The Infinite Actuary Exam 1/P Online Seminar A.1 Fundamentals of Probability Solutions Last updated February 23, 2013 1. Let A be those buying auto insurance and H those who buy homeowners insurance. Then P[A] = 2P[H ] 0.7 = P[A ∪ H ] = P[A] + P[H ] - P[AH ] 0.7 = 2P[H ] + P[H ] - 0.2 P[H ]=0.3 P[HA 0 ] = P[H ] - P[AH ]=0.3 - 0.2= 0.1 The Venn diagram is 0.4 A 0.2 H 0.1 2. Let F be those who buy fire insurance and T those who buy theft insurance. We want P[F ∪ T ] - P[F ∩ T ] = (P[F ] + P[T ] - P[FT ]) - P[FT ] =0.4+0.3 - 0.2 - 0.2 =0.3 The Venn diagram is 0.2 F 0.2 T 0.1 3.(A ∪ B) 0 = A 0 B 0 , so we have 0.3 = P[AB 0 ] and 0.4 = P[A 0 B 0 ]. Adding gives us 0.7 = P[B 0 ], and thus P[B]=1 - 0.7= 0.3 0.3 A B 0.4 4. Let L denote the event that the trip requires lab work, and S the event that it results in a referral to a specialist. We are given: P[(L ∪ S ) 0 ]=0.35 P[L ∪ S ]=1 - 0.35 = 0.65 P[S ]=0.30 P[L]=0.40 c 2013 The Infinite Actuary, LLC p. 1 A.1 Solutions
Transcript
The Infinite Actuary Exam 1/P Online SeminarA.1 Fundamentals of Probability SolutionsLast updated February 23, 2013
1. Let A be those buying auto insurance and H those who buy homeowners insurance. Then
P[A] = 2P[H]
0.7 = P[A ∪H] = P[A] + P[H]− P[AH]
0.7 = 2P[H] + P[H]− 0.2
P[H] = 0.3
P[HA′] = P[H]− P[AH] = 0.3− 0.2 = 0.1
The Venn diagram is
0.4
A
0.2
H
0.1
2. Let F be those who buy fire insurance and T those who buy theft insurance. We want
P[F ∪ T ]− P[F ∩ T ] = (P[F ] + P[T ]− P[FT ])− P[FT ]
= 0.4 + 0.3− 0.2− 0.2
= 0.3
The Venn diagram is
0.2
F
0.2
T
0.1
3. (A∪B)′ = A′B′, so we have 0.3 = P[AB′] and 0.4 = P[A′B′]. Adding gives us 0.7 = P[B′], and
thus P[B] = 1− 0.7 = 0.3
0.3
A B
0.4
4. Let L denote the event that the trip requires lab work, and S the event that it results in areferral to a specialist. We are given:
5. We are given 0.7 = P[A∪B] = P[A]+P[B]−P[AB] and 0.9 = P[A∪B′] = P[A]+P[B′]−P[AB′].Summing gives us 1.6 = 2P [A] + P [B] + P [B′] − (P[AB] + P[AB′]). But P[B] + P[B′] = 1 and
P[AB] + P[AB′] = P[A], so the right hand side is P[A] + 1. Subtracting 1 gives us P[A] = 0.6 .Or, 0.7 = P[A ∪B] so 1− 0.7 = 0.3 = P[A′B′], and P[A ∪B′] = 0.9 so 1− 0.9 = 0.1 = P[A′B].
Adding those gives us 0.4 = P[A′] and P[A] = 1− 0.4 = 0.6 This method is easier to follow if you
draw the Venn diagram:
A B
0.1
0.3
6. Let T denote the event that a patient visits a physical therapist, and C the event that a patientvisits a chiropractor. We are given:
P[T ∩ C] = 0.22
P[(T ∪ C)′] = 0.12 P[T ∪ C] = 1− 0.12 = 0.88
P[C] = 0.14 + P[T ]
and we want P[T ]
P[T ∪ C] = P[C] + P[T ]− P[T ∩ C]
0.88 = 0.14 + P[T ] + P[T ]− 0.22
P[T ] = 0.48
7. A′ ∪B′ means that it is either not in A or not in B, which includes everything except AB (i.e.,
A′ ∪B′ = (A ∩B)′), so P[A′ ∪B′] = 1− P[AB] = 1− 0.2 = 0.8
8. Let S be the event that a student took the SAT and A the event that a student took the ACT.We know P[AS] = 0.3 and P[S] = 0.75, so P[A′S] = 0.75−0.3 = 0.45 and P[AS ′] = 0.4−0.3 = 0.1.
10. There are 3,000 young policy holders. 1,320 are male, leaving 1,680 young female policyholders. There are 1,400 young married policyholders, 600 of whom are male, so there are 800young female married policyholders, leaving 1,680−800 = 880 young female single policyholders.
11. Using the fact that R and H are mutually exclusive, and then given properties (i), (iv), (v)and (ii) gives the following Venn diagram:
H
R
A
0.35− x
x
0.29
0.11
y
0.17
1 = 0.35− x + 0.11 + 0.29 + x + y + 0.17
1 = 0.92 + y, 0.08 = y
2(x + y) = 0.46− x by given (iii)
2x + 0.16 = 0.46− x
x = 0.10
12. P[K] = 3P[T ]
P[Q] =1
4(P[K] + P[T ])
=1
4(3P[T ] + P[T ]) = P[T ]
1 = P[K] + P[Q] + P[T ]
= 3P[T ] + P[T ] + P[T ]
P[T ] = 0.2
P[K ∪Q] = P[K] + P[Q] = 4P[T ] = 0.80
13. The Venn diagram we get is
C N
A
4
3
12
3
25 7
H18
Subtracting the pieces we have solved for from 100 gives
100− 25− 3− 7− 2− 4− 1− 3− 18 = 37
14. Let G denote those who read the Globe, H those who read the Herald, and P those who read
We want x + y, and using that everything sums to 1 gives
1 = (0.25− 0.09− x) + x + 0.06 + 0.03 + (0.29− 0.09− y) + y
+ 0.25 + (0.4− 0.3− x− y)
= 1.07− x− y
x + y = 0.07
15. In general, all we know about probabilities is that they can range from 0 to 1. To getrestrictions on P[A ∩B] we want an equation that combines P[A ∩B], along with the things thatwe know, namely P[A] and P[B], and gives us a probability. In particular, we can say that
0 ≤ P[A ∪B] ≤ 1
P[A ∪B] = P[A] + P[B]− P[AB]
0 ≤ 0.7 + 0.6− P[AB] ≤ 1
−1.3 ≤ −P[AB] ≤ −0.3
1.3 ≥ P[AB] ≥ 0.3
That gives us a useful lower bound of P[AB] ≥ 0.3Note that this bound can be obtained if P[AB] = 0.3, P[AB′] = 0.4, P[A′B] = 0.3 and
P[A′B′] = 0.
16. The problem with the approach we used in the previous problem is that it gave the upperbound of P[AB] ≤ 1.3, which we already knew since it is a probability and hence no more than 1.But we also know that AB ⊂ A and AB ⊂ B so P[AB] ≤ P[A] and P[AB] ≤ P[B]. That givesus P[AB] ≤ 0.6 and P[AB] ≤ 0.7. Taking the most restrictive of those gives us the maximum
possible value of P[AB] of 0.6This upper bound can be realized if B ⊂ A.
17. Let C, T and S denote the event of being a car, train, and spaceship respectively. We are toldP[C] = 2P[T ], so P[T ] = (0.5)P[C]. Likewise, P[S] = 2(P[C] + P[T ]) = 2(1 + 0.5)P[C] = 3P[C].Since the total probability is 1, and these events are disjoint, we have 1 = P[T ] + P[C] + P[S] =
4.5P[C], so P[C] = 2/9
18. You could do it by explicitly listing all the possibilities. Alternatively, there are 5 numbers inthe set that are divisible by 5, 12 that are even, and 2 (10 and 20) that are both, so P[even] = 12/25,P[divisible by 5] = 5/25 and P[both] = 2/25, so P[even or divisible by 5, but not both] = (12 +
From (iii), we know that L and S are disjoint, hence they are non-overlapping sets in our Venndiagram. Likewise, F and G are also disjoint.
From (v) and (vi), let x be the number who take French and Latin, so 2x is the number whotake Spanish and French and x is the number who take German and Latin.
Let y be the number who take Latin only. (iv) tells us that the number who take German onlyis 2y, and (vii) tells us that the total number who take German and Spanish equals the numberwho take Latin only, which is 2x + y from our diagram.
We now see that 3x + 3y take German, so from (ii), we also have 3x + 3y who take French.3x take French plus another language, so 3y take French only. The number who take Spanish is2(x + 2y + 2x + y) = 6x + 6y, so 6x + 6y − 2x− (2x + y) = 2x + 5y take Spanish only.
Finally, twice as many take German as Latin, so 2(2x+y) = (3x+ 3y) and so x = y. There are20x who take at least one language, and 7x who take more than 1, so P[more than 1 language] =
From (i), we can draw T and E as disjoint, but need to keepsome overlap between Fire and Flood. Since no one has 3 coverages, that overlap is outside ofboth T and E, hence the basic shape of our Venn diagram.
Let x be T only, so 2x is fire only by (iii). Let y be flood and T , so flood and E is also y (by iv).Let z be fire and E, so 2z is fire and T by (v). Finally, 0.35 = P[E] so P[E only] = 0.35− y − z
We can then get the system of equations
0.53 = 2x + x + 0.1 + 0.35− y − z, 0.08 = 3x− y − z
0.35 = 2z + 2x + z + 0.05, 0.30 = 3z + 2x
1 = 3x + 2z + y + 0.5, 0.50 = 3x + y + 2z
from (ii), (vii), and the fact that all the probabilities sum to 1. Solving gives x = 0.09
23. One approach is to draw a Venn diagram for each combination and compare. It turns out thatthe (i) and (ii) are the same as the original expression, so the answer is A and the Venn diagram
for those is
A B
C
Expression (iii) is different since it also includes everything outside of A ∪B ∪ C.
24. P[A ∪ B] − P[A ∩ B] = (P[A] + P[B] − P[AB]) − P[AB] = 1.6 − 2P[AB] so it is maximizedwhen P[AB] is minimized. Since 1 ≥ P[A ∪ B] = P[A] + P[B] − P[AB] = 1.6 − P[AB], we have
P[AB] ≥ 0.6 and so the minimum of our quantity is 1.6− 2 · 0.6 = 0.4
25. One approach to problems like this is to let one die be red, one blue (so we can tell themapart), list all 36 cases, and count.
Using our probability rules, there are 11 ways to get at least one 6, and 3 ways to roll an 8without a 6 (red 3 and blue 5, or both 4, or red 5 and blue 3). Since those are disjoint, we can
sum the probabilities for a final answer of (11/36) + (3/36) = 14/36 .
The Infinite Actuary Exam 1/P Online SeminarA.2 Conditional Probability SolutionsLast updated February 6, 2013
1. Note that there are 6 ways to get a 7 (1 and 6, 2 and 5, 3 and 4, 4 and 3, 5 and 2, or 6 and 1)and 5 ways to get an 8 (2 and 6, 3 and 5, 4 and 4, 5 and 3, or 6 and 2). There are 6 · 6 = 36 totalcombinations for 2 dice, which gives
P[X = 8 | X = 7 or 8] =P[X = 8]
P[X = 7 or 8]
=5/36
(5 + 6)/36=
5
11
2. Let H denote those with health coverage and D those with dental. Then
P[HD] = P[H] · P[D | H] = 0.5 · 0.6 = 0.3
P[H ′D] = 0.15
P[D] = P[HD] + P[H ′D] = 0.30 + 0.15 = 0.45
3.
A0.5
Ac
B
0.4
Bc
0.1
0.3
0.4
0.2
P[B | A] =P[AB]
P[A]=
0.1
0.5= 0.2
P[A | B] = 0.05 + P[B | A]
P[AB]
P[B]=
0.1
P[B]= 0.05 + 0.20 = 0.25
P[B] = 0.4
4. Drawing a grid gives
Insure1 car
Insure morethan 1 car 70%
Sports Car20%
No SC
10.5%
9.5% 20.5%
The probability of insuring more than one car and a sports car is 0.7 · 0.15 = 0.105, andsince 20% insure a sports car, the probability of insuring exactly one car that is a sports car is0.20 − 0.105 = 0.095, leaving 0.3 − 0.095 = 0.205 as the chance of insuring exactly one car thatis not a sports car.
5. Let C denote those purchasing collision coverage, and D those with disability coverage. Then
11. The probability of rolling two 4s and the first roll being a 4 is the same as the probability ofrolling two 4s (the second condition is redundant), so
P[roll two 4s | 1st roll is a 4] =P[roll two 4s]
P[1st roll is a 4]
=
1
3· 1
6· 1
6+
1
3· 6
6· 6
6+
1
3· 2
6· 2
61
3· 1
6+
1
3· 6
6+
1
3· 2
6
=41
54= 0.76
12. A conditional probability is still just a probability, so all of our basic laws of probability stillhold. P[A ∪ B | C] = P[A | C] + P[B | C] − P[AB | C] ≤ P[A | C] + P[B | C] = 0.10, so theinequality in E is backwards. But A′B′ is the complement of A∪B, so P[A′B′ | C] = 1−P[A∪B |C] ≥ 1− 0.10 = 0.90 and the answer is B
Note that C is false since the inequality is backwards, and A and D need not hold because wecan’t just multiply probabilities without knowing things about the independence of A and B.
