ) ( )4

BC HH P BC P B P C= = = so B and C are independent.

(b) 1

{ }, ( )4

ABC HH P ABC= = but 1

( ) ( ) ( )8

P A P B P C = is not the same as P(ABC).


4.81 (a)

( | ) ( )( )

( | ) ( ) ( | ) ( )

(0.96)(0.014)0.577

(0.96)(0.014) (0.01)(0.986)

P A B P BP B A

P A B P B P A PB B=

+

= =+

(b) This is the probability computed in (a). He has about a 58% chance of having the disease.

4.82 (a) By the rule of total probability,

( ) ( ) ( )

( ) ( )

Incomplete Incomplete Carol did repair Carol did repair

Incomplete Karl did repair Karl did repair

(0.04)(0.60) (0.06)(0.40) 0.048

P P P

P P

= ⋅

+ ⋅

= + =

(b) By Bayes’ Formula, ( )

( ) ( )

( ) ( ) ( ) ( )

Karl did repair|Repair incomplete

Incomplete Karl did repair Karl did repair

Incomplete Karl did repair Karl did repair Incomplete Carol did repair Carol did repair

(0.06)(0.40)

(0.06)(0.40) (0.04)(

P

P P

P P P P

⋅=

⋅ + ⋅

=+

0.500.60)

=

4.83 (a) 6 6 5 4

203 3 2 1

× ×= =

× ×

(b) 10 10 9 8 7

2104 4 3 2 1

× × ×= =

× × ×

(c) 22 22 21

2312 2 1

×= =

×

(d) 22 22

23120 2

= =

(see part c)

(e) 30 30 29 28

40603 3 2 1

× ×= =

× ×

(f) 30 30

406027 3

= =

(see part e)

105 4.84 (a) { , } { , } { , } { , }

{ , } { , } { , }

{ , } { , }

{ , }

a b a c a d a e

b c b d b e

c d c e

d e

Number of samples 10=

(b) { , , } { , , } { , , }

{ , , } { , , } { , , }

{ , , } { , , } { , , }

{ , , }

a b c a b d a b e

a c d a c e a d e

b c d b c e b d e

c d e

Number of samples 10=

4.85 (a) The number of possible selections of 4 persons out of 10 persons is

10 10 9 8 7210

4 4 3 2 1

× × ×= =

× × ×

(b) The number of possible selections of 2 men out of 6 men is

6 6 515

2 2 1

×= =

×

and the number of possible selections of 2 women out of 4 women is

4 4 36

2 2 1

×= =

×

Since the men can be selected in 15 ways, and for each selection of men, there

are 6 ways the women can be selected, the number of possible selections of two men and two women is 15 6 90× = .

4.86 The number of ways 4 positions can be chosen out of 11 positions is

11 11 10 9 8330

4 4 3 2 1

× × ×= =

× × ×

4.87 (a) The number of possible selections of 5 children out of 11 is 11

4625

=

.

(b) The number of selections of 2 out of the 4 young males is 4

62

=

.

The number of selections of 3 out of the 7 young females is 7

353

=

.

Each of the 6 choices of the young males can accompany each of the 35 choices of the young females, so the number of selections of 2 males and 3 female

students is 4 7

6 35 2102 3

× = × =

.


4.88 Of the 11 people, 8 can do the work and 3 cannot.

(a) Number of possible pairs 11

552

= =

(b) Number of pairs where both are able to work 8

282

= =

Number of pairs where one cannot do the work and one can 3 8

241 1

= =

Hence, the number of pairs such that 0 or 1 people cannot do the work

28 24 52= + = .

(c) The number of pairs such that neither can do the work 3

32

= =

.

The required probability is

3

2 30.055

11 55

2

= =

.

4.89 The number of possible samples of 5 jurors out of 17 is 17

61885

=

. Under

random sampling, these 6188 possible samples are equally likely. The number of

possible samples where all 5 jurors selected are males is 10

2525

=

. The jury

selection shows discrimination since

P(no female members)

10

5 2520.041

17 6188

5

= = =

.

