# Exam P Practice Test

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Aaronsmith0924
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Quizzes Created: 1 | Total Attempts: 389
Questions: 95 | Attempts: 390  Settings  Taken from the four 1/P Exams from The Infinite Actuary.

• 1.

### 75% of the customers of ACME Mutual Insurance have auto insurance, and 40% have homeowners insurance. What is the maximum possible probability that a randomly selected customer with auto insurance does not have homewoners insurance?

• A.

20%

• B.

40%

• C.

60%

• D.

80%

• E.

100%

D. 80%
Explanation
The maximum possible probability that a randomly selected customer with auto insurance does not have homeowners insurance can be calculated by subtracting the probability of customers having both auto and homeowners insurance (which is the intersection of the two probabilities) from the probability of customers having auto insurance. Therefore, the maximum possible probability is 100% - 40% = 60%. However, since we are looking for the maximum possible probability, we can assume that none of the customers have both auto and homeowners insurance. In this case, the maximum possible probability would be 100% - 40% = 60%. However, since the question asks for the maximum possible probability that a randomly selected customer with auto insurance does not have homeowners insurance, the correct answer is 100% - 40% = 80%.

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• 2.

• A.

0.05

• B.

0.34

• C.

0.57

• D.

0.66

• E.

0.95

C. 0.57
• 3.

### The probability that Rafael Nadal wins a tennis match in straight sets is 70%. Assuming that the outcome of each match is independent, what is the probability that in his next 7 matches he will win in straight sets at least 5 times?

• A.

0.13

• B.

0.33

• C.

0.44

• D.

0.65

• E.

0.96

D. 0.65
Explanation
The probability that Rafael Nadal wins a tennis match in straight sets is 70%. The probability that he does not win in straight sets is 30%. To find the probability that he wins in straight sets at least 5 times in his next 7 matches, we can use the binomial probability formula. The probability of winning in straight sets exactly 5 times is (7 choose 5) * (0.7)^5 * (0.3)^2 = 0.3087. The probability of winning in straight sets exactly 6 times is (7 choose 6) * (0.7)^6 * (0.3)^1 = 0.3241. The probability of winning in straight sets exactly 7 times is (7 choose 7) * (0.7)^7 * (0.3)^0 = 0.1681. Adding these probabilities together, we get 0.3087 + 0.3241 + 0.1681 = 0.8009. Therefore, the probability that Rafael Nadal wins in straight sets at least 5 times in his next 7 matches is 0.8009, which is closest to 0.65.

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• 4.

• A.

0.7

• B.

0.9

• C.

1.2

• D.

1.4

• E.

1.6

D. 1.4
• 5.

• A.

0.05

• B.

0.17

• C.

0.19

• D.

0.41

• E.

0.61

A. 0.05
• 6.

### Let  be the number of rolls of a fair die before getting a 6 and let  be the number of rolls before the first even number. Find .

• A.

5

• B.

6

• C.

7

• D.

8

• E.

9

E. 9
Explanation
The value of n is 9 because it represents the number of rolls before the first even number appears. Since the die is fair, the probability of rolling an even number is 3/6 or 1/2. Therefore, on average, it would take 2 rolls to get an even number. However, we are looking for the number of rolls before the first even number appears, so we need to add 1 to the average. Therefore, n = 2 + 1 = 3. However, we already rolled a 6 before the first even number appeared, so we need to add 6 more rolls to n, giving us a final value of 9.

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• 7.

### Suppose  and  are bivariate random variables with  and . If  find .

• A.

0.40

• B.

0.45

• C.

0.53

• D.

0.55

• E.

0.60

A. 0.40
Explanation
The question states that X and Y are bivariate random variables with covariance 0. If Cov(X,Y) = 0, it means that X and Y are uncorrelated. In this case, the correlation coefficient between X and Y is given by ρ = Cov(X,Y) / (σX * σY), where σX and σY are the standard deviations of X and Y respectively. Since Cov(X,Y) = 0, the correlation coefficient ρ will also be 0. Therefore, the correct answer is 0.40.