13. Since C and D are disjoint, P[C ∪D] = P[C] + P[D].Since A and B are complements, we have
24. Let L denote those with liability coverage, and P those with property coverage. ThenP[L ∪ P ] = P[L ∩ P ′] + P[P ] = 0.01 + 0.10 = 0.11, so P[(L ∪ P )′] = 1− 0.11 = 0.89
25. Let S denote a smoker, and D the event that someone died. Then
P[S | D] =P[SD]
P[D]
=P[SD]
P[SD] + P[ScD]
=0.10 · 0.05
0.10 · 0.05 + 0.90 · 0.01
= 0.357
26. The fastest way to do this is to use the complement. P[R ∪ S ∪ T ] = 1 − P[R′ ∩ S ′ ∩ T ′] =
1− P[R′] · P[S ′] · P[T ′] = 1− (2/3)3 =19
27
27. P[makes first] = 0.44
P[makes 2nd | makes first] = 0.55
P[makes both] = 0.44 · 0.55
= 0.242
28. Let A be the event that he made the previous shot, and B the event that he makes the currentshot. Then
P[A] = P[B] = 0.44
P[B | A] = 0.55
and we want P[B | A′]. Note that
P[B] = P[A]P[B | A] + P[A′]P[B | A′]
0.44 = 0.44 · 0.55 + 0.56 · P[B | A′]
0.354 = P[B | A′]
Another way to think about it is to suppose he took 1,000 shots over the course of a year. Hemade 44% of them, or 440 shots, leaving 560 misses. He makes 55% of the shots when he made the
previous shot. There were 440 times when he made the previous shot, he made 55% of those, or 242.That likewise means there were 440-242 = 198 times he made a short after missing the perviousshot. There are 560 total misses, 198 were followed by a make, so P[Make | missed previous shot] =35.35%
29.
A0.4
Ac0.6
B
0.6
Bc
0.4
0.3
0.3
0.1
0.3
P[AB] = P[A]P[B | A] = 0.4 · 0.75 = 0.3
30. See the previous problem for the Venn diagram. We get
P[A | B] =P[AB]
P[B]=
0.3
0.6=
1
2
P[A | B′] =P[AB′]
P[B′]=
1
4
31.P[1st is blue | 2nd is blue] =
P[both are blue]
P[2nd is blue]
=
7
12· 4
87
12· 4
8+
5
12· 3
8
=28
43= 0.65
32. We want the probability of the shaded set:
A B
C
The complement is easier to find: we want 1 − P[ABC ′] = 1 − P[A] ·P[B] · P[C ′] = 1− 0.5 · 0.6 · (1− 0.1) = 0.73
33. Let A be the event that he breaks 80 in 2007, and B the event that he breaks 80 in 2008.Then P[A] = 0.4,P[B] = 0.6 and P[AB] = 0.3. We want
7. First, let’s make a table of the possible values and their probabilities. As we aren’t initiallygiven µX , we will first find it and then fill in that column of the table.
8. Again, let’s make a table of the possible values. Because we want to find the median, we willwant a column for F (x) = P[X ≤ x] since the median is the smallest x that makes F (x) > 1/2
The median m(x) is 5 since that is the first time P[X ≤ x] > 1/2. Next, we need to find themedian of |X − 5|. There are 4 possible values of |X − 5|. It is 0 with probability 1/4, 1 with
probability 3/8, 3 with probability 1/8, and 5 with probability 1/4. P[|X−5| ≤ 0] =1
4<
1
2, while
P[|X − 5| ≤ 1] =1
4+
3
8=
5
8>
1
2, so the first time P[|X − 5| ≤ t] > 1/2 is when t = 1, giving us
an answer of 1
9. Let x denote the value of the 6th face, and Y the outcome of a roll. Then
The Infinite Actuary Exam 1/P Online SeminarA.4 Combinatorics SolutionsLast updated August 2, 2012
1. There are two cases, based on the color that is transferred, and the answer is
7
10·
(5
2
)(
10
2
) +3
10·
(4
2
)(
10
2
) =44
225
2. There are a total of 3n counters in the urn, so there are
(3n
2
)ways to choose the two counters.
There are 3 ·(n
2
)ways for them to be the same color, and n ·
(3
2
)ways for them to be the same
number. There is no overlap between those possibilities as if they are the same color they cannotbe the same number.
Combining this gives us
3 ·(n
2
)+ n ·
(3
2
)(
3n
2
) =
3n(n− 1)
2+ 3n
3n(3n− 1)
2
=(n− 1) + 2
3n− 1=
n + 1
3n− 1
3. We want 3 aces and therefore 10 non-aces, for an answer of(4
3
)·(
48
10
)(
52
13
)4. He will plant 7 white bushes in a row if a) the 2 red bushes are both in the first 3, or b) thered bushes are both in the last 3 or c) the red bushes are 1st and 9th, 1st and 10th, or 2nd and
10th. There are
(3
2
)= 3 ways for a) to occur, and also the same number of ways for b) to occur.
c) also had 3 cases, so there are 9 ways that work. There are
(10
2
)ways to choose where to put
the red bushes for a final answer of
3 · 3(10
2
) =9
45=
1
5
5. We will keep on rolling until we get either a 7 or an 8. All we care about is the result of thelast roll, and since the rolls are independent and identically distributed, we can ignore all of therolls until then. The fastest way to do so is to condition on rolling a 7 or an 8. Then we want
7. Let N be the number of employee’s who achieve high performance and get a bonus. The totalcost of the bonuses is therefore CN . First, let’s figure out the distribution of N .
P[N = 0] = (0.98)20 = 0.668
P[N = 1] =
(20
1
)(0.98)19(0.02)1 = 0.272
P[N ≤ 1] = 0.668 + 0.272 = 0.940
P[N = 2] =
(20
2
)(0.98)18(0.02)2 = .053
P[N ≤ 2] = 0.993
So there is a better than 99% chance that no more than 2 people will get a bonus. In other words,if C is chosen so that we can pay 2 bonuses, then we have more than a 99% chance of being safeand less than a 1% chance of the fund being inadequate.
But if we have 120 in the fund and need to cover 2 bonuses, then we can give 120/2 = 60 for
each of them, so C = 60 is the maximum possible payout.
8. There are 3 basic cases to have at least 2 more high risk losses than low risk losses, namely
Case I : All 4 are high risk
Case II : 3 are high risk, 1 is not
Case III : 2 are high risk, and 2 are moderate risk
Note that in Case II, the 1 who is not high risk can be either low risk or moderate risk. In caseIII, on the other hand, the 2 who are not high risk must be moderate risk since if either of themwere low risk, the number of high risk - number of low risk would be at most 1.
P[Case I] = (0.2)4
P[Case II] =
(4
3
)(0.2)3 (0.8)1
P[Case III] =
(4
2
)(0.2)2 (0.3)2
Answer = sum of 3 cases = 0.0488
9. Let X be the number of hurricanes in 20 years, so X has a binomial distribution with n = 20and p = 0.05. Then
10. Let N1 be the number of people in group 1 who complete the study, and N2 the number ingroup 2 who complete the study. For either one of the groups,
11. There are a total of 30 · 0.2 + 50 · 0.8 = 10 defective modems in the total inventory of 80.Because we are sampling without replacement, we want a hypergeometric distribution and not abinomial distribution, and our answer is(
10
2
)·(
70
3
)(
80
5
) = 0.102
12. Suppose that all 3 items are chosen from the first 3 rows. There would then be 5 ways tochoose the column for the first row, 4 for the column in the 2nd row, and 3 for the column in thethird row, or 60 ways.
But we don’t know that all 3 items come from the first 3 rows, just that they come from 3 of
the 6 different rows, giving
(6
3
)ways to choose the rows and(
6
3
)· 60 = 20 · 60 = 1200
total ways to do it.
13. We will have at least 3 kings if at least one of the three new cards is a king. When talkingabout “at least one” it is often easier to use a complement.
P[at least one more king] = 1− P[no more kings]
= 1−
(45
3
)(
47
3
) =135
1,081
14. There are
(52
5
)ways to choose 5 cards, so that is our denominator. For the numerator,
)ways to choose one non-king/queen from the 52− 4− 4
non-king/queens, for a final answer of (3
1
)·(
4
2
)·(
44
1
)(
52
5
)
15. The range is 3 if we draw 1, (2 or 3), 4, or if we draw 2, (3 or 4), 5. So there are 2 · 2 = 4 ways
for the range to be 3, and
(5
3
)= 10 ways to choose the sample, for a final answer of 4/10 = 2/5
16. The probability of a specific person not responding is 0.5 · 0.6 = 0.3. Let N be the numberwho don’t respond. Then N is a binomial with n = 4 and p = 0.3, so
P[N ≥ 3] = 0.34 + 4 · 0.33 · 0.7
17. For the outcomes to be decreasing, they must all be different. There are
(6
3
)ways to pick
3 different values, and the probability of any one specific sequence occurring is (1/6)3, so the
probability is
(6
3
)· (1/6)3 of having a decreasing sequence of rolls.
That ignores the requirement that the product is even. The only way for the product to be
odd is to have the sequence be 5, 3, 1, so there are
(6
3
)− 1 = 19 ways for the product to be even,
giving us a final answer of 19 · (1/6)3 = 19/216
18. Let W be the number of white balls drawn. To have 1 red and 2 white means that we needW = 3− 1 = 2, so we want
P[W = 2 | W ≥ 2] =P[W = 2,W ≥ 2]
P[W ≥ 2]
=P[W = 2]
P[W = 2] + P[W = 3]
=
(4
1
)·(
6
2
)(
4
1
)·(
6
2
)+
(4
0
)·(
6
3
)=
4 · 15
4 · 15 + 20=
3
4
19. If the committee has all 3 women, then the probability of the chairperson being a woman is
25. Because we are sampling with replacement, each drawing is independent and so we get amultinomial distribution. (
5
1
)·(
4
2
)4
14
(8
14
)2(2
14
)2
=960
16,807
26. In one round, the probability of losing is1
2· 1
2=
1
4. The chance of exactly 3 losses in 5 plays
is thus (5
3
)·(
1
4
)3
·(
3
4
)2
= 10 · 32
45
27. The chance that there will be the same number of heads as tails is
(2k
k
)·(12
)2k. If the number
of heads and tails are different, it is equally likely that there are more heads than tails as it is thatthere are more tails than heads. So
P[More tails] =1
2·[1−
(2k
k
)· 1
22k
]=
1
2−(
2k
k
)· 1
22k+1
and the answer is D
28. We need the Fire to win games 6 and 7. But for the final total to be 4 games to 3, that meansthat the Fire only won 2 of the first 5.
P[Win 2 of first 5] · P[Win 6 and 7] =
(5
2
)· 1
25· 1
22=
10
128=
5
64
29. There are 6 total pairs of shoes in the closet, so we have 6 · 2 = 12 total shoes. Since we seethe phrase “at least” I want to consider using the complement: P[at least one matching pair] =1− P[no matching pair] and to have no matching pair, we get
P[No matching pair] =12
12· 10
11· 8
10· 6
9=
16
33
To see where that equation comes from, there are 12 choices for the first shoe that we pick, noneof which form a pair, so all 12 are okay for us. After we have picked one shoe, there are 11 shoesleft, 10 of which don’t match the first one. For the next shoe, there are 10 shoes left, 1 matchesthe first, one matches the second, so 8 don’t match either. Finally, for the last shoe there are 9shoes left, 3 of which match our first 3 and 9-3=6 don’t match any of our first 3.
Once we have that equation, we see that
P[At least 1 pair] = 1− 16
33=
17
33
On a side note, it turns out to be somewhat hard to find the probability of exactly 1 matchingpair, but it is easy to find the chance of exactly 2 matching pairs:
There are
(12
4
)ways to choose 4 shoes from the closet. Name the 6 pairs A, B, C, D, E, F .
For any specific two pairs, such as A and B, there are 4 shoes between the two pairs and
We could now find P[exactly 1 pair] = P[at least 1 pair]− P[2 pairs] =17
33− 1
33=
16
33
30. Method 1: First, let’s find the probability that we have exactly 2 suits, and those suits arehearts (H) and spades (S). There are 4 ways for that to happen: we could have 4 spades and 1heart, 3 spades and 2 hearts, 2 spades and 3 hearts, or 1 spade and 4 hearts. So
P[at least 1 S and at least 1H] =
(13
4
)·(
13
1
)+
(13
3
)·(
13
2
)+
(13
2
)·(
13
3
)+
(13
1
)·(
13
4
)(
52
5
)=
63,206
2,598,960= 0.02432
But that isn’t what we want, as we don’t know which 2 suits we want. Instead, there are
(4
2
)= 6
ways to choose the two suits, giving
P[Exactly 2 suits] = 6 · 0.02432 = 0.146
Method 2: Again, let’s first find the probability that the 2 suits are hearts and spades, and
then multiply by 6. There are
(26
5
)ways for all cards to be hearts and/or spades, but that also
includes the possibility that all 5 are hearts or all 5 are spades. So to have both hearts and spades,we get
P[at least 1 S and at least 1H] =
(26
5
)−(
13
5
)−(
13
5
)(
52
5
)=
63,206
2,598,960
P[Exactly 2 suits] = 6 · 63,206
2,598,960= 0.146
Method 3: Finally, we could try to draw a tree diagram, although this will be pretty compli-cated. After we look at the first card, we have exactly 1 suit in our hand. After looking at thesecond card, we might have 2 cards from the same suit, or we might have 2 different suits. Then
The Infinite Actuary Exam 1/P Online SeminarA.5 Key Discrete Distributions SolutionsLast updated March 19, 2012
1. Since x is the number of businesses examined prior to finding a failing business, x is a geometricstarting at 0. p = P[x = 0] = 0.10, so the answer is A
2. Let N be the number of claims. Taking complements, P[N > 1] = 1 − p0 − p1. We don’tknow p0 and p1, but we do have a recursive connection between the pn that we can use to writeeverything in terms of p0. We can then solve for p0 using the fact that the probabilities sum to 1:
p1 =p05, p2 =
p15
=p052
p3 =p25
=p052· 1
5=p053
pn =p05n
1 =∞∑n=0
pn =(p0 +
p05
+p052
+ · · ·)
1 =p0
1− 15
=5
4p0, p0 =
4
5
and from our recursion relation, p1 = (4/5) · (1/5) = 4/25 so
P[N > 1] = 1− p0 − p1 = 1− 4
5− 4
5· 1
5= 0.04
Alternatively, the distribution is a geometric: As before, pn =p05n
= p0
(1
5
)nBut any time P[X = n] = c (blah)n then X is a geometric random variable. Here, P[X = 0] =
p0 > 0, so we have a geometric starting at 0, and if X is a geometric starting at 0 then
P[X = n] = (1− p)n p =
(1
5
)np0
so 1− p =1
5, p =
4
5p = p0
p1 = (1− p) p =1
5· 4
5=
4
25
1− p0 − p1 = 1− 4
5− 4
25= 0.04
3. Let B denote the benefit amount. The chances of seeing the possible values of B are:
P[B = 4,000] = 0.4
P[B = 3,000] = (0.6)(0.4) = 0.24
P[B = 2,000] = P[fail in 3rd year] = P[Survive first two years, fail in next]
4. There are a few different ways to think about this. One is that there are 25 = 32 equally likelyoutcome from flipping a coin 5 times, and so we want #{Ways to get 3 consecutive tails}/32.