4.90 Denote

A = at least one of your cards will be drawn, so that

A = none of your cards will be drawn.

It is convenient to first calculate ( )P A . The number of possible samples of 4 cards

out of 5000 is

5000 5000 4999 4998 4997

4 4 3 2 1

× × ×=

× × × .

107

These are equally likely under random sampling. For A to occur, the 4 cards must come from the 4998 cards that are not yours. The number of such samples is

4998 4998 4997 4996 4995

4 4 3 2 1

× × ×=

× × ×

Therefore,

4998

4 4996 4995( ) 0.998

5000 5000 4999

4

P A

× = = =

×

.

The required probability is ( ) 1 0.998 0.002P A = − = .

4.91 The batch contains 4 defective and 16 good alternators. The number of possible

samples of size 3 is 20

11403

=

and these are equally likely.

(a)

16

3 560( ) 0.491

20 1140

3

P A

= = =

(b)

4 16

2 1 6 16( ) 0.084

20 1140

3

P B

×

× = = =

4.92 1 2 3 1 2 1 3 1 2( ) ( ) ( ) ( | ) ( | )P A P G G G P G P G G P G G G= =

16 15 14 3360

0.49120 19 18 6840

= × × = =

The event B is the union of the following incompatible events:

1 2 3 1 2 3 1 2 3D D G D G D G D D

We calculate

1 2 3 1 2 1 3 1 2( ) ( ) ( | ) ( | )P D D G P D P D D P G D D=

4 3 16 192

0.028120 19 18 6840

= × × = =

The same result holds for 1 2 3( )P D G D and 1 2 3( )P G D D . Thus,

192( ) 3 0.084

6840P B = × = .

4.93 No. The states with fewer seniors get more representation in this process than what

the random selection would permit.


4.94

3 hard 4 easy → 3 selected at random

7

The number of possible samples 7 7 6 5

353 3 2 1

× ×= = =

× × .

These are equally likely.

The number of possible samples containing only the hard problems 3

13

= =

.

Hence, the required probability

3

3 10.0286

7 35

3

= = =

.

4.95 (a) The number of possible selections of 3 plots out of 9 is 9

843

=

, and the 84

selections are equally likely. One row can be chosen in 3

31

=

ways, and

within that row, 3 plots can be chosen in 3

13

=

way. Therefore, the number of

choices such that the three plots are in the same row is 3 3

3 1 31 3

× = × =

.

Hence, the required probability 3

0.03684

= = .

(b) The number of ways one plot can be selected from row 1 is 3

31

=

. Likewise,

a plot can be selected from row 2 in 3 ways, and from row 3 in 3 ways. The number of possible selections of 3 plots, one in each row, is 3 3 3 27× × = , so

the required probability 27

0.32184

= = .

109 4.96

10 bushy 7 lean → 4 randomly selected

17

Number of possible selections of 4 trees out of 17 trees is 17

23804

=

Number of selections of 2 bushy trees and 2 lean trees is

10 745 21 945

2 2

× = × =

Hence, the required probability 945

0.3972380

= = .

4.97

Row A Row B 4 bushy 6 bushy 4 lean → 2 3 lean → 2

8 9

(a) There are 8 trees in row A from which 2 trees can be selected in 8

282

=

ways.

Of these 28 equally likely selections, there are 4

62

=

selections in which both

trees are bushy. Therefore,

P[2 bushy trees selected in row A] 6

28= .

Similarly,

P[2 lean trees selected in row B]

3

2 3

9 36

2

= =

.

By independence of the two selections, the required probability is therefore

6 30.018

28 36= × = .