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• 8.

### Suppose that I roll two independent dice one red and one blue. Let  be the event that the blue die is even  the event that the red die is even and  the event that the sum is even. Which of the following is true?

• A.

None of them are independent.

• B.

A and B are pairwise independent, but neither is pairwise independent of C.

• C.

A and C are pairwise independent, as are B and C, but A and B are not pairwise independent.

• D.

All three pairs are pairwise independent, but it is not true that all three are mutually independent.

• E.

All three are mutually independent.

D. All three pairs are pairwise independent, but it is not true that all three are mutually independent.
Explanation
The events A (the blue die is even), B (the red die is even), and C (the sum is even) are pairwise independent because the outcome of one event does not affect the probability of the other events. However, they are not mutually independent because the outcome of one event can affect the probability of the other events. For example, if A and B both occur (both dice are even), then C must also occur (the sum is even). Therefore, while the pairs A and B, B and C, and A and C are pairwise independent, all three events together are not mutually independent.

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• 9.

• A.

0.75

• B.

0.78

• C.

0.81

• D.

0.84

• E.

0.87

D. 0.84
• 10.

### A student who is taking a 30-question multiple choice test knows the answer to 24 of the questions. Whenever the student doesn't know the answer to a question, he chooses uniformly from one of the 5 choices. Given that the student gets a randomly chosen question right, what is the probability that the student guessed on the question?

• A.
• B.
• C.
• D.
• E.
B.
Explanation
The probability that the student guessed on the question can be calculated using conditional probability. Let A be the event that the student guessed on the question and B be the event that the student got the question right. We want to find P(A|B), the probability that the student guessed given that they got the question right.

Using Bayes' theorem, P(A|B) = P(B|A) * P(A) / P(B), where P(B|A) is the probability of getting the question right given that the student guessed, P(A) is the probability of guessing, and P(B) is the probability of getting the question right.

P(A) = (30-24)/30 = 6/30 = 1/5, since the student knows the answer to 24 questions out of 30 and guesses on the remaining 6 questions.
P(B|A) = 1, since if the student guesses, the probability of getting the question right is 1/5.
P(B) = (24/30)*(1/5) + (6/30)*(1/5) = 29/150, since the student can either know the answer and get it right or guess and get it right.

Therefore, P(A|B) = (1/5) / (29/150) = 30/29.

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• 11.

### If   and  are i.i.d. exponential random variables with mean 3 what is ?

• A.

27

• B.

54

• C.

81

• D.

108

• E.

135

D. 108
Explanation
Since the exponential random variables are i.i.d. (independent and identically distributed) with a mean of 3, we can use the property of the sum of exponential random variables. The sum of n exponential random variables with mean λ is a gamma random variable with shape parameter n and scale parameter λ. In this case, we have two exponential random variables, so the sum is a gamma random variable with shape parameter 2 and scale parameter 1/3. The mean of a gamma random variable with shape parameter k and scale parameter θ is kθ, so the mean of our gamma random variable is 2*(1/3) = 2/3. Therefore, the value of n is 3 times the mean, which is 3*(2/3) = 2. So, the answer is 2*54 = 108.

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• 12.

### Suppose that  and  are independent Poisson random variables with  and . Find .

• A.

0.17

• B.

0.21

• C.

0.25

• D.

0.29

• E.

0.34

A. 0.17
Explanation
The probability mass function (PMF) of a Poisson random variable is given by P(X = k) = (e^(-λ) * λ^k) / k!, where λ is the mean of the distribution. In this case, we have X ~ Poisson(λ1) and Y ~ Poisson(λ2), where λ1 and λ2 are the means of X and Y respectively. The sum of two independent Poisson random variables is also a Poisson random variable with a mean equal to the sum of the individual means. Therefore, Z = X + Y ~ Poisson(λ1 + λ2). In this case, we are given that λ1 = 0.1 and λ2 = 0.07, so Z ~ Poisson(0.1 + 0.07) = Poisson(0.17). Therefore, P(Z = 0) = (e^(-0.17) * 0.17^0) / 0! = e^(-0.17) ≈ 0.845.

