exam p practice exam by jared nakamura

Exam P Practice Exam© by Jared Nakamura, March 28, 2015

This is an original sample exam to be used freely by those preparing for Exam 1/P.

This exam is based off the Society of Actuaries’ September 2014 syllabus for Exam 1/P.

This exam consists of 30 questions with 5 answer choices each to be taken in under 3 hours.

The pages after question 30 of this exam contain the solutions.

This exam may or may not reflect the format, content, or difficulty of the actual exam.

Question 1 begins on the next page.

1. Let X be a binomial random variable such that 16*P(X = 1) = 4*Var(X) = E(X).

What is E(X)?

A. 1.5

B. 2

C. 2.25

D. 2.5

E. 3.25

2. Let X1, X2, and X3 be independent identically distributed normal variables with mean μ and standard deviation σ.

What is the probability that the second largest of these variables is between μ - σ and μ + σ?

A. 0.483

B. 0.683

C. 0.717

D. 0.779

E. 0.865

3. The chief security officer of a large business is having two meetings, one after the other, with lengths of time modeled by random variables X and Y. The joint density function fX,Y(x, y) is uniform over the region

0.5|x - 10| + 0.2|y - 12| ≤ 1.

What is the expected length of time from the beginning of the first meeting to the end of the second meeting?

A. 11

B. 21

C. 22

D. 23

E. 80

4. An urn contains 10 balls: 4 red and 6 blue. A number n is chosen randomly and uniformly between 1 and 10 inclusive, then n balls are drawn from the urn without replacement.

What is the probability that all red balls are drawn?

A. 0.19

B. 0.22

C. 0.35

D. 0.50

E. 0.60

5. In the next 50 years, the probability that a certain house experiences damage from an earthquake is 0.6, given the house experiences damage from a landslide. The probability that the house experiences damage from a landslide is 0.5, given the house experiences damage from an earthquake.

What is the probability that the house experiences damage from just a landslide given the house experiences damage from either an earthquake or a landslide in the next 50 years?

A. 0.20

B. 0.25

C. 0.30

D. 0.38

E. 0.62

6. An actuary models the damage amount of a particular class of automobile accidents as an exponential variable with mean 400. Suppose policyholders A and B hold policies under this class of accidents with deductibles of 100 and 200 respectively.

If both policyholders get into an accident which falls under this particular class, what is the probability that the payout to policyholder A exceeds the payout to policyholder B?

A. 0.54

B. 0.64

C. 0.74

D. 0.84

E. 0.94

7. Suppose the amount of a surgical claim is uniformly distributed on the interval [30, 90]. Given the surgical claim amount a, the size of the actual payout by the insurance company is uniformly distributed on the interval [0, a]. Finally, the actual payout is reduced by ten percent for fees and given to the policyholder as benefit.

What is the variance of the final benefit?

A. 97

B. 128

C. 155

D. 324

E. 400

8. The joint moment generating function for random variables X and Y is

MX,Y(t1, t2) = (2t1t2 - 2t1 - t2 + 1)-1.

What is the correlation coefficient between X and Y?

A. -2

B. -1

C. 0

D. 1

E. 2

9. Assuming the claim for a certain type of umbrella policy is greater than the deductible, an actuary determines the proportion of the claim that exceeds the deductible as payout follows the distribution

fX(x) = cx-1/3, 0 < x < 1,fX(x) = 0 elsewhere,

where c is an appropriate constant. Suppose three such claims are made.

Assuming these claims are independent, what is the probability that the payout of the maximum of these claims is greater than its deductible?

A. 0.125

B. 0.25

C. 0.5

D. 0.75

E. 0.875

10. The sum of two independent identically distributed random variables X and Y has moment generating function

MX+Y(t) = 0.25(4 - 4t + t2)/(1 - 2t + t2).

What is P(X < 1)?

A. 1 - 0.5e-1

B. 0.5 - e-1

C. 0.5e-1

D. 1 - e-0.5

E. e-0.5

11. The joint cumulative distribution function for two random variables X and Y is

FX,Y(x, y) = 3xy/(1 + 2y) , 0 < x < 1, 0 < y < 1.

What is P(X > 0.5 | X + Y = 1)?

A. 0.25

B. 0.50

C. 0.60

D. 0.75

E. 0.91

12. Let X1, X2, …, X75 be 75 independent identically distributed Poisson variables with mean 3.

What is the probability that the average of these variables exceeds 3 given that each of these variables is at least 1?

A. 0.80

B. 0.90

C. 0.94

D. 0.97

E. 0.99

13. Let X1, X2, X3, X4, and X5 be five random variables such that the variance of the sum of any three of them is 300,000 and the variance of the sum of all five of them is 400,000.

What is the sum of the variances of all Xi?

A. 100,000

B. 400,000

C. 500,000

D. 600,000

E. 900,000

14. An insurance company decides to use a rating system to describe the intensity of a thunderstorm. The rating of a thunderstorm is modeled by a random variable X supported over all positive integers. An actuary determines the following:

1) P(X = a)*P(X = a + 2) = P(X = a + 1)2 for all positive integers a.

2) P(X = 1) = 0.75.

3) The expected damage for a given rating a is equal to 90a2.

What is the expected damage from a single thunderstorm?