We have 3 consecutive tails if we get:
TTTHH TTTHT TTTTH TTTTT
HTTTH HTTTT
HHTTT THTTT
so we have 8 ways that work, for an answer of 8/32 = 1/4
In practice, I don’t want to list all the possible ways and would like to find a shortcut, especiallyif the numbers are larger! There are 3 different types of cases that give us 3 consecutive tails:
1. We could have the first 3 all be tails (probability (1/2)3 = 1/8). That is also the first row ofour complete list, which had 4 cases with probability 1/32 each, for a total of 4/32 = 1/8.
2. The first flip could be heads, followed by 3 tails. That has probability (1/2)4 = 1/16 becausewe are specifying the results of the first 4 flips. That is also the second row of cases, whichhad 2 cases (giving us the same 2/32 = 1/16).
3. The last 3 flips could be tails. To avoid overlap with our previous cases, we need the 2nd flipto be heads. That specifies the last 4 flips, so that has probability (1/2)4 = 1/16, and is also,as you may have guessed, the third row of outcomes (so 2/32 = 1/16).
Summing those 3 cases gives us1
8+
1
16+
1
16=
1
4
5. Let X be the number of rolls until we get a 6. So X is a geometric on {1, 2, . . . } and
P[X = 5 | X ≥ 2] =P[X = 5, X ≥ 2]
P[X ≥ 2]=
P[X = 5]
P[X ≥ 2]
=
(5
6
)4
· 1
6
1− P[X = 0]− P[X = 1]
=
(5
6
)4
· 1
6
1− 0− 1
6
=
(5
6
)3
· 1
6= 0.096
An alternative approach is that in order to have X = 5 given X ≥ 2, we need the 2nd, 3rd, and4th rolls to not be 6, but the 5th roll to be a 6, which gives
6. The number of rolls until the first 3 is a geometric on {1, 2, ...} with mean 1/p = 6 and variance(1− p)/p2 = 30.
The number of rolls after the first 3 until the second is likewise a geometric with the sameparameters, as is the number of rolls after the second 3 and until the third, etc.
So N is the sum of 5 independent geometrics, and hence E[N ] = 5 · 6 = 30 and Var[N ] =5 · 30 = 150, so
CV(N) =
√150
30= 0.408
Alternatively, the number of non-3s before the 5th 3 is a negative binomial with r = 5 andp = 1/6, and thus has mean r(1 − p)/p = 25 and variance r(1 − p)/p2 = 150. But that variableis N − 5, so E[N ] = 25 + 5 = 30 and Var[N ] = 150 (the variances are the same since adding orsubtracting a constant doesn’t change the variance) which again gives us 0.408 for the Coefficientof Variation.
7. A lot of people have trouble understanding what the SOA meant by their phrasing, so let’sconsider an example. Suppose that January is the first month of the year, and that there areaccidents in January, March, April, and June, but not in February or May. Then June is the 4thmonth with an accident, and we have had no accidents in two months, so that is not a case that wewant to include in computing our probability. On the other hand, if we have accidents in January,April, June, and September, but none in February, March, May, July, or August, then we havehad 5 months with no accidents prior to our 4th month with an accident (and in particular havehad at least 4 months with no accidents), so that is in the event that we care about. Note that weare not requiring the months with accidents to be consecutive.
Let N denote the number of months with no accident before the fourth month with an accident.We want to find P[N ≥ 4] = 1 − P[N < 4]. In order to have N = k, we need month k + 4 to bethe fourth month with an accident, meaning that the first k + 3 months have k months withoutaccidents and 3 months with an accident, giving us
8. Let N denote the number of days of rain, so N ∼ Poisson (0.6). Let X be the amount of thepayment. We want to find SD(X), so first let’s make a table listing the possible payment amounts
and probabilities.x n P[X = x]0 0 P[N = 0] = e−λ = e−0.6
1,000 1 P[N = 1] = λ e−λ = 0.6e−0.6
2,000 at least 2 P[N ≥ 2] = 1− P[N < 2]= 1− e−0.6 − 0.6 e−0.6
E(X) = 0 · e−.6 + 1,000(.6 e−.6
)+ 2,000
(1− e−.6 − .6 e−.6
)= 573
E(X2) = 02 · e−.6 + 1,0002(.6 e−.6
)+ 2,0002
(1− 1.6 e−.6
)= 816,893
Var(X) = 816,893− 5732 = 488,461
SD(X) =√
488,461 ≈ 699
9. Let X be the number of tests completed, so X is a geometric starting at 1, and E[X] = 1/p,where p is the probability of “success” (which in this case means that the person has high bloodpressure).
1
p= 12.5
p =1
12.5P[X = 6] = (1− p)5 p
=
(11.5
12.5
)5
· 1
12.5= 0.053
10. The point of this problem is the fact that the person made a claim in 1997 means he is morelikely than a random selected policyholder to be high risk, and we have to take that into account.
P[θ = 0.6 | 1 claim in 1997] =P[θ = 0.6, 1 claim in 1997]
P[1 claim in 1997]
=(0.1) (0.6) e−0.6
(0.1) (0.6) e−0.6 + (0.9) (0.1) e−0.1
= 0.288
P[θ = 0.1 | 1 claim in 1997] = 1− 0.288
E[# claims in 1998 | 1 claim in 1997] = P[θ = 0.6 | 1 claim in 1997] · 0.6+ P[θ = 0.1 | 1 claim in 1997] · 0.1= (0.288) (0.6) + (0.712) (0.1)
= 0.244
11. In 12 hours, the mean number of power surges is 1. In 24 hours, the mean number is therefore2. The total number in a 24 hours period is still Poisson (because the sum of independent Poissonsis Poisson), and
12. Given that Y = 2, we know that the first roll is 1, 2, 3, 4, or 5. So there are 5 possible
outcomes for roll 1, one of which is a 5. Therefore P[X = 1 | Y = 2] = 1/5
Alternatively, P[X = 1, Y = 2] =1
6· 1
6and P[Y = 2] =
5
6· 1
6, and dividing gives us P[X = 1 |
Y = 2] = 1/5.
13. If Y = 2 then the 2nd roll is a 6. If X = 2 then the 2nd roll is a 5. We can’t have both atonce, so the answer is 0
14. We are given that the first rolls is not 6, leaving 5 possibilities. We also want the first roll tonot be a 5 (probability 4/5). We are given that the second roll is a 6 (and in particular not a 5).We aren’t given what happens on the third roll, but need it to be a 5 (probability 1/6), giving us
an answer of4
5· 1
6=
2
15
Alternatively, P[X = 3, Y = 2] =4
6· 1
6· 1
6and P[Y = 2] =
5
6· 1
6, and dividing gives us
P[X = 1 | Y = 3] = 4/30 = 2/15.
15. Given that Y = 2, the first roll is not 6, the 2nd is a 6, and the rolls after that can be anything.X = k requires that the first not be a 5 (probability 4/5 given that it isn’t a six), the second notbe a 5 (probability 1 since we know it is a 6), rolls 2 through k − 1 not be a 5 (probability 5/6
each) and the kth be a 5, so we get4
5· 1 ·
(5
6
)(k−1)−2
· 1
6= E
16. The sum of independent Poisson variables is Poisson, so X + Y ∼ Poisson(3 + 2 = 5) and
More generally (this is not important for the exam) if X and Y are independent Poissonvariables with means λ and µ then given that X + Y = n, the conditional distribution of X is abinomial with parameters n = X + Y and p = λ/(λ+ µ).
18. There are 8 possible cases for us to consider: 7 claims in week 1 and 0 in week 2, 6 in week 1and 1 in week 2, 5 in week 1 and 2 in week 2, etc. For each case, if there are x claims in week 1and 7− x in week 2, the probability is
2−(x+1) · 2−(7−x+1) = 2−9
and so our final answer is 8 · 2−9 = 2−6 = 1/64
19. Let X be the number of chips in the first cookie and Y the number of chips in the second.We could either list out 4 cases and sum their probabilities (X = 0, Y = 3 or X = 1, Y = 2 orX = 2, Y = 1 or X = 3, Y = 0). A faster approach is to use the fact that the sum of independentPoissons is Poisson, so X + Y ∼ Poisson(4), and so
P[X + Y = 3] = e−4 · 43
3!=
32
3e−4
20. Recall that ex =∑∞
n=0xn
n!. Then
P[X = N = n] = P[X = n] · P[N = n] by independence
= P[n tails, then 1 head] · P[N = n]
=
(1
2
)n+1
· e−1 · 1n
n!=e−1
2·(1/2)n
n!
P[X = N ] =∞∑n=0
P[X = N = n]
=∞∑n=0
e−1
2·(1/2)n
n!=e−1
2
∞∑n=0
(1/2)n
n!
=e−1
2· e1/2 =
e−1/2
2= 0.30
See video solution for how to approximate this using the calculator.
21. Since the sum of independent Poisson variables is Poisson, the total number of accidents over
13 weeks is a Poisson with mean 13 · 25 = 325, and thus coefficient of variation
√325
325= 0.055
22. We need to have 0 accidents in the first 3 days, and then at least one accident on day4. P[No accidents in first 3 days] = e−3·(1/2) = e−1.5, and P[At least one accident on day 4] =1− e−1/2, for a final answer of
e−1.5[1− e−0.5
]= 0.088
23. Let Ni be the number of storms in month i. Then Ni is a binomial with n = 4 and 1 =E[Ni] = np = 4p so p = 1/4. We are given that there are no tropical storms in the first month, sowe want to find
24. The number of tails, T , that comes up before the first heads is a Geometric(p) starting at 0.
E[T ] = 3 =1− pp
p =1
4
Var[T ] =1− pp2
= 12
SD[T ] =√
12
25. Since the standard deviation is 4, the variance is 16. For a Poisson, the mean equals thevariance, giving us an answer of 16
26. Because we are sampling with replacement, each draw is independent. That means that thenumber of white balls drawn before the fourth red ball is a negative binomial with r = 4 andp = 0.4, so
The Infinite Actuary Exam 1/P Online SeminarA.6 Discrete Deductibles SolutionsLast updated June 21, 2013
1. We want to find the expected payment, but first we need to find K and thus distribution ofloss. The sum of the probabilities of all possible events is 1, so
1 =K
1+
K
2+
K
3+
K
4+
K
5= 2.2833K
K ≈ 0.43796
So taking into account the deductible, the expected payment is
E[payment] = 0.05
(0 · K
1+ 0 · K
2+ (3− 2) · K
3+ (4− 2)
K
4+ (5− 2)
K
5
)= 0.031
2. The tour operator receives 21 · 50 if 20 or fewer tourists show up, but only 21 · 50 − 100 if all21 show up. So the expected revenue is
5. Let N denote the number of snowstorms. The insurance company pays
Payment =
{10,000(N − 1) N ≥ 1
0 N = 0
We can view that as saying that the payment is 10,000 per storm, with a deductible of 10,000.Then using our methods for finding the expected payment with a deductible,
E[Total loss] = E[Payment] + E[Uncovered Loss]
10,000E[N ] = E[Payment] + 10,000 · P[N ≥ 1]
10,000 · 1.5 = E[Payment] + 10,000(1− P[N = 0])
E[Payment] = 5,000 + 10,000 · e−1.5 = 7,231
For a more algebraic approach:
E[Payment] =∞∑n=1
10,000(n− 1)P[N = n]
=
[∞∑n=0
10,000(n− 1)P[N = n]
]− 10,000(0− 1)P[N = 0]
where the last line comes from adding and subtracting the n = 0 term
= 10,000 · E[N − 1]− 10,000 · (−1)e−1.5
= 10,000(1.5− 1) + 10,000e−1.5 = 7,231
6. There are 20 bulbs in the shipment, 6 of which are defective. We are testing 5 bulbs, and the
total expected damage in those 5 test bulbs is 5 · 6
20· 10 = 15. The uncovered loss is 10 if any of
the tested bulbs are defective. Let N be the number of tested bulbs that are defective, so
8. The loss amount will exceed the policy limit only if there are either 4 months with storms (foran excess of 5,000) or if there are 5 months with storms (and an excess of 15,000). So
E[Total Losses] =15,000
E[Uncovered Loss] =5,000 ·(
5
4
)0.34 · 0.7 + 15,000 · 0.35 = 178
E[Payment] =15,000− 178 = 14,822
9. The number of months with storms ranges from 0 to 5, with the following resulting payments:
15. Let N be the number of snowstorms. If N = 0 then there is no loss and therefore no payment.If N = 1 then the loss is 10,000 and since that is less than the deductible, there is still no payment.If N = 2 then the loss is 20,000 and the payment is 5,000. So
The Infinite Actuary Exam 1/P Online SeminarA.7 Discrete Reveiw SolutionsLast updated April 6, 2013
1. When we have only two sets, I often prefer drawing a 2x2 grid to a traditional Venn diagram.That is especially true when doing conditional problems, or when dealing with complements (thelatter of which applies here). But you can do it either way.
