Economics 202

Spring 2009

Homework 2

Your homework is due at the beginning of class onMonday, 2/16/08. Homework must be legible, stapled, and must have your name on it. No late homework will be accepted under any circumstances.

Chapter 4 problems
4.2

a. P(Brown) = 310/982 = 0.3157

b. P(YZ-99) = 375/982 = 0.3819

c. P(YZ-99 AND Brown) = 205/982 = 0.2088

d. No, it is possible to choose a White YZ-99 product, so they are not mutually exclusive (or, they have a positive joint probability, so they are not exclusive).

4.12

a. There are nine events:

AA,AB,AC,BA,BB,BC,CA,CB,CC

There are two events in which the first package goes through A and the second does not, so the probability is 2/9 = 0.2222

b. There are four events in which neither package travels through A, so the probability is 4/9 = 0.4444.

4.20
a. P(all three pick same color) = P(3 Brown) + P(3Gray) + P(3 Red)

= 1/27 + 1/27 + 1/27 = 3/27 = 1/9 (=.1111)

b. Using the complement rule, P (not all select same color) = 1 – P(all select same color) = 8/9 (=.8888)

4.26

a. P(Color down)P(Color Down) = (.2)(.2) = 0.04

b. Since color copiers can make black and white copies, all the copiers would have to be down in order to not be able to make a BW copy, and the color copiers would have to both be down to not make a color copy. So the probability of being able to make both copies is just the probability that one or more color copiers will be up = 1 -.04 = .96.

c. The probability that all 5 copiers are down = (.1)(.1)(1.)(.2)(.2) = .00004, so the probability that they can meet demand is .99996, which is fine by the manager’s standard.

d. P(all 5 running) = (.9)(.9)(.9)(.8)(.8) = 0.46656

P(all 6 running) = (.9)(.9)(.9)(.9)(.8)(.8) = 0.41990

Having more machines makes it less likely that all will be running.

4.38 (you need to get the data from the CD that comes with the book)

An automobile company is trying to determine which colors of cars are preferred by its customers. It surveys customers and gets the following data about sex of customer and color preference.

Male / Female
White / 148 / 141
Red / 285 / 92
Blue / 321 / 129
Black / 246 / 138

a. Create a table that shows the joint and marginal probabilities of each event.

(below)

b. Given that a customer is a woman, what is the probability that she will prefer a red car?

P(red|woman) = .063/.333 = .189

c. What is the probability that the customer is either a woman or prefers a white car?

P(woman OR white car) = .333 + 0.19 - .094 = 0.429

d. What is the probability that a man will prefer a blue car?

P(man AND blue) = .214

Or, you could read this as the probability that, given that the customer is a man, he will prefer a blue car, which would be

P(blue|man) = .214/.666 = .321

e. What is the probability that a customer will prefer a blue car?

P(blue) = 0.30

f. What is the probability that a customer is a woman and prefers a red car?

P(woman AND red) = .063

The following are the joint and marginal probabilities.

Male / Female
White / 148 / 141 / 289
Red / 285 / 92 / 377
Blue / 321 / 129 / 450
Black / 246 / 138 / 384
1,000 / 500 / 1500
Male / Female
White / .098 / .094 / 0.19
Red / .19 / .063 / 0.25
Blue / .214 / .086 / 0.3
Black / .164 / .092 / 0.256
0.666 / .333 / 1.00