A. 160

B. 200

C. 320

D. 720

E. 2520

15. Suppose A, B, and C are pairwise independent events such that P(A ∩ B ∩ C) = 0 and P(A) = P(B) = P(C). What is the maximum value of P(A ∪ B ∪ C)?

A. 0

B. 1/2

C. 2/3

D. 3/4

E. 1

16. A bivariate normal variable in X and Y satisfies P(X > Y + 50) = 0.9 and P(X > Y + 72) = 0.8. What is the difference between E(X) and E(Y)?

A. 28

B. 50

C. 108

D. 114

E. 138

17. The cumulative distribution function for X is

FX(x) = 0, x < 1,

FX(x) = k*ln(x), 1 < x < 3,

FX(x) = 1, x > 3,

where k is an appropriate constant. Let Y = ln(X).

What is the variance of Y?

A. 0.10

B. 0.20

C. 0.30

D. 0.40

E. 0.50

18. Let X and Y be two independent continuous variables with means 500 and 200 and standard deviations 8 and 6 respectively.

What is the smallest possible value for P(275 < X - Y < 325)?

A. 0.00

B. 0.04

C. 0.16

D. 0.69

E. 0.84

19. The perks given to various policyholders depend on their level of membership. A membership level can be any integer between 1 and 10 inclusive. A policyholder with a given level of membership also has all lower levels of membership, and all policyholders have a membership level. The probability that a policyholder has a level of membership given he/she has the next lower membership is constant over all such levels, and the probability that a randomly chosen policyholder has a level 10 membership is 0.001.

What is the probability that a randomly chosen policyholder has a level 3 membership?

A. 0.100

B. 0.126

C. 0.215

D. 0.250

E. 0.933

20. The joint probability density of random variables X and Y is

fX,Y(x, y) = 4/3 + x + y, 0 < x, 0 < y, x + y < 1,

fX,Y(x, y) = 0 elsewhere.

What is P(2Y2 > X)?

A. 0.426

B. 0.488

C. 0.512

D. 0.574

E. 0.831

21. Within a given month, n claims fell below the deductible. If 2n claims are chosen at random from all claims made during the month, it is expected that 4 of those claims fall below the deductible. If n + 6 claims are chosen at random from all claims made during the month, it is expected that 3 of those claims fall below the deductible.

How many claims were made during the month?

A. 12

B. 18

C. 72

D. 162

E. 288

22. The number n of the first customer to make a purchase at grocery store A follows a geometric distribution with mean μ, supported only on all positive integers n. The number m of the first customer to make a purchase at grocery store B follows an independent identical distribution.

What is the probability that m and n are the same on any given day?

A. (2μ - 1)-1

B. μ-1

C. (2μ - 1)-2

D. μ(2μ - 1)-2

E. (2μ - 1)μ-2

23. The joint probability distribution function for random variables X and Y is

fX,Y(x, y) = k + xy/2, 1 < x, y < 2,

fX,Y(x, y) = 0 elsewhere,

where k is an appropriate constant.

What is the conditional distribution of X given Y = y over the region 1 < x < 2, 1 < y < 2?

A. (6x + 1)/8

B. (6y + 1)/8

C. (1 + 4xy)/8

D. (4xy - 1)/(6x - 1)

E. (4xy - 1)/(6y - 1)

24. Suppose X and Y are two random variables such that Var(X) = Var(Y) = 100 and Cov(X, Y) = 10. Let W = 2X - Y and Z = X + 2Y.

What is Cov(W, Z)?

A. 0

B. 30

C. 50

D. 430

E. 450

25. The joint probability density function fX,Y(x, y) for random variables X and Y is uniform over the region 0 < x < 1, 0 < y < 2 - x. Let Z = Y / X.

What is the cumulative distribution function FZ(z) for Z when z > 1?

A. z/3

B. z/(z + 2)

C. 1/(2z + 1)

D. (3z - 1)/(3z + 3)

E. 1/3 + (2z - 2)/(3z - 1)

26. Let A, B, and C be mutually independent events such that P(C) > 0. You are given the following:

1) P(A ∩ B ∩ C) = 0.4P((A ∩ B) | C)

2) P((A ∩ C) ∪ (B ∩ C)) = 0.5P(C | A)

What is P((A ∪ B) | C)?

A. 0.2

B. 0.3

C. 0.4

D. 0.5

E. 0.8

27. Let W, X, Y, and Z be independent identically distributed variables with a uniform distribution among the integers 1 through 20 inclusive.

What is P(W = 9 | W + X + Y + Z = 17)?

A. 1/70

B. 2/105

C. 7/360

D. 1/40

E. 3/80

28. A couple receives a 5-year life insurance policy with a one-time premium of 5,000. The death benefit is 50,000 received if the husband or wife does not survive at least 5 years. The expected excess of premiums over claims is 1,000. Assume the survival probabilities of the husband and wife are independent and the probability that the husband does not survive at least 5 years is twice the probability that the wife does not survive at least 5 years.

What is the probability that the husband survives at least 5 years?

A. 0.027

B. 0.054

C. 0.892

D. 0.948

E. 0.973

29. A device has three components, A, B, and C. The device fails if any component fails. The lifetimes of A, B, and C are independent exponential variables with means 1.4, 1.6, and 2 years.

What is the probability that the lifetime of the device is less than one year?