(b) Let 0 1,A A , and 2A respectively denote the events of getting exactly 0, 1, or 2

bushy trees in row A, and let 0 1 2, ,B B B denote the corresponding events for

row B. Then

[Exactly 2 bushy] 2 0 1 1 0 2A B A B A B= ∪ ∪ (union of mutually exclusive events)


2 0 2 0

4 3

2 2 6 3( ) ( ) ( ) 0.018

8 9 28 36

2 2

P A B P A P B

= = × = × =

(see part (a))

1 1 1 1

4 4 6 3

1 1 1 1 16 18( ) ( ) ( ) 0.286

8 9 28 36

2 2

P A B P A P B

= = × = × =

0 2 0 2

4 6

2 2 6 15( ) ( ) ( ) 0.089

8 9 28 36

2 2

P A B P A P B

= = × = × =

Adding these probabilities we get P[exactly 2 bushy] 0.393= . 4.98 (a) No – Students who enjoy outdoor activities would be over-represented. (b) No – Studious students would be over-represented. (c) No – Students taking statistics courses would be over-represented. 4.99

5 below thirty 6 over thirty → 4 randomly selected

11

(a) The number of possible choices of 4 persons out of 11 is 11

3304

=

.

(b) The number of choices of 3 persons below thirty and 1 over thirty is

5 610 6 60

3 1

× = × =

.

So, the required probability 60

0.182330

= = .

4.100 Males Females 2 below thirty 3 below thirty 4 above thirty → 2 2 above thirty → 2

6 5

(a) 6 5

15 10 1502 2

× = × =

possible selections

(b) The number of selections such that both selected females are below 30 and

both males are over 30 is 3 43 6 18

2 2

× = × =

. So, the required probability

180.12

150= = .

111 4.101

6 yellow 5 red → 4

11

The number of possible samples of 4 bulbs out of 11 is 11

3304

=

, and all

choices are equally likely. (a) The number of ways 2 red and 2 yellow bulbs can be selected is

5 610 15 150

2 2

× = × =

So, P[exactly 2 red] 150

0.455330

= = .

(b) We calculate

P[2 red] 150

330= (done in part (a))

P[3 red]

5 6

3 1 10 6 60

330 330 330

×

× = = =

P[4 red]

5

4 5

330 330

= =

Adding these probabilities, we obtain

P[at least 2 red] 150 60 5 215

0.652330 330

+ += = =

(c) P[all 4 red]

5

4 5

330 330

= =

P[all 4 yellow]

6

4 15

330 330

= =

Adding these probabilities, we obtain

P[all 4 of the same color] 20

0.061330

= = .


4.102 (a) The number of possible selections of 3 files from 8 is 8

563

=

each of which

is equally likely. (b) Label the folders A through H in alphabetical order. The selections that

correspond to the three folders being adjacent are

{ , , } { , , } { , , }

{ , , } { , , } { , , }

A B C B C D C D E

D E F E F G F G H

So, there are 6 such selections. Thus, the required probability is

P[all adjacent] 6

0.10756

= = .

4.103 (a) {1, 2, ..., 24}S =

(b) { : 0}S p p= > , p is tire pressure in psi.

(c) {0, 1, 2, ..., 50}S =

(d) { : 0}S t t= ≥ , t is time in days.

4.104 (a) and (c) are discrete while (b) and (d) are continuous. 4.105 (a) {23, 24}A =

(b) { : 0 28}A p p= < ≤

(c) Since 0.25 50 12.5, {0, 1, ..., 12}A× = =

(d) { : 0 500.5}A t t= ≤ <

4.106 (a) The total exceeds 1. (b) 1.1 is not permissible for a probability. (c) The probability of both A’s cannot be more than the probability of one A. 4.107 { : 0 100}S p p= ≤ < where p = percentage of alcohol in blood.

{ : .10 100}A p p= < <

113 4.108 Denote player 1’s winning and loss in a set by W and L, respectively. (a) Championship ends in 3 sets { , }WWW LLL= , two elementary outcomes.

Ends in 4 sets { , , , , , }.WWLW WLWW LWWW LLWL LWLL WLLL=

(b) 1

( ) ( )2

P W P L= = . Assuming the results of different sets to be independent,

the outcome WWLW has the probability 1 1 1 1 1

2 2 2 2 16× × × = and similarly for

the other outcomes. Hence, P(ends in 4 sets) 6

16= .

4.109 The sample space is listed by means of a tree diagram which is given in Figure

4.4.

