president universityerwin sitompulpbst 6/1 dr.-ing. erwin sitompul president university lecture 6...

President University Erwin Sitompul PBST 6/1

Dr.-Ing. Erwin SitompulPresident University

Lecture 6

Probability and Statistics

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Chapter 5

Some Discrete Probability Distributions

Chapter 5 Some Discrete Probability Distributions


Introduction Often, the observations generated by different statistical

experiments have the same general type of behavior. The discrete random variables associated with these experiments

can be described by essentially the same probability distribution in a single formula.

Chapter 5.1 Introduction

In fact, one needs only a handful of important probability distributions to describe many of the discrete random variables encountered in practice.

In this chapter, we are going to present these commonly used distributions with various examples.


Discrete Uniform Distribution If the random variable X assumes the values x1, x2, ..., xk, with

equal probabilities, then the discrete uniform distribution is given by

Chapter 5.2 Discrete Uniform Distribution

1 2

1( ; ) , , , , kf x k x x x x

k

When a light bulb is selected at random from a box that contains a 40-watt bulb, a 60-watt bulb, a 75-watt bulb, and a 100-watt bulb, each element of the sample space S = {40, 60, 75, 100} occurs with probability 1/4.

Therefore, we have a uniform distribution, with1

( ;4) , 40,60,75,1004

f x x


Discrete Uniform DistributionWhen a dice is tossed, each element of the sample space S = {1, 2, 3,

4, 5, 6} occurs with probability 1/6.

Therefore, we have an uniform distribution with

Chapter 5.2 Discrete Uniform Distribution

1( ;6) , 1,2,3,4,5,6

6f x x


Bernoulli Process An experiment often consists of repeated trials, each with two

possible outcomes that may be labeled success or failure. We may choose to define either outcome as a success.

The process is referred to as a Bernoulli process. Each trial is called a Bernoulli trial.

Chapter 5.3 Binomial and Multinomial Distributions

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 classified 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.


Bernoulli ProcessChapter 5.3 Binomial and Multinomial Distributions

Consider the set of Bernoulli trials where three items are selected at random from a manufacturing process, inspected, and classified defective or non-defective. A defective item is designated a success. The number of successes is a random variable X assuming integer values from 0 to 3.

The items are selected independently from a process and we shall assume that it produces 25% defectives.

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

The probability of the outcome NDN can be calculated as

3 1 3 9

4 4 4 64


Bernoulli ProcessChapter 5.3 Binomial and Multinomial Distributions

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 denoted by b(x; n, p).

14

9( 2) (2) (2 : 3, )

64P X f b

The probabilities for the other possible outcomes can also be calculated to result the probability distribution of X


Binomial Distribution |Binomial Distribution| A Bernoulli trial can result in a success

with probability p and a failure with probability q = 1 – p. Then the probability distribution of the binomial random variable X, the number of successes in n independent trials, is

( : , ) , 0,1, 2,...,x n xn xb x n p C p q x n


The probability that a certain kind of component will survive a given shock test is 3/4. Find the probability that exactly 2 of the next 4 components tested will survive.

3

4p

2 2

4 2

3 3 12 : 4,

4 4 4b C

The mean and variance of the binomial distribution b(x; n, p) are2np npq

54

256


Binomial and Multinomial DistributionsChapter 5.3 Binomial and Multinomial Distributions

The probability that a patient recovers from a rare blood disease is 0.4. If 15 people are known to have contracted this disease, what is the probability that (a) at least 10 survive, (b) from 3 to 8 survive, and (c) exactly 5 survive?

Let X be the number of people that survive. Table A.1 gives help.

(a)

(b)

( 10) 1 ( 10)P X P X 9

0

1 ( ;15,0.4)x

b x

1 0.9662 0.033815

10

( ;15,0.4)x

b x

8

3

(3 8) ( ;15,0.4)x

P X b x

8 2

0 0

( ;15,0.4) ( ;15,0.4)x x

b x b x

0.9050 0.0271 0.8779

(c) ( 5) (5;15,0.4)P X b 0.18595 1015 5 (0.4) (0.6)C

?Can you calculate manually?


Table A.1 Binomial Probability SumsChapter 5.3 Binomial and Multinomial Distributions



A large chain retailer purchase a certain kind of electronic device from a manufacturer. The manufacturer indicates that the defective rate of the device is 3%.(a) The inspector of the retailer randomly picks 20 items from a

shipment. What is the probability that there will be at least one defective item among these 20?

(b) Suppose that the retailer receives 10 shipments in a month and the inspector randomly tests 20 devices per shipment. What is the probability that there will be 3 shipments containing at least one defective device?

Let X be the number of defective devices among the 20 items.

