4.1 Six of the first 10 digits on line 131 correspond to “heads,” so the proportion of heads is 60%. Although the average number of heads in 10 tosses is five, the actual outcome is random and can vary from sample to sample.

4.3 S={all numbers from 0 to 1,440}.

4.5 0.83. Adding three probabilities and subtracting that result from 1 is slightly easier than adding the five probabilities of interest.

4.7 P(3 or 8 or more)=0.222.

4.9 For each outcome (1, 2, 3, 4, 5, and 6), the probability is 1/6.

4.11 If we know the first card was not an ace, we know there are 4 aces still in the deck and 51 cards left, so the probability of the second card being an ace is 4/51. Because the outcome of the first card changes the probability for the second card, these outcomes are not independent.

4.13 S={0, 1, 2, 3} P(X=0)=1/8, P(X=1)=3/8, P(X=2)=3/8, P(X=3)=1/8.

4.15 P(X<15)=0.8413.

4.17 Use Applet. Answers will vary.

4.19 μW=17.

4.21 0.3333.

4.23 11/48.

4.25 The addition rule for disjoint events.

4.1

1. Yes.
2. No.
3. Yes.
4. No.

4.3

1. The probability is the number of times you would see a certain number when you roll the six-sided die.
2. Probability would not apply because we cannot predict the number of times we would use a certain number at the end of our personal phone number.
3. The probability is the number of tens in the card deck out of the 52 cards.
4. Probability would not apply here because we can predict when a birth date will be.

4.5

1. No.
2. Yes.
3. No.
4. Yes.

4.7 The probability is 0.518. The number of rolls and estimated probability will vary.

4.9

1. S={0, 1, 2, 3,⋯} people.
2. S={any amount $0 and up}.
3. S={any amount from 0 second and up}.
4. S={Yes, No}.

4.11

1. 1/4 or 2/8; Rule 3.
2. 3/4 or 6/8; Rule 4.
3. No; Rule 2.
4. No; Rule 1.
5. 0.77; Rule 4.
6. 1; Rule 2.

4.13

1. Equally likely because each side is either even or odd, and it is a fair die.
2. The events are not equally likely in general. For many intersections the straight through traffic will be most likely; some intersections may prohibit left turns, etc.
3. Not equally likely because a lot of factors go into the home team winning or losing.
4. Not equally likely for same reason as (c).

4.15 There are five possible outcomes: S={link1, link2, link3, link4, leave}.

4.17

1. 0.44.
2. 0.86.

4.19

1. Legitimate.
2. Legitimate.
3. Not legitimate because they must all add up to 1.

4.21

1. P(some education beyond high school but no degree)=0.28.
2. P(at least high school)=0.88.

4.23 Possible types: A+, A−, B+, B−, AB+, AB−, O+, O−. Probabilities: 0.353, 0.067, 0.092, 0.018, 0.025, 0.005, 0.370, 0.070.

4.25

1. P(win)=0.006.
2. P(win)=0.003.

4.27 P(at least one is O-negative)=0.353.

4.29 Observe that P(A and Bc)=P(A)−P(A)P(B)=P(A)(1−P(B)).

4.31

1. Either A or O.
2. P(O)=0.25 and P(A)=0.75.

4.33

1. 0.25.
2. 0.015625; 0.140625.

4.35

4.39

1. P(X≤2)=0.12.
2. P(X=3 or 4)=0.32.
3. P(X=7)=0.

4.41

1. P(X<0.2)=0.2.
2. P(X≥0.7)=0.3.
3. P(0.4<X<0.8)=0.3.
4. P(X=0.7)=0.

4.43

1. P(X≥22)=0.3085.
2. P(X<22)=0.6915.
3. P(22<X<24)=0.1498.
4. P(X>40)=0.

4.45 μH=2.52, σH=1.41, μF=3.11, σF=1.26. The most important difference is the fact that a single person cannot be a family. This makes the family mean much larger than the household mean and also shrinks the standard deviation. Otherwise, the distribution among sizes 2 through 7 has similar patterns.

4.47

1. 0.8507+0.1448+0.0045=1. It is a legitimate discrete distribution.

3. P(X≥1)=0.1493

4.49

4.51

1. P(X≥0.40)=0.6.
2. P(X=0.40)=0.
3. P(0.40<X<1.40)=0.6.
4. P(0.22≤X≤0.25 or 0.42≤X≤0.45)=0.06.
5. 0.7.

4.53

1. Area=0.5bh=0.5(2)(1)=1.
2. P(Y<1)=0.5.
3. P(Y>1.5)=0.125.
4. P(Y>0.5)=0.875.

4.55

1. P(0.52≤p^≤0.60)=0.9652.
2. P(p^≥0.72)=this is basically 0.

4.57 μx=−0.1.

