McGraw-Hill/Irwin

Copyright © 2009 by The McGraw-Hill Companies, Inc. All Rights Reserved.

Chapter 4 and 5

Probability and Discrete Random Variables

4-2

The Concept of Probability

• An experiment is any process of observation with an uncertain outcomeThe 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

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

Complement

• The complement (Ā) of an event A is the set of all sample space outcomes not in A

• P(Ā) = 1 – P(A)

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

4-6

Discrete Random Variables

5.1 Two Types of Random Variables

5.2 Discrete Probability Distributions

5.3 The Binomial Distribution

5.4 The Poisson Distribution

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

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

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)

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…

1all

x

xp

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) 0 p(0) = 0.03 1 p(1) = 0.20 2 p(2) = 0.50 3 p(3) = 0.20 4 p(4) = 0.05 5 p(5) = 0.02

1.00

4-12

Example 5.3 Continued

• What is the chance that two radios will be sold in a week?

– p(x = 2) = 0.50

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 rulefor the mutuallyexclusive values ofthe random variable.

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

xAll

X xpx

4-15

Example 5.3: Number of RadiosSold 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, µX

• On average, expect to sell 2.1 radios per week

Radios, x Probability, p(x) x p(x) 0 p(0) = 0.03 0 0.03 = 0.00

1 p(1) = 0.20 1 0.20 = 0.20

2 p(2) = 0.50 2 0.50 = 1.00

3 p(3) = 0.20 3 0.20 = 0.60

4 p(4) = 0.05 4 0.05 = 0.20

5 p(5) = 0.02 5 0.02 = 0.10 1.00 2.10

xAll

X xpx

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:

xAll

XX xpx 22

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

2XX

4-18

Example 5.7: Number of RadiosSold at Sound City in a Week

Radios, x Probability, p(x) (x - X)2 p(x)

0 p(0) = 0.03 (0 – 2.1)2 (0.03) = 0.1323

1 p(1) = 0.20 (1 – 2.1)2 (0.20) = 0.2420

2 p(2) = 0.50 (2 – 2.1)2 (0.50) = 0.0050

3 p(3) = 0.20 (3 – 2.1)2 (0.20) = 0.1620

4 p(4) = 0.05 (4 – 2.1)2 (0.05) = 0.1805

5 p(5) = 0.02 (5 – 2.1)2 (0.02) = 0.1682

1.00 0.8900

xAll

XX xpx 22

4-19

Example 5.7 Continued

• Variance equals 0.8900

• Standard deviation is the square root of the variance

• Standard deviation equals 0.9434

4-20

Calculation the mean and variance

• Roll a die. Suppose all outcomes are equally possible.

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

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

x-nxqp

x-nx

n =xp

!!

!

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 = n•p

– Variance 2x = n•p•q

– Standard deviation x = square root n•p•qnpqX

4-24

Example 5.10: Incidence of Nauseaafter 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