(a)

(b)

( 1) 1 ( 1)P X P X

0.4562p

1 ( 0)P X 0 2020 01 (0.03) (1 0.03)C 0.4562

( 3) (3;10,0.4562)P Y b 3 10 310 3(0.4562) (1 0.4562)C 0.1602



It is conjectured that an impurity exists in 30% of all drinking wells in a certain rural community. In order to gain some insight on this problem, it is determined that some tests should be made. It is too expensive to test all of the many wells in the area so 10 were randomly selected for testing.(a) Using the binomial distribution, what is the probability that

exactly three wells have the impurity assuming that the conjecture is correct?

(b) What is the probability that more than three wells are impure?

(a) 3 10 3

10 3( 3) (0.3) (1 0.3)P X C 0.2668

(b) ( 3) 1 ( 3)P X P X

1 0.0282 0.1211 0.2335 0.2668

0.3504

3

0

1 ( ;10,0.3)x

b x

?Try also to use Table A.1 to find this value



Consider the previous “drinking wells” example. The “30% are impure” is merely a conjecture put forth by the area water board.Suppose 10 wells are randomly selected and 6 are found to contain the impurity. What does this imply about the conjecture? Use a probability statement.

6 10 610 6( 6) (0.3) (1 0.3)P X C 0.0368

Should the 30% impurity conjecture is true, there is only 3.68% chance that it stands after 6 wells are found contaminated.

The investigation suggests that the impurity problem is much more severe than 30%.


Binomial and Multinomial Distributions The binomial experiment becomes a multinomial experiment if

we let each trial have more than 2 possible outcomes.

1 21 2 1 2 1 2

1 2

!( , ,..., ; , ,..., , ) ...

! !... !kxx x

k k kk

nf x x x p p p n p p p

x x x


1

k

ii

x n

1

1k

ii

p

|Multinomial Distribution| If a given trial can result in the k outcomes E1, E2, ..., Ek with probabilities p1, p2, .., pk, then the probability distribution of the random variables X1, X2, ..., Xk, representing the number of occurrence for E1, E2, ..., Ek in n independent trials is

with

and



The complexity of arrivals and departures into an airport are such that computer simulation is often used to model the “ideal” conditions. For a certain airport containing three runways it is known that in the ideal setting the probabilities that the individual runways are accessed by a randomly arriving commercial jet are 2/9, 1/6, and 11/18 for runway 1, runway 2, and runway 3, respectively.If there are 6 randomly arriving airplanes, what is the probability that 2 airplanes will do the landing in runway 1, 1 in runway 2, and 3 in runway 3?

2 1 32 1 11 6! 2 1 11

2,1,3; , , ,69 6 18 2!1!3! 9 6 18

f

0.1127

?What is the probability that 2 airplanes will do the landing in runway 1?


Hypergeometric Distribution As opposed to the binomial distribution, the hypergeometric

distribution is based on the sampling done without replacement. The independence among trials is not required.

Applications for the hypergeometric distribution are found in many areas, with heavy uses in acceptance sampling, electronic sampling, and quality assurance.

Chapter 5.4 Hypergeometric Distribution

The experiment where the hypergeometric distribution applies must possess the following two properties:1. A random sample of size n is selected without replacement from

N items2. k of the N items may be classified as successes and N–k are

classified as failures.

The number X of successes of a hypergeometric experiment is called a hypergeometric random variable.

The hypergeometric distribution of such variable is denoted by h(x;

N, n, k)


Hypergeometric DistributionA particular part that is used as an injection device is sold in lots of 10. The producer feels that the lot is deemed acceptable if no more that one defective is in the lot. Some lots are sampled and the sampling plan involves random sampling and testing 3 of the parts out of 10. If none of the 3 are defective, the lot is accepted. Give comment on the utility of this plan.


2 0 8 3

10 3

( 0)C C

P XC

3 0 7 3

10 3

( 0)C C

P XC

In case there are 2 defectives, there is still a chance of 46.7% that the lot is accepted.

0.467

0.291 In case there are 3 defectives,

there is still a chance of 29.1% that the lot is accepted.

As conclusion, a plan to do this kind of quality control is faulty. Unacceptable lot can still be accepted with high probability.

3 samples are not enough. The sample size must be increased.


Hypergeometric Distribution The probability distribution of the hypergeometric random variable

X, the number of successes in a random sample of size n selected from N items of which k are labeled success and N–k labeled failure, is


( ; , , ) , 0,1,2,...,k x N k n x

N n

C Ch x N n k x n

C

The mean and variance of the hypergeometric distribution h(x;N,n,k) are

2 11

nk N n k kn

N N N N

and


Hypergeometric DistributionLots of 40 components each are called unacceptable if they contain as many as 3 defectives or more. The procedure for sampling the lot is to select 5 components at random and to reject the lot if a defective is found. (a) What is the probability that exactly 1 defective is found in the

sample if there are 3 defectives in the entire lot?(b) Find the mean and variance of the random variable and use

Chebyshev’s theorem to interpret the interval μ ± 2σ.


