Course 1 May 2000 Multiple Choice Exams - Society of Actuaries

33. Answer: B
Let Y = positive test result
D = disease is present (and ~D = not D)
Using Baye’s theorem:
PY [ | DPD ] [ ] (0.95)(0.01)
P[D|Y] =
PY [ | DPD ] [ ] + PY [ |~ DP ] [~ D ] = (0.95)(0.01) + (0.005)(0.99)
= 0.657 .
34. Answer: D
⎧ y for 1< y≤10
Let W denote claim payments. Then W = ⎨
⎩10 for y ≥10
10
∞
2 2 2 10 10 ∞
It follows that E[W] = y dy+ 10 dy =− −
3 3 1 2
y y y y 10
∫ ∫ = 2 – 2/10 + 1/10 = 1.9 .
1 10
35. Answer: E
Let X J , X K , and X L represent annual losses for cities J, K, and L, respectively. Then
X = X J + X K + X L and due to independence
xt
( xJ + xK + xL)
t xt J xt K xt L
M(t) = E⎡e E⎡e ⎤
⎣
⎤
⎦ = = E⎡e ⎤E⎡ ⎣
e ⎤
⎦
E⎡ ⎣
e ⎤
⎣ ⎦ ⎣ ⎦
⎦
= M J (t) M K (t) M L (t) = (1 – 2t) –3 (1 – 2t) –2.5 (1 – 2t) –4.5 = (1 – 2t) –10
Therefore,
M¢(t) = 20(1 – 2t) –11
M≤(t) = 440(1 – 2t) –12
M≤¢(t) = 10,560(1 – 2t) –13
E[X 3 ] = M≤¢(0) = 10,560
36. Answer: A
1 11 1 1 1 ⎛1⎞
p k = pk− 1
= pk−2 = ⋅ ⋅ pk−3 = ... = ⎜ ⎟ p0
k ≥0
5 55 5 5 5 ⎝5⎠
∞ ∞
⎛1⎞
p0
5
1 = ∑ pk
= ∑⎜
⎟ p = = p
1
k= 0 k=
0⎝5⎠ 1−
4
5
k
0 0
p 0 = 4/5 .
Therefore, P[N > 1] = 1 – P[N £1] = 1 – (4/5 + 4/5 ◊ 1/5) = 1 – 24/25 = 1/25 = 0.04 .
k
Course 1 12 May 2000