EPS2

Part 1: Background

Parts 1 and 2 of this homework assignment were copied and slightly modied from pages 143 - 146 of Prob-

ability and Statistics for Engineers and Scientists, 8th edition, by Walpole, Myers, Myers, and Ye, Pearson

Prentice Hall, 2007.

An experiment often consists of repeated trials, each with two possible outcomes that may be labeled

success or failure. The most obvious application deals with the testing of items as they come o an

assembly line, where each test or trial may indicate a defective or a nondefective item. We may choose to

dene either outcome as a success. The process is referred to as a Bernoulli process. Each trial is called

a Bernoulli trial.

Bernoulli Process

Strictly speaking, the Bernoulli process must possess the following properties:

1. The experiment consists of n repeated trials.

2. Each trial results in an outcome that may be classied as a success or a failure.

3. The probability of success, denoted by p, remains constant from trial to trial.

4. The repeated trials are independent.

Consider the set of Bernoulli trials where three items are selected at random from a manufacturing process,

inspected, and classied as defective or nondefective. A defective item is designated a success. The number

of successes in an arbitrary trial is a random variable X assuming integer values zero through 3, where x = 0

means no defects and x = 3 means 3 defects. Let D denote defective and N denote nondefective. The eight

possible outcomes and the corresponding values of X are

Outcome x

NNN 0

NND 1

NDN 1

NDD 2

DNN 1

DND 2

DDN 2

DDD 3

Assuming the defective probability rate is 25% (i.e. P(D)=0.25 or P(D) = 1

4 ) and the non-defective

probability rate is then P(N)= 1 - P(D) = 0.75, or P(N)=3

4 , then the probability of each outcome will be

determined by the product of the probability of the individual trials. For example, the probability of having

the rst item nondefective, the second item defective, and the third item nondefective is given by:

P(NDN) = P(N)P(D)P(N) =

3

4

1

4

3

4

= 9

64

:

Similar calculations yield the probabilities for the other possible outcomes. colorWe can now construct what

is known as a probability distribution of variable, X. The probability distribution of X is therefore

x 0 1 2 3

f(x)

27

64

27

64

9

64

1

64

1EPS2: Homework 2 Due Friday, September 18, 2015, 5:00 p.m.

The number X of successes in n Bernoulli trials is called a binomial random variable. The probability

distribution of this discrete random variable is called the binomial distribution, and its values will be

denoted by b(x; n; p) since they depend on the number of trials and the probability of a success on a given

trial (the number of trials in the above example is 3 and the probability of success, or nding a defect, for

each trial is 1

4 ). Thus, given the probability distribution on X for the above example, the probability of a

given X is

P(X = 0) = f(0) = b

0; 3;

1

4

= 27

64

P(X = 1) = f(1) = b

1; 3;

1

4

= 27

64

P(X = 2) = f(2) = b

2; 3;

1

4

= 9

64

P(X = 3) = f(3) = b

3; 3;

1

4

= 1

64

:

Binomial Distribution

A Bernoulli trial can result in a success with probability p and a failure with probability q = 1