A. 0.16

B. 0.48

C. 0.52

D. 0.84

E. 0.99

30. The value of the damage done to a particular structure due to a fire is determined to be a random variable X which is exponentially distributed such that E(X | X > 120) = 140. The claim for damages is subject to a deductible of 20 and a payout limit of 60. Let the final payout be modeled by a random variable Y.

What is the moment generating function MY(t) for Y?

A. 1 + 20t(e60t - 4 - e-1)/(20t - 1)

B. 1/(1 - 20t) + e60t - 4 - e-4

C. (e-1 - e-4)/(1 - 20t) + 1 + e-4 - e-1

D. (20te60t - 3 - 1)/(20t - 1)

E. 20(te60t - 4 - 1)/(20t - 1)

Solutions

1. C2. E3. C4. B5. B6. A7. D8. C9. D10. A11. D12. A13. D14. B15. D16. D17. A18. E19. C20. A21. C22. A23. E24. B25. D26. D27. E28. D29. D30. A

The solutions are given difficulties roughly based on the following measures:

Difficulty 1 requires basic understanding or key formulae. The math is never messy. They are among the easiest questions on a real exam. ~2 minutes for a proficient exam taker.

Difficulty 2 requires intermediate understanding and some work. They are among the easier questions on a real exam. ~4 minutes for a proficient exam taker.

Difficulty 3 requires advanced understanding, a moderate level of work, and/or some small mathematical insight. They can be linear but long, or short but insightful. These are average questions on a real exam. ~6 minutes for a proficient exam taker.

Difficulty 4 requires thorough understanding, lots of work, and/or insight. They can be linear but ugly, or short but greatly insightful. These questions are among the harder questions on a real exam. ~8 minutes for a proficient exam taker.

Difficulty 5 requires mastery, tons of work, and/or great insight (usually all three). They are never linear or short, and sometimes combine multiple syllabus sections. These questions are equal to the hardest questions expected on a real exam. ~10 minutes for a proficient exam taker.

Solution to 1. (Difficulty 3)

We begin this question by writing down the formulae for the probability mass function at 1, the variance, and the expected value for a binomial distribution with parameters n and p:

P(X = 1) = nC1 * p1 * (1 - p)n - 1 = np(1 - p)n - 1

Var(X) = np(1 - p)

E(X) = np

Substitute the formulae into what we are given in the problem statement:

4*Var(X) = E(X) ⇔ 4np(1 - p) = np ⇔ 4(1 - p) = 1 ⇔ p = 3/4

16*P(X = 1) = E(X) ⇔ 16np(1 - p)n - 1 = np ⇔ 16(1/4)n - 1 = 1 ⇔ n = 3

Finally, E(X) = np = (3)(3/4) = 2.25


The probability that a normal variable is within one standard deviation from the mean is around 68 percent. This means that the probability that a normal variable is above or below one standard deviation above or below the mean respectively is around 1/2*(100 - 68) = 16 percent. Separate the standard normal into three regions: 1) the left tail, 2) the middle region within one standard deviation from the mean, and 3) the right tail.

Randomly choose standard normal variable A from these three regions, then normal variable B , then normal variable C. There are four possibilities where the middle variable is in region 2:

• all three variables in region 2• smallest variable in 1, other two in 2• largest variable in 3, other two in 2• smallest variable in 1, middle in 2, largest in 3

The second and third choices can happen three different ways depending on which of A, B, or C are in the tail region. The fourth choice can happen six different ways, or 3! ways. Therefore:

P(all three variables in region 2) = 0.683 = 0.314

P(smallest variable in 1, other two in 2) = 3 * 0.16 * 0.682 = 0.222

P(largest variable in 3, other two in 2) = 3 * 0.16 * 0.682 = 0.222

P(smallest variable in 1, middle in 2, largest in 3) = 6 * 0.162 * 0.68 = 0.104

Adding these up, we get 0.862, which is roughly 0.865.


This is a quick problem which requires insight. In short, the region is symmetric in the x and y directions about the point (10, 12), so the line x + y = c is just as much in the region as the same line reflected about x + y = 22. Therefore, the marginal distribution of x + y is symmetric about 22, so E(X + Y) is 22.

Remarks: To see that the region is symmetric, substitute (20 - x) for x or (24 - y) for y. Notice these transformations reflect the region about x = 10 and y = 12 respectively and the inequality remains the same. Then, assume y is greater than or equal to 12 and x is greater than or equal to 12 to eliminate the absolute value signs. The boundary to the inequality will end up being a line, and the line reflected about x = 10, y = 12, and both x = 10 and y = 12 will yield a diamond-shaped region.

Unfortunately, if you solve this problem by integrating the piecewise analytic marginal distribution of x and y, it could take a while.


Draw all ten balls and line up the ten balls in the order they are drawn. Uniformly choose n from among 1 through 10 and put a partition after the nth ball. Assuming four of the balls to the left of the partition are red, then there are n choose 4 ways this can happen. Without the assumption, there were 10 choose 4 ways to draw the ten balls. Therefore, given that the integer n was chosen, the probability that all four red balls are drawn is nC4 / 10C4. The probability that a particular n was chosen is ten percent because the distribution is uniform. Therefore, the probability that all four red balls are drawn is:

(sum from b = 1 to 10) [0.1 * nC4 / 10C4] = 0.1 / 10C4 * (sum from b = 1 to 10)[bC4]

= 0.1 / 210 * (0 + 0 + 0 + 1 + 5 + 15 + 35 + 70 + 126 + 210)

= 0.22

Remarks: Those who have studied the free SOA sample problems may recognize the first sentence of this question, question number 4, as being identical to the first sentence of question number 4 in the free SOA sample problems.