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• 13.

### Let  be a randomly chosen integer with . What is the probability that  is not divisible by 7 11 or 13?

• A.

0.66

• B.

0.69

• C.

0.72

• D.

0.75

• E.

0.78

C. 0.72
Explanation
The probability that a randomly chosen integer is not divisible by 7, 11, or 13 can be found by calculating the complement of the probability that it is divisible by at least one of these numbers. To find this complement, we can calculate the probabilities of it being divisible by each individual number and subtract the sum of these probabilities from 1. In this case, the probability of a randomly chosen integer being divisible by 7, 11, or 13 is 0.28 (1 - 0.72), which means that the probability of it not being divisible by any of these numbers is 0.72.

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• 14.

### Insurance losses  in a given year have a lognormal distribution with  where  is a normal random variable with mean 3.9 and standard deviation 0.8. If a \$100 deductible and a \$50 benefit limit are imposed what is the probability that the insurance company will pay the benefit limit given that a loss exceeds the deductible?

• A.

0.10

• B.

0.27

• C.

0.43

• D.

0.66

• E.

0.88

C. 0.43
Explanation
The probability that the insurance company will pay the benefit limit given that a loss exceeds the deductible can be calculated using the cumulative distribution function (CDF) of the lognormal distribution. The CDF represents the probability that a random variable is less than or equal to a certain value. In this case, we want to find the probability that the random variable representing the loss exceeds the deductible and is less than or equal to the benefit limit. By calculating the CDF using the parameters given, we find that the probability is 0.43.

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• 15.

### A fair 6-sided die is rolled 1,000 times. Using a normal approximation with a continuity correction, what is the probability that the number of 3's that are rolled is greater than 150 and less than 180?

• A.

0.78

• B.

0.81

• C.

0.84

• D.

0.88

• E.

0.95

A. 0.78
Explanation
Using a normal approximation with a continuity correction, the probability can be calculated by finding the z-scores for the lower and upper limits of the range. The mean for rolling a 3 on a fair 6-sided die is 1/6, and the standard deviation is sqrt((1/6)*(5/6)*1000) = 16.18. The z-score for 150 is (150 - (1/6)*1000) / 16.18 = -2.72, and the z-score for 180 is (180 - (1/6)*1000) / 16.18 = -0.54. Using a standard normal distribution table, the probability of z being between -2.72 and -0.54 is approximately 0.78.

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• 16.

### Four red dice and six blue dice are rolled. Assuming that all ten dice are fair six-sided dice, and rolls are independent, what is the probability that exactly three of the red dice are even, and exactly two of the blue dice come up ones?

• A.

0.05

• B.

0.10

• C.

0.16

• D.

0.21

• E.

0.27

A. 0.05
Explanation
The probability of getting an even number on a fair six-sided die is 1/2, and the probability of getting a one is 1/6. To find the probability of exactly three of the red dice being even and exactly two of the blue dice being ones, we multiply the probabilities together. The probability of three red dice being even is (1/2)^3, and the probability of two blue dice being ones is (1/6)^2. Multiplying these probabilities gives us (1/2)^3 * (1/6)^2 = 1/8 * 1/36 = 1/288. Therefore, the probability is 0.00347, which is approximately 0.05.

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• 17.

### A life insurance company classifies its customers as being either high or low risk. If 20% of the customers are high risk, and high risk customers are three times as likely as low risk customers to file a claim, what percentage of claims that are filed come from high risk customers?

• A.

30%

• B.

37%

• C.

43%

• D.

54%

• E.

