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Math 408, Spring 2005
Final Exam Solutions
- Assume A and B are independent events with P (A) = 0.2 and P (B) = 0.3. Let C be the event that neither A nor B occurs, let D be the event that exactly one of A or B occurs.
(a) Find P (C). Solution. P (C) = P (A′^ ∩ B′) = P (A′)P (B′) = (1 − P (A))(1 − P (B)) =
8 · 0 .7 = 0. 56
56 (b) Find P (D). Solution. P (D) = P (A ∪ B \ A ∩ B) = P (A ∪ B) − P (A)P (B) = P (A) + P (B) − 2 P (A ∩ B) = 0. 38
38 (c) Find P (A|D). Solution. P (A|D) = P (A ∩ D)/P (D) = P (A \ A ∩ B)/P (D) = (0. 2 − 0. 2 · 0 .3)/ 0 .38 = 7/ 19 7 19 = 0.^368 (d) Are C and D independent? Justify your answer! Solution. C and D are not independent since P (C∩D) = 0, but P (C)P (D) =
56 · 0. 38 6 = 0. NO
[2.4-9, Variant] Suppose A, B, and C are mutually independent events with prob- abilities P (A) = 0.5, P (B) = 0.8, and P (C) = 0.3. Find the probability that at least one of these events occurs.
(^93) Solution. The probability in question is P (A ∪ B ∪ C). Using the formula for
the probability of a union of three events, and the independence of the three sets, we get
P (A ∪ B ∪ C) = P (A) + P (B) + P (C) − P (A ∩ B) − P (A ∩ C) − P (B ∩ C)
- P (A ∩ B ∩ C) = 0.5 + 0.8 + 0. 3 − (0. 5 · 0 .8) − (0. 5 · 0 .3) − (0. 8 · 0 .3)
- (0. 5 · 0. 8 · 0 .3) = 0. 93.
- How many ways are there to seat 10 people, consisting of 5 couples, in a row of seats (10 seats wide) if all couples are to get adjacent seats? Solution. 5!2^5 = 3840 (or 10 · 8 · 6 · 4 · 2 = 3840) 3840
- [Actuarial Exam Problem #27] The probability that a randomly chosen male has a circulation problem is 0.25. Males who have a circulation problem are twice as likely to be smokers as those who do not have a circulation problem. What is the conditional probability that a male has a circulation problem, given that he is a smoker? Solution. Let C denote the event “has a circulation problem” and S the event
- 4 “is a smoker”. We are given that P (C) = 0.25 and P (S|C) = 2P (S|C′), and we need to compute P (C|S). By Bayes’ Rule,
P (C|S) =
P (S|C)P (C)
P (S|C)P (C) + P (S|C′)P (C′)
P (S|C)0. 25
P (S|C)0.25 + (1/2)P (S|C)(1 − 0 .25)
- [Similar to Actuarial Exam Problem #57] Suppose a random variable X has mo- ment generating function
M (t) =
2 + et 3
Calculate the variance of X. 2 Solution. We use the formula Var X = E(X^2 ) − E(X)^2 = M ′′(0) − M ′(0)^2. Computing the derivatives of M (t), we get
M ′(t) = 9
2 + et 3
et 3
M ′′(t) = 3 · 8
2 + et 3
e^2 t 3
2 + et 3
et,
M ′(0) = 3
2 + e^0 3
e^0 = 3 ,
M ′′(0) = 8
so Var(X) = 11 − 32 = 2.
- [Actuarial Exam Problem #46] A device that continuously measures and records seismic activity is placed in a remote region. The time, T , to failure of this device is exponentially distributed with mean 3 years. Since the device will not be monitored during its first two years of service, the time to discovery of its failure is X = max(T, 2). Determine E(X). Solution. First note that since T has exponential distribution with mean 3, the
- 54 density of T is f (t) = (1/3)e−t/^3 for t ≥ 0. Since X = max(T, 2) equals 2 if T ≤ 2, and T if T ≥, it follows that (using integration by parts for the second integral)
E(X) =
0
e−t/^3 dt +
2
t ·
e−t/^3 dt
[
−e−t/^3
] 2
0 +^
[
−e−t/^3 t
]∞
2
e−t/^3 dt
= 2(1 − e−^2 /^3 ) − (−e−^2 /^3 2) +
e−^2 /^3 = 2 + 3e−^2 /^3 = 3. 54.