Let E and L represent the event that the house experiences damage from an earthquake and landslide respectively. Then we want to find P(L \ E | E ∪ L). We can rewrite this as:

P(L \ E | E ∪ L) = P((L \ E) ∩ (E ∪ L))/ P(E ∪ L)

Because L \ E is a subset of E ∪ L, (L \ E) ∩ (E ∪ L) = (L \ E):

P(L \ E | E ∪ L) = P(L \ E) / P(E ∪ L)

From the addition rule, P(L \ E) = P(L) - P(E ∩ L). Also, from the inclusion-exclusion principle, P(E ∪ L) = P(E) + P(L) - P(E ∩ L). Therefore:

P(L \ E | E ∪ L) = (P(L) - P(E ∩ L))/(P(E) + P(L) - P(E ∩ L))

From the problem statement:

P(E | L) = 0.6 = P(E ∩ L) / P(L), which implies that P(L) = P(E ∩ L) / 0.6.

P(L | E) = 0.5 = P(E ∩ L) / P(E), which implies that P(E) = P(E ∩ L) / 0.5.

We can now substitute P(E ∩ L) / 0.6 for P(L) and P(E ∩ L) / 0.5 for P(E) and get P(E ∩ L) in the numerator and denominator, which will cancel out:

P(L \ E | E ∪ L) = (P(L) - P(E ∩ L)) / (P(E) + P(L) - P(E ∩ L))

= (P(E ∩ L) / 0.6 - P(E ∩ L)) / (P(E ∩ L) / 0.5 + P(E ∩ L) / 0.6 - P(E ∩ L))

= (1 / 0.6 - 1) / (1 / 0.5 + 1 / 0.6 - 1)

= 0.25

Remarks: My favorite way to solve problems like this is to use the partition method. Draw the Venn Diagram and label the outside region (E’ ∩ L’) a, the E but not in L region (E ∩ L’) b, the E and L region (E ∩ L) c, and the L but not in E region (E’ ∩ L) d. Then we have the equations:

c / (c + d) = 0.5

c / (c + b) = 0.6

and we want to solve for d / (b + c + d). Solve the above two equations for d and b, substitute into what we want to solve, and the c’s will cancel.


Let A and B be the damage to policyholders A and B respectively. There are four cases: 1) A and B are below deductible, 2) A is above and B is below deductible, 3) A is below and B is above deductible, and 4) A and B are above deductible. We are interested in 2 and 4 since the payout to policyholder A can be greater than the payout to policyholder B only in these two cases.

Note that A and B are distributed as f(x) = e-x / 400 / 400.

Case 2) The probability that A is above deductible is e-100 / 400. This is a result of integrating f(x) from 100 to infinity.

The probability that B is below deductible is 1 - e-200 / 400. This is a result of integrating f(x) from 0 to 200.

The probability that A is above and B is below deductible is e-100 / 400 * (1 - e-200 / 400) = 0.3064

Case 4) The probability that A is above deductible is e-100 / 400.

The probability that B is above deductible is e-200 / 400.

Here is the key insight: Assuming A and B are above deductible, the distribution of their payouts are identical. This is due to the memoryless exponential distribution along with the idea that the payout after a deductible is a shift to the left by the amount of the deductible. Because their distributions are identical exponential distributions, the probability that the payout of A is greater than the payout of B given that both exceed the deductible is 0.5. Therefore, the probability that A and B are above deductible and payout A is greater than payout B is:

e-100 / 400 * e-200 / 400 * 0.5 = 0.2362

Adding Case 2 and Case 4 probabilities, we get:

P(A pays more than B) = 0.3064 + 0.2362 = 0.54

Remarks: Without the key insight, solving this problem using convolutions is hell.


This is a direct application of the law of total variance. Let A be the surgical claim amount random variable a, and let B be the actual payout random variable. Then:

Var(B) = E(Var(B | A)) + Var(E(B | A))

= E(A2 / 12) + Var(A / 2)

= E(A2) / 12 + Var(A) / 4

= (Var(A) + E(A)2) / 12 + Var(A) / 4

= ((90 - 30)2 / 12 + (60)2) / 12 + (90 - 30)2 / 48

= 400

If B is reduced by ten percent, then B is multiplied by 0.9. Therefore, the variance of the final benefit is:

Var(0.9B) = 0.81 * Var(B) = 324


We can rearrange the denominator of the joint moment generating function:

2t1t2 - 2t1 - t2 + 1 = 2t1 * (t2 - 1) - (t2 - 1) = (2t1 - 1) * (t2 - 1)

Therefore, the joint moment generating function is the product of a function of the moment generating variable of X and a function of the moment generating variable of Y. Whenever this is the case, the underlying variables are independent. Therefore, the correlation coefficient is 0.

Remarks: In fact, X is an exponential distribution with mean 1/2 and Y is an exponential distribution with mean 1.


Integrating f(x) from 0 to 1 and setting it equal to 1:

(integral from x = 0 to 1) [c * x-1/3] = 3/2 * c * (1)2/3 - 3/2 * c * (0)2/3 = 3/2 * c = 1

We get that the normalizing constant c equals 2/3.