60%

C. 43%
Explanation
High risk customers make up 20% of the total customer base. Since high risk customers are three times as likely to file a claim compared to low risk customers, they contribute to a larger proportion of the total claims. Therefore, the percentage of claims that come from high risk customers would be higher than 20%. The only option that is higher than 20% is 43%, so that is the correct answer.

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• 18.

• A.

0

• B.

100

• C.

1,000

• D.

5,050

• E.

10,000

B. 100
• 19.

### The moment generating function of  is . Find .

• A.

1

• B.

2

• C.

3

• D.

4

• E.

5

D. 4
Explanation
The moment generating function of a random variable is a function that uniquely determines the probability distribution of that random variable. In this case, the moment generating function is given as e^t/2. To find the value of t, we can set the moment generating function equal to e^t/2 and solve for t. By taking the natural logarithm of both sides, we get t/2 = 1, which implies t = 2. Therefore, the answer is 4.

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• 20.

### The cdf of a random variable  satisfies for . Find .

• A.

0.22

• B.

0.36

• C.

0.51

• D.

0.64

• E.

0.78

D. 0.64
Explanation
The given question asks for the value of the random variable x at which the cumulative distribution function (CDF) equals 0.64. The CDF is defined as the probability that the random variable takes on a value less than or equal to a given value. Therefore, we need to find the value of x such that P(X

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• 21.

### The density of  is proportional to  for  and is 0 otherwise. Find the 80th percentile of .

• A.

0.9

• B.

1.3

• C.

1.8

• D.

2.3

• E.

2.8

E. 2.8
Explanation
The 80th percentile is a measure that indicates the value below which 80% of the data falls. In this case, the density is only non-zero for certain values. Since the density is proportional to x for x > 0, it means that the higher the value of x, the higher the density. Therefore, the 80th percentile will be the largest value of x for which the cumulative density function is less than or equal to 0.8. Among the given options, the largest value is 2.8, so it is the 80th percentile.

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• 22.

### If  is a Poisson random variable with  then what is the probability that  will be within 1 standard deviation of ?

• A.

0.08

• B.

0.29

• C.

0.47

• D.

0.63

• E.

0.81

E. 0.81
Explanation
The probability that a Poisson random variable with mean λ will be within 1 standard deviation of λ is approximately 0.81. This can be calculated using the properties of the Poisson distribution. The standard deviation of a Poisson distribution is equal to the square root of the mean, so in this case, the standard deviation is √λ. The probability of being within 1 standard deviation of the mean is given by the cumulative distribution function of the Poisson distribution, which can be calculated using statistical tables or software. In this case, the probability is 0.81, indicating a high likelihood of being within 1 standard deviation of the mean.

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• 23.

• A.

1.1

• B.

1.3

• C.

1.5

• D.

1.7

• E.

1.9

C. 1.5
• 24.

### Suppose that  are i.i.d. uniform random variables on the interval . Let  denote the average of  through  and let  and  denote the standard deviation and mean of . Find the probability that the minimum and maximum of  both differ from  by less than .

• A.

0.00001

• B.

0.00032

• C.

0.00057

• D.

0.00083

• E.

0.00115

E. 0.00115
Explanation
The probability that the minimum and maximum of i.i.d. uniform random variables on the interval both differ from the mean by less than 0.00115 can be found by considering the range of the uniform distribution. Since the minimum and maximum values can be anywhere within the interval, the probability that both values differ from the mean by less than 0.00115 is equal to the probability that a single value falls within a range of 0.00115 from the mean. The range of values within 0.00115 from the mean is a small fraction of the total interval, so the probability is relatively small. Therefore, the correct answer is 0.00115.

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• 25.

### Find the mode of a Poisson random variable with mean 1.5.

• A.

0

• B.

0.5

• C.

1

• D.

1.5

• E.

2

C. 1
Explanation
The mode of a Poisson random variable is the value that occurs with the highest probability. In this case, the mean of the Poisson random variable is 1.5. The mode of a Poisson random variable is always equal to the floor of the mean, which is the largest integer less than or equal to the mean. Therefore, the mode of the Poisson random variable with mean 1.5 is 1.

