Answers to Selected Exercises

1.

  1. P(L′) = P(S)
  2. P(M OR S)
  3. P(F AND L)
  4. P(M|L)
  5. P(L|M)
  6. P(S|F)
  7. P(F|L)
  8. P(F OR L)
  9. P(M AND S)
  10. P(F)

3. P(N) = [latex]\frac{{15}}{{42}}=\frac{{5}}{{14}}=0.36[/latex]

5. P(C) = [latex]\frac{{5}}{{42}}=0.12[/latex]

7. P(G) = [latex]\frac{{20}}{{150}}=\frac{{2}}{{15}}=0.13[/latex]

9. P(R) = [latex]\frac{{22}}{{150}}=\frac{{11}}{{75}}=0.15[/latex]

11. P(O) = [latex]\frac{{{22}-{38}-{20}-{28}-{26}}}{{150}}=\frac{{16}}{{150}}=\frac{{8}}{{75}}=0.11[/latex]

13. P(E) = [latex]\frac{{47}}{{194}}=0.24[/latex]

15. P(N) = [latex]\frac{{23}}{{194}}=0.12[/latex]

17. P(S)=[latex]\frac{{12}}{{194}}={{6}}{{97}}=0.06[/latex]

19. [latex]\frac{{13}}{{52}}={{1}}{{4}}=0.25[/latex]

21. [latex]\frac{{3}}{{6}}={{1}}{{2}}=0.5[/latex]

23. P(R) = [latex]\frac{{4}}{{8}}=0.5[/latex]

25. P(O or H)

27. P(H | I)

29. P(N | O)

31. P(I or N)

33. P(I)

35. The likelihood that an event will occur given that another event has already occurred.

37. 1

39. the probability of landing on an even number or a multiple of three

Independent and Mutually Exclusive Events — Practice

41. P(J) = 0.3
43. P(Q AND R) = P(Q)P(R)0.1 = (0.4)P(R)P(R) = 0.25

Two Basic Rules of Probability — Practice

45. 0.376
47. C|L means, given the person chosen is a Latino Californian, the person is a registered voter who prefers life in prison without parole for a person convicted of first degree murder.
49. L AND C is the event that the person chosen is a Latino California registered voter who prefers life without parole over the death penalty for a person convicted of first degree murder.
51. 0.6492
53. No, because P(L AND C) does not equal 0.

 Contingency Tables — Practice

55. P(musician is a male AND had private instruction) = [latex]\frac{15}{130} = [latex]\frac{3}{26} = 0.12[/latex]
57. The events are not mutually exclusive. It is possible to be a female musician who learned music in school.

Tree and Venn Diagrams — Practice

58.This is a tree diagram with two branches. The first branch, labeled Cancer, shows two lines: 0.4567 C and 0.5433 C'. The second branch is labeled False Positive. From C, there are two lines: 0 P and 1 P'. From C', there are two lines: 0.51 P and 0.49 P'.

60. [latex]\frac{35,065}{100,450}[/latex]

62. To pick one person from the study who is Japanese American AND smokes 21 to 30 cigarettes per day means that the person has to meet both criteria: both Japanese American and smokes 21 to 30 cigarettes. The sample space should include everyone in the study. The probability is [latex]\frac{4,715}{100,450}[/latex].

64. To pick one person from the study who is Japanese American given that person smokes 21-30 cigarettes per day, means that the person must fulfill both criteria and the sample space is reduced to those who smoke 21-30 cigarettes per day. The probability is [latex]\frac{4715}{15,273}[/latex].

Terminology — Homework

67.

  1. You can’t calculate the joint probability knowing the probability of both events occurring, which is not in the information given; the probabilities should be multiplied, not added; and probability is never greater than 100%
  2. A home run by definition is a successful hit, so he has to have at least as many successful hits as home runs.

Independent and Mutually Exclusive Events — Homework

69. 0
71. 0.3571
73. 0.2142
75. Physician (83.7)
77. 83.7 − 79.6 = 4.1
79. P(Occupation < 81.3) = 0.5

Two Basic Rules of Probability — Homework

81.

  1. The Forum Research surveyed 1,046 Torontonians.
  2. 58%
  3. 42% of 1,046 = 439 (rounding to the nearest integer)
  4. 0.57
  5. 0.60.

83.