If we want the payout to be greater than the deductible, then the proportion of the claim that exceeds the deductible has to be greater than one half. However, because we want the probability that any of the three claims exceeds the deductible, this direct route will be infeasible. Instead, we compute the complement, the probability that all three claims are below the deductible.

The probability that the payout is less than the deductible is:

(integral from x = 0 to 1/2) [2/3 * x-1/3] dx = 2/3 * (integral from x = 0 to 1/2) [x-1/3] dx

= 2/3 * (3/2 * (1/2)2/3 - 3/2 * (0)2/3) = (1/2)2/3

The probability that all three claims have payouts less than the deductible is therefore ((1/2)2/3)3 = 1/4.

Finally, the probability that the maximum of these claims exceeds the deductible is 1 - 1/4 = 0.75.


Because X and Y are independent and identically distributed, the moment generating function of their sum is the square of either moment generating function, as they are equivalent. We can see that:

MX + Y(t) = 0.25(4 - 4t + t2)/(1 - 2t + t2) = (0.5(2 - t)/(1 - t))2

Therefore, using polynomial division:

MX(t) = 0.5(2 - t)/(1 - t) = 0.5 + 0.5/(1 - t)

This is the moment generating function of a random variable which is 0 half the time and which is exponential with mean 1 the other half of the time, otherwise known as the moment generating function of X. Consider two cases:

1) Given that X is 0, the probability that X is less than 1 is 1.

2) Given that X is exponential, the probability that X is less than 1 is the integral from x = 0 to 1 of the exponential distribution with mean 1, which is 1 - e-1.

Therefore, the probability that X is less than 1 is:

P(X < 1 | X = 0) * P(X = 0) + P(X < 1 | X > 0) * P(X > 0) = 0.5(1) + 0.5(1 - e-1) = 1 - 0.5e-1


There are many simple mistakes to be made in this problem.

Note that the expression given is the joint cumulative distribution, so we need to take the derivative with respect to x and y to get the probability distribution. Upon doing this, we see that the joint probability density function is:

fX, Y = 6/(1+2y)2

Substitute 1 - x in for y to get f(x | y = 1 - x) = 6/(3 - 2x)2. In order to normalize this conditional distribution, divide f(x | y = 1 - x) by the integral from x = 0 to 1 of f(x | y = 1 - x). The definite integral is:

(integral from x = 0 to 1) [6/(3 - 2x)2] dx = 3/(3-2(1)) - 3/(3 - 2(0)) = 2

The conditional distribution is thus 3/(3-2x)2. To find P(X > 0.5), integrate the conditional distribution from 0.5 to 1 to get:

(integral from x = 0 to 1) [3/(3 - 2x)2] dx = 3/(2(3-2(1))) - 3/(2(3-2(0.5))) = 0.75


Using the probability mass function of the Poisson variable with mean 3, the probability that it equals 0 is 30 / 0! * e-3 = e-3. Likewise, the probability that it is greater than 0 is 1 - e-3. Therefore, we have the following equation:

e-3 * E(X | X = 0) + (1 - e-3) * E(X | X > 0) = E(X)

Knowing that E(X | X = 0) = 0 and E(X) = 3, we have E(X | X > 0) = 3/(1 - e-3) = 3.157.

The variance of the Poisson variable is equal to the mean, which is 3. The general formula for the variance is E(X2) - E(X)2, and since E(X) = 3, it must be that E(X2) = 12. Therefore, we have the following equation:

e-3 * E(X2 | X = 0) + (1 - e-3) * E(X2 | X > 0) = E(X2)

Knowing that E(X2 | X = 0) = 0 and E(X2) = 12, we have E(X2 | X > 0) = 12/(1 - e-3).

Therefore:

Var(X | X > 0) = E(X2 | X > 0) - E(X | X > 0)2 = 12/(1 - e-3) - 9/(1 - e-3)2 = 2.661

The average of 75 independent variables with mean 3.157 and variance 2.661 has mean 3.157 and variance 2.661/75 = 0.0355. By the central limit theorem, this is approximately normal with mean 3.157 and standard deviation 0.03551/2 = 0.188. Using the Z table with a z score of (3.157 - 3) / 0.188 = 0.835, we see that the probability the average is greater than 3 is 0.80.


Three of the five random variables can be chosen 5C3 = 10 ways.

Consider the sum of the variances over all such threesomes. The variance of one particular variable, for example Var(X1), appears in 4C2 = 6 of these variances. The covariance of two different variables, for example Cov(X1, X2), appears in 3C1 = 3 of these variances. Note also that the covariance of two different variables in the formula for the variance of the sum of variables is always doubled. Therefore:

6 * (sum over all variances) + 3 * 2 * (sum over all covariances) = 10 * 300,000

We are given the variance of the sum of all variables. The variance of the sum of variables is equal to the sum of the variance of the variables plus two times the sum of the covariances of each pair of different variables. Therefore:

(sum over all variances) + 2 * (sum over all covariances) = 400,000

Let x equal (sum over all variances) and let y equal (sum over all covariances). Then we need to solve for x in the following system:

6x + 6y = 3,000,000

x + 2y = 400,000

Solving for x, we get x = 600,000.

