Introduction Discrete random variables take on only a finite

or countable number of values. Three discrete probability distributions serve as

models for a large number of practical applications:

The binomial random variable

The Poisson random variable

The hypergeometric random variable

The binomial random variable

The Poisson random variable

The hypergeometric random variable

Chapter 5: Several Useful Discrete Distributions

The Binomial Experiment

Ex: A coin-tossing experiment is a simple example of a binomial random variable, with n tosses and x number of heads

• The experiment consists of n identical trials.• Each trial results in one of two outcomes, success (S) or

failure (F).• The probability of success on a single trial is p and remains

constant from trial to trial. The probability of failure is q = 1 – p.

• The trials are independent.• We are interested in x, the number of successes in n

trials.

The Binomial Probability Distribution

For a binomial experiment with n trials and probability p of success on a given trial, the probability of k successes in n trials is

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)!(!

! Recall

.,...2,1,0for )!(!

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nk

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)!(!

! Recall

.,...2,1,0for )!(!

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knk

nC

nkqpknk

nqpCkxP

nk

knkknknk

The Mean and Standard Deviation

For a binomial experiment with n trials and probability p of success on a given trial, the measures of centre and spread are:

npq

npq

np

:deviation Standard

:Variance

:Mean2

npq

npq

np

:deviation Standard

:Variance

:Mean2

n = p = x =success =

ExampleA cancerous tumour that is irradiated will die 80% of the time. A doctor treats 5 patients by irradiating their tumours. What is the probability that exactly 3 patients will have their tumours disappear?

333)3( nn qpCxP

5 .8cure # of cures

353 )2(.)8(.!2!3

!5

2048.)2(.)8(.10 23

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Example

What is the probability that more than 3 patients are cured?

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45454)3( qpCqpCxP

0514 )2(.)8(.!0!5

!5)2(.)8(.

!1!4

!5

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Cumulative Probability Tables

You can use the cumulative probability tables to find probabilities for selected binomial distributions.

Find the table for the correct value of n.

Find the column for the correct value of p.

The row marked “k” gives the cumulative probability, P(x k) = P(x = 0) +…+ P(x = k)

Find the table for the correct value of n.

Find the column for the correct value of p.

The row marked “k” gives the cumulative probability, P(x k) = P(x = 0) +…+ P(x = k)

Examplek p = .80

0 .000

1 .007

2 .058

3 .263

4 .672

5 1.000

What is the probability that exactly 3 patients are cured?

P(x = 3) = P(x 3) – P(x 2)= .263 - .058= .205

P(x = 3) = P(x 3) – P(x 2)= .263 - .058= .205 Check from formula:

P(x = 3) = .2048

AppletApplet

Examplek p = .80

0 .000

1 .007

2 .058

3 .263

4 .672

5 1.000

What is the probability that more than 3 patients are cured?

P(x > 3) = 1 - P(x 3)= 1 - .263 = .737

P(x > 3) = 1 - P(x 3)= 1 - .263 = .737

Check from formula: P(x > 3) = .7373

AppletApplet

Example Would it be unusual

to find that none of the patients are cured?

• The value x = 0 lies

49.489.

40

x

z 49.489.

40

x

z

more than 4 standard deviations below the mean. Very unusual.

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4)8(.5

npq

np

:deviation Standard

:Mean

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npq

np

:deviation Standard

:Mean

The Poisson Random Variable The Poisson random variable x is a model for data

that represents the number of occurrences of a specified event in a given unit of time or space.

It is a special approximation to the Binomial Distribution for which n is large and the probability of success (p) is small. ( and ) 50n

Examples:

• The number of traffic accidents at a given intersection during a given time period.

• The number of radioactive decays in a certain time.

5np

The Poisson Probability Distribution x is the number of events that occur in a period

of time or space during which an average of such events can be expected to occur. The probability of k occurrences of this event is

For values of k = 0, 1, 2, … The mean and standard deviation of the Poisson random variable are

Mean:

Standard deviation:

For values of k = 0, 1, 2, … The mean and standard deviation of the Poisson random variable are

Mean:

Standard deviation:

!)(

k

ekxP

k

ExampleThe average number of traffic accidents on a certain section of highway is two per week. Find the probability of exactly one accident during a one-week period.

