Exam 1
Winter 2011
Name: _____________________________ This is a closed book exam. You may use a calculator and the
formulas handed out along with the exam. Show your work so I can see how you arrived at your answers. (This is
particularly important if I am to be able to give part credit.) Turn in this exam with your answers. However, don't
write your answers on the exam itself; leave them on the pages with your work. Also turn in the formulas; put them
on the formula pile.
1.
Security Bank makes 70% of its loans to men and 30% to women. For a man there is a 20% chance he will
default on the loan, but for a woman there is only a 10% chance she will default on the loan.
a.
(10 points) What percent of the bank's loans ends up in default?
b.
(5 points) Suppose a certain loan ends up in default. What is the probability the loan was made to a man?
2.
You work for a market research firm and you are doing a telephone survey about consumer attitudes toward
buying a new car. You estimate that for any given person you call there is a 30% chance that they will
participate in the survery and whether any person agrees to participate is independent of whether all the other
people agree to participate. Calculate the numerical value for your answer to each of the following.
a. (10 points) What is the probability that exactly 3 of 7 people you call agree to participate in the survey?
b. (10 points) What is the probability that at least 3 of 7 people you call agree to participate in the survey?
3.
During the winter in Detroit, there is a 20% chance that on any day it will snow and whether it snows on any
day is independent of any other day. Ann and Bob decide to play the following game. If it snows tomorrow,
Bob pays Ann a dollar. If it doesn't snow tomorrow, but it snows the next day, Ann pays Bob two dollars. If it
doesn't snow tomorrow or the next day, but it snows the day after, then Bob pays Ann three dollars. In general,
let N be the first day after today that it snows, with tomorrow being N = 1, the day after N = 2, and so on. If N
is odd, then Bob pays Ann N dollars. If N is even Ann pays Bob N dollars.
a. (6 points) Find the probability that N = n for n = 1, 2, 3, …
b. (8 points) Find the probability that Bob ends up paying Ann something. The formula sheet should be
helpful for evaluating infinite sums.
c. (11 points) Let X be the amount Ann ends up winning on the game. Thus X = 1 if it snows tomorrow,
X = - 2 if it doesn't snow tomorrow, but it snows the next day, etc. So X = N if N is odd, but X = - N if N is
even. Find the expected value of X. The formula sheet should be helpful for evaluating infinite sums.
4.
The joint probability mass function fXY(x,y) of random variables X and Y is given by the following.
|
x = -1 |
x= 0 |
x= 1 |
a.
b.
c.
5.
y = -1
0.1
0.05
0.25
y=0
0.2
0.15
0.05
y=1
0.1
0.05
0.05
(7 points) Find the conditional probability mass function fY|X(y|x) of Y given X.
(6 points) Are X and Y independent? Explain why or why not?.
(7 points) Find the probability mass function of the product Z = XY.
Tom operates a hardware store and one of the things he sells is snowblowers. Tom needs to decide how may
snowblowers to order for the next winter season. It costs him $200 to purchase each one wholesale. Then he
sells them retail for $400 each. Any that he can’t sell by the end of the winter he is able to get rid of for $100
each. Let D be the number of snowblowers he would be able to sell at $400 if he had them available. Suppose
1
Pr{ D = k } = 20, for k = 0 1 2 … 19. Let y be the number of snowblowers he orders and py be the expected
' ' ' '
profit if he orders y snowblowers. Assume he can only order snowblowers wholesale before the winter season
starts.
a. (4 points) Suppose Tom would be able to sell k snow blowers where k is greater than or equal to y. What is
his profit? Your answer should be some formula that can involve y and k.
b. (4 points) Suppose Tom is able to sell k snow blowers where k is less than or equal to y. What is his profit?
Your answer should be some formula that can involve y and k.
c. (12 points) Find a formula for py. For this part you can express your answer using summation notation
where k is the summation variable.
1.
2.
Solutions
a. Given: Pr{M} = 0.7, Pr{D|M} = 0.2, Pr{D|W} = 0.1. So Pr{DM} = Pr{D|M} Pr{M} = (0.2)(0.7) = 0.14
and Pr{DW} = Pr{D|W} Pr{W} = (0.1)(0.3) = 0.03. So Pr{D} = Pr{DM} + Pr{DW} = 0.14 + 0.03 =
0.17.
b. Pr{M|D} = Pr{DM}/Pr{D} = 0.14/0.17  0.82.
7
a. Let N = number that participate. Pr{N = 3} = 3 (0.3)3(0.7)4 = (35)(0.027)(0.2401)  0.2269
7
7
b. Pr{N  3} = 1 - Pr{N = 0, 1 or 2} = 1 – [(0.7)7 + 1 (0.3)1(0.7)6 + 2 (0.3)2(0.7)5] = 1 – 0.0824 – 0.2471 –
0.3176  0.3529
3.
a. Pr{N = n} =
1 4n-1
5 5
b. Pr{Bob pays Ann} = Pr{N is odd} = Pr{N = 1} + Pr{N = 3} + Pr{N = 5} +  =
=
1 1 42 1 44 
+
+
+
5 5 5 5 5
1
4 2
4 4
1
16
16 2
1
1
 5
[1 +   +   +  ] = [1 +   +   +  ] =   
5
5
5 5
25 25
5 1 - 16/25 = 9
1
1 4
1 4 2
- (2)    + (3)    5
5 5
5 5
3
2
3
2
3
1
4
1
4
4
4
1
4
4
4
(4)    +  = [1 – (2)  + (3)  - (4)  +  ] = = [1 + (2)-  + (3)-  + (4)-  +  ] =
5
5
5 5
5
5
5
 5
 5
 5
1
1
1
1
5
 