A 0.65
A′ 0.35
B B′
0.15 1− 0.5
(A′ ∪ B)′ = (A ∩ B′) so P[A ∩ B′] = 1 − 0.5 = 0.5. We then have P[A] = P[AB] + P[AB′] =
0.15 + 0.5 = 0.65 and P[A′] = 1− 0.65 = 0.35
2. P[A ∪B] = P[A] + P[B]− P[AB]
0.8 = 0.6 + 0.5− P[AB]
P[AB] = 0.3
3. P[A ∩ B is a probability, so 0 ≤ P[A ∩ B] ≤ 1. Also, P[A ∩ B] ≤ min{P[A],P[B]} = 0.3. If wesolve P[A ∪B] = P[A] + P[B]− P[AB] for P[AB] we get P[AB] = P[A] + P[B]− P[A ∪B] so
0 ≤ P[A ∩B] ≤ 0.3
0 ≤ P[A] + P[B]− P[A ∪B] ≤ 0.3
0 ≤ 0.7− P[A ∪B] ≤ 0.3
0.4 ≤ P[A ∪B] ≤ 0.7
From this, we get a minimum possible value of 0.4 which is realized if A ⊂ B.
4. From the previous part, we had a maximum possible value of 0.7 which is realized if A and Bare mutually exclusive.
5. Let L denote people who need lab work, and S those who need a specialist. Then
8. Let A, H, and R denote those who have auto, homeowners, and renters insurance, respectively.Since P[HR] = 0, we know that those two sets don’t overlap. To fill in the Venn diagram, Iuse properties (i) then (v) then (ii) then (iv) to obtain the numbers that are filled in. Then letx = P[R∩A′] and use the fact that the probabilities sum to 1 to see that P[A∩R′∩H ′] = 0.53−x.That gives
9. Let M denote people who insure more than 1 car, and S those who insure a sports car.
P[M ′ ∩ S ′] = 1− P[(M ′ ∩ S ′)′]= 1− P[M ∪ S]
= 1− (P[M ] + P[S]− P[MS])
= 1− 0.64− 0.20 + 0.12
= 0.28
M
M ′
S S ′
0.12 0.52 0.64
0.08 0.28
10. Let M denote people who insure more than 1 car, and S those who insure a sports car. Thedifference between this problem and the previous one is that here we are given P[S | M ] instead
of P[MS] in condition (iv). P[MS] = P[M ] · P[S |M ] = 0.64 · 0.15 = 0.096 so P[M ′S ′] = 0.256
M
M ′
S S ′
0.096 0.544 0.64
0.104 0.256
11. Let g(k) be the payment for k days in the hospital.
hurricanes is a negative binomial with r = 2 and p = 0.4, so P[N = n] =
(n+ (2− 1)
2− 1
)· 0.6n · 0.42
Using the MultiView, we see that this is maximized when n = 1. So the mode of N is 1, but thevariable that we care about is N+2 (since we want to add on the 2 hurricanes that cause damage),
and so that has mode 1 + 2 = 3
16. E[total bonus] = 400E[bonus for 1 low risk] + 600E[bonus for 1 high risk]
21. Let A be the event that the first component fails, and B the event that the second componentfails. Then we are given that
0.28 = P[A ∪B]
0.28 = P[A] + P[B]− P[A ∩B]
0.28 = P[A] + 2P[A]− P[A] · (2P[A])
2(P[A])2 − 3P[A] + 0.28 = 0
and solving the quadratic gives P[A] = 0.1 or P[A] = 1.4. The second is impossible since 0 ≤P[A] ≤ 1, so the answer is P[A] = 0.1
22. For one specific toss, the probability of no 7 or 11 is 1 − 6
36− 3
36=
28
36. All 10 tosses are
independent, so to find the probability of that happening all 10 times, we raise that probability to
the 10th power, giving
(28
36
)10
23. First, let’s find the probability of selecting box k. We are given that P[box 1] = c, P[box 2] = 2cand P[box 3] = 3c. Since we must choose one of the boxes, we have 1 = c+ 2c+ 3c so c = 1/6.
Summing over all of our cases gives a final answer of
2 · 1
6· 1
5· 4
4+ 2 · 2
6· 2
5· 3
4+ 2 · 3
6· 3
5· 2
4=
17
30
24. P[X = 1 | X ≤ 1] =P[X = 1, X ≤ 1]
P[X ≤ 1]
0.8 =P[X = 1]
P[X ≤ 1]=
λe−λ
e−λ + λe−λ
0.8 =λ
1 + λ0.8 + 0.8λ = λ
λ = 4
25. The mode is 0 since the probability is maximized at 0. The median is likewise 0 since theprobability of 0 accidents is greater than 1/2.
The mean is 0 · 0.55 + 1 · 0.2 + 2 · 0.1 + · · · > 0, so the mean is different.That means that (i) and (ii) are both false, but (iii) is correct, so the answer is C
26. Since we are selecting with replacement, each draw is independent and we have a multinomialdistribution. That gives an answer of
6!
2!2!2!
(10
20
)2(5
20
)2(5
20
)2
= 90 ·(
1
2
)10
=45
512
27. Here we are sampling without replacement, so the draws are not independent and we wantthe hypergeometric distribution, not the binomial distribution. There are two cases that we want:namely 2 blue socks, or 2 white socks, and so we get(
6
2
)+
(4
2
)(
10
2
) =
6 · 52
+4 · 3
210 · 9
2
=7
15
28. On each roll, the chance of rolling a seven is 6/36 = 1/6. In order for it to take exactly 10rolls to observe 8 sevens, we need the 10th roll to be a seven (probability 1/6) and for 7 of the first9 rolls to be sevens, giving (
9
7
)(1
6
)7 (5
6
)2
· 1
6=
36 · 52
610=
52
68
29. Let N be the number of variables in our sample that are equal to 6. Then N is a binomialrandom variable with n = 3 and p = 0.6. The distribution of X and N is therefore as follows:
31. Let A be the event that the red die is 1, 2, or 3 and let B be the event that the sum is 11.There are two ways that B can occur: either the red die is 5 and the green die is 6, or the red dieis 6 and the green die is 5; in particular, A ∩B = ∅.
P[A ∪B] = P[A] + P[B]− P[AB]
=3
6+
2
36− 0
36
=20
36=
5
9
32.
1/2 T
1/2 H
2 dice sum to 65/36
one roll equals 61/6
1
2· 5
36+
1
2· 6
36=
11
72
33. Since we are drawing with replacement, we have independent draws and hence a multinomialdistribution, so the answer is
37. If we were drawing with replacement, then each draw would be independent and we wouldhave a negative binomial distribution with r = 2 and k = 2. But since we are drawing withoutreplacement, that doesn’t apply. We have 3 cases to consider:
Real, Fake, Fake, RealFake, Real, Fake, RealFake, Fake, Real, Real
Probability of Case 1:10
35· 25
34· ·24
33· 9
32
Probability of Case 2:25
35· 10
34· 24
33· 9
32
Probability of Case 3:25
35· 24
34· 10
33· 9
32
Note that all 3 cases are equally likely. This gives
Total Probability: 3 · 10 · 9 · 25 · 24
35 · 34 · 33 · 32=
675
5,236
38. P[at least one occurs] = 1− P[none of R, S, or T occurs]
= 1−(
P[not R])(
P[not S])(
P[not T ])
= 1−(
2
3
)3
=19
27
39. Since the coins are independent, we simply multiply the probability of the gold coin beingheads 3 times by the probability that the silver coin is heads 6 times, giving(
5
3
)(0.3)3 (0.7)2
(10
6
)(0.1)6 (0.9)4
40. The sum of independent Poissons is Poisson, so W ∼ Poisson(3 + 1 + 4) and
P[W ≤ 1] = e−8 + 8 e−8 = 9 e−8
41. Since tails is twice as likely as heads, P[H]+P[T ] = P[H]+2P[H] = 3P[H] = 1 so P[H] = 1/3and P[T ] = 2/3.
For the third head to be on the fifth trial, we need the fifth trial to be heads, and two of thefirst 4 to also be heads, giving an answer of(
4
2
)·(
1
3
)2
·(
2
3
)2
· 1
3= 6 · 4
35=
8
81
Note that we have a negative binomial distribution with r = 3 (because we want 3 heads) andk = 2 (because we have two tails).
43. (A ∩ B)′ = A′ ∪ B′, so (A ∩ B)′ ∪ C = A′ ∪ B′ ∪ C. Since B and C are mutually exclusive,C ⊂ B′, so A′ ∪B′ ∪ C = A′ ∪B′. But A and B are independent, so
P[A′ ∪B′] = P[A′] + P[B′]− P[A′B′] =3
4+
5
6− 3
4· 5
6=
23
24
44. X has a binomial distribution with parameters n and p, so
E[X] = np
SD[X] =√npq
np =√np(1− p)
n2p2 = np(1− p)np = 1− p
n =1− pp
Since n must be an integer, we need to find an answer choice such that (1− p)/p is an integer. Ifp = 0.1 then n = 9 then that holds, so A is possible. If p = 0.15 then n = 5.67, so B is wrong.Likewise 0.7/0.3 is not an integer, nor is 0.6/0.4 or 0.4/0.6, so C, D and E don’t work either.
The answer is therefore A
45. In one week, the expected number of accidents is 2·3+3·4+2·1 = 20. The sum of independentPoissons is Poisson, so the number of accidents in a week is a Poisson(20) random variable, andthe answer is thus
e−20 · 2018
18!= 0.084
46. Let G be the event that the driver is good and N the number of claims. The complement ofgood is bad, so G′ is the event that the driver is bad. Then
48. P[S ∪ T ] = P[S] + P[T ]− P[S ∩ T ] = P[S] + P[T ]− P[S] · P[T ] by independence
1
4= P[S] + P[S]− (P[S])2 since P[S] = P[T ]
(P[S])2 − 2P[S] +1
4= 0
P[S] =2±
√4− 4 · 1 · 1
4
2
P[S] =2−√
3
2
Note that we discard the solution to the quadratic that is greater than 1 as it isn’t possible.
49.P[first is red | 2nd is red] =
P[both are red]
P[2nd is red]
=
6
9· 5
96
9· 5
9+
3
9· 7
9
=30
51=
10
17
50. There are three possible cases: 1st accident is minor, second is also minor, 1st is minor, 2ndis moderate, or 1st is moderate and 2nd is minor. That gives 0.5 · 0.5 + 0.5 · 0.4 + 0.4 · 0.5 = 0.65
You can also think of it as P[1 moderate, 1 minor] + P[2 minor]/ There are two ways to have1 moderate and 1 minor leaving 2 · 0.4 · 0.5 + 0.52 = 0.65
51. For the next two problems, let us pretend that one die is blue and the other is red. There are4 ways for the sum to be 5 (red=1 and blue=4, or red=2 and blue =3, or red=3 and blue=2, orred=4 and blue=1). 2 of those ways involve a 3, giving an answer of 2/4=1/2. Alternatively,
P[at least one 3 | sum is 5] =P[at least one 3 and sum is 5]
P[sum is 5]
=2/36
4/36=
1
2
52. There are 5 ways to get a sum of 6 (red=1 and blue=5, or red=2 and blue=4, or red=blue=3,or red=4 and blue=2, or red=5 and blue=1). Only one of those 5 ways involves a 3, so the answer
is 1/5
53. Let N be the number of passengers who arrive, so N ∼ Binom(32, 0.9). Then
You could also do it as a sum of 3 cases: one die is 4 and 2 are less than 4, two dice are 4 and oneis less than 4, and all three are 4, giving
P[max ≤ 4] =
(3
1
)·(
1
6
)·(
3
6
)2
+
(3
2
)·(
1
6
)2
·(
3
6
)+
(3
3
)·(
1
6
)3
=27 + 9 + 1
216=
37
216
55. After two rounds, there are 3 cases:
P[both in after 2 rounds] = (0.8)2(0.8)2 = 0.642
P[exactly 1 in after 2 rounds] = 2(0.8)2(1− 0.82) = 2(0.64)(0.36)
P[both out after 2 rounds] = [1− 0.82][1− 0.82] = 0.362
If both are in after 2 rounds, then for both to be out by round 5 they have to both lose in oneof the next 2 rounds, which has probability 0.362. If exactly one is in after 2 rounds, then theyare both out by round 5 if that person loses in one of the next 2 rounds, which has probability1− 0.82 = 0.36. So
P[both out by round 5 | ≥ 1 in after 2 rounds] =0.642 · 0.362 + 2(0.64)(0.36)(0.36)
0.642 + 2(0.64)(0.36)= 0.2516
56. Let N be the number of defects. Since at least 99 of the sampled items are not defective, weare given that N = 0 or N = 1. If N = 0 then the first item is not defective because there are nodefects. If N = 1, then exactly 1 of the 100 items is defective, and it is equally likely to be eachof the 100 items, so the chance that the first item is not defective given N = 1 is 99/100.