Remarks: For clarity,

(sum over all variances) = Var(X1) + Var(X2) + Var(X3) + Var(X4) + Var(X5)

(sum over all covariances) = Cov(X1, X2) + Cov(X1, X3) + Cov(X1, X4) + Cov(X1, X5) + Cov(X2, X3) + Cov(X2, X4) + Cov(X2, X5) + Cov(X3, X4) + Cov(X3, X5) + Cov(X4, X5)


Rearrange the equation given in 1) to get:

P(X = a) / P(X = a + 1) = P(X = a + 1) / P(X = a + 2)

Therefore, the ratio between successive probabilities is constant. This is a property of the geometric distribution. Because P(X = 1) = 0.75, the parameter p is 0.75 and P(X = k) = 0.75 * 0.25k - 1.

Therefore, the expected damage from a single thunderstorm is the weighted sum:

(sum over all positive integers k) [90 * k2 * 0.75 * 0.25k - 1] = 90 * E(X2)

= 90 * (Var(X) + E(X)2)

= 90 * ((1 - 0.75)/0.752 + (1/0.75)2)

= 200


Let P(A) = P(B) = P(C) = x.

Note that because A, B, and C are pairwise independent:

P(A ∩ B) = P(A)P(B) = x2

P(A ∩ C) = P(A)P(C) = x2

P(B ∩ C) = P(B)P(C) = x2

From the inclusion-exclusion principle:

P(A ∪ B ∪ C) = P(A) + P(B) + P(C) - P(A ∩ B) - P(A ∩ C) - P(B ∩ C) + P(A ∩ B ∩ C)

= x + x + x - x2 - x2 - x2 + 0

= 3x - 3x2

The vertex of this parabola occurs at (1/2, 3/4), which we can obtain by setting the derivative of the above quadratic function to zero. Therefore, the maximum value of P(A ∪ B ∪ C) is 3/4.


Consider the random variable Z = X - Y. Because X and Y are bivariate normal, Z must be normally distributed. Additionally, E(X) - E(Y) = E(X - Y) = E(Z) and:

P(X > Y + 50) = P(Z > 50) = 0.9

P(X > Y + 72) = P(Z > 72) = 0.8

Let N(x) be the cumulative normal and let μ and σ be the mean and standard deviation of Z respectively. Then the following equations are true:

N((μ - 50) / σ) = 0.9

N((μ - 72) / σ) = 0.8

Using the z-tables, the values of z at N(z) = 0.9 and N(z) = 0.8 are 1.28 and 0.84 respectively. Therefore:

(μ - 50) / σ = 1.2

(μ - 72) / σ = 0.84

Divide the two equations to get:

(μ - 50) / (μ - 72) = 1.28 / 0.84 ⇔ μ = 114


Because the mapping between Y and X is one-to-one, we can directly substitute y = ln(x) into the cumulative distribution function and boundaries to get:

FY(y) = k * y, ln(1) = 0 < y < ln(3)

We get the probability density function by taking the derivative:

fY(y) = k, 0 < y < ln(3)

This is a uniform distribution over the interval [0, ln(3)]. Therefore, the variance is:

(ln(3) - 0)2 / 12 = 0.1

Remarks: In general, a direct substitution does not work if the function given is the probability density function or if the mapping is not one-to-one. Neither of these cases occur here, so we can directly substitute into the cumulative distribution function and the boundaries. Nevertheless, we can still use the general formula to check our answer. We first find the probability density function:

fX(x) = d(k * ln(x))/dx = k/x

Then we find the inverse mapping between Y and X:

g(x) = ln(x)

g-1(x) = ex

Using the general formula for a univariate transformation:

fY(y) = |d(g-1(y))/dy| * fX(g-1(y))

= |d(ey)/dy| * fX(ey)

= |ey| * k/ey

= k


The key to this problem is using Chebyshev’s inequality, which does not usually appear on the exam but has appeared occasionally.

Let Z = X - Y. Then:

E(Z) = μ = 500 - 200 = 300

Var(Z) = 82 + 62 = 100

σ(Z) = 1001/2 = 10

By Chebyshev’s inequality, P(μ - kσ < Z < μ + kσ) is at least 1 - 1/k2.

For this problem, k is 2.5 because 275 = 300 - 2.5(10) and 325 = 300 + 2.5(10).

Therefore, the smallest possible value for P(275 < Z < 325) is 1 - 1/2.52 = 0.84.


Note from the problem statement that all policyholders have membership level 1.

Also from the problem statement, P(policyholder has level n | policyholder has level n - 1) = k for some constant k and any integer n in {2, 3, …, 10}. From the definition of conditional probability:

P(A | B) = P(A ∩ B) / P(B). If B is contained in A, as in this problem, then P(A ∩ B) = P(A). Therefore:

P(policyholder has level n) / P(policyholder has level n - 1) = k

where k is some positive constant. This can be rearranged to:

P(policyholder has level n) = k * P(policyholder has level n - 1)

Iterating from n = 2, we get:

P(policyholder has level n) = kn - 1

We are given P(policyholder has level 10) = 0.001 = k9, so k = 0.0011/9.

Finally, P(policyholder has level 3) = (0.0011/9)3 - 1 = 0.215


This problem gets messy and can quickly discourage those who believe they are on the wrong track.

The boundary of the probability density function is a right triangle with vertices (0,0), (0,1), and (1,0).