  1. P(Betting on two line that touch each other on the table) = [latex]\frac{{6}}{{38}}[/latex]
  2. P(Betting on three numbers in a line) = [latex]\frac{{3}}{{38}}[/latex]
  3. P(Bettting on one number) = [latex]\frac{{1}}{{38}}[/latex]
  4. P(Betting on four number that touch each other to form a square) = [latex]\frac{{4}}{{38}}[/latex]
  5. P(Betting on two number that touch each other on the table ) = [latex]\frac{{2}}{{38}}[/latex]
  6. P(Betting on 0-00-1-2-3) = [latex]\frac{{5}}{{38}}[/latex]
  7. P(Betting on 0-1-2; or 0-00-2; or 00-2-3) = [latex]\frac{{3}}{{38}}[/latex]

85.

  1. {G1, G2, G3, G4, G5, Y1, Y2, Y3}
  2. [latex]\frac{{5}}{{8}}[/latex]
  3. [latex]\frac{{2}}{{3}}[/latex]
  4. [latex]\frac{{2}}{{8}}[/latex]
  5. [latex]\frac{{6}}{{8}}[/latex]
  6. No, because P(G AND E) does not equal 0.

87.

NOTE

The coin toss is independent of the card picked first.

  1. {(G,H) (G,T) (B,H) (B,T) (R,H) (R,T)}
  2. P(A) = P(blue)P(head) = ([latex]\frac{{3}}{{10}}[/latex])([latex]\frac{{1}}{{2}}[/latex])=[latex]\frac{{3}}{{20}}[/latex]
  3. Yes, A and B are mutually exclusive because they cannot happen at the same time; you cannot pick a card that is both blue and also (red or green). P(A AND B) = 0
  4. No, A and C are not mutually exclusive because they can occur at the same time. In fact, C includes all of the outcomes of A; if the card chosen is blue it is also (red or blue). P(A AND C) = P(A) = 320

89.

  1. S = {(HHH), (HHT), (HTH), (HTT), (THH), (THT), (TTH), (TTT)}
  2. [latex]\frac{{4}}{{8}}[/latex]
  3. Yes, because if A has occurred, it is impossible to obtain two tails. In other words, P(A AND B) = 0.

91.

  1. If Y and Z are independent, then P(Y AND Z) = P(Y)P(Z), so P(Y OR Z) = P(Y) + P(Z) – P(Y)P(Z).
  2. 0.5

93.

1. iii 2. i 3. iv 4. ii

95.

  1. P(R) = 0.44
  2. P(R|E) = 0.56
  3. P(R|O) = 0.31
  4. No, whether the money is returned is not independent of which class the money was placed in. There are several ways to justify this mathematically, but one is that the money placed in economics classes is not returned at the same overall rate; P(R|E) ≠ P(R).
  5. No, this study definitely does not support that notion; in fact, it suggests the opposite. The money placed in the economics classrooms was returned at a higher rate than the money place in all classes collectively; P(R|E) > P(R).
 97.

  1. P(type O OR Rh-) = P(type O) + P(Rh-) – P(type O AND Rh-)

    0.52 = 0.43 + 0.15 – P(type O AND Rh-); solve to find P(type O AND Rh-) = 0.06

    6% of people have type O, Rh- blood

  2. P(NOT(type O AND Rh-)) = 1 – P(type O AND Rh-) = 1 – 0.06 = 0.94

    94% of people do not have type O, Rh- blood

99.

  1. Let C = be the event that the cookie contains chocolate. Let N = the event that the cookie contains nuts.
  2. P(C OR N) = P(C) + P(N) – P(C AND N) = 0.36 + 0.12 – 0.08 = 0.40
  3. P(NEITHER chocolate NOR nuts) = 1 – P(C OR N) = 1 – 0.40 = 0.60

 Contingency Tables — Homework

101. 0

103. [latex]\frac{{10}}{{67}}[/latex]

105.[latex]\frac{{10}}{{34}}[/latex]

107. d

109.

a.

Race and Sex 1–14 15–24 25–64 over 64 TOTAL
white, male 210 3,360 13,610 4,870 22,050
white, female 80 580 3,380 890 4,930
black, male 10 460 1,060 140 1,670
black, female 0 40 270 20 330
all others 100
TOTALS 310 4,650 18,780 6,020 29,760

b.