P[N = 0 | N ≤ 1] =0.95100
0.95100 + 100(0.05)(0.95)99=
0.95
5.95
P[N = 1 | N ≤ 1] =100(0.05)(0.95)99
0.95100 + 100(0.05)(0.95)99=
5.00
5.95
Note that we could have also said P[N = 1 | N ≤ 1] = 1− P[N = 0 | N ≤ 1]
Note: later on we will see another approach to this, namely that X has a Gamma distributionwith θ = 1 and α = 4.
6.0.75 =
∫ t
−∞f(x)dx =
∫ t
−∞ex−2dx = ex−2
∣∣∣t−∞
0.75 = et−2
t = 2 + ln(3/4)
7. The only requirements to be a cdf that are likely to be tested is that it be 0 at −∞ and 1, aswell as increasing. (It must also be right continuous, but that has never been tested). All of ourchoices are 0 at −∞ and 1 at ∞, so we only need to check whether each one is increasing. Todo that, we need to verify that the derivative, when defined, is non-negative, and where it is notdifferentiable that we either have continuity or a jump up.
For (i), F ′(x) = 0 if x < 1 or x > 2, and F ′(x) = 2x− 2 > 0 if 1 < x < 2. At 1, F (x) goes from0 to 12 − 2 + 1 = 0, so there is no jump. At 2, F (x) goes from 22 − 2 · 2 + 1 = 1 to 1, so there isagain no jump. Choice (i) is therefore a valid cdf.
For (ii), F ′(x) = 2x− 2 for 0 < x < 1 +√
2, and 2x− 2 < 0 for 0 < x < 1. Since F ′(x) can benegative, we do not have a valid cdf.
For (iii), we have a jump from 0 to 1/2 at 2. That is a jump up, so it is ok. At 1 +√
6/2we go from 1 to 1, so there is no jump. When differentiable, F ′(x) is either 0 or 2x − 2 > 0 for2 < x < 1 +
Because of the absolute value, we will need to break into cases: |x − 1| = x − 1 if x > 1 and|x− 1| = −(x− 1) = 1− x if x < 1
P[X < 1.5 | X > 0.5] =
1∫0.5
2(1− x)3 dx+1.5∫1
2(x− 1)3 dx
1∫0.5
2(1− x)3 dx+2∫1
2(x− 1)3 dx
=−12
(1− x)4∣∣10.5
+ 12(x− 1)4
∣∣1.51
−12
(1− x)4∣∣10.5
+ 12(x− 1)4
∣∣21
=12
(12
)4+ 1
2
(12
)412
(12
)4+ 1
2(1)4
=2
17
20. P[X > 1] =
∫ θ
1
f(x) dx =
∫ θ
1
4x3
θ4dx
3
4=x4
θ4
∣∣∣∣θ1
= 1− 1
θ4
θ =√
2
21. P[|X − 1/2| > 1/4] = P[X < 1/4] + P[X > 3/4]
= 1− P[1/4 ≤ X ≤ 3/4]
= 1−∫ 3/4
1/4
6x(1− x)dx
= 1−(3x2 − 2x3
) ∣∣∣3/41/4
= 0.3125
22. The mode is either an endpoint (x = 0 or x = 1), or a critical value where f ′(x) = 0. At theendpoints, f(0) = f(1) = 0, while between the density is positive, so the mode must be a criticalvalue.
f(x) = cx2(1− x) = cx2 − cx3
f ′(x) = 2cx− 3cx2
0 = 2x− 3x2
x = 0 or x = 2/3
Since we have already ruled out x = 0, the mode must therefore be 2/3
Note that we didn’t have to find c since it doesn’t affect the value of x that maximizes f(x).
To find P[Y > y], note that if y < 2 then P[Y > y] = P[X > y] so for 0.6 < y < 2,
P[Y > y] =
∞∫y
2.5 (0.6)2.5
x3.5dx =
−(0.6)2.5
x2.5
∣∣∣∣∞y
=0.62.5
y2.5
so E[Y ] = 0.6 +
2∫0.6
0.62.5
y2.5dy = 0.6− 0.62.5
1.5
1
y1.5
∣∣∣∣20.6
= 0.934
Or: Y is a mixed distribution. Either Y = 2 (when X ≥ 2) or 0.6 < y < 2 and fY (y) =2.5(0.6)2.5
y3.5
E[Y ] =
2∫0.6
yfY (y) dy
+ 2 · P[Y = 2]
=
2∫0.6
2.5(0.6)2.5
y2.5dy + 2
∞∫2
2.5 (0.6)2.5
x3.5dx
=−2.5
1.5
(0.6)2.5
y1.5
∣∣∣∣20.6
+−2 (.6)2.5
x2.5
∣∣∣∣∞2
= 0.934
6. The payment has a mixed distribution, with a density up to 10, and then a positive probabilityof being equal to 10, so
E[Y ] =
10∫1
y · f(y) dy + 10 · P[Y ≥ 10]
=
10∫1
2
y2dy + 10
∞∫10
2
y3dy
=−2
y
∣∣∣∣10
1
+ 10−1
y2
∣∣∣∣∞10
=
(−2
10+ 2
)+
1
10= 1.9
7. There are 3 cases: either there is no damage, partial damage, or total damage. In the partialdamage case with a damage of X, the payment amount is 0 when X < 1, (X − 1) when X > 1and our conditional density is 0.5003e−x/2 so
12. Y has a mixed distribution. Either the machine fails in the first 4 years, during which timefY (x) = fX(x) = 1/5, or the machine doesn’t fail in the first 4 years and is replaced at time 4 (soP[Y = 4] = P[X ≥ 4] = 1/5. To find the variance, we will compute
Var(Y ) = E[Y 2]− (E[Y ])2
E[Y ] =
4∫0
y · 1
5dy + 4 · P[Y = 4]
=y2
10
∣∣∣∣40
+ 4
5∫4
1
5dy
=8
5+ 4 · 1
5=
12
5
E[Y 2]
=
4∫0
y2 · 1
5dy + 42 · P[Y = 4]
=y3
15
∣∣∣∣40
+ 42 · 1
5=
112
15
Var[Y ] =112
15−(
12
5
)2
= 1.7
13. The value after 3 years of use is 100(0.5)X , and so we want
14. The man purchases the life insurance policy on his 40th birthday, so we know that he livedto be 40. The policy pays 5,000 if he dies before turning 50, and 0 otherwise, so the expectedpayment is
5,000 · P[T < 50 | T > 40] = 5,000 · P[40 < T < 50]
26. Since our density involves |x+ 1|, we want to break things into two cases: one when x+ 1 > 0,i.e., x > −1, and one when x+ 1 < 0, i.e., when x < −1. Doing so gives
E[X] =
−1∫−∞
x f(x) dx+
∞∫−1
x f(x) dx
=
−1∫−∞
x
2· ex+1 dx+
∞∫−1
x
2· e−x−1 dx
And for both of these integrals we will use integration by parts. Let u = x/2 and dv = ex+1dx sodu = dx/2 and v = ex+1 to get∫
x
2ex+1dx =
x
2ex+1 −
∫ex+1dx
2
=x
2ex+1 − 1
2ex+1
−1∫−∞
x
2ex+1dx =
x
2ex+1 − 1
2ex+1
∣∣∣∣−1
−∞
=−1
2− 1
2= −1
and for the other one, again let u = x/2 but now set dv = e−x−1dx so v = −e−x−1. Then∫x
2e−x−1 dx =
x
2
(−e−x−1
)−∫−e−x−1dx
2
=−x2e−x−1 − e−x−1
2∞∫−1
x
2e−x−1 dx =
−x2e−x−1 − e−x−1
2
∣∣∣∣∞−1
= (0− 0)−(
+1
2e0 − 1
2
)= 0
So E[X] = −1 + 0 = −1
Alternatively, f(x) =1
2e−(x+1) is symmetric about x = −1, so E[X] = −1
27. We are given that E[X] = µ = 3 and that E [X2] = 13. That gives us that Var[X] = 13−32 = 4so σ = 2. Plugging into Chebyshev’s inequality,
31. The area is length · width = 2w · w = 2w2, so the expected area is∫ 1
0
2w2f(w) dw =
∫ 1
0
2 · 3w4 dw =6
5= 1.200
32. M(0) = 1 since M(t) is an MGF. Likewise, in order to be MGFs, the other functions mustalso be 0 at t = 0, But (iii) fails that as 3e0 ·M(0) = 3 so (iii) is not an MGF. The other two are:M(0) ·M(0) = 1 and e0 ·M(0) = 1, so they are probably MGFs, meaning the answer is probably
C (there are other conditions that an MGF must satisfy, but they are too complicated to verify
under exam conditions.) To verify, note that (i) is the MGF of X1 + 3X2 where X1 and X2 are iidvariables with the same distribution as X, and (ii) is the MGF of X + 3.
33. M(0) = 1
M ′(0) = E[X] =
∫ 1
0
2x2 dx =2
3
M ′′(0) = E[X2] =
∫ 1
0
2x3 dx =2
4
M(0) +M ′(0) +M ′′(0) =13
6
34.E[Payment] =
∫ 2
1
(x− 1)f(x) dx
=
∫ 2
1
3x3
8− 3x2
8dx
=
(3
4
16
8− 8
8
)−(
3
4
1
8− 1
8
)= 0.53
35.E[Payment] =
∫ ∞2
(3
x3− 6
x4
)dx
=3
8− 6
24=
1
8
36. Let t be the 75th percentile of losses that result in positive payments. To result in a payment,the loss must exceed 3, so being the 75th percentile means that
0.75 = P[X ≤ t | X > 3]
0.25 = P[X > t | X > 3] =P[X > t]
P[X > 3]
0.25 · P[X > 3] = P[X > t]
0.25 · 1
32=
1
t2
t = 6
Comments: You could also have used P[X ≤ t | X > 3] = P[3 < X ≤ t]/P[X > t]. Also, we usedthe fact that P[X > x] = 1/x2.
The Infinite Actuary Exam 1/P Online SeminarB.3 Key Continuous Distributions SolutionsLast updated January 11, 2012
1. Let X be the loss amount, so X is uniform on (0, c). The payment is X − 5 if X > 5, and 0otherwise. Given X > 5, the payment is uniform on (0, c− 5) so
8 = E[payment] = E[X − 5|X > 5] · P[X > 5]
=c− 5
2· c− 5
c16c = c2 − 10c+ 25
0 = c2 − 26c+ 25
c = 25
Note: c = 1 is also a solution to the quadratic, but not to our problem as we need c > 5 in orderfor it to be possible for a payment to be made.
2. Let X denote the loss amount. The expected payment is E[X − 20 | X > 20] · P[X > 20] =40 · 0.8 = 32, and the expected loss amount is 50, so the expected uncovered loss is 50− 32 = 18.
Alternatively, the uncovered loss Y is a mixed distribution that has density 1/100 for 0 < y < 20and a probability of 0.8 of being equal to 20, so
E[Y ] =
∫ 20
0
0.01dy + 20 · 0.8 = 2 + 16 = 18
3. Let T be the amount of time spent in the hospital, and Y the payment. The payment is $100per day, rounded up, so Y is a discrete random variable that can be 100, 200, or 300. Making atable, we get
t y P[Y = y]0 < t < 1 100 1/31 < t < 2 200 1/32 < t < 3 300 1/3
and we want E[Y ] =1
3· 100 +
1
3· 200 +
1
3· 300 = 200
4. E[X] =1 + a
2
Var[X] =(a− 1)2
121 + a
2= 6 · (a− 1)2
121 + a = a2 − 2a+ 1
a2 − 3a = 0
a = 3
Again, the quadratic has an extraneous solution (a = 0) that contradicts the assumptions of ourproblem so we discard it.
5. Let X denote the loss amount, so the payment is X − 500 if X > 500 and 0 otherwise. Then
E[Payment] = E[X − 500 | X > 500] · P[X > 500]
=0 + 4500
2· 4500
5000= 2,025
6. Saying that we want the average value over the interval means that we want E[c(T )] where Tis chosen uniformly from the interval. The interval length is 2, so f(t) = 1/2.
E[c(T )] =
∫ 2
0
c(t) f(t) dt =
2∫0
t cos (t2)
2· 1
2dt
To do this integral, substitute u = t2, so du = 2tdt. Then tdt/4 = du/8, so we are now integratingcos(u)du/8. The limits also change: when t = 0 we have u = 0, so the lower limit remains 0, butwhen t = 2 we get u = 4 so the upper limit becomes 4, giving
E[c(T )] =
4∫0
cos (u)
8du
=1
8sinu
∣∣∣∣40
=1
8sin 4
This problem is from an older syllabus. It is theoretically still testable, but in practice you rarelysee trig functions on the exam today.
7. We want to find P[T ≤ 80] and are given that T is an exponential with P[T ≤ 50] = 0.3.Plugging into the formula for the cdf of an exponential gives
P[T > t] = e−λt = e−t/θ
P[T ≤ t] = 1− e−λt
0.3 = 1− e−λ·50
e−λ·50 = 0.7
−λ · 50 = ln(0.7)
λ = 0.00713
P[T ≤ 80] = 1− e−λ·80 = 0.4349
8. Let T denote the lifetime of one printer, and X the payment amount for that printer. We want100E[X]. Since the payment is either 0, 100, or 200, we have a discrete payment amount and cansimply list the possible payments and their probabilities to find E[X].