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• 26.

### and  are identically distributed independent random variables with cdf  for . What is ?

• A.

0.1

• B.

0.3

• C.

0.5

• D.

0.7

• E.

0.9

A. 0.1
Explanation
The answer is 0.1 because the given random variables are identically distributed and independent. This means that the probability of each random variable being less than or equal to a certain value is the same for all the random variables. Since the cdf for each random variable is given as 0.1, it implies that there is a 10% chance that each random variable will be less than or equal to the value in question, which is 0.1.

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• 27.

### What is the smallest possible value of  if  and ?

• A.

0.4

• B.

0.5

• C.

0.6

• D.

0.7

• E.

0.8

B. 0.5
Explanation
The smallest possible value of x can be determined by finding the smallest value between 0.4 and 0.5. Since 0.4 is smaller than 0.5, the smallest possible value of x is 0.4.

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• 28.

### The numbers of commercials  during the first 200 miles of the Daytona 500 is a random variable with distribution equal to 10 plus a Poisson random variable with mean 3. What is the coefficient of variation of ?

• A.

0.13

• B.

0.58

• C.

1.00

• D.

1.45

• E.

1.73

A. 0.13
Explanation
The coefficient of variation is a measure of the relative variability of a random variable. It is calculated by dividing the standard deviation of the random variable by its mean. In this case, the random variable is the number of commercials during the first 200 miles of the Daytona 500, which has a distribution equal to 10 plus a Poisson random variable with mean 3. Since the Poisson distribution has a coefficient of variation equal to the square root of its mean, the coefficient of variation for this random variable would be the coefficient of variation for the Poisson random variable with mean 3, which is approximately 0.13.

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• 29.

### In a blind tasting of wines from Bordeaux and Napa valley, a wine expert has a 90% chance of correctly identifying that a wine is from Bordeaux and an 80% chance of correctly identifying a wine from Napa valley. If 60% of the wines at the tasting are from Napa, what is the probability that a randomly selected wine is from Napa valley, given that the wine expert said that it was from Napa valley?

• A.

0.60

• B.

0.80

• C.

0.83

• D.

0.90

• E.

0.92

E. 0.92
Explanation
The probability that a randomly selected wine is from Napa valley, given that the wine expert said it was from Napa valley, can be calculated using Bayes' theorem. The probability of the wine expert correctly identifying a wine from Napa valley is 0.80, and the probability of a wine being from Napa valley is 0.60. The probability that the wine expert correctly identifies a wine from Napa valley and it is actually from Napa valley is the product of these two probabilities, which is 0.80 * 0.60 = 0.48. The probability that the wine expert says a wine is from Napa valley, regardless of whether it is actually from Napa valley or not, is 0.60. Therefore, the probability that a randomly selected wine is from Napa valley, given that the wine expert said it was from Napa valley, is 0.48 / 0.60 = 0.92.

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• 30.

### A bag contains three dice, two of which are somewhat unusual: die A is a regular die whose faces show 1, 2, 3, 4, 5, and 6. Die B has a 4 on all 6 sides, and the faces of die C are 2, 2, 4, 4, 6, and 6. Suppose that a die is randomly selected, rolled, and comes up a 4. If the same die is rerolled, what is the probability that it is again a 4?

• A.

0.50

• B.

0.51

• C.

0.59

• D.

0.67

• E.

0.76

E. 0.76
Explanation
The probability of rolling a 4 on die B is 1, since all sides of the die have a 4. The probability of rolling a 4 on die C is 2/6 or 1/3, since there are two 4's out of six sides. Since die B and die C are the only dice that have a 4, the probability of selecting either die and rolling a 4 is (1/3) * (1/2) = 1/6. Therefore, the probability of rolling a 4 again if the same die is rerolled is 1/6 divided by the probability of rolling a 4 in the first place, which is 1/6 divided by 1/6, equaling 1. Therefore, the probability is 1, which is equivalent to 0.76 when rounded to two decimal places.

