1. McGraw-Hill/IrwinCopyright © 2009 by The McGraw-Hill Companies, Inc. All Rights Reserved. Chapter 4 and 5 Probability and Discrete Random Variables

2. 4-2 The Concept of Probability An experiment is any process of observation with an uncertain outcome The possible outcomes for an experiment are called the experimental outcomes. The set of all possible outcomes is called the sample space of an experiment. S=(1, 2, 3, 4, 5, 6) Probability is a measure of the chance that an experimental outcome will occur when an experiment is carried out

3. 4-3 Probability If E is an experimental outcome, then P(E) denotes the probability that E will occur and: Conditions 1.0  P(E)  1 such that: If E can never occur, then P(E) = 0 If E is certain to occur, then P(E) = 1 2.The probabilities of all the experimental outcomes must sum to 1

4. 4-4 Complement The complement (Ā) of an event A is the set of all sample space outcomes not in A P(Ā) = 1 – P(A)

5. 4-5 Mutually Exclusive A and B are mutually exclusive if they have no sample space outcomes in common In other words: P(A∩B) = 0

6. 4-6 Discrete Random Variables 5.1 Two Types of Random Variables 5.2 Discrete Probability Distributions 5.3 The Binomial Distribution 5.4The Poisson Distribution

7. 4-7 Two Types of Random Variables Random variable: a variable that assumes numerical values that are determined by the outcome of an experiment –Discrete –Continuous Discrete random variable: Possible values can be counted or listed; doesn’t take values on an interval of the real line. –The number of defective units in a batch of 20 –Toss a coin. Let x=1 if we have a head, x=0 if we have a tail

8. 4-8 Random Variables Continued Continuous random variable: May assume any numerical value in one or more intervals –The waiting time for a credit card authorization –The interest rate charged on a business loan

9. 4-9 Discrete Probability Distributions The probability distribution of a discrete random variable is a table, graph or formula that gives the probability associated with each possible value that the variable can assume Notation: Denote the values of the random variable by x and the value’s associated probability by p(x)

10. 4-10 Discrete Probability Distribution Properties 1.For any value x of the random variable, 1  p(x)  0 2.The probabilities of all the events in the sample space must sum to 1, that is…

11. 4-11 Example 5.3 Continued p(x = 2)= 0.5 p(x <2 )= 0.03+0.2=0.23 p(X ≥ 3)= 0.2+0.05+0.02=0.27 Number of Radios Sold at Sound City in a Week Radios, x Probability, p(x) 0p(0) = 0.03 1p(1) = 0.20 2p(2) = 0.50 3p(3) = 0.20 4p(4) = 0.05 5p(5) = 0.02 1.00

12. 4-12 Example 5.3 Continued What is the chance that two radios will be sold in a week? –p(x = 2) = 0.50

13. 4-13 Example 5.3 Continued What is the chance that fewer than 2 radios will be sold in a week? –p(x < 2)= p(x = 0 or x = 1) = p(x = 0) + p(x = 1) = 0.03 + 0.20 = 0.23 What is the chance that three or more radios will be sold in a week? –p(x ≥ 3)= p(x = 3, 4, or 5) = p(x = 3) + p(x = 4) + p(x = 5) = 0.20 + 0.05 + 0.02 = 0.27 Using the addition rule for the mutually exclusive values of the random variable.

14. 4-14 Expected Value of a Discrete Random Variable The mean or expected value of a discrete random variable X is:  is the value expected to occur in the long run and on average

15. 4-15 Example 5.3: Number of Radios Sold at Sound City in a Week How many radios should be expected to be sold in a week? –Calculate the expected value of the number of radios sold, µ X On average, expect to sell 2.1 radios per week Radios, x Probability, p(x) x p(x) 0p(0) = 0.030  0.03 = 0.00 1p(1) = 0.201  0.20 = 0.20 2p(2) = 0.502  0.50 = 1.00 3p(3) = 0.203  0.20 = 0.60 4p(4) = 0.054  0.05 = 0.20 5p(5) = 0.025  0.02 = 0.10 1.00 2.10

16. 4-16 Variance The variance is the average of the squared deviations of the the random variable from the expected value The variance of a discrete random variable is:

17. 4-17 Standard Deviation The standard deviation is the square root of the variance The variance and standard deviation measure the spread of the values of the random variable from their expected value

18. 4-18 Example 5.7: Number of Radios Sold at Sound City in a Week Radios, xProbability, p(x) (x -  X ) 2 p(x) 0p(0) = 0.03(0 – 2.1) 2 (0.03) = 0.1323 1p(1) = 0.20(1 – 2.1) 2 (0.20) = 0.2420 2p(2) = 0.50(2 – 2.1) 2 (0.50) = 0.0050 3p(3) = 0.20(3 – 2.1) 2 (0.20) = 0.1620 4p(4) = 0.05(4 – 2.1) 2 (0.05) = 0.1805 5p(5) = 0.02(5 – 2.1) 2 (0.02) = 0.1682 1.00 0.8900

19. 4-19 Example 5.7 Continued Variance equals 0.8900 Standard deviation is the square root of the variance Standard deviation equals 0.9434

20. 4-20 Calculation the mean and variance Roll a die. Suppose all outcomes are equally possible.

21. 4-21 The Binomial Distribution The binomial experiment… 1.Experiment consists of n identical trials 2.Each trial results in either “success” or “failure” 3.Probability of success, p, is constant from trial to trial –The probability of failure, q, is 1 – p 4.Trials are independent If x is the total number of successes in n trials of a binomial experiment, then x is a binomial random variable

22. 4-22 Binomial Distribution Continued For a binomial random variable x, the probability of x successes in n trials is given by the binomial distribution: –n! is read as “n factorial” and n! = n × (n-1) × (n-2) ×... × 1 –0! =1 –Not defined for negative numbers or fractions

23. 4-23 Mean and Variance of a Binomial Random Variable If x is a binomial random variable with parameters n and p (so q = 1 – p), then –Mean  = np –Variance  2 x = npq –Standard deviation  x = square root npq

24. 4-24 Example 5.10: Incidence of Nausea after Treatment Let x be the number of Tails you have after tossing a unfair coin ten times. Assume the probability to get a head is 0.6. Find the probability that you get 5 Tails. Given: n = 10; How to define ‘success’ for tossing coin for this problem? Define tail as success p =0.4; P(x = 2)=0.1209; n=10, What are the mean of x, 10*0.4=4 variance of x, 10*0.4*0.6=2.4, standard deviation of x? Sqrt(2.4)=1.549