The region 2y2 > x is to the left of a sideways parabola opening to the right. This boundary intersects the triangle at the origin and also on the hypotenuse at (1/2, 1/2). The region of interest is therefore bound by the y-axis, the line y = 1 - x, and the curve y = (x / 2)1/2.

To find the probability for this region, perform a double integral in the y-direction and then in the x-direction. The integral setup should be:

(integral from x = 0 to 1/2) (integral from y = (x / 2)1/2 to 1 - x) [4/3 + x + y] dydx

= (integral from x = 0 to 1.2) [11/6 - 19x/12 - x2/2 - 4x1/2/(3*21/2) - x3/2/(21/2)] dx

= 0.426


Because we are choosing a number of claims from a finite set and computing expected values, this problem will use the hypergeometric distribution. The population size is N, the number of successes is n, and either 2n or n + 6 claims are chosen at random with expected success rate 4 and 3 respectively. From the expression for the mean of the hypergeometric function:

4 = 2n(n/N)

3 = (n + 6)(n/N)

Dividing the first equation by the second, we get:

4/3 = 2n/(n + 6) ⇔ n = 12

Therefore:

4 = 2(12)(12/N) = 288/N ⇔ N = 72


The probability mass function of the geometric distribution is p*(1 - p)k - 1. Because p = 1/μ, the distribution for the number of the first customer becomes:

f(k) = (1 - 1/μ)k - 1/μ

If stores A and B have independent distributions A and B identical to the distribution above, then the probability that they are equal is the sum of the probabilities that both equal k, where k is every positive integer. Due to independence, the probability that both equal k is equal to the probability that A equals k times the probability that B equals k, which is:

((1 - 1/μ)k - 1/μ)2 = (1 - 1/μ)2k - 2/μ2

The sum over all positive integers k of these probabilities is thus:

(sum over all k > 0) [(1 - 1/μ)2k - 2/μ2] = (1/μ2) * (sum over all k > 0) [((1 - 1/μ)2)k - 1]

= (1/μ2) * (sum over all k > -1) [((1 - 1/μ)2)k]

= (1/μ2) * 1 / (1 - (1 - 1/μ)2) (using the formula for an infinite geometric sum)

= (1/μ2) * 1 / (2/μ - 1/μ2)

= (2μ - 1)-1


To find k, integrate the joint probability function from x = 1 to 2 and from y = 1 to 2 and set the result equal to 1:

1 = (integral from x = 1 to 2) (integral from y = 1 to 2) [k + xy/2] dydx

= (integral from x = 1 to 2) [k + 3x/4] dx

= k + 9/8

Solving for k, we get k = -1/8 and therefore fX, Y(x, y) = xy/2 - 1/8.

To get the conditional distribution, treat y as a constant. In a conditional distribution, it is necessary to normalize the result, so we need to divide fX, Y(x, y) by the integral from x = 1 to 2 of fX, Y(x, y).

If we integrate xy/2 - 1/8 from x = 1 to 2, we get 3y/4 - 1/8. Therefore, the conditional distribution is:

(xy/2 - 1/8) / (3y/4 - 1/8) = (4xy - 1) / (6y - 1)


Cov(W, Z)

= Cov(2X - Y, X + 2Y)

= Cov(2X, X) + Cov(2X, 2Y) + Cov(-Y, X) + Cov(-Y, 2Y)

= 2Cov(X, X) + 4Cov(X, Y) - Cov(Y, X) - 2Cov(Y, Y)

= 2Var(X) + 3Cov(X, Y) - 2Var(Y)

= 200 + 30 - 200

= 30


Drawing the region out, the region is a trapezoid with vertices (0, 0), (1, 0), (1, 1), and (0, 2). The area of this trapezoid is 3/2, therefore the value of the joint probability density function is 2/3.

Because Z = Y / X, Z * X = Y. Therefore, if Z is constant, Y is directly proportional to X, and the line describing this conditional distribution goes through the origin and has slope z. Assuming z > 1, this line intersects the boundary of the trapezoid at the line y = 2 - x.

Next, the value of the cumulative distribution function of Z at a constant z value is the probability that Y / X is less than or equal to z, which is the volume of the joint probability density function in the region y / x < z, and which in turn is the volume of the joint probability density function in the region y < zx.

This region is confined to the interior of the trapezoid and is to the right of the line y = zx. Because z > 1, this region is somewhat complicated to compute its volume. Instead, we know that the volume of the entire joint probability density function is 1, so we just need to find 1 minus the volume of the region to the left of the line y = zx. This region is a triangular prism with height 2/3 and with base vertices (0, 0), (0, 2), and a point along y = 2 - x such that y = zx.

Setting 2 - x = zx, we see that x = 2/(1 + z) is the x-coorinate of the intersection point. Therefore, the area of the triangular base is:

1/2 * b * h = 1/2 * 2 * (2/(1 + z))

= 2/(1 + z)

The volume of the triangular prism is just the height times the area of the triangular base, which is:

2/3 * 2/(1 + z) = 4/(3 + 3z)

Finally, because we found the complement of the region we wanted to find, the cumulative distribution function of Z is:

1 - 4/(3 + 3z) = (3z - 1)/(3z + 3)


The fact that A, B, and C are mutually independent allows us to get rid of all the conditionals and treat a probability of intersections as the product of probabilities. Apply the inclusion-exclusion principle to the second equation given in the problem statement:

0.5P(C) = P((A ∩ C) ∪ (B ∩ C))

0.5P(C) = P(A ∩ C) + P(B ∩ C) - P((A ∩ C) ∩ (B ∩ C))

0.5P(C) = P(A ∩ C) + P(B ∩ C) - P(A ∩ B ∩ C)

0.5P(C) = P(A)P(C) + P(B)P(C) - P(A)P(B)P(C)

0.5 = P(A) + P(B) - P(A)P(B)

Applying the inclusion-exclusion principle in reverse:

0.5 = P(A ∪ B)

Because A and B are independent of C, it must be that:

P(A ∪ B) = P(A ∪ B | C) = 0.5

Remarks: The first equation in the problem does not need to be used, but it does show that P(C) = 0.4.