(a) S consists of 12 elementary outcomes which are equally likely because of

random selection. Of the 12 elementary outcomes, 6 correspond to even numbers so

P[even number] 6

0.512

= =

(b) There are 9 elementary outcomes where the number is larger than 20, so the

required probability is 9

0.7512

= .

(c) There are 2 elementary outcomes, namely {23, 24}, for which the number is

between 22 and 30, so the required probability is 2

0.16712

= .


4.110 (a) The possible assignments of ‘a’ are listed by the pairs of plot numbers

{(1, 2), (1,3), (1, 4), (2,3), (2, 4), (3,4)}S =

(b) Under random assignment, the elementary outcomes are equally likely so each

has probability 1/ 6 .

(i) [Same column] {(1,3), (2, 4)}= , Probability 2 1

6 3= =

(ii) [different rows and different columns] {(1, 4), (2,3)}= , Probability 1

3= .

4.111 Consider the plot selected from each row to be assigned to variety ‘a’. We list the

sample space by drawing a tree diagram which is given in the figure below.

1 2 3 4{ , , , }S e e e e= and the elements are equally likely.

[same column] {(1, 3), (2, 4)}= , Probability 2 1

4 2= = .

4.112 Denote the events A = 1 appears in the first toss B = 1 appears in the second toss The the event “1 appears at least once in the two tosses” can also be described as

“either A or B or both occur”. Referring to the sample space of Exercise 4.21, this event has the composition

{(1,1), (1, 2), (1,3), (1, 4), (1,5), (1,6),

(2,1), (3,1), (4,1), (5,1), (6,1)}

Hence the correct probability is 11

36. Indeed

1 1( ) , ( )

6 6P A P B= = , and

1 1 12

6 6 36+ = . The outcome (1,1) belongs to each of A and B, and the incorrect

answer has resulted from counting (1,1) twice.

115 4.113 There are 9 letters which are equally likely to be selected.

(a) Since there are 4 vowels, the probability of getting a vowel is 4

9.

(b) Among the 9 letters there is one T and one V, so the probability of getting a T

or V is 2

9.

4.114 (a) No. Heavy air traffic occurs on Mondays and Fridays. (b) Yes. The temperature has nothing to do with the day of the week. (c) No. Ozone lower on weekends when not as much automobile exhaust. (d) No. December almost always has highest sales. 4.115 (a) {125, 152, 251, 215, 512, 521}S =

(b) P(less than 400) 4 2

6 3= = .

(c) P(even number) 2 1

6 3= = .

4.116 Let us label the 11 flowers as follows:

roses that will open 1 2 3, ,e e e

roses that will not open 4 5,e e

tulips that will open 6 7,e e

tulips that will not open 8 9 10 11, , ,e e e e

(a) The Venn diagram is given in Figure 4.5.

(b) Since [not open] 4 5 8 9 10 11{ , , , , , }e e e e e e= ,

P(not open) 6

0.54511

= = .


4.117 (a) Either a faulty transmission or faulty brakes. (b) Transmission, brakes and exhaust system all faulty. (c) No faults with the transmission, brakes or the exhaust system. (d) Either the transmission is not faulty or the brakes are not faulty.

4.118 (a) AB (b) AB (c) ( ) ( )AB AB∪

4.119 Because ( ) ( ) ( )P AB P A P A B≤ ≤ ∪ , we have ( ) 0.30, ( ) 0.1,P A P AB= = and

( ) 0.50P A B∪ = .

4.120 ( ) 0.4 0.1 0.18 0.68,P A = + + = ( ) 0.02P AB = ,

( ) 0.02 0.4 0.18 0.60P B C∪ = + + = , ( ) 0P BC = .

4.121 (a) ABC (b) A B C∪ ∪ (c) ABC (d) ABC 4.122 (a) The Venn diagram is given in Figure 4.6.