  

5 (1 - (- 4/5))2 = 5 (1 + 4/5)2 = 81  0.0617
a. fY|X(y|x) = fXY(x,y)/fX(x). So we need to compute fX(x). These are the row sums of the joint distribution table.
c. E{X} = (1)Pr{N = 1} – (2)Pr{N = 2} + (3)Pr{N = 3} – (4)Pr{N = 4} + -  =
4.
|
y = -1
y=0
y=1
fX(x)
x = -1 |
0.1
0.2
0.1 | 0.4
x= 0 |
0.05
0.15
0.05 | 0.25
x= 1 |
0.25
0.05
0.05 | 0.35
Now we divide each row by its row sum to get fY|X(y|x).
b.
c.
5.
a.
b.
|
y = -1
y=0
y=1
x = -1 |
1/4
1/2
1/4
x= 0 |
1/5
3/5
1/5
x= 1 |
5/7
1/7
1/7
X and Y are independent if and only if fY|X(y|x) = fY(y) for all x and y. This means fY|X(y|x) should not depend
on x. However, we can see from the table in part a that fY|X(y|x) does depend on x. So X and Y are not
independent.
The possible values of Z are –1, 0 and 1. For each of these three values fZ(z) = Pr{Z = z} = Pr{XY = z}. So
fZ(-1) = Pr{XY = -1} = Pr{(X,Y) = (-1,1) or (1,-1)} = 0.1 + 0.25 = 0.35.
fZ(0) = Pr{XY = 0} = Pr{(X,Y) = (-1,0), (0,-1), (0,0) (0,1) or (1,0)} = 0.2 + 0.05 + 0.15 + 0.05 + 0.05 = 0.5.
fZ(1) = Pr{XY = 1} = Pr{(X,Y) = (-1,-1) or (1,1)} = 0.1 + 0.05 = 0.15.
Since k  y, Tom sells only the y that he stocks. So the profit is 200y.
Since k  y, Tom sells k of the ones he orders for retail and (y – k) for $100. So the profit is 200k - 100(y - k)
= 300k – 100y.
y
c. py =

k=0
1
(300k - 100y)   +
20
19
 (200y) 201 
k = y+1