X T P[X = x]200 t ≤ 1 P[T ≤ 1] = 1− e−1/2 = 0.3934100 1 < t ≤ 2 P[T ≤ 2]− P[T ≤ 1] = e−1/2 − e−2/2 = 0.23870 t > 2 P[T > 2] = e−2/2 = 0.3679
9. Let T be the failure time and X the discovery time. Since the machine isn’t checked for thefirst two years, P[X > x] = 1 for x < 2. After two years, a failure is discovered when it occurs, soP[X > x] = P[T > x] = e−x/3 for x > 2. Using the survival function method,
E[X] =
∞∫0
P[X > x] dx
=
2∫0
1 dx+
∞∫2
e−x/3 dx
= 2 +(−3 e−x/3
) ∣∣∣∞2
= 2 + 3 e−2/3
Alternatively, X has a mixed distribution with a positive probability of being equal to 2 (X = 2if T < 2) and density (1/3)e−x/3 for x > 2, so
E[X] = 2 · P[X = 2] +
∞∫2
x · f(x) dx
That expression is also what we get if you view X as g(T ) with g(T ) = 2 when T ≤ 2 and g(T ) = Tfor T > 2 and integrate g(t) · f(t)dt. To compute this integral, we can use tabular integration
11. First, let’s ignore the maximum benefit and find the median loss. We can recognize the densityof f(x) as that of an exponential random variable. For an exponential, f(x) = λe−λx, and we havece−0.004x, so c = λ = 0.004.
For the median loss,
0.5 = P[X ≤ t] = 1− e−0.004t
e−0.004 t = 0.5
t =ln(0.5)
−0.004= 173
So the median loss amount is 173. Since 173 is less than the maximum benefit of 250, a loss of173 results in a benefit of 173 as well, and the median benefit is 173
12. Let X be the failure time.
0.5 = P[X ≤ 4] = 1− e−4λ
λ =ln(0.5)
−4P[X ≥ 5] = P[X > 5] since X is continuous
= e−λ·5
= e−5 ln(0.5)/(−4) = 0.42
13. Since we are looking only at those losses that exceed the deductible, we have a conditionalstatement. Conditioning and having an exponential distribution is a hint that we probably wantto use the memoryless property, which says that the distribution of X− c, given that X > c, is thesame as the original random variable. If you prefer, the distribution of the payment, given thatthere is a payment, is the same as the distribution of the original loss.
Here, we want the 95th percentile of losses that exceed the deductible, so we are conditioningon X > 100. First, note that the 95th percentile of the loss (with no conditioning) is
0.95 = P[X ≤ x] = 1− e−x/300
− ln(0.05) =x
300x = 898.8
The memoryless property means that the 95th percentile of the original loss distribution is equalto the 95th percentile of the payment given that the loss exceeds the deductible, so 899 is also the95th percentile of the payment given loss exceeds the deductible.
The loss amount is the payment plus the deductible, so the 95th percentile of the loss, giventhat X > 100, is 899 + 100 = 999
If you prefer not to use the memoryless property, then
14. Since we care about the effect of a deductible on payment amounts, we want to use thememoryless property. Initially, our loss amounts are exponential with mean θ. After we impose adeductible of d, our expected payment is 0.9θ. But the expected payment amount given that theloss exceeds the deductible is θ by the memoryless property, by which in symbols I mean
15. Since the density f(x) = xe−x, then X is a Gamma random variable with α = 2 and θ = 1.P[X > 1] = e−1/1 + (1/1)e−1/1 = 2e−1 = 0.736. If there are 100 possible claims, each resulting in a
payment with probability 0.736, then the expected number of payments is 73.6
Or, since we are conditioning an exponential, we can use the memoryless property:
P[T > 45 | T > 30] = P[T − 30 > 45− 30 | T > 30]
= P[T − 30 > 15 | T > 30]
= P[T > 15] = e−15/15 = e−1
21. We are summing 3 iid exponential random variables, and thus have a Gamma distributionwith α = 3 and θ = 1/4. So we want to find the Gamma cdf at 1/2. The easiest way to do it isto just know the Gamma cdf:
α = 1 : P[X ≤ t] = 1− e−t/θ
α = 2 : P[X ≤ t] = 1− e−t/θ −(t
θ
)e−t/θ
α = 3 : P[X ≤ t] = 1− e−t/θ −(t
θ
)e−t/θ − (t/θ)2
2!· e−t/θ
To remember these, the α = 1 case is just the CDF of an exponential, and then for larger α, eachtime α goes up by 1, we get one more term that is the next term of the probability distribution ofa Poisson(t/θ) random variable.
and then you can evaluate that integral using tabular integration or integration by parts.
22. If Y is exponential with mean θ then E(Y ) = θ and Var(Y ) = θ2.A Gamma is a sum of α exponentials, so if X is Gamma (α, θ) then E[X] = αθ and Var[X] = αθ2.For us that means that
4 = E[X] = αθ
8 = Var[X] = αθ2
2 = θ, 2 = α
fX (x) =xα−1e−x/θ
θα(α− 1)!=xe−x/2
4
P[X ≤ 4] =
4∫0
x
4e−x/2 dx
=
(−x2e−x/2 − e−x/2
)∣∣∣∣40
= 1− 3 e−2
Or if you prefer the memorization / connection the Poisson route
F (x) = 1− e−x/θ −(xθ
)e−x/θ
F (4) = 1− e−4/2 − 2e−4/2 = 1− 3e−2
23. If X is a Gamma(α, θ) random variable then MX(t) =
(1
1− θt
)α. We are given that
MX(t) = (1− 3t)−4 so α = 4 and θ = 3. That means that
)6. Let X denote the loss amount, and Y the reimbursement.
G(115) = P[Y ≤ 115 | Y > 0] = P[X ≤ x | X > 20]
where x is the loss amount that results in a payment of 115. A loss of 20 results in a payment of0, a loss of 120 results in a payment of 100, and a loss of 120 + t results in a payment of 100 + 0.5t.That gives
100 + 0.5t = 115
t = 30
120 + t = 150 = x
G(115) = P[X ≤ 150 | X > 20] =P[20 < X ≤ 150]
P[X > 20]
G(115) =e−0.2 − e−1.5
e−0.2= 0.727
Note: for the last step, since X has an exponential distribution we could have used the memorylessproperty: P[X ≤ 150 | X > 20] = P[X ≤ 130] = 1− e−1.3
8. ln(X) is increasing, so Y = −2 ln(X) is decreasing. If y = −2 ln(x) then x = e−y/2, so
P[Y ≤ y] = P[−2 ln(X) ≤ y]
= P[X ≥ e−y/2] = 1− e−y/2
fY (y) =1
2e−y/2
Or, using the density formula,
y = g(x) = −2 ln(x)
x = g−1(y) = e−y/2
fX(x) =1
1− 0= 1 because X is uniform(0,1)
fY (y) = fX(g−1(y)
)·∣∣∣∣ ddy g−1(y)
∣∣∣∣= 1 ·
∣∣∣∣ ddy e−y/2∣∣∣∣ =
∣∣∣∣−1
2e−y/2
∣∣∣∣fY (y) =
1
2e−y/2
9. Y is an exponential with mean 1/θ = 2, θ = 1/2, so
P[Z > z] = P[√Y > z] = P[Y > z2]
= e−z2/(1/2) = e−2z
2
10. Since Y = X2, we potentially have X = ±√Y . One question, though, is whether or not we
have both roots. If we knew X > 0, then we would only have the positive root. But −2 < X < 3.That means that we have 2 roots when 0 < Y < 4, as they correspond to −2 < X < 2, but weonly have one root when 4 < Y < 9 (as that means that 2 < X < 3).
So we have two cases: one for fY (y) when 0 < y < 4, and one for 4 < y < 9. If 0 < y < 4 then
With the cdf method, we still have 2 cases based on whether or not y < 4. If 0 < y < 4 then
P[Y ≤ y] = P[X2 ≤ y] = P[−√y ≤ X ≤ √y]
=
√y − (−√y)
5=
2√y
5
fY (y) =1
5√y
while if 4 < y < 9 then
P[Y ≤ y] = P[X2 ≤ y] = P[−2 ≤ X ≤ √y]
=
√y − (−2)
5=
√y + 2
5
fY (y) =1
10√y
and combining the two cases gives the same answer as before. For the cdf method, if you havetrouble following the equations, try plugging in specific values for y such as 3 or 7 and see whatyou get.
+ E[Payment | 5,000 < X ≤ 25,000] · P[5,000 < X ≤ 25,000]
+ E[Payment | 25,000 < X] · P[25,000 < X]
= 0 + E[Uniform(0, 20,000)] · 25− 5
50+ 20,000 · 50− 25
50
= 0 + 10,000 · 0.4 + 20,000 · 0.5 = 14,000
22. We don’t know that the distribution is symmetric (in fact, it probably isn’t since there cannotbe a negative number of crimes but could potentially be a large positive number). So
The Infinite Actuary Exam 1/P Online SeminarC.1 MultiVariate Distributions SolutionsLast updated February 25, 2013
1.
1 2
X
2
Y
The device fails either if X < 1 (blue area) or if X > 1 but Y < 1 (green area). The probabilityof this occurring is
2∫0
1∫0
f(x, y) dx dy +
1∫0
2∫1
f(x, y) dx dy =
2∫0
x2
16+xy
8
∣∣∣∣10
dy +
1∫0
x2
16+xy
8
∣∣∣∣21
dy
=
2∫0
1
16+y
8dy +
1∫0
3
16+y
8dy
=
(1
16+y2
16
)∣∣∣∣20
+
(3
16+y2
16
)∣∣∣∣10
=2
16+
4
16+
3
16+
1
16=
5
8
= 0.625
2. Let G denote the first claim time for a good driver, and B the first time for a bad driver.
P[G < 3, B < 2] = P[G < 3]P[B < 2]
=(1− e−3/6
) (1− e−2/3
)= 1− e−2/3 − e−1/2 + e−7/6
3. We want x > 20 and y > 20. The inner integral in the answer choices is dy, so we start aty = 20 (the smallest possible y). x < 50 − y implies that y < 50 − x, so the upper limits of ourinner integral are 50− x.
Combining y > 20 and x < 50− y gives x < 50− 20 = 30 so the outer limits are 20 < x < 30,and the answer is B
8. We want P[X < 0.2] and have 0 < x < 1, 0 < y < 1− x, so our picture is
0.2 1X
Y1
x+ y = 1
y = 1− x
0.2∫0
1−x∫0
f(x, y) dy dx =
0.2∫0
1−x∫0
6[1− (x+ y)] dy dx
=
0.2∫0
6
(y − xy − y2
2
)∣∣∣∣1−x0
dx
=
0.2∫0
6
[1− x− x+ x2 − (1− x)2
2
]dx
=
0.2∫0
6− 12x+ 6x2 − 3 (1− 2x+ x2) dx
=
0.2∫0
3− 6x+ 3x2 dx =
0.2∫0
3(x− 1)2 dx
= (x− 1)3∣∣0.20
= (−0.8)3 − (−1)3 = 0.488
9. We want x2 < y < x, so our picture isx
1
y
1
y = x2
x =√y
y = x
For any specific y, the rangeof x is from y to
√y so
fY (y) =
√y∫
y
f(x, y) dx =
√y∫
y
15 y dx = 15 y (√y − y) = 15y3/2(1− y1/2)
10. fX,Y (x, y) = fX (x) · fY |X (y | X = x) = 1 · 1 = 1. Since this is constant, (X, Y ) are uniformon the set of their possible values (which has total area 1/f(x, y) = 1).
18. We are given that X and Y are uniformly chosen from a disk with radius 2/√π. The area of
that disk is πr2 = π · 4
π= 4.
X
x2 + y2 = r2
y = ±√r2 − x2
= ±√
4
π− x2
Y(x,
√4
π− x2
)
(x,−
√4
π− x2
)
x 2√π
f (x, y) =1
area=
1
4
fX (x) =
∫1
4dy =
1
4· 2√
4
π− x2 =
√1
π− x2
4
19. P[(X, Y ) = (x, y)] =16− 4x− 4 y + xy
36
=(4− x)
6· (4− y)
6= P[X = x] · P[Y = y]
Since the probabilities factor and the support is a rectangle, X and Y are independent. The factorsare both (4− variable)/6 so the marginal distributions match as well. The answer is therefore D
20. We want P[X < 1, Y < 1]/P[Y < 1], which is the probability of being in the dark shaded
Since x ranges from 1 to y, to find the marginal of Y we integrate f(x, y) from 1 to y, i.e.,
fY (y) =
∫ y
1
f(x, y) dx =
∫ y
1
12
25(x+ y2)dx
=
[12
25· x
2
2+
12
25· xy2
]∣∣∣∣y1
=6
25
[(y2 − 1) + 2y2(y − 1)
]=
6
25
[2y3 − y2 − 1
]24.
x
y
1 2
1
From our picture, we see that the top boundary is somewhat complicated (it is y = x if x < 1 andy = 1 if 1 < x < 2) so we would rather have our inner integral be dx to avoid having to split intomultiple cases.
The Infinite Actuary Exam 1/P Online SeminarC.2 MultiVariate Moments SolutionsLast updated April 2, 2012
1.
t1
t2t2 = t1
L (L,L)
f(t1, t2) =1
area=
1
L2/2=
2
L2
E[T 21 + T 2
2 ] =
L∫0
t2∫0
(t21 + t22) ·2
L2dt1 dt2
=
L∫0
(t313
+ t22t1
)∣∣∣∣t20
· 2
L2dt2
=
∫ L
0
8
3L2t32 dt2 =
2 t423L2
∣∣∣∣L0
=2
3· L
4
L2=
2L2
3
2.P[Y = 1] = P[X1 = X2 = X3 = 1] =
(2
3
)3
P[Y = 0] = 1− P[Y = 1] = 1−(
2
3
)3
M(t) = E[et y]
= et·0P[Y = 0] + et·1P[Y = 1]
= 1 · 19
27+ et
8
27
Note that we worked with P[Y = 1] because that was easier to find than P[Y = 0]. In particular,P[Y = 0] is not equal to (P[X = 0])3 since that is P[X1 = X2 = X3 = 0] but Y can also be 0 ifX1 = 1 and X2 = X3 = 0 or if X1 = 0 and X2 = X3 = 1 or ...