!)1(

k

exP

k

2707.2!1

2 221

ee

Cumulative Probability Tables

You can use the cumulative probability tables to find probabilities for selected Poisson distributions.

Find the column for the correct value of .

The row marked “k” gives the cumulative probability, P(x k) = P(x = 0) +…+ P(x = k)

Find the column for the correct value of .

The row marked “k” gives the cumulative probability, P(x k) = P(x = 0) +…+ P(x = k)

Example

k = 2

0 .135

1 .406

2 .677

3 .857

4 .947

5 .983

6 .995

7 .999

8 1.000

What is the probability that there is exactly 1 accident? P(x = 1) = P(x 1) – P(x 0)

= .406 - .135= .271

P(x = 1) = P(x 1) – P(x 0)= .406 - .135= .271

Check from formula: P(x = 1) = .2707

ExampleWhat is the probability that 8 or more accidents happen?

P(x 8) = 1 - P(x < 8)= 1 – P(x 7) = 1 - .999 = .001

P(x 8) = 1 - P(x < 8)= 1 – P(x 7) = 1 - .999 = .001

k = 2

0 .135

1 .406

2 .677

3 .857

4 .947

5 .983

6 .995

7 .999

8 1.000

This would be very unusual (small probability) since x = 8 lies

standard deviations above the mean.

This would be very unusual (small probability) since x = 8 lies

standard deviations above the mean.

24.4414.1

28

x

z

The Hypergeometric Probability Distribution Example: a bowl contains M red candies and N-

M blue candies. Select n candies from the bowl and record x, the number of red candies selected, where red candies are a success.

M= successes N-M = failures n= total number x= number selected

The probability of exactly k successes in n trials is

Nn

NMkn

Mk

C

CCkxP

)(

The Mean and VarianceThe mean and variance of the hypergeometric random variable x resemble the mean and variance of the binomial random variable:

1 :Variance

:Mean

2

N

nN

N

MN

N

Mn

N

Mn

1 :Variance

:Mean

2

N

nN

N

MN

N

Mn

N

Mn

ExampleA group of 8 drugs used to treat Alzheimer’s disease contains 2 drugs that are not effective. A researcher randomly selects four drugs to test on her subject. What is the probability that all four drugs work? What is the mean and variance for the number of drugs that work?

84

20

64)4(C

CCxP

N = 8 M = 6 n = 4

70

15

)1)(2)(3(4/)5)(6)(7(8

)1(2/)5(6

Success = effective drug

38

64

N

Mn

4286.7

4

8

2

8

64

12

N

nN

N

MN

N

Mn

Key ConceptsI. The Binomial Random Variable

1. Five characteristics: n identical independent trials, each resulting in either success S or failure F; probability of success is p and remains constant from trial to trial; and x is the number of successes in n trials.

2. Calculating binomial probabilities

a. Formula:

b. Cumulative binomial tables

3. Mean of the binomial random variable: np

4. Variance and standard deviation: 2 npq and

knknk qpCkxP )(

npq npq

Key ConceptsII. The Poisson Random Variable

1. The number of events that occur in a period of time or space, during which an average of such events are expected to occur

2. Calculating Poisson probabilities

a. Formula:b. Cumulative Poisson tables

c. Individual and cumulative probabilities using Minitab

3. Mean of the Poisson random variable: E(x) 4. Variance and standard deviation: 2 and

5. Binomial probabilities can be approximated with Poisson probabilities when np 7, using np.

!)(

k

ekxP

k

!)(

k

ekxP

k

Key ConceptsIII. The Hypergeometric Random Variable

1. The number of successes in a sample of size n from a finite population containing M successes and N M failures2. Formula for the probability of k successes in n trials:

3. Mean of the hypergeometric random variable:

4. Variance and standard deviation:

N

Mn

N

Mn

12

N

nN

N

MN

N

Mn

12

N

nN

N

MN

N

Mn

Nn

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n

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