(Note: For parts (b) and (c), we add the relevant entries of the probability

table or work with the Venn diagram.) (b) ( ) 0.05 0.10 0.15P AB = + =

( ) 0.10 0.17 0.27P AC = + =

( ) 0.05 0.05 0.18 0.20 0.48P C = + + + =

(c)

B B A 0.15 0.22

A 0.35 0.28

117 4.123 (a) ( ) 0.05 0.20 0.25P BC = + = (see Figure 4.6 above.)

(b) ( ) ( ) ( ) ( )

( ) 0.05 0.10 0.20 0.15 0.50

( ) 0.05 0.05 0.20 0.18 0.48

( ) 0.25

P B C P B P C P BC

P B

P C

P BC

∪ = + −

= + + + =

= + + + =

=

(Addition law)

Therefore, ( ) 0.50 0.48 0.25 0.73P B C∪ = + − =

Alternatively,

( ) ( ) ( )

0.50 0.23 0.73

P B C P B P BC∪ = +

= + =

(See Figure 4.6 above.)

(c) ( ) 0.10 0.15 0.25P BC = + =

(d) ( ) 0.17 0.15 0.18 0.50P ABC ABC ABC∪ ∪ = + + =

4.124 The Venn diagram is given in Figure 4.7. To mark probabilities in the Venn

diagram, we start with the intersection of all three events i.e., ABC , fill in the

given probability 0.12. Then note that the sets ABC and ABC together make up the intersection AB . Since ( ) 0.17P AB = and ( ) 0.12P ABC = , the difference

must be ( ) 0.17 0.12 0.05P ABC = − = . We continue this process, and in the final

step, obtain ( ) 1 ( ) 1 0.88 0.12P ABC P A B C= − ∪ ∪ = − = .

Probability Table:

B B C C C C

A 0.12 0.05 0.21 0.13

A 0.08 0.20 0.09 0.12


4.125 The probabilities can be determined either by using the Venn diagram or the

probability table presented in the solution of Exercise 4.124.

(a) ( ) 0.05 0.20 0.25P BC = + =

(by summing all probabilities in B but outside of C) (b) ( ) ( ) ( ) ( )

0.51 0.45 0.17

0.79

P A B P A P B P AB∪ = + −

= + −

=

(Addition law)

(c) P[exactly two of the three events occur] ( ) ( ) ( )P ABC P ABC P ABC= + +

0.05 0.21 0.08= + + (see Figure 4.7) 0.34= .

4.126 (a) 2 1

( )6 3

P A = =

(b) 3 1

( )6 2

P B = =

(c) 1

( )6

P AB =

(d) ( ) 1/ 6 1

( | )( ) 3/ 6 3

P ABP A B

P B= = =

(e) 2 3 1 4 2

( ) ( ) ( ) ( ) .6 6 6 6 3

P A B P A P B P AB∪ = + − = + − = =

4.127 ( ) 0.4

( | ) 0.8( ) 0.5

P ABP A B

P B= = =

( ) ( ) 0.8 0.5 0.40 ( )P A P B P AB= × = = , so A and B are independent.

4.128 (a) The Venn diagram is

119 (b) ( ) ( ) ( | ),

20.3 0.2

3

P AB P B P A B=

= × =

Multiplication law

( ) ( ) ( ) 0.25 0.1 0.15P AC P C P AC= − = − =

( ) ( ) ( ) 0.3 0.2 0.1P BA P B P AB= − = − =

( ) ( ) ( ) ( ) 0.6 0.2 0.15 0.25P ABC P A P AB P AC= − − = − − =

( ) 1 ( ) 1 0.8 0.2P ABC P A B C= − ∪ ∪ = − =

(c) ( ) ( ) ( ) 0.25 0.1 0.1 0.45P ABC P AB P AC+ + = + + =

4.129 (a) ( ) 0.15P AC =

( ) ( ) 0.6 0.25 0.15P A P C = × =

Because ( ) ( ) ( )P AC P A P C= , the events A and C are independent.

