Math/Stat 304
Solutions: TEST II
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Part I (Each question is worth 10 pts.)
1. [10 pts] [Actuarial sample question] Albertine’s car dealership in BetaVille
sells 0, 1, or 2 luxury cars on any given day. When selling a car, the dealer also
tries to persuade the customer to buy an extended warranty for the car. Let X
denote the number of luxury cars sold in a given day, and let Y denote the
number of extended warranties sold.
Let px,y(x, y) be the joint pmf of X and Y.
Assume that
px,y (0, 0) = 1/6
px,y (1, 0) = 1/12
px,y (1, 1) = 1/6
px,y (2, 0) = 1/12
px,y (2, 1) = 1/3
px,y (2, 2) = 1/6
Find the variance of X. Express your answer to the nearest hundredth.
Solution: First find the marginal pmf of X.
pX(0) = pX,Y(0, 0) = 1/6
pX(1) = pX,Y(1, 0) + pX,Y(1, 1) = 1/12 + 1/6 = ¼
pX(2) = pX,Y(2, 0) + pX,Y(2, 1) + pX,Y(2, 2) = 1/12 + 1/3 + 1/6 = 7/12
Hence E(X) = 0(1/6) + 1(1/4) + 2(7/12) = 17/12 = 1.4167 and
E(X2) = 02(1/6) + 12(1/4) + 22(7/12) = 31/12 = 2.5833
Finally, Var(X) = E(X2) – (E(X))2 = 2.5833 – (1.4167)2 = 0.5763 ≈ 0.58
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2. [10 pts] If X is a binomial random variable with expected value 6 and variance
2.4, find P(X = 5).
Solution: If X ~ binomial(n, p) then E(X) = np and var(X) = np(1 – p).
So np = 6 and np(1 – n) = 2.4. Hence 6(1 – p) = 2.4, from which we obtain
p = 0.6 and n = 10.
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Thus P(X = 10) = ( ) (0.6)5 (0.4)5 .
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3. [10 pts] Consider the following experiment. An unfair coin (probability of a
head is 1/4) is flipped 100 times; then another unfair coin (probability of head is
2/3) is tossed until a tail appears. Let X be the total number of heads that occur
during the course of the experiment.
Find E(X) and Var(X).
Solution: Let X1 = number of heads obtained from the 100 flips of the first unfair
coin.
Let X2 = number of heads obtained by flipping the second unfair coin until a head
appears.
Note that X = X1 + X2. Hence E(X) = E(X1) + E(X2) = 100(1/4) + 1/(2/3) = 26.5
Using independence,
var(X) = var(X1) + var(X2) = 100 (1/4)(1 – 1/4) + (1 – 2/3)/(2/3)2 = 19.5
Here we have used the facts that X1 is binomial(100, 1/4) and X2 is geometric with
parameter p = 2/3.
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Part II Answer any 3 of the 4 questions below. You will be given extra credit
for answering all 4. (Each question is worth 20 pts.)
1. The AlphaVille Cereal Company has a new marketing scheme. Each time you
purchase a box of cereal, you will find contained inside the box one small coin
bearing the likeness of one of five individuals: Kolmogorov, Poisson, Gauss,
Bayes, or Bernoulli. Assume that these five coins are equally likely to appear
in any given box of cereal, and that the coin in any box of cereal is independent
of the content of any other box. Let X = the number of boxes Albertine
purchases to obtain a complete set of coins (a complete set of coins is one in
which every one of the five probabilists is included). Find the expected value
of X.
Solution: Let X1 be the number of flips required to find a new type of coin; of
course, P(X1) = 1. Let X2 = number of flips required to find the next new medal.
Note that X2 is a geometric r.v. with p = 4/5. Then E(X2) = 1/p = 5/4.
Similarly let X3, X4, X5 denote the geometric r.v. that, respectively, count the
number of flips necessary for the next new medal. So the parameters for X3, X4, X5
are, respectively, 3/5, 2/5 and 1/5. Now X = X1 + X2 + X3 + X4 + X5.
Thus E(X) = E(X1 + X2 + X3 + X4 + X5) = 1 + 5/4 + 5/3 + 5/2 + 5/1 = 137/12 ≈
𝟏𝟏. 𝟒 𝒃𝒐𝒙𝒆𝒔.
2.
Lucky and Pozzo need to purchase new smartphone cases. The phone store
has a stock of only one red case, 4 blue cases and 3 green cases. The sale
