Binomial vs. Geometric

Chapter 8

Binomial and Geometric Distributions

Binomial vs. Geometric The Binomial Setting The Geometric Setting

1. Each observation falls into

one of two categories.

2. The probability of success

is the same for each

observation.

3. The observations are all

independent.

4. There is a fixed number n

of observations.

4. The variable of interest is the

number of trials required to

obtain the 1st success.

1. Each observation falls into

one of two categories.

2. The probability of success

is the same for each

observation.

3. The observations are all

independent.

Binomial vs. Geometric The Binomial Setting The Geometric Setting

1. Each observation falls into

one of two categories.

2. The probability of success

is the same for each

observation.

3. The observations are all

independent.

4. There is a fixed number n

of observations.

4. The variable of interest is the

number of trials required to

obtain the 1st success.

1. Each observation falls into

one of two categories.

2. The probability of success

is the same for each

observation.

3. The observations are all

independent.

Are Random Variables and Binomial Distributions Linked?

X

P(X)

X = number of people who purchase electric hot tub

0 1 2 3

GGG (.6)(.6)(.6)

.216

EGG

GEG

GGE

(.4)(.6)(.6)

(.6)(.4)(.6)

(.6)(.6)(.4)

.432

EEG

GEE

EGE

(.4)(.4)(.6)

(.6)(.4)(.4)

(.4)(.6)(.4)

.288

EEE (.4)(.4)(.4)

.064

Combinations

Formula: n

k

n

k n k

FHGIKJ

!

! !a fPractice:

16

4. FHGIKJ

6

4 6 4

!

! !a f6 5 4 3 2 1

4 3 2 1 2 1a f6 5

2 115

28

5. FHGIKJ

8

5 3

!

! !

8 7 6 5

5 3 2 1

!

!56

Developing the Formula X

P X( )

Outcomes Probability Rewritten

0

OcOcOc (. )(. )(. )75 75 75

.4219

1

OOcOc

OcOOc

OcOcO

(. )(. )(. )25 75 75

.4219

2

OOOc

OOcO OcOO

(. )(. )(. )25 25 75

.1406

3

OOO (. )(. )(. )25 25 25

.0156

1 .253a f

3 . .25 752a f a f1

3 . .25 751 2a f a f

.750a f

1 25 750 3

. .a f a f3

0

FHGIKJ

3

1

FHGIKJ

3

2

FHGIKJ

3

3

FHGIKJ

Developing the Formula

Rewritten

1 .253a f

3 . .25 752a f a f1

3 . .25 751 2a f a f

.750a f

1 25 750 3

. .a f a f3

0

FHGIKJ

3

1

FHGIKJ

3

2

FHGIKJ

3

3

FHGIKJ

n = # of observations

p = probablity of success

k = given value of variable

P X k( )n

k

FHGIKJ pk

1 pn ka f

Working with probability distributions

State the distribution to be used

Define the variable

State important numbers

Binomial: n & p

Geometric: p

Twenty-five percent of the customers entering a grocery store between 5 p.m. and 7 p.m. use an express checkout. Consider five randomly selected customers, and let X denote the number among the five who use the express checkout.

binomial

X = # of people use express

n = 5 p = .25

What is the probability that two used express checkout?

binomial

X = # of people use express

n = 5 p = .25

2P X5

2

2.25

3.75 .2637

What is the probability that at least four used express checkout?

binomial

X = # of people use express

n = 5 p = .25

4P X5

4

4.25

1.75

.0156

55.25

5

“Do you believe your children will have a higher standard of living than you have?” This question was asked to a national sample of American adults with children in a Time/CNN poll (1/29,96). Assume that the true percentage of all American adults who believe their children will have a higher standard of living is .60. Let X represent the number who believe their children will have a higher standard of living

from a random sample of 8 American adults.

binomial

X = # of people who believe…

n = 8 p = .60

Interpret P(X = 3) and find the numerical answer.

binomial

X = # of people who believe

n = 8 p = .60

3P X8

3

3.6

5.4 .1239

The probability that 3 of the people from the

random sample of 8 believe their children will

have a higher standard of living.

Find the probability that none of the parents believe their children will have a higher standard.

binomial

X = # of people who believe

n = 8 p = .60

0P X8

0

0.6

8.4 .00066

Binomial vs. Geometric The Binomial Setting The Geometric Setting

1. Each observation falls into

one of two categories.

2. The probability of success

is the same for each

observation.

3. The observations are all

independent.

4. There is a fixed number n

of observations.

4. The variable of interest is the

number of trials required to

obtain the 1st success.

1. Each observation falls into

one of two categories.

2. The probability of success

is the same for each

observation.

3. The observations are all

independent.

Developing the Geometric Formula

X Probability

1 16

25

61

6d id i3 5

65

61

6d id id i4

56

56

56

16d id id id i

P X na f 11

pna f p

The Mean and Standard Deviation of a Geometric Random

Variable If X is a geometric random variable with probability of success p on each trial, the expected value of the random variable (the expected number of trials to get the first success) is

1

p

12

p

p

Suppose we have data that suggest that 3% of a company’s hard disc drives are defective. You have been asked to determine the probability that the first defective hard drive is the fifth unit tested.

geometric

X = # of disc drives till defective

p = .03

5P X4

.97 .03 .0266

A basketball player makes 80% of her free throws. We put her on the free throw line and ask her to shoot free throws until she misses one. Let X = the number of free throws the player takes until she misses.

geometric

X = # of free throws till miss

p = .20

What is the probability that she will make 5 shots before she misses?

geometric

X = # of free throws till miss

p = .20

6P X5

.80 .20 .0655

What is the probability that she will miss 5 shots before she makes one?

geometric

Y = # of free throws till make

p = .80

6P Y5

.20 .80 .00026

What is the probability that she will make at most 5 shots before she misses?

geometric

X = # of free throws till miss

p = .20

6P X .20 .80 .20

.7379

2.80 .20

3.80 .20

4.80 .20

5.80 .20