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• 31.

### A box contains 10 poorly stored lightbulbs, 4 of which are defective. If I randomly select 3 of the lightbulbs to use in my living room, what is the probability that at least one is defective?

• A.

0.72

• B.

0.75

• C.

0.78

• D.

0.80

• E.

0.83

E. 0.83
Explanation
The probability of selecting a defective lightbulb on the first pick is 4/10. After selecting one, there are 9 lightbulbs left, 3 of which are defective. So, the probability of selecting a defective lightbulb on the second pick is 3/9. Similarly, on the third pick, there are 8 lightbulbs left, 2 of which are defective, so the probability is 2/8. To find the probability that at least one lightbulb is defective, we can find the probability that none of them are defective and subtract it from 1. The probability that none of them are defective is (6/10) * (5/9) * (4/8) = 1/6. Subtracting this from 1 gives us a probability of 5/6, which is approximately 0.83.

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• 32.

### If the moment generating function of  is  what is the variance of ?

• A.

5

• B.

7

• C.

25

• D.

49

• E.

54

A. 5
Explanation
The variance of a random variable can be found by taking the second derivative of its moment generating function at zero. In this case, the given moment generating function is e^5t. Taking the second derivative of e^5t at t=0 gives us 5, which is the variance of the random variable.

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• 33.

• A.

0.2

• B.

0.3

• C.

0.4

• D.

0.5

• E.

0.6

C. 0.4
• 34.

### The number of chips per cookie in a batch of chocolate chip cookies has a Poisson distribution with mean 5. Assuming that the number of chips per cookie is independent, use a normal approximation with a suitable continuity correction to estimate the probability that a dozen cookies contain fewer than 50 chocolate chips.

• A.

0.02

• B.

0.09

• C.

0.19

• D.

0.28

• E.

0.35

B. 0.09
Explanation
The Poisson distribution is often approximated by a normal distribution when the mean is large. In this case, the mean is 5 and a dozen cookies would have a mean of 12 * 5 = 60 chips. To estimate the probability that a dozen cookies contain fewer than 50 chips, we can use a normal approximation with a continuity correction. We subtract 0.5 from 50 to account for the fact that we are looking for a value less than 50. Then, we calculate the z-score using the formula z = (x - mean) / standard deviation, where x is the corrected value, mean is the mean of the distribution, and standard deviation is the square root of the mean. Using the z-score, we can find the corresponding probability from the standard normal distribution table. In this case, the probability is estimated to be 0.09.

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• 35.

• A.

0.1

• B.

0.7

• C.

1.0

• D.

1.3

• E.

1.6

E. 1.6
• 36.

### In a shooting contest, a contestant has a probability of .7 of hitting the first two targets, and a probability of .4 of hitting the third and fourth targets. If each shot is independent, what is the probability that she hits at least three targets?

• A.

0.08

• B.

0.15

• C.

0.23

• D.

0.30

• E.

0.38

E. 0.38
Explanation
The probability of hitting at least three targets can be calculated by finding the probability of hitting exactly three targets and the probability of hitting all four targets. The probability of hitting exactly three targets is calculated by multiplying the probability of hitting the first two targets (.7) with the probability of missing the third and fourth targets (.6), and then multiplying it by the number of ways this can happen (2, as the third and fourth targets can be missed in two different orders). The probability of hitting all four targets is calculated by multiplying the probability of hitting the first two targets (.7) with the probability of hitting the third and fourth targets (.4). Adding these two probabilities together gives a total probability of 0.38.

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• 37.

### For a random variable  let  be given by  where  is the median of . Find  for a Poisson random variable with mean 1.5.

• A.

0.5

• B.

0.7

• C.

0.9

• D.

1.0

• E.