The number of ways four positive integers w, x, y, and z can add up to 17 is equal to the number of ways four nonnegative integers (w - 1), (x - 1), (y - 1), and (z - 1) can add up to 17 - 4 = 13.

Imagine 13 balls in a row and add three partitions with each partition being between two balls or on either end. Then the integers can be represented by the numbers of balls in each section between partitions or on the ends, for example:

The number of ways this can happen is the number of ways 13 identical balls and 3 identical partitions can be ordered, which is 16C3 = 560.

If W = 9, then the number of ways the remaining three positive integers x, y, and z can add up to 17 - 9 = 8 is equal to the number of ways three nonnegative integers (x - 1), (y - 1), and (z - 1) can add up to 8 - 3 = 5. We use 2 partitions in this case, so the number of ways 5 identical balls and 2 identical partitions can be ordered is 7C2 = 21.

Therefore:

P(W = 9 | W + X + Y + Z = 17) = P(W = 9 ∩ W + X + Y + Z = 17) / P(W + X + Y + Z = 17)

= 21/560

= 3/80

Remarks: Admittedly, this problem lies more in the realm of combinatorics than exam P, but hopefully the idea of marking balls and using partitions can clarify how to solve simpler combinatorics problems which might appear on exam P.


Let P(H) represent the probability that the husband dies and P(W) represent the probability that the wife dies within five years. Then we are given:

P(H) = 2 * P(W) ⇔ P(W) = 0.5 * P(H)

50,000 * P(H ∪ W) + 1,000 = 5,000 ⇔ P(H ∪ W) = 0.08

From the inclusion-exclusion principle:

0.08 = P(H ∪ W) = P(H) + P(W) - P(H ∩ W) = 1.5 * P(H) - P(H ∩ W)

Because H and W are independent:

P(H ∩ W) = P(H)P(W) = 0.5 * P(H)2

We substitute this result:

0.08 = 1.5 * P(H) - 0.5 * P(H)2 ⇔ 0.5P(H)2 - 1.5P(H) + 0.08 = 0

By the quadratic formula, P(H) = 0.052 or 2.95. Probabilities are between 0 and 1, so P(H) = 0.052.

Finally, we find the complement by subtracting the result from 1:

P(husband survives 5 years) = 1 - 0.052 = 0.948

Remarks: This problem is similar to SOA sample problem 113.


Let LA, LB, and LC be the lifetimes of components A, B, and C respectively.

The probability that the lifetime of the device is less than one year is P(LA < 1 ∪ LB < 1 ∪ LC < 1). Its complement is much easier to find, so we compute:

P(LA < 1 ∪ LB < 1 ∪ LC < 1) = 1 - P(LA > 1 ∩ LB > 1 ∩ LC > 1)

= 1 - P(LA > 1) * P(LB > 1) * P(LC > 1)

= 1 - (e-1/1.4)(e-1/1.6)(e-1/2)

= 0.84


Because E(X | X > 120) = 140, E(X) = 140 - 120 = 20 due to the memoryless property of the exponential.

The damage due to fire thus has an exponential distribution with mean 20.

The payout only with the deductible of 20 is 0 with probability 1 - e-20/20 = 1 - e-1 and x with a probability distribution function of e-(x+20)/20/20 = e-x/20/(20e) for 0 < x.

The payout with deductible and payout limit is therefore 0 with probability 1 - e-1, x with a probability distribution function of e-x/20/(20e) for 0 < x < 60, and 60 with probability e-60/20/e = e-4.

This is a mixed distribution, so the moment generating function E(eXt) will involve weighted sums and integrals. In particular, there is a point mass at x = 0, a continuous distribution where 0 < x < 60, and a point pass at x = 60. Therefore:

E(eXt) = (1 - e-1) * e0t + (integral from x = 0 to 60) [ext * e-x/20/(20e)] + e-4 * e60t

= (1 - e-1) + ((integral from x = 0 to 60) [e(t - 1/20)x])/(20e) + e60t - 4

= (1 - e-1) + [e(t - 1/20)60 /(t - 1/20) - 1/(t - 1/20)]/(20e) + e60t - 4

= (1 - e-1) + (e60t - 4 - e-1)/(20t - 1) + e60t - 4

= 1 + (-20te-1 + e-1 + e60t - 4 - e-1 + 20te60t - 4 - e60t - 4)/(20t - 1)

= 1 + 20t(e60t - 4 - e-1)/(20t - 1).

Remarks: At this point I must have decided to throw the most complicated problem I could think of at you just to test your mastery of exam P. Hopefully it worked, but if not, we know that:

P(Question 30 appearing on the real exam) < r, where r is any positive real number.

exam p practice exam by jared nakamura

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