4.59 As the sample size gets larger, the standard deviation decreases. The mean for 1000 will be much closer to μ than the mean for 2 (or 100) observations.

4.61

1. μZ=−215.
2. μZ=252.
3. μZ=60.
4. μZ=−20.
5. μZ=60.

4.63 σ2=0.890, σ=0.943.

4.65

1. σZ2=576, σZ=24.
2. σZ2=1089, σZ=33.
3. σZ2=13, σZ=3.61.
4. σZ2=13, σZ=3.61.
5. σZ2=52, σZ=7.21.

4.67

1. σZ2=576, σZ=24.
2. σZ2=1089, σZ=33.
3. σZ2=17.8, σZ=4.22.
4. σZ2=8.2, σZ=2.86.
5. σZ2=32.8, σZ=5.73.

4.69

1. The balance point is 1.
2. X1 and X2 each have a mean 0.5. (The square will balance at its center.) Indeed, μY=1=μX1+μX2.

4.71 If D is the result for rolling a single four-sided die, μD=2.5 and σD2=1.25. Then, for the sum I=D1+D2+1, we have μI=6, σI2=2.5, σI=1.5811.

4.73 Let ρ=1, then σ(α+Y)2=σX2+σY2+2(1)σXσY2=σX2+2σXσY+σY2=(σX+σY)(σX+σY)=(σX+σY)2. So, σ(X+Y)=σX+σY.

4.75 To convert from centimeters to inches, we divide by 2.54. μin=176.8/2.54=69.61 and σin=7.2/2.54=2.83.

4.77 Total loss for 5 policies has μ5=$1,500, σ5=$894.43. The average loss for 5 policies has μT/5=$300, σT/5=$178.89. Total loss for 20 policies μ20=$6,000 σ20=$1,788.85. The average loss for 20 policies has μT/20=$300, σT/20=$89.44.

4.79

1. Cannot have negative probabilities, and probability of 1 is certain.
2. P(A)+P(Ac)=P(A)+1−P(A)=1.
3. P(A or B)=P(A)+P(B).
4. P(Ac)=1−P(A).
5. P(A and B)=P(A)P(B).

4.81 P(AUBUC)=0.6.

4.83 P(B|A)=0.5.

4.85 P(A and B)=0.02.

4.87 P(B|A)=0.255.

4.89 Not independent; P(AC|CC)=0.7451≠P(AC)=0.48.

4.93 P(F|A)=0.3913.

4.97 P(A or B)=0.317.

4.99

1. P(A and B)=0.082.
2. P(AC and B)=0.179.
3. P(A and BC)=0.056.
4. P(AC and BC)=0.683.

4.101

1. P(AC|BC)=0.5625.
2. Not independent. 0.72≠0.5625.

4.103 P(at least one offer)=0.9.

4.105 P(B|C)=0.33. P(C|B)=0.2.

4.107

1. If Beth is Aa, then P(Child is non-albino|Beth is Aa)=1/2. P(Child is non-albino|Beth is AA)=1
2. P(Beth is Aa|Child is Aa)=1/2.

4.109 The value will be 4 60% of the time.

4.111

1. μ=5.4, σ=1.96.
2. μy=4μx−2=19.6. σy=7.84.
3. We used the rules about a linear function of a random variable and the fact that the standard deviation is the square root of the variance.

4.113

1. Neither.
2. Independent; the outcome of the first roll will not affect the outcome of the second.
3. Independent; the outcomes of the two rolls do not affect each other.
4. Neither.

4.115

1. μ=5, σ=0.7746.

• (b – c) Answers will vary.

4.117 P(B|A)=P(B and A)/P(A)=P(both heads)/P(A)=0.25/.5=0.5=P(B). Therefore, A and B are independent.

4.119 The probability of a No is 0.5(0.7)=35%, which is half of the actual percent who have not plagiarized. If the probability of plagiarism is 0.2, the probability of a No would be 0.5(0.8)=40%, again half of the actual percent who have not plagiarized. With 39% No answers, we estimate twice as much, 78%, to not have plagiarized, which means we estimate that 22% of the population have indeed plagiarized a paper.

4.121

1. All the probabilities are between 0 and 1 and sum to 1.
2. 0.61.
3. 0.39; 0.39.

4.123 For two-year, 58.11% are public, 41.89% are private. For four-year, 80.50% are public, 19.50% are private. A much larger percent of four-year institutions are public than for two-year institutions. For public, 26.50% are two-year, 73.50% are four-year. For private, 51.76% are two-year, 48.24% are four-year. For public institutions, a much larger percent are four-year versus two-year, whereas for private institutions, they are split about half and half, two-year and four-year.