3 1 37 4

40 5

(1;40,5,3)C C

hC

0.3011(a) Again, this method of testing is

not acceptable, since it detects a bad lot (with 3 defectives) only about 30% of the time

(b) nk

N

2 11

N n k kn

N N N

(5)(3) 30.375

40 8

40 5 3 3(5) 1

40 1 40 40

0.3113

In at least 3/4 of the time, the number of defectives will be between –0.741 and 1.491 components

0.558


Hypergeometric Distribution If the number of sample n is small compared to the sample size N,

the nature of the N items changes very little in each draw, although without replacement.

In this case, where n/N ≤ 0.05, the value of binomial distribution can be used to approximate the value of hypergeometric distribution.


A manufacturer of automobile tires reports that among a shipment of 5000 sent to a local distributor, 1000 are slightly blemished. If one purchase 10 of these tires at random from the distributor, what is the probability that exactly 3 are blemished?

1000 3 4000 7

5000 10

(3;5000,10,1000)C C

hC

1,5

p 3 10 3

10 3

1 1 43;10,

5 5 5b C

Exact hypergeometric probability

Approximation using binomial distribution

0.2015

0.2013


Negative Binomial Distribution Consider an experiment where the properties are the same as

those listed for a binomial experiment, with the exception that the trials will be repeated until a fixed number of successes occur.

We are interested in the probability that the kth success occurs on the xth trial.

This kind of experiment is called negative binomial experiment.

Chapter 5.5 Negative Binomial and Geometric Distributions

The number X of trials to produce k successes in a negative binomial experiment is called a negative binomial random variable, and its probability distribution is called the negative binomial distribution.

*1 2( ; , ) , , 1, 2,...k x k

x kb x k p C p q x k k k

|Negative Binomial Distribution| If repeated independent trials can result in a success with probability p and a failure with probability q = 1–p, then the probability distribution of the random variable X, the number of the trial on which the kth success occurs, is


Negative Binomial DistributionChapter 5.5 Negative Binomial and Geometric Distributions

In an NBA (National Basketball Association) championship series, the team which wins four games out of seven will be the winner. Suppose that team A has probability 0.55 of winning over the team B and both teams A and B face each other in the championship games. (a) What is the probability that team A will win the series in six

games?(b) What is the probability that team A will win the series?

(a)

(b) * * * *

( )(4;4,0.55) (5;4,0.55) (6;4,0.55) (7;4,0.55)

team A wins the championship seriesPb b b b

* 4 6 26 1 4 2(6;4,0.55) (0.55) (0.45)b C 0.1853

0.0915 0.1647 0.1853 0.1668 0.6083


Negative Binomial DistributionChapter 5.5 Negative Binomial and Geometric Distributions

In an NBA (National Basketball Association) championship series, the team which wins four games out of seven will be the winner. Suppose that team A has probability 0.55 of winning over the team B and both teams A and B face each other in the championship games. (c) If both teams face each other in a regional playoff series and the

winner is decided by winning three out of five games, what is the probability that team A will win a playoff?

(c) * * *

( )(3;3,0.55) (4;3,0.55) (5;3,0.55)

team A wins the reginal seriesPb b b

0.1664 0.2246 0.2021 0.5931


Geometric Distribution If we consider the special case of the negative binomial distribution

where k = 1, we have a probability distribution for the number of trials required for a single success.

Chapter 5.5 Negative Binomial and Geometric Distributions

If repeated independent trials can result in a success with probability p and a failure with probability q = 1–p, then the probability distribution of the random variable X, the number of the trial on which the first success occurs, is

1( ; ) , 1, 2,3,...xg x p pq x

The mean and variance of a random variable following the geometric distribution are

22

1 1 p

p p and


Geometric DistributionChapter 5.5 Negative Binomial and Geometric Distributions

In a certain manufacturing process it is known that, on the average, 1 in every 100 items is defective. What is the probability that the fifth item inspected is the first defective item found?

4(5;0.01) (0.01)(0.99)g

At “busy time” a telephone exchange is very near capacity, so callers have difficulty placing their calls. It may be of interest to know the number of attempts necessary in order to gain a connection.Suppose that we let p = 0.05 be the probability of a connection during busy time. We are interested in knowing the probability that 5 attempts are necessary for a successful call.

4( 5) (5;0.05) (0.05)(0.95)P X g

0.0096

0.041


Poisson Distribution and Poisson Process Experiments yielding numerical values of a random variable X, the

number of outcomes occurring during a given time interval or in a specified region, are called Poisson experiments.