- [Actuarial Exam Problem #76] Claim amounts for wind damage to insured homes are independent random variables with common density function f (x) = 3x−^4 for x > 1, and f (x) = 0 otherwise, where x is the amount of a claim in thousands. Suppose 3 such claims are made. What is the expected value of the largest of the three claims? Solution. Let X 1 , X 2 , X 3 denote the three claims, and Y = max(X 1 , X 2 , X 3 ) 2025 the largest of these. We need to compute E(Y ). This requires first computing the c.d.f. F (y) and the p.d.f. f (y) = F ′(y) of Y. By the maximum trick,
F (y) = P (max(X 1 , X 2 , X 3 ) ≤ y) = P (X 1 ≤ y)P (X 2 ≤ y)P (X 3 ≤ y)
=
(∫ (^) y
1
3 x−^4 dx
y^3
so f (y) = F ′(y) = 3
y^3
y^4
y^4
y^3
for y ≥ 1. Therefore
E(Y ) =
1
yf (y)dy =
1
y^3
y^3
dy
1
y^3
y^6
y^9
dy
Since the units are thousands of dollars, the answer is 2025.
- [5.3-15] Let X have uniform distribution on the interval [0, 2], and given X = x, let Y have uniform distribution on the interval [0, x^2 ].
(a) Find the joint density f (x, y) of X and Y. (Be sure to specify the range!) Solution. We are given that fX (x) = 1/2, 0 ≤ x ≤ 2 (uniform distribution on [0, 2]), and g(y|x) = 1/x^2 , 0 ≤ y ≤ x^2 (uniform distribution on [0, x^2 ]). Therefore
f (x, y) = g(y|x)fX (x) =
x^2
(a)
2 x^2
, 0 ≤ x ≤ 2 , 0 ≤ y ≤ x^2 ,
(b) Find the marginal density fY (y) of Y. (Be sure to specify the range!) Solution.
fY (y) =
f (x, y)dx =
x=√y
2 x^2
dx =
[
2 x
]x=
x=√y
(b)
y
, 0 ≤ y ≤ 4
(c) Find E(XY ). Solution. We have
E(XY ) =
xyf (x, y)dydx =
x=
∫ (^) x 2
y=
(xy)
2 x^2
dydx
x=
x
(x^2 )^2 2
dx =
0
x^3 dx =
- [Actuarial Exam Problem #83] A company manufactures a brand of light bulb with a lifetime in months that is normally distributed with mean 3 and variance
- A consumer buys a number of these bulbs with the intention of replacing them successively as they burn out. The light bulbs have independent lifetimes. What is the smallest number of bulbs to be purchased so that the succession of light bulbs produces light for at least 40 months with probability 0.9772? (^16) Solution. Let n be the (unknown) number of light bulbs to be purchased,
X 1 ,... , Xn their respective lifetimes, and S =
∑n i=1 Xi^ the total lifetime of all^ n bulbs. We need to choose n minimal so that P (S ≥ 40) ≥ 0 .9772. Now, by the CLT, P (S ≥ 40) ≈ P
S − 3 n 1
n
40 − 3 n 1
n
40 − 3 n 1
n
This is equal to 0.9772 when (40− 3 n)/
n = −2, or equivalently, 3n− 2
n−40 = 0. Setting x =
n, the latter equation becomes 3x^2 − 2 x − 40 = 0. Solving (ignoring the negative solution) gives x = (1/6)(2 +
4 + 480 = 4, so n = x^2 = 16 is the number sought.
- [6.3-12] Assume the math scores on the SAT test are normally distributed with mean 500 and standard deviation 60, and the verbal scores are normally distributed with mean 450 and standard deviation 80. If two students who took both tests are chosen at random, what is the probability that the first student’s math score exceeds the second student’s verbal score?
- (^6915) Solution. Let X and Y denote the scores of the two students. Then X − Y is
N (500 − 450 , 602 + 80^2 ) = N (50, 1002 ), so
P (X > Y ) = P (X − Y > 0) = P
X − Y − 50
= P (Z > − 1 /2)
- [Example 6.3-2] Let X 1 , X 2 , X 3 , X 4 be a random sample of size 4 from the nor- mal distribution N (76. 4 , 383), and let X be the sample mean and S^2 the sample variance. Determine a such that P (S^2 ≤ a) = 0.90. Solution. We know that S^2 (n − 1)/σ^2 = (3/383)S^2 = 0. 0078 S^2 has χ^2 (3) a = 798 distribution, and from the chi-square table the 90-th percentile of a χ^2 (3) is 6.251. Thus,
0 .90 = P ((3/383)S^2 ≤ 6 .251) = P (S^2 ≤ 6 .251(383/3)) = P (S^2 ≤ 798).
so a = 798.
- [6.1-1(b)] Suppose X and Y are independent, each having Poisson distribution with means 2 and 3, respectively. Let Z = X + Y. Find P (X + Y = 1).
- (^0336) Solution.
P (X + Y = 1) = P (X = 0, Y = 1) + P (X = 1, Y = 0)
= (e−^220 /0!)(e−^331 /1!) + (e−^330 /0!)(e−^221 /1!)
= 5e−^5 = 0. 0336.