3. The new total benefit is (X + 100) + 1.1Y , and
6. Finding the covariance from scratch is tedious as it often requires 3 (in this case 4) doubleintegrals, so we often want to look for a shortcut. Here, the shortcut is to notice that the domainis rectangular, and that f(x, y) factors as a function of X (namely kx) times a function of Y
(namely 1), so X and Y are independent and thus Cov (X, Y ) = 0 The long way to do theproblem is as follows: First, we need to find k.
8. The fact that the random variable Y is described in terms of the variable X (namely as beinguniform on (0, X) if we know X) suggests that we probably want to use double expectation to findany moments involving Y .
E(X) =0 + 12
2= 6
E[Y |X = x] =0 + x
2=x
2
E[Y ] = E[E[Y |X]
]= E
(X
2
)=
1
2E(X) = 3
E[XY |X] = X E[Y |X] = X · X2
=X2
2
E[XY ] = E[E[XY |X]
]= E
(X2
2
)=
1
2· E[X2]
And to find E[X2], we use the fact that we know the mean and variance of a uniform:
Var(X) =(12− 0)2
12= 12
E(X) =0 + 12
2= 6
E[X2] = Var(X) + (EX)2 = 12 + 36 = 48
E[XY ] =1
2· 48 = 24
Cov(X, Y ) = E(XY )− (E[X])(E[Y ])
= 24− 6 · 3 = 6
If we didn’t want to use double expectation, then we could have set
10. The total premium paid is 500 + 500 = 1,000. Since we are conditioning on the husbandsurviving, the possible total claims are either 0 (if the wife survives) or 10,000 (if the wife dies).The unconditional probabilities that we are given are:
H dies
H lives0.96 0.01
0.025 0.005
W lives W dies
With no conditioning
P[W lives|H lives] =0.96
0.97, P[W dies|H lives] =
0.01
0.97E[total premiums− total claims|H lives] = E[1,000− total claims|H lives]
Given X = 1, there are two possible values for Y , namely 0 and 1, so we can use our usual formulafor the variance of a Bernoulli random variable
Var(Y | X = 1) = P[Y = 1 | X = 1] · P[Y = 0 | X = 1]
=0.125
0.175· 0.05
0.175= 0.204
12. The formula for the joint density of X and Y doesn’t depend on y (so long as it is a possiblevalue), so given that X = x, Y is uniform on x < y < x+ 1. Therefore
Var(Y | X = x) =(x+ 1− x)2
12=
1
12
13. Let X denote the number of tornadoes in county P , and Y the number in county Q. We aregiven that X = 0, and P[X = 0] = 0.12 + 0.06 + 0.05 + 0.02 = 0.25 so 1/P[X = 0] = 4. Thatmeans that the conditional distribution of Y is given by
To evaluate the last two integrals, we could do integration by parts (or equivalently tabularintegration). Alternatively, if Y is an exponential random variable with mean 1/λ then∫ ∞
0
λe−λydy = E[Y ] =1
λ∫ ∞0
e−λydy =1
λ
∫ ∞0
λe−λydy
=1
λE[Y ] =
1
λ2
and we are using that with λ = 2 and λ = 3 for our two integrals. A final way of looking at it isto use the Gamma trick from the calculus review:∫ ∞
0
xa · e−bxdx =a!
ba+1
Here, we have a = 1 and b = 2 or 3 respectively.
16. Again, since we are looking for the covariance, our hope is that we can find a shortcut. (Sadly,that won’t always be possible on the exam.)
so the density factors. Therefore X and Y are independent, and thus Cov(X, Y ) = 0 .
17. There are two approaches to this. The shorter approach is to notice that our original region isa rectangle, and the joint density factors as f(x, y) = ce−xe−2y, so X and Y are independent. TheY factor of the joint density is that of an exponential random variable with mean 1/2, so that isour marginal distribution of Y . Combining independence with the memoryless property,
Var(Y | X > 3, Y > 3) = Var(Y | Y > 3)
= Var(Y − 3 | Y > 3)
= Var(Y )
=
(1
2
)2
= 0.25
Alternatively, we could find the conditional joint density:
f(x, y | X > 3, Y > 3) =f(x, y)
P[X > 3, Y > 3]x > 3, y > 3
=2e−x−2y
e−3 · e−6= e−(x−3) · 2e−2(y−3)
from which we can either again conclude that the conditional marginal density of Y is a shiftedexponential, or we could do a couple of double integrals to find E[Y ] or E[Y 2].
18. We are given that N1 = 2, and
p(n1, n2 | N1 = 2) =p(2, n2)
P[N1 = 2]
=
3
4
(1
4
)2−1
e−2(1− e−2)n2−1
P[N1 = 2]
= c · (1− e−2)n2
for n2 = 1, 2, . . . . That is the distribution of a geometric random variable starting at 1. The ratioin a geometric is 1 − p = 1 − e−2 so p = e−2. The expected value of a geometric starting at 1 is
1/p, so E[N2 | N1 = 2] = 1/p = e2
19. We can see that we want to use double expectation or the law of total variation since if weknew λ, the problem would be easy. So we can either work with the raw moments:
20. There are a few ideas going on here. First, the MGF of X + Y is the product of the MGFs ofX and Y by independence, so we want to factor M(t). Second, the MGF is a sum of terms of theform cie
tai , with the ci summing to 1, suggesting that we have a discrete distribution.
22. Let N be the number of hurricanes that hit. If we knew N , then we would have the sum of Nindependent exponentials and could find the variance. So we want to use either double expectationor the law of total variation. The law of total variation is easier since we can find the variance ofa sum more easily than the second moment.
23. Let N be the number of accidents and S the total unreimbursed amount. The reimbursementfor an accident is 70% of the loss amount, so the unreimbursed part is 30%. You could scale theexponential (e.g., say that the unreimbursed loss is exponential with mean 0.3 · 0.8 = 0.24) butthat approach wouldn’t necessarily work with some other distributions.
A more general approach is to let X be a loss and Y the unreimbursed part (so Y = 0.3X).Then E[Y ] = E[0.3X] = 0.3E[X] = 0.24 and Var(Y ) = 0.32Var(X) = 0.242, so
Var(S) = E[Var(S | N)] + Var[E(S | N)]
= E[0.242 ·N ] + Var[0.24 ·N ]
= 0.242(E[N ] + Var[N ])
= 0.242(0.25 · 3 + 0.25 · 0.75 · 3)
= 0.0756
24. fY |X
(y∣∣∣X =
1
3
)=
f(13, y)∫ 1
0f(13, y)dy
=13
+ y∫ 1
013
+ y dy=
13
+ y13
+ 12
=6
5
(1
3+ y
)E
[Y∣∣∣X =
1
3
]=
∫ 1
0
y · 6
5
(1
3+ y
)dy
=
∫ 1
0
(2
5y +
6
5y2)dy
=
(y2
5+
2 y3
5
)∣∣∣∣10
=3
5
25. E[k (X2 − Y 2) + Y 2
]= k E(X2)− k E(Y 2) + E(Y 2)
σ2 = k [12 + σ2]− k [22 + σ2] + (22 + σ2)
σ2 = k + kσ2 − 4k − kσ2 + 4 + σ2
σ2 = 4− 3k + σ2
4 = 3k
k =4
3
26. The marginal density of Y is that of an exponential with mean 1, so E[Y ] = 1 and E[Y 2] =Var(Y ) + (EY )2 = 1 + 1 = 2.
We have two basic approaches. One is to use double expectation, which we can only do withthe raw moments of X, and the other is to use the law of total variation. Using double expectation:
27. We are given that the support of (X, Y ) is the shaded triangle:
X
Y Y = X
1
x
The joint density of X and Y doesn’t depend on Y , so given X = x, Y is uniform over thepossible values, i.e., given X = x, Y is uniform on (x, 1) and thus
30. D is true since that is the definition of the second moment. A is false because we are missingthe marginals and also don’t know that X and Y are independent. B is false because if we aredoing a single integral to find E[X], we want the marginal of X, not the joint density. C is falsebecause we should use the marginals, not joint densities, and also we don’t know that X and Yare independent. E is false because we want a moment of Y , so we should integrate dy using themarginal of Y , not X.
31.
E[X1] =∂
∂t1M(t1, t2)
∣∣∣∣(0,0)
=(0.1et1 + 0.4et1+t2
) ∣∣∣(0,0)
= 0.5
E[X2] =∂
∂t2M(t1, t2)
∣∣∣∣(0,0)
=(0.2et2 + 0.4et1+t2
) ∣∣∣(0,0)
= 0.6
E[2X1 −X2] = 2 · 0.5− 0.6 = 0.4
32. Since the domain is not a rectangle, X and Y are not independent so we cannot just say thatthe answer is 0. Fortunately they gave us E[X] and E[Y ] so we only have to find E[XY ]
Since the joint density is constant, (X, Y ) are uniform on the large triangle. Given that X < 0.6,every point in the shaded triangle is still equally likely, so the conditional distribution is uniformand hence the conditional joint density is 1/(shaded area) = 1/(0.5 · 0.62) = 50/9. Alternatively,
The Infinite Actuary Exam 1/P Online SeminarC.3 Sums and Transformations SolutionsLast updated July 23, 2012
1.
X1
Y
1
2
y = 1− x
P [loss ≥ 1] = 1− P [loss < 1]
= 1−1∫
0
1−x∫0
2x + 2− y
4dy dx
= 1−1∫
0
2x + 2
4· (1− x)− (1− x)2
8dx
= 1− 1
8
1∫0
(4x + 4)(1− x)− (1− x)2 dx
= 1− 1
8
1∫0
3 + 2 x− 5x2 dx
= 1− 1
8
[3 + 1− 5
3
]= 0.7083
2. Let X denote the loss with deductible 1, and Y the loss with deductible 2. If X ≤ 1 then thatloss results in 0 payment, so if X ≤ 1 then the total payment is at most 5 when Y ≤ 5 + 2 = 7.Likewise, if Y ≤ 2 then that loss results in 0 payment, so if Y ≤ 2 then the total payment is at most5 if X ≤ 5+1 = 6. Finally, if X > 1 and Y > 2 then the total payment is X−1+Y −2 = X+Y −3so we want X + Y ≤ 5 + 3 = 8. That gives the following region:
3. Since the joint density is constant over the possible values, we have a uniform distribution andthe density is 1/area. So first let’s draw a picture and find the area.
If we want to use the change of variable approach, let g(C,P ) = (X,P ) (the second coordinate isarbitrary and is to make finding the inverse easier). To find the inverse, we want to solve for Cand P in terms of X and P . But this is just
6. We want P[X + Y ≤ 60], or the chance of being in the shaded area.
x
y
60
60x + y = 60
60∫0
60−x∫0
f(x, y) dy dx =1
800
60∫0
60−x∫0
e−x/40e−y/20 dy dx
7.
x
y
3
1
1 3
(2, 3)
(3, 2)y = 5− x
P[Sum > 5] =Area of Triangle
Total Area=
12· 1 · 12 · 2
=1
8
8. Since our joint distribution is discrete, we can just list the possible outcomes.
P
[X +
Y
2≤ 5
]= 1− P
[X +
Y
2> 5
]= 1− (P[X = 4, Y = 4] + P[X = 2, Y = 8] + P[X = 4, Y = 8])
= 1− 4
24 · 4− 8
24 · 2− 8
24 · 4
= 1− 1
24− 4
24− 2
24=
17
24
9. First, we need to find c, then we need to integrate over the region we care about. The range ofpossible values is the rectangle, and the shaded area is the region that we care about.
We have two choices: we could either integrate the joint density along the green line to directlyfind fZ(z), or we could do a double integral over the blue region to find the cdf of Z and thendifferentiate.
13. Because we have a discrete distribution, we again want to sum over possible cases that work,namely (X, Y ) = (1, 1) or (X, Y ) = (2, 2). Summing those probabilities gives us 0.1 + 0.2 = 0.3
21. To find the range where the new joint density is positive, we know that Y < 1 so U = XY <X < 1. Since W = X, that means that U < W < 1. We also know that X < Y , so U > X2 = W 2,giving us 0 < W 2 < U < W < 1. That reduces the answer choices to A and D. As for the formulafor the density,
The Infinite Actuary Exam 1/P Online SeminarC.4 Bivariate Normal Solutions
1. Unfortunately neither A nor B is normally distributed because there is a positive probabilityof no claim being made. So before we can use what we know about normal distributions, we haveto condition on a payment being made for both A and B.
I will use A > 0 and B > 0 to denote the case when A and B have claims (that is technicallywrong as there is a very small probability of the normal variables being negative, but it simplifiesthe notation).
P[B > A] = P[B > A,A > 0] + P[B > A,A = 0]
= P[A > 0] · P[B > A | A > 0] + P[A = 0] · P[B > A | A = 0]
6. For standard normals, E[Z1 | Z2 = z] = ρ · z. For general normals, E[X | Y = y] is E[X] plusρ · SD(x) · z, where z = (y − E[Y ])/SD(Y ) is how many standard deviations y is above the mean.
E(Y ) = −1, Var(Y ) = 4, SD(Y ) = 2
1− (−1)
2= 1
i.e., 1 is one SD above the mean of Y
E[X | Y = 1] = E[X] + ρ · SD(X) · 1
= 1 +−1
2
√5− 12 = 1− 1
2· 2 = 0
7. The conditional variance of Y is 0 if ρ = ±1 and is Var(Y ) if ρ = 0. In general, the formula forthe conditional variance is arguably the simplest formula that meets those requirements, namely
10. Let X denote the father’s height, and Y denote the son’s height. Since both are adult males,we have E[X] = E[Y ] = 68 and SD[X] = SD[Y ] = 2.5.
