Binomial vs. Geometric

Chapter 8Binomial and Geometric

Distributions

Binomial vs. GeometricThe 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?

RECALL: Suppose that each of three randomly selected customers purchasing a hot tub at a certain store chooses either an electric (E) or a gas (G) model. Assume that these customers make their choices independently of one another and that 40% of all customers select an electric model. The number among the three customers who purchase an electric hot tub is a random variable. What is the probability distribution?

Are Random Variables and Binomial Distributions Linked?

XP(X)

X = number of people who purchase electric hot tub

0 1 2 3

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

.216

EGGGEGGGE

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

.432

EEGGEEEGE

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

.288

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

.064

Combinations

Formula:n

kn

k n kFHGIKJ

!

! !a fPractice:

16

4. FHGIKJ

64 6 4

!! !a f

6 5 4 3 2 14 3 2 1 2 1a f

6 52 1

15

28

5. FHGIKJ

85 3

!! !

8 7 6 55 3 2 1

!!

56

Developing the Formula

Ex. 8.1 Blood Type; pg. 440 – Pract. Of Stats 2nd EdBlood type is inherited. If both parents carry genes for the O and A blood types, each child has a probability 0.25 of getting two O genes, thus having blood type O. Let X = the number of O blood types among 3 children. Different children inherit genes independently of each other.Since X has the binomial distribution with n = 3 and p = 0.25. We say that X is B(3, 0.25)

Developing the FormulaX

P X( )

Outcomes Probability Rewritten

0

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

.42191

OOcOc

OcOOc

OcOcO

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

.42192

OOOc

OOcOOcOO

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

.14063

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

.0156

1 .25 3af3 . .25 752afaf1

3 . .25 751 2afaf

.75 0af

1 25 750 3. .afaf3

0FHGIKJ

3

1FHGIKJ

3

2FHGIKJ

3

3FHGIKJ

Developing the Formula

Rewritten

1 .25 3af3 . .25 752afaf1

3 . .25 751 2afaf

.75 0af

1 25 750 3. .afaf3

0FHGIKJ

3

1FHGIKJ

3

2FHGIKJ

3

3FHGIKJ

n = # of observationsp = probablity of successk = given value of variable

P X k( ) n

kFHGIKJ pk 1 p n ka f

Be able to recognize the formula even though you can set up a Prob. Distribution without it.

Working with probability distributions

State the distribution to be usedState important numbers

Binomial: n & pGeometric: p

Define the variableWrite proper probability notation

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.

binomialX = # of people use express

n = 5 p = .25

What is the probability that two used the express checkout?

1. Binomial, n = 5, p = .252. X = # of people use express3. P(X = 2)

Get started with the required components:

Calculator option #1:

nCr = n choose r = 5 math prb 3:nCr 2 enter

1. Define the situation: Binomial, with n = 5 and p = 0.252. Identify X: let X = # of people use express3. Write proper probability notation that the question is asking: P( X = 2)

2P X 5

2

2.25 3

.75 .2637

Calculator option #2:

2nd vars 0:binompdf(5, 0.25, 2) = .2637

REQUIRED COMPONENTS 1, 2, 3

What is the probability that no more than three used express checkout?

1. Binomial, n = 5, p = .252. X = # of people use express3. P(X ≤3)

Calculator option:

P(X ≤3) = binomcdf(5, 0.25, 3) = .9844

P(X ≤3)=P(X=0)+P(X=1)+P(X=2)+P(X=3)

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

1. Binomial, n = 5, p = .252. X = # of people use express3. P(X ≥ 4)

4P X 5

4

4.25 1

.75 .0156 55.25

5

Calculator option:

P(X 4) = 1-P(X<4) 1-binomcdf(5, 0.25, 3) = .0156

“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. binomialX = # of people who believe…

n = 8 p = .60

Interpret P(X = 3) and find the numerical

answer.

binomialX = # of people who believe

n = 8 p = .60

The probability that 3 of the people from the random sample of 8 believe their children willhave a higher standard of living.

Interpret P(X = 3) and find the numerical

answer.

binomialX = # of people who believe

n = 8 p = .60

3P X 8

3

3.6 5

.4 .1239

P(X=3) = binomialpdf(8, .6, 3) = .1239

or

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

•binomial, n = 8, p = 0.60•X = # of people who believe•P(X = 0 )

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

binomialX = # of people who believe

n = 8 p = .60

0P X 8

0

0.6 8

.4 .00066

or

P(X=0) = binomialpdf(8, .6, 0) = .00066

Binomial vs. GeometricThe 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

2 56

16didi

3 56

56

16didid i

4 56

56

56

16didid id i

P X n a f 1 1 p na f p

The Mean and Standard Deviation of a Geometric

Random VariableIf 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

1p

1

2p

pX

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•Find P(X = 5)

p = .03

5P X 4.97 .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? (When is the 1st success?)

geometricX = # of free throws till miss

p = .20

6P X 5.80 .20 .0655

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

geometricY = # of free throws till make

p = .80

6P Y 5.20 .80 .00026

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

geometricX = # 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

Or 1 – P(makes all 6 shots) = 1 – (.806) = .7379