(b) ( ) ( ) 0.15P ABC P AC= = (see the Venn diagram in Exercise 4.128)

( ) 0.25 0.15 0.40P AB = + =

( ) 0.15 0.1 0.25P C = + =

( ) ( ) 0.40 0.25 0.1 ( )P AB P C P ABC= × = ≠ , so the events AB and C are not

independent. 4.130 (a) Given that the person is male, the probability that he is a moderate to heavy

drinker is 0.29. (b) By the law of total probability,

( ) ( | ) ( ) ( | ) ( )

(0.12)(0.50) (0.29)(0.50) 0.205

P A P A B P B P A B P B= +

= + =

(c) By Bayes’ Formula,

( | ) ( ) (0.12)(0.50)( | ) 0.293

0.205( | ) ( ) ( | ) ( )

P A B P BP B A

P A B P B P A B P B= = =

+

4.131 (a) ( )

( | )( )

P ABP A B

P B=

From the probability table in Exercise 4.122, we get

( ) 0.05 0.17 0.22

( ) 0.05 0.17 0.18 0.10 0.50

P AB

P B

= + =

= + + + =

So, 0.22

( | ) 0.440.50

P A B = = .


(b) ( ) 0.05

( | ) 0.50( ) 0.10

P ABCP B AC

P AC= = =

(c) From the probability table, we find ( ) 0.37, ( ) 0.48. ( ) 0.10P A P C P AC= = = .

Since ( ) ( ) 0.37 0.48 0.178 ( )P A P C P AC= × = ≠ , the events A and C are

dependent. 4.132 In order to be successful Mr. Hope must receive the good pill in each of the two

trials. The probability that he gets the good pill in the first trial 1

1( )

2P G = , and

that in the second trial 2

1( )

2P G = .

Because the trials are independent, 1 2 1 2

1 1 1( ) ( ) ( )

2 2 4P G G P G P G= = × = . Thus,

the probability of Mr. Hope’s success is only 1

4.

4.133 (a) 1 2 3 4{ , , , }S e e e e= where 1e corresponds to the selection sequence 11, 2e to

12, 3e to 21, and 4e to 22.

1

10 9 90( )

15 14 210P e = × = (Because the probability of a #1 marble in the first

draw is 10

15, and the conditional probability of a #1 marble in the second draw

given that a #1 marble appears in the first draw, is 9

14.)

Similarly, we calculate

2

10 5 50( )

15 14 210P e = × =

3

5 10 50( )

15 14 210P e = × =

4

5 4 20( )

15 14 210P e = × =

(b) [Even number] 2 4{ , }e e= , Probability 50 20 70 1

210 210 210 3= + = =

(c) [Larger than 15] 3 4{ , }e e= , Probability 50 20 70 1

210 210 210 3= + = = .

121 4.134 (a) In the total pool of the product, the proportion of defectives and from Line 1 is 0.2 0.1 0.020× =

proportion of defectives and from Line 2 is 0.5 0.05 0.025× =

proportion of defectives and from Line 3 is 0.3 0.06 0.018× =

Total 0.063= The pool has a proportion 0.063 or 6.3% defective items. Alternatively, to cast in terms of probability calculation, let us denote by

1 2, ,L L and 3L the events that a randomly selected item comes from Line 1,

Line 2, and Line 3, respectively. Let D denote the event that the item is defective. The specified percentages represent the following probabilities.

1 1( ) 0.2, ( | ) 0.1P L P D L= =

2 2( ) 0.5, ( | ) 0.05P L P D L= =

3 3( ) 0.3, ( | ) 0.06P L P D L= =

By the multiplication law of probability, we have

1 1 1( ) ( ) ( | ) 0.2 0.1 0.020P DL P L P D L= = × =

2 2 2( ) ( ) ( | ) 0.5 0.05 0.025P DL P L P D L= = × =

3 3 3( ) ( ) ( | ) 0.3 0.06 0.018P DL P L P D L= = × =

Since 1 2 3D DL DL DL= ∪ ∪ , a union of incompatible events, we obtain

( ) 0.020 0.025 0.018 0.063P D = + + = or 6.3%.

(b) 11

( ) 0.020( | ) 0.317

( ) 0.063

P L DP L D

P D= = = .

4.135 (a) & (b) The tree diagram is given below.