prices of the red, blue and green cases are $40, $30 and $10, respectively.
Lucky enters the store and impulsively purchases one of the cases at random.
Then Pozzo enters the store and also randomly chooses a case. Let X = cost of
Lucky’s case and Y = cost of Pozzo’s case.
(a) Find E(X).
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Solution: Let pX(x) donote the pmf of X.
Then pX (10) = 3/8; pX (30) = 4/8; pX(40) = 1/8.
Thus E(X) = 10(3/8) + 30(4/8) + 40(1/8) = 190/8 = 95/4 = $ 23.75
(b) Find E(X + 2Y + 1)
Solution: Note that Y has the same distribution as X. Now, since E is linear,
E(X + 2Y + 1) = E(X) + 2E(Y) + 1 = 3E(X) + 1 = 3(95/4) + 1 = 289/4 = $ 72.25
(c) Let pX|Y(x|y) denote the conditional pmf of X given Y. Find pX|Y(x|40) for
x = 10, 30, 40
Solution: Since
pX|Y(x|40) = pX,Y(x,40) / pY(40) = py|X(40 | x) pX(x) / pY(40) = 8 py|X(40 | x) pX(x)
we find that
pX|Y(10|40) = py|X(40 | 10) pX(10) = 8(1/7)(3/8) = 3/7,
pX|Y(30|40) = py|X(40 | 30) pX(30) = 8(1/7)(4/8) = 4/7,
pX|Y(40|40) = py|X(40 | 40) pX(10) = 8(0) = 0.
(d) Find E(X| Y = 40)
Solution: Clearly E(X| Y = 40) = 10 pX|Y(10|40) + 30 pX|Y(30|40) = 10(3/7) +
30(4/7) = 150/7 = $ 21.43
(e) Are X and Y independent? Justify your answer.
Solution: No, X and Y are not independent; for example,
pX,Y(40, 40) = 0, yet pX(40) pY(40) = (1/8)(1/8) = 1/64.
That is, P(X = 40, Y = 40) ≠ P(X = 40) P(Y = 40).
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3. Suppose that a fair three-sided die is thrown three times, with outcomes X1,
X2, and X3, respectively. Let Y be the maximum of the three values.
(a) Find the pmf of Y.
Solution: Let pY(y) denote the pmf of Y.
The possible values of Y lie in the set {1, 2, 3}.
The equiprobable sample space, , consists of all ordered triples from the set
{1, 2, 3}.
Now, if Y = 1, then the outcome must have been (1, 1, 1). Thus P(Y = 1) = 1/27.
Next, if Y = 2, the possible outcomes are
(2, 1, 1), (1, 2, 1), (1, 1, 2)
(1, 2, 2), (2, 1, 2), (2, 2, 1)
(2, 2, 2)
Hence P(Y = 2) = 7/27.
Finally P(Y = 3) = 1 – P(Y = 1) – P(Y = 2) = 1 – 1/27 – 7/27 = 19/27.
Summarizing, pY(1) = 1/27, pY(2) = 7/27, pY(3) = 19/27.
(b) Find E(Y).
Solution: E(Y) = 1(1/27) + 2(7/27) + 3(19/27) = 72/27 = 8/3 = 2.67
(c) Find E(Y| X1 < X3).
Solution: Consider the event A = “X1 < X3”. This implies that the possible
outcomes are:
(1, 1, 2), (1, 2, 2), (1, 3, 2)
(1, 1, 3), (1, 2, 3), (1, 3, 3)
(2, 1, 3), (2, 2, 3), (2, 3, 3)
Thus P(A) = 9/27 = 1/3.
So pY|A(1|A) = 0, pY|A(2|A) = 2/9, pY|A(3|A) = 7/9.
Hence E(Y|A) = 0 pY|A(1|A) + 2 pY|A(2|A) + 3pY|A(3|A) = 2(2/9) + 3(7/9) = 25/9 =
2.78
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4.
Albertine ponders her daily commute. She leaves home at 7 am and drives to
work. If all the traffic lights are green, her trip takes 40 minutes. Albertine’s
route includes 5 traffic lights, each of which is red with probability 1/3 (assume
that yellow is not an option), independent of every other light. Each red traffic
light that she encounters adds 4 minutes to her commute (for slowing, stopping,
and returning to speed). Let X be Albertine’s commute time (in minutes).
(a) Find the pmf of X.
Solution: The values of X are 40, 44, 48, 52, 56, 60. The number of lights that are
red during Albertine’s commute is given by a binomial random variable, D with
parameters n = 5 and p = 1/3. Albertine’s delay due to traffic lights is 4 times the
number of lights that are red.
Now X = 40 + 4D; so
(2 / 3) 5 if x  40

 5  (1 / 3)( 2 / 3) 4 if x  44
1 
 
 5 
2
3
  (1 / 3) (2 / 3) if x  48
 2 
p X ( x)  
 5  (1 / 3) 3 (2 / 3) 2 if x  52
 3 

 5 
4
1
  (1 / 3) (2 / 3) if x  56
4
 
(1 / 3) 5 if x  60

(b) Find E(X).
Solution: E(X) = 40 + 4 E(D) = 40 + 4 (5)(1/3) = 140/3 = 46.67 minutes
(c) Find Var(X).
Solution: var(X) = var(40 + 4 D) = var(4 D) = 16 var(D) =
16 np(1 – p) = 16 (5) (1/3) ( 1 – 1/3) = 160/9 = 17.78 minutes2.