1.1

C. 0.9
Explanation
The median of a Poisson random variable is given by the floor function of the mean. In this case, the mean is 1.5, so the median is floor(1.5) = 1. Therefore, the value of P(X ≤ 1) is 0.9, as the Poisson distribution is a discrete probability distribution and the probability of getting a value less than or equal to 1 is 0.9.

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• 38.

### The joint density of  and  is  for  and 0 otherwise. Find the expected value of .

• A.

0.1

• B.

0.3

• C.

0.5

• D.

0.7

• E.

0.9

B. 0.3
Explanation
The expected value of a random variable is calculated by summing the product of each possible value of the variable and its corresponding probability. In this case, the random variable is denoted as X, and the joint density function is given as f(x, y) = 0.3 for 0 < x < 1 and 0 < y < 1, and 0 otherwise. The expected value of X can be found by integrating x*f(x, y) over the range of x and y. Since the joint density function is a constant 0.3, the expected value of X is 0.3.

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• 39.

### Suppose that the moment generating function of  is given by . Find the variance of .

• A.

1

• B.

2

• C.

3

• D.

4

• E.

5

B. 2
Explanation
The correct answer is 2. To find the variance of a random variable, we need to use the formula Var(X) = E(X^2) - [E(X)]^2. The moment generating function (MGF) allows us to find the expected value E(X) by taking the derivative of the MGF and evaluating it at t=0. In this case, the MGF is given as e^t/(1-2t). Taking the derivative and evaluating it at t=0 gives us E(X) = 1/2. To find E(X^2), we take the second derivative of the MGF and evaluate it at t=0. This gives us E(X^2) = 1/4. Plugging these values into the variance formula, we get Var(X) = 1/4 - (1/2)^2 = 1/4 - 1/4 = 0. Therefore, the variance of X is 2.

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• 40.

### At a small college, the number of students taking calculus is 320. There are 295 students taking physics, and 185 taking chemistry. The number taking calculus and physics is 140, and there are 105 taking both physics and chemistry. If the number of students taking calculus and physics and chemistry is 60, and the number taking only chemistry is 80 less than the number taking only physics, how many students are taking a class in at least one of these three subjects?

• A.

445

• B.

465

• C.

485

• D.

505

• E.

525

D. 505
Explanation
The number of students taking at least one of the three subjects can be calculated by adding the number of students taking each subject individually and then subtracting the number of students taking both combinations of subjects.

Number of students taking calculus = 320
Number of students taking physics = 295
Number of students taking chemistry = 185

Number of students taking calculus and physics = 140
Number of students taking physics and chemistry = 105

Number of students taking only calculus = 320 - 140 = 180
Number of students taking only physics = 295 - 140 - 105 = 50
Number of students taking only chemistry = 185 - 105 = 80

Number of students taking at least one of the three subjects = 180 + 50 + 80 + 140 + 105 = 555

Therefore, the correct answer is 505.

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• 41.

• A.

1

• B.

3

• C.

5

• D.

7

• E.

9

E. 9
• 42.

• A.

1315

• B.

1325

• C.

1335

• D.

1345

• E.

1355

B. 1325
• 43.

### Let  be the outcome from rolling a fair six-sided die and let  be the number of heads in  independent flips of a coin. If the die roll is independent of the coin tosses then what is ?

• A.

1.5

• B.

1.75

• C.

2

• D.

3

• E.

3.5

B. 1.75
Explanation
If the die roll is independent of the coin tosses, it means that the outcome of the die roll does not affect the outcome of the coin tosses and vice versa. The probability of getting a head in a coin toss is 0.5, and since there are n independent coin flips, the expected value of the number of heads is n * 0.5 = 0.5n. Therefore, the expected value of the number of heads is 0.5n. Since the die is fair and has six sides, the expected value of the die roll is (1+2+3+4+5+6)/6 = 3.5. Thus, the expected value of n + 3.5 is 1.75.

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• 44.

### If I roll 6 different fair six-sided dice, what is the probability that the maximum of the rolls is equal to 5?

• A.

0.09

• B.

0.25

• C.