NFND

NN


(i) The elementary outcome probabilities are contained within the above diagram. (ii) ( ) ( ) ( ) ( ) ( ) 0.64P FF P FD P FN P DF P NF+ + + + =

(iii) ( ) ( ) ( ) ( ) 0.49P FF P FD P DF P DD+ + + =

4.136 (a) The Venn diagram is given below.

The possible outcomes are all contained within the above diagram. All outcomes

are equally likely, so 2 1

( ) ( )6 3

P A P B= = = and ( ) 0P AB = .

(b) 2 2 0 4 2

( ) ( ) ( ) ( )6 6 6 6 3

P A B P A P B P AB∪ = + − = + − = =

(c) 1 1 1

( ) ( ) ( )3 3 9

P A P B P AB= × = = so the events A and B are independent.

(d) 2 1

( )6 3

P AB = = .

4.137 There are 11 11 10 9

1653 3 2 1

× ×= =

× × ways to select three trucks at random. If the

company’s argument is correct, that there are exactly three noncompliant trucks, then only one selection can consist of three noncompliant vehicles. The

probability of selecting all three noncompliant trucks is 1

0.0061165

= , a small

probability. It is quite unlikely a random selection would produce the indicated sample so we can question the veracity of the company’s claim.

4.138 (a) P(appears only once) 14376

0.0163884,647

= =

(b) P(appears exactly twice) 4343

0.0049884,647

= =

(c) P(appears more than twice) 1= − P(appears 2 or fewer times) 1 0.0163 0.0049 0.9788= − − = .

123 4.139 Denote G = Good standing, D = Illegal deduction.

11 G 7 D → 4

18 (a) The number of possible selections of 4 returns out of 18 is

183060

4

=

, all equally likely.

P[Sample has all 4 G’s]

11

4 330.108

18 3060

4

= = =

.

(b) P[at least 2 D’s] [2 ' ] [3 ' ] [4 ' ]P D s P D s P D s= + +

7 11

2 2 21 55 1155[2 ' ]

18 3060 3060

4

P D s

× = = =

7 11

3 1 35 11 385[3 ' ]

18 3060 3060

4

P D s

× = = =

7

4 35[4 ' ]

18 3060

4

P D s

= =

Adding these we obtain P[at least 2 D’s] =

1155 385 35 1575

0.5153060 3060 3060 3060

+ + = = .


4.140 (a) 1 2 1 2 1 2 1 2{ , , , }S R R R G G R G G=

(b) 1

6( ) 0.6

10P G = =

(c) Given that a green ball appears in the first draw, the urn contains 4 red balls and 9 green balls prior to drawing the second ball. The probability that a red

ball will be drawn from this mix is 4

13, so 2 1

4( | )

13P R G = .

(d) 2 1 2 1 2G R G G G= ∪

1 2 1 2 1

4 6 24( ) ( ) ( | )

10 13 130P R G P R P G R= = × =

1 2 1 2 1

6 9 54( ) ( ) ( | )

10 13 130P G G P G P G G= = × =

2 1 2 1 2

24 54 78( ) ( ) ( ) 0.6

130 130 130P G P R G P G G= + = + = = .

4.141 (a) P(no common birthday) 365 364 363 364 363

1 0.992365 365 365 365 365

× ×= = × × =

× ×.

Hence, P(at least two have the same birthday) 1 0.992 0.008= − = . (b) For each person there are 365 possible birthdays. Hence, for N persons, the

number of possible birthdays is

365 365 ... 365× × × (N factors) (365)N= .

In order that the birthdays of N persons to be all different, the first person can have any of the 365 days, the second person can have any of the remaining

365 1 364− = days, the third person can have any of the remaining

365 2 363− = days, ……………………….. the Nth person can have any of the remaining 365 1N− + days.

Hence, the number of ways N persons can have all different birthdays is 365 364 ... (365 1)N× × × − + . Therefore,

( )365 364 365 1[No common birthday]

365N

NP

× × × − +=

…

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MAT 540 Statistical Concepts for Research