1. © 2008 Pearson Addison-Wesley. All rights reserved 12-5-1 Probability Level 8 The key idea of probability at Level 8 is investigating chance situations using probability concepts and distributions. Investigating chance situations using concepts such as randomness, probabilities of combined events and mutually exclusive events, independence, conditional probabilities and expected values and standard deviations of discrete random variables, and probability distributions including the Poisson, binomial and normal distributions.

2. © 2008 Pearson Addison-Wesley. All rights reserved 12-5-2 Chapter 1 Random variables A random variable, X, is a numerical measure of the outcomes of an experiment Random variables can be discrete or continuous

3. © 2008 Pearson Addison-Wesley. All rights reserved 12-5-3 Chapter 1 Random variables Discrete random variables have a countable number of outcomes –Examples: Dead/alive, treatment/placebo, dice, counts, etc. Continuous random variables have an infinite continuum of possible values. –Examples: blood pressure, weight, the speed of a car, the real numbers from 1 to 6.

4. © 2008 Pearson Addison-Wesley. All rights reserved 12-5-4 Chapter 1 Expected Value The Expected Value of a random variable is a kind of theoretical average that it should take. It is an extremely useful concept for good decision making! The Expected Value is often referred to as the population mean (symbol µ ) Expected Value = E(X) = µ The symbol for Expected Value is E(X)

5. © 2008 Pearson Addison-Wesley. All rights reserved 12-5-5 Expected Value It is an extremely useful concept for good decision making and is used in Games and Gambling Investments Business and Insurance

6. © 2008 Pearson Addison-Wesley. All rights reserved 12-5-6 Expected Value It is an extremely useful concept for good decision making and is used in Games and Gambling Investments Business and Insurance

7. © 2008 Pearson Addison-Wesley. All rights reserved 12-5-7 To work out the expected value of a random variable from a probability table MULTIPLY each possible value of X by its probability Then ADD these products Discrete case:

8. © 2008 Pearson Addison-Wesley. All rights reserved 12-5-8 Expected Value example 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. x (hrs)0123 P(x)P(x)0.30.40.20.1 Calculate a “weighted average” by multiplying each possible time value by its probability and then adding the products.

9. © 2008 Pearson Addison-Wesley. All rights reserved 12-5-9 Expected Value 1.1 hours is the expected value (or the mathematical expectation) of the quantity of time spent playing electronic games.

10. © 2008 Pearson Addison-Wesley. All rights reserved 12-5-10 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.

11. © 2008 Pearson Addison-Wesley. All rights reserved 12-5-11 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. In context this means that if a large number of three child families were studied the “average” number of girls in these families would be very close to1.5.

12. © 2008 Pearson Addison-Wesley. All rights reserved 12-5-12 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. Die OutcomePayoff NetP(x)P(x)

13. © 2008 Pearson Addison-Wesley. All rights reserved 12-5-13 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

14. © 2008 Pearson Addison-Wesley. All rights reserved 12-5-14 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.

15. © 2008 Pearson Addison-Wesley. All rights reserved 12-5-15 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.

16. © 2008 Pearson Addison-Wesley. All rights reserved 12-5-16 Investments Expected value can be a useful tool for evaluating investment opportunities.

17. © 2008 Pearson Addison-Wesley. All rights reserved 12-5-17 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 $800.51000.2 $1300.3 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.

18. © 2008 Pearson Addison-Wesley. All rights reserved 12-5-18 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.

19. © 2008 Pearson Addison-Wesley. All rights reserved 12-5-19 Business and Insurance Expected value can be used to help make decisions in various areas of business, including insurance.

20. © 2008 Pearson Addison-Wesley. All rights reserved 12-5-20 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?

21. © 2008 Pearson Addison-Wesley. All rights reserved 12-5-21 Example: Expected Lumber Revenue Income xP(x)P(x) $9500.25$2375 $9000.60$5400 $8500.15$1275 Solution The expected revenue from sales can be found below. Expected revenue: E(x) = $9050

22. © 2008 Pearson Addison-Wesley. All rights reserved 12-5-22 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.