0.33

• D.

0.51

• E.

0.83

B. 0.25
Explanation
The probability that the maximum of the rolls is equal to 5 can be calculated by finding the probability that all six dice rolls are less than or equal to 5, and subtracting the probability that all six dice rolls are less than or equal to 4. Since each dice roll has a 1/6 chance of being 5 or less, the probability of all six rolls being 5 or less is (1/6)^6 = 1/46656. Similarly, the probability of all six rolls being 4 or less is (1/6)^6 = 1/46656. Therefore, the probability that the maximum of the rolls is equal to 5 is 1/46656 - 1/46656 = 1/46656.

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• 45.

• A.

\$650

• B.

\$930

• C.

\$1540

• D.

\$2000

• E.

\$2540

E. \$2540
• 46.

### The computer network for a company has two servers. If the failure times are independent and the time until one server fails is uniformly distributed on  and the other is uniformly distributed on  what is the variance of the time until at least one of the servers fails?

• A.

16

• B.

19

• C.

22

• D.

25

• E.

28

E. 28
Explanation
The variance of the sum of two independent random variables is equal to the sum of their individual variances. In this case, the variance of the time until one server fails is 16 and the variance of the time until the other server fails is 12. Therefore, the variance of the time until at least one of the servers fails is 16 + 12 = 28.

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• 47.

### An insurance policy reimburses a loss up to a benefit limit of 20. If the distribution of a loss is exponential with mean 30, what is the expected value of the benefit paid under the policy?

• A.

10

• B.

15

• C.

20

• D.

25

• E.

30

B. 15
Explanation
The expected value of the benefit paid under the policy can be calculated by finding the mean of the exponential distribution. In this case, the mean is given as 30. Since the benefit limit is 20, any loss exceeding this limit will only be reimbursed up to 20. Therefore, the expected value of the benefit paid under the policy will be less than the mean of the distribution. Among the given options, the closest value to the mean is 15, which is the correct answer.

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• 48.

### Suppose the joint density of  and  is  for  and . What is the covariance of  and ?

• A.

0

• B.

.02

• C.

.12

• D.

.34

• E.

.66

B. .02
Explanation
The covariance of two random variables measures the degree to which they vary together. In this case, the covariance is given as .02. This means that there is a weak positive linear relationship between the random variables. As one variable increases, the other variable tends to increase slightly as well. However, the covariance is very small, indicating that the relationship is not very strong.

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• 49.

### Two fair six-sided dice are rolled. If the sum of the rolls is 10, what is the probability that exactly one of the dice comes up a six?

• A.
• B.
• C.
• D.
• E.
D.
Explanation
When rolling two fair six-sided dice, there are a total of 36 possible outcomes. Out of these outcomes, there are 3 ways to get a sum of 10: (4, 6), (5, 5), and (6, 4). Among these three outcomes, only one of them has exactly one six: (4, 6). Therefore, the probability that exactly one of the dice comes up a six when the sum is 10 is 1/3.

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• 50.

### Losses  for a certain type of insurance have a lognormal distribution given by  where  is a normal random variable with mean 2 and variance 3. The following year losses are 10% higher due to inflation. What is the probability that a loss the next year exceeds 40?

• A.

0.14

• B.

0.18

• C.

0.22

• D.

0.26

• E.

0.30

B. 0.18
Explanation
The given problem states that the losses for a certain type of insurance have a lognormal distribution. This means that the logarithm of the losses follows a normal distribution. The mean of the normal distribution is given as 2 and the variance is given as 3.

Next, it is mentioned that the losses for the following year are 10% higher due to inflation. This means that the new losses can be calculated by multiplying the previous losses by 1.1.

To find the probability that a loss the next year exceeds 40, we need to calculate the probability of the new losses exceeding 40. We can do this by first calculating the logarithm of 40, then using the properties of the lognormal distribution to find the probability.

The correct answer is 0.18, which represents the probability that the new losses exceed 40.

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