Var[Y | X = 72] = (1− ρ2) · 2.52
2 = (1− ρ2) · 2.52
ρ2 = 0.68, ρ = 0.8246
E[Y | X = 72] = ρ · z · SD[Y ] + E[Y ]
= 0.8246 · 72− 68
2.5· 2.5 + 68
= 71.3
11. Since X and Y are normal, the 60th percentile of Y is µY + 0.2533σY = 0.2533σY and the70th percentile of X is µX + 0.5244σX = 0.5244σX . Those are equal, so σY = (0.5244/0.2533)σX
The 70th percentile of Y is 0.5244σY = (0.52442/0.2533)σX = 1.0854σX , and Φ(1.0854) =0.8611, so that is the 86th percentile of X.
Calculation note: I used a computer to compute all of the numbers to a greater precision thanyou are given on the tables. On the exam, you can either round to 2 decimal places, and thenchoose the answer choice closest to what you end up with (doing so here gives you 0.8599) or youcould do linear interpolation, which will take longer but typically give you something closer to the“exact” answer.
12. Let N denote the number of shutdowns, and S the total cost. Then
4. In order for exactly one of the two variables to exceed 1/2, there are two cases: Either X > 1/2and Y ≤ 1/2, or X ≤ 1/2 and Y > 1/2. This means we want
P
[X >
1
2, Y ≤ 1
2
]+ P
[X ≤ 1
2, Y >
1
2
]= P
[X >
1
2
]P
[Y ≤ 1
2
]+ P
[X ≤ 1
2
]P
[Y >
1
2
]= 2 ·
(1− F
(1
2
))F
(1
2
)
F
(1
2
)=
1/2∫0
2− 2 t dt
= 2t− t2∣∣∣1/20
= 1− 1
4=
3
4
Answer = 2
(1− 3
4
)(3
4
)= 2 · 1
4· 3
4=
6
16
5. For one of the variables,
P[X ≤ 1] =
∫ 1
0
√2− x dx
=√
2− 1
2
To have exactly 2 of the 3 variables exceed 1, we also need exactly 1 less than 1, so
15. In order for only the median to be between 2 and 3, we need two variables to be at most 2,one variable to be between 2 and 3, and two variables to be at least 3. The probability of that is(using a multinomial distribution)
5!
2!1!2!(P[X ≤ 2])2P[2 < X < 3](P[X ≥ 3])2
For the components of that,
F (x) =
∫ x
0
f(t)dt =
∫ x
0
t
8dt =
x2
16
P[X ≤ 2] = F (2) =4
16
P[2 < X ≤ 3] = F (3)− F (2) =9− 4
16=
5
16
P[X > 3] = 1− F (3) =16− 9
16=
7
16
Answer =5!
2!1!2!
(4
16
)25
16
(7
16
)2
= 0.112
16. Let T denote the max.
P[T ≤ t] = (P[X ≤ t])3 = (t2)3 = t6
fT (t) = 6t5
E[T ] =
∫ 1
0
t · 6t5dt =6
7
17. 1
4= P[at least one Xi > d]
1
4= 1− P[both Xi ≤ d]
1
4= 1−
(d
100
)2
d = 50√
3
18. The first machine fails at time min{X, Y } and the second fails at time max{X, Y }, so we want
The Infinite Actuary Exam 1/P Online SeminarC.6 Multivariate Review SolutionsLast updated April 9, 2012
1.
x
y
1 3
3
1
The device fails during its first hour if either X < 1 or if Y > 1. There are two key cases: X < 1(blue region), and X > 1, Y < 1 (green region), giving∫ 1
0
∫ 3
0
f(x, y) dy dx +
∫ 3
1
∫ 1
0
f(x, y) dy dx =
∫ 1
0
(x + y)2
2 · 27
∣∣∣∣30
dx +
∫ 3
1
(x + y)2
2 · 27
∣∣∣∣10
dx
=
∫ 1
0
(3 + x)2
54− x2
54dx +
∫ 3
1
(1 + x)2
54− x2
54dx
=43 − 33
162− 13
162+
43 − 22
162− 33 − 13
162
=11
27= 0.41
2. Our picture is almost the same as before: either the first component fails in the first half hour(blue area) or the first component survives for the first half hour but the second fails in the firsthalf hour (green area),
4. Let N be the number of people hospitalized and Z the total cost. If N = 0 or 1 then the totalcost is less than 1. If N = 2, with costs X and Y for each, then X and Y are uniformly distributedover the following square:
x
y
The total cost is less than 1 if and only if (X, Y ) are in the shaded region, which takes up halfof the square, so P[Z < 1 | N = 2] = 1/2. That gives us
P[Z < 1] =P[N = 0] + P[N = 1] +1
2· P[N = 2]
= 0.72 + 2 · 0.3 · 0.7 + 0.5 · 0.32 = 0.955
E[N | Z < 1] = 0 · P[N = 0 | Z < 1] + 1 · P[N = 1 | Z < 1] + 2 · P[N = 2 | Z < 1]
= 0 + 1 · 2 · 0.3 · 0.70.955
+ 2 · 0.5 · 0.32
0.955= 0.534
5. Let H denote the event that the coin comes up heads, and T a tails. Then
E[X] =E[X | H] · P[H] + E[X | T ] · P[T ]
=1 + 6
2· 1
2+
1 + 4
2· 1
2= 3
6. Let H and T again denote heads and tails. From before, we know that E[X] = 3. Then
7. Let X be the running time for the assembling machine, and Y the running time for the packingmachine. Then f(x, y) = x + y for 0 < x < 1 and 0 < y < 1. We want P[Y < 2X], so we want
x
y
1
1
P[Y < 2X] =
∫ 1
0
∫ 1
y/2
(x + y) dx dy
=
∫ 1
0
12 − (y/2)2
2+ y ·
(1− y
2
)dy =
∫ 1
0
1
2+ y − 5y2
8dy
=1
2+
1
2− 5
24=
19
24
8.
x
y
1
1
0.5
0.5
The unshaded region is less complicated, so we will use the complement.
Originally our distribution is over the largest triangle. We are given that X ≤ 1/2, so we are giventhat we are in one of the shaded regions, and we want the probability of being in the green shaded
We are given that we are in the blue triangle, and want the conditional density of y. If it werea discrete situation, we would want P[Y = y,X ≤ 1/2]/P [X ≤ 1/2]. In a joint continuous case,P[X ≤ 1/2] still makes sense, so we still want to use that. The numerator has to become a densitysince P[Y = y] = 0, and since we want X ≤ 1/2, we want to integrate f(x, y) along the green lineand divide by P[X ≤ 1/2]. From the previous problem, we know that P[X ≤ 1/2] = 1/8
Bernoulli 0 or 1 p, x = 1 p p(1− p) pet + q1− p, x = 0
Sum of Bernoullis
Binomial number of successes
(n
x
)px(1− p)n−x np np(1− p) (pet + q)
n
in n trials
Geometric Number of trials p(1− p)x−1 1
p
1− pp2
pet
1− qeton {1, 2, . . . } until first success
Geometric Number of failures p(1− p)x 1
p− 1
1− pp2
p
1− qeton {0, 1, . . . } before first success
Negative Sum of Geometric on 0, . . .
Binomial Number of failure
(x+ r − 1
r − 1
)pr(1− p)n−r r(1− p)
p
r(1− p)p2
(p
1− qet
)r
until r-th succss
Poisson Sum of Poissons e−λ · λn
n!λ λ eλ(e
t−1)
is Poisson
Bernoulli Variance Shortcut: If X is a random variable that can only take on 2 values, with P[X = a] = p and P[X = b] = 1 − p,then Var[X] = (b− a)2p(1− p)
Continuous Distributions
One trick to remembering the Gamma distribution is to recall that if α is an integer, then a Gamma(α, θ) distribution is the sum ofα exponentials with mean θ.
Densities are only given for the region on which they are positive, and are 0 otherwise.
Distribution range density cdf EX VarX MGF
Uniform on (0, 1) 0 < x < 1 1 x for 0 < x < 11
2
1
12
et − 1
t
Uniform on (a, b) a < x < b1
b− ax− ab− a
a+ b
2
(b− a)2
12
etb − eta
t(b− a)Exponential
with “rate” λ 0 < x <∞ λe−λx =1
θe−x/θ 1− e−λx = 1− e−x/θ θ =
1
λθ2 =
1
λ21
1− θt=
λ
λ− tand mean θ =
1
λ
Gamma 0 < x <∞ xα−1
θα(α− 1)!e−x/θ see below αθ αθ2
(1
1− θt
)α
Standard Normal −∞ < x <∞ 1√2πe−x
2/2 Φ(x) 0 1 et2/2
Normal(µ, σ2) −∞ < x <∞ 1
σ√
2πe−(x−µ)2/(2σ2) Φ
(x− µσ
)µ σ2 etµ+(t2σ2)/2
For α = 1, a Gamma is an exponential with mean θ and so has CDF P[X ≤ t] = 1− e−t/θ
For α = 2, a Gamma has cdf P[X ≤ t] = 1− e−t/θ −(t
θ
)e−t/θ
For α = 3, a Gamma has cdf P[X ≤ t] = 1− e−t/θ −(t
θ
)e−t/θ −
((t/θ)2
2
)e−t/θ
Note that each time we increase α by 1, we subtract another term from the cdf that looks that the probability distribution of aPoisson. In general, if α is an integer, the cdf of a Gamma(α, θ) is given by P[X ≤ t] = P[X < t] = 1 − P[N ≤ α − 1] = P[N ≥ α]where N is a Poisson random variable with mean λ = t/θ.
Fundamentals and conditional probabilities
P[Ac] = 1− P[A] P[A ∪B] = P[A] + P[B]− P[AB]
P[A | B] =P[AB]
P[B]P[AB] = P[B]P[A | B]
A and B are mutually exclusive if P[AB] = 0.A and B are independent if P[AB] = P[A]P[B], which also implies that P[A | B] = P[A].Three events A,B and C are independent if every possible combination factors, e.g., P[ABC] = P[A]P[B]P[C],P[AcC] = P[Ac]P[C], etc.
If A1, A2, . . . , An are mutually exclusive and∑
P[Ai] = 1 (i.e., the Ai are a way of breaking things down into allpossible cases), then
P[B] =∑i
P[B | Ai]P[Ai] “Law of Total Probability”
P[Aj | B] =P[AjB]
P[B]=
P[B | Aj ]P[Aj ]∑i P[B | Ai]P[Ai]
Bayes’ Theorem
CDFs and densities:
F (x) = P[X ≤ x]
If F (x) is continuous, then f(x) =d
dxF (x) is the density and P[a ≤ X ≤ b] =
∫ b
a
f(x)dx
If F (x) has a jump at a, i.e. F (a) > limx↑a
F (x), then the size of the jump is the probability of X = a, i.e.,
P[X = a] = F (a)− F (a−).
Moments:
E[X] =∑x
x · P[X = x] +
∫ ∞−∞
xf(x)dx
(for discrete random variables, only the first part applies, for purely continuous, only the second, for mixed distri-butions use both)
If X ≥ 0, then E[X] =
∫ ∞0
P[X > x]dx.
If X ≥ 0 and only takes on integer values, E[X] =∑∞
k=0 P[X > k].
E[X2] =∑x
x2 · P[X = x] +
∫x2f(x)dx
E[g(X)] =∑x
g(x)P[X = x] +
∫g(x)f(x)dx
Var[X] = E[(X − µX)
2]
= E[X2]− (E[X])2
Var[aX] = a2Var[X]
SD[X] =√
Var[X]
Cov(X,Y ) = E[XY ]− E[X] · E[Y ]
Cov(X,X) = Var[X]
Corr(X,Y ) =Cov(X,Y )
SD(X)SD(Y )
Coefficient of Variation of X =SD(X)
E(X)
Covariance is bilinear, meaning that it distributes as you might expect. This means that Cov(aX + bY, cZ) =acCov(X,Z) + bcCov(Y,Z),Var(aX + bY ) = a2VarX + 2abCov(X,Y ) + b2VarY .
Moment generating functions:
MX(t) = E[etX]
MX(0) = E[e0]
= 1
d
dtMX(t)
∣∣∣∣0
= E[X]
d(n)
dt(n)MX(t)
∣∣∣∣0
= E[Xn]
MaX+b(t) = E[eatX+bt] = ebtMX(at)
Normal, Gamma, and Binomial are the most common mgf’s used on the exam.If X and Y are independent, then MX+Y (t) = MX(t)MY (t).
Joint moment generating functions:
MX1,X2(t1, t2) = Eet1X1+t2X2
E[Xm1 X
n2 ] =
∂(m)
∂x(m)1
∂(n)
∂x(n)1
M(t1, t2)
∣∣∣∣∣(0,0)
MX1(t1) = Eet1X1+0X2 = MX1,X2(t1, 0)
Convolutions and transformations
If Z = X + Y , then fZ(z) =
∫ ∞−∞
fX,Y (x, z − x)dx =
∫ ∞−∞
fX(x)fY (z − x) The last equality (convolution formula)
only holds if X and Y are indepedent.You can also find fZ(z) by finding the CDF and differentiating.
If Y = g(X) and g is one-to-one, then fY (y) = fX [g−1(y)]
∣∣∣∣ ddy g−1(y)
∣∣∣∣ If g is two-to-one (or more), then you can
use this with each “inverse” and add them up.
If (Y1, Y2) = g(X1, X2), then fY1,Y2(y1, y2) = fX1,X2
