Chapter 12 Probability © 2008 Pearson Addison-Wesley. All rights reserved

© 2008 Pearson Addison-Wesley. All rights reserved Chapter 12: Probability 12.1 Basic Concepts 12.2 Events Involving “Not” and “Or” 12.3 Conditional Probability; Events Involving “And” 12.4 Binomial Probability 12.5Expected Value

© 2008 Pearson Addison-Wesley. All rights reserved Chapter 1 Section 12-5 Expected Value

© 2008 Pearson Addison-Wesley. All rights reserved Expected Value Games and Gambling Investments Business and Insurance

© 2008 Pearson Addison-Wesley. All rights reserved Expected Value Children in third grade were surveyed and told to pick the number of hours that they play electronic games each day. The probability distribution is given below. # of Hours xProbability P(x)

© 2008 Pearson Addison-Wesley. All rights reserved Expected Value Compute a “weighted average” by multiplying each possible time value by its probability and then adding the products. 1.1 hours is the expected value (or the mathematical expectation) of the quantity of time spent playing electronic games.

© 2008 Pearson Addison-Wesley. All rights reserved Expected Value If a random variable x can have any of the values x 1, x 2, x 3,…, x n, and the corresponding probabilities of these values occurring are P(x 1 ), P(x 2 ), P(x 3 ), …, P(x n ), then the expected value of x is given by

© 2008 Pearson Addison-Wesley. All rights reserved Example: Finding Expected Value Find the expected number of boys for a three-child family. Assume girls and boys are equally likely. Solution # BoysProbabilityProduct xP(x)P(x) 01/80 13/8 2 6/8 31/83/8 S = {ggg, ggb, gbg, bgg, gbb, bgb, bbg, bbb} The probability distribution is on the right.

© 2008 Pearson Addison-Wesley. All rights reserved Example: Finding Expected Value Solution (continued) The expected value is the sum of the third column: So the expected number of boys is 1.5.

© 2008 Pearson Addison-Wesley. All rights reserved Example: Finding Expected Winnings A player pays $3 to play the following game: He rolls a die and receives $7 if he tosses a 6 and $1 for anything else. Find the player’s expected net winnings for the game.

© 2008 Pearson Addison-Wesley. All rights reserved Example: Finding Expected Winnings Die OutcomePayoff NetP(x)P(x) 1, 2, 3, 4, or 5$1–$25/6–$10/6 6$7$41/6$4/6 Solution The information for the game is displayed below. Expected value: E(x) = –$6/6 = –$1.00

© 2008 Pearson Addison-Wesley. All rights reserved Games and Gambling A game in which the expected net winnings are zero is called a fair game. A game with negative expected winnings is unfair against the player. A game with positive expected net winnings is unfair in favor of the player.

© 2008 Pearson Addison-Wesley. All rights reserved Example: Finding the Cost for a Fair Game What should the game in the previous example cost so that it is a fair game? Solution Because the cost of $3 resulted in a net loss of $1, we can conclude that the $3 cost was $1 too high. A fair cost to play the game would be $3 – $1 = $2.

© 2008 Pearson Addison-Wesley. All rights reserved Investments Expected value can be a useful tool for evaluating investment opportunities.

© 2008 Pearson Addison-Wesley. All rights reserved Example: Expected Investment Profits Company ABCCompany PDQ Profit or Loss x Probability P(x) Profit or Loss x Probability P(x) –$400.2$600.8 $ $ Mark is going to invest in the stock of one of the two companies below. Based on his research, a $6000 investment could give the following returns.

© 2008 Pearson Addison-Wesley. All rights reserved Example: Expected Investment Profits Solution ABC: –$400(.2) + $800(.5) + $1300(.3) = $710 PDQ: $600(.8) + $1000(.2) = $680 Find the expected profit (or loss) for each of the two stocks.

© 2008 Pearson Addison-Wesley. All rights reserved Business and Insurance Expected value can be used to help make decisions in various areas of business, including insurance.

© 2008 Pearson Addison-Wesley. All rights reserved Example: Expected Lumber Revenue A lumber wholesaler is planning on purchasing a load of lumber. He calculates that the probabilities of reselling the load for $9500, $9000, or $8500 are.25,.60, and.15, respectfully. In order to ensure an expected profit of at least $2500, how much can he afford to pay for the load?

© 2008 Pearson Addison-Wesley. All rights reserved Example: Expected Lumber Revenue Income xP(x)P(x) $ $2375 $ $5400 $ $1275 Solution The expected revenue from sales can be found below. Expected revenue: E(x) = $9050

© 2008 Pearson Addison-Wesley. All rights reserved Example: Expected Lumber Revenue Solution (continued) profit = revenue – cost or cost = profit – revenue To have an expected profit of $2500, he can pay up to $9050 – $2500 = $6550.