The time interval may be given in any length, such as minute, day, week, and month.

The specified region may be a line segment, an area, a volume, or a piece of material

Chapter 5.6 Poisson Distribution and Poisson Process


Properties of Poisson Process A Poisson experiment is derived from the Poisson process and

possesses the following properties:1. The number of outcomes occurring in one time interval or

specified region is independent of the number that occurs in any other disjoint time interval or region of space.

2. The probability that a single outcome will occur during a very short time interval or in a small region is proportional to the length of the time interval or the size of the region and does not depend on the number of outcomes occurring outside this time interval or region.

3. The probability that more than one outcome will occur in such a short time interval or fall in such a small region is negligible



Poisson Distribution and Poisson Process |Poisson Distribution| The probability distribution of the Poisson

random variable X, representing the number of outcomes occurring in a given time interval or specified region denoted by t, is


( )( ; ) , 0,1, 2,...

!

t xe tp x t x

x

where λ is the average number of outcomes per unit time or region, and e = 2.71828.... (natural number).

The mean and variance of the Poisson distribution p(x;λt) both have the value λt.


Poisson Distribution and Poisson ProcessChapter 5.6 Poisson Distribution and Poisson Process

During a laboratory experiment the average number of radioactive particles passing through a counter in 1 millisecond is 4. What is the probability that 6 particles enter the counter in a given millisecond?

4 6(4)(6;4)

6!

ep

Ten is the average number of oil tankers arriving each day at a certain port city. The facilities at the port can handle at most 15 tankers per day. What is the probability that on a given day tankers have to be turned away?

6, 4x t

15

0

( 15) 1 ( 15) 1 ( ;10)x

P X P X p x

10 1 10 2 10 15(10) (10) (10)

11! 2! 15!

e e e

1 0.9513 0.0487

0.1042

Table A.2 gives help


Table A.2 Poisson Probability SumsChapter 5.6 Poisson Distribution and Poisson Process


Poisson Distribution As a Limit of BinomialChapter 5.6 Poisson Distribution and Poisson Process

It should be clear from the three properties of the Poisson process that the Poisson distribution relates to the binomial distribution.

In the case of the binomial, if n is quite large and p is small, the conditions begin to simulate the continuous space or time region implications of the Poisson process.

Let X be a binomial random variable with probability distribution b(x;n,p). When n ® ∞ and p ® 0, and μ = np remains constant,

( ; , ) ( , )b x n p p x

Poisson distribution can be taken as a limiting form of the binomial distribution when n ® ∞ and p ® 0, and np remains constant.

If the conditions are fulfilled, the Poisson distribution can be used with μ = np, to approximate binomial distribution.


0.2707


In a certain industrial facility accidents occur infrequently. It is known that the probability of an accident on any given day is 0.005 and accidents are independent of each other.(a) What is the probability that in any given period of 400 days there

will be an accident on one day?(b) What is the probability that there are at most three days with an

accident?

1, (0.005)(400) 2x t

(a)

(b)

2 1(2)(1;2)

1!

ep

23 3

0 0

(2)( 3) ( ;2)

!

x

x x

eP X p x

x

0.8571

Considered as Poisson process

Considered as Bernoulli process

1 399400 1(1;400,0.005) (0.005) (0.095)b C 0.2707

3

0

( 3) ( ;400,0.005)x

P X b x

0.8571



In a manufacturing process where glass products are produced, defects or bubbles occur, occasionally rendering the piece undesirable for marketing. It is known that, on average, 1 in every 1000 of these items produced has one or more bubbles. What is the probability that a random sample of 8000 will yield fewer than 7 items possessing bubbles?

6

0

( 7) ( ;8000,0.001)x

P X b x

(8000)(0.001) 8

6

0

( ;8)x

p x

0.3134

Actually a problem for Binomial Distribution

Solved by approximation using Poisson Distribution


Homework 6Probability and Statistics

1. A communications system consists of n components, each of which will, independently, function with probability p. The total system will be able to operate effectively if at least one-half of its components function. For what values of p is a 5-component system more likely to operate effectively than a 3-component system?

(Ro.E5.1c s144)

2. It has been established that the number of defective stereos produced daily at a certain plant is Poisson distributed with mean 4. Over a 2-day span, what is the probability that the number of defective stereos does not exceed 3?

(Ro.E5.2f s+10)

3. The probability of hitting a target is 1/5 and ten shots are fired independently.(a) What is the probability of the target being hit at least twice?(b) Find the conditional probability that the target is hit at least twice,

assuming that at least one hit is scored.

(Fe.VI.10.5-6 s16.9)

president universityerwin sitompulpbst 6/1 dr.-ing. erwin sitompul president university lecture 6...

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