Race and Sex 1–14 15–24 25–64 over 64 TOTAL
white, male 210 3,360 13,610 4,870 22,050
white, female 80 580 3,380 890 4,930
black, male 10 460 1,060 140 1,670
black, female 0 40 270 20 330
all others 10 210 460 100 780
TOTALS 310 4,650 18,780 6,020 29,760

c. [latex]\frac{{22050}}{{29760}}[/latex]

d. [latex]\frac{{330}}{{29760}}[/latex]

e. [latex]\frac{{2000}}{{29760}}[/latex]

f. [latex]\frac{{23720}}{{29760}}[/latex]

g. [latex]\frac{{5010}}{{6020}}[/latex]

111. b

113.

a. [latex]\frac{{26}}{{106}}[/latex]

b. [latex]\frac{{33}}{{106}}[/latex]

c. [latex]\frac{{21}}{{106}}[/latex]

d. [latex]\left(\frac{{26}}{{106}}\right)+\left(\frac{{33}}{{106}}\right)-\left(\frac{{21}}{{106}}\right)=\left(\frac{{38}}{{106}}\right)[/latex]

e. [latex]\frac{{21}}{{33}}[/latex]

Tree and Venn Diagrams — Homework

115. a

Extra Practice

118.

 

a. P(C) = 0.4567

b. not enough information

c. not enough information

d. No, because over half (0.51) of men have at least one false positive text

120.

 

  1. P(J OR K) = P(J) + P(K) − P(J AND K); 0.45 = 0.18 + 0.37 – P(J AND K); solve to find P(J AND K) = 0.10
  2. P(NOT (J AND K)) = 1 – P(J AND K) = 1 – 0.10 = 0.90
  3. P(NOT (J OR K)) = 1 – P(J OR K) = 1 – 0.45 = 0.55

 

 

121.

a.

This is a tree diagram with branches showing probabilities of each draw. The first branch shows two lines: 5/8 Green and 3/8 Yellow. The second branch has a set of two lines (5/8 Green and 3/8 Yellow) for each line of the first branch.

b. P(GG) =[latex]\left(\frac{{5}}{{8}}\right)\left(\frac{{5}}{{8}}\right)=\frac{{25}}{{64}}[/latex]
c. P(at least one green) = P(GG) + P(GY) + P(YG) = [latex]\left(\frac{{25}}{{64}}\right)+\left(\frac{{15}}{{64}}\right)+\left(\frac{{15}}{{64}}\right)=\frac{{55}}{{64}}[/latex]
d. P(G|G) = 58. 
e. Yes, they are independent because the first card is placed back in the bag before the second card is drawn; the composition of cards in the bag remains the same from draw one to draw two.123.

a.
<20 20–64 >64 Total
Female 0.0244 0.3954 0.0661 0.486
Male 0.0259 0.4186 0.0695 0.514
Totals 0.0503 0.8140 0.1356 1

b. P(F) = 0.486

c. P(>64|F) = 0.1361

d. P(>64 and F) = P(F) P(>64|F) = (0.486)(0.1361) = 0.0661

e. P(>64|F) is the percentage of female drivers who are 65 or older and P(>64 and F) is the percentage of drivers who are female and 65 or older.

f. P(>64) = P(>64 and F) + P(>64 and M) = 0.1356

g. No, being female and 65 or older are not mutually exclusive because they can occur at the same time P(>64 andF) = 0.0661.

125.

  1. Car, Truck or Van Walk Public Transportation Other Total
    Alone 0.7318
    Not Alone 0.1332
    Totals 0.8650 0.0390 0.0530 0.0430 1
  2. If we assume that all walkers are alone and that none from the other two groups travel alone (which is a big assumption) we have: P(Alone) = 0.7318 + 0.0390 = 0.7708.
  3. Make the same assumptions as in (b) we have: (0.7708)(1,000) = 771
  4. (0.1332)(1,000) = 133

127.

Homosexual/Bisexual IV Drug User* Heterosexual Contact Other Total
Female 0 70 136 49 255
Male 2,146 463 60 135 2,804
Totals 2,146 533 196 184 3,059
  1. [latex]\frac{{255}}{{3059}}[/latex]
  2. [latex]\frac{{196}}{{3059}}[/latex]
  3. [latex]\frac{{718}}{{3059}}[/latex]
  4. 0
  5. [latex]\frac{{463}}{{3059}}[/latex]
  6. [latex]\frac{{136}}{{196}}[/latex]