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Chapter 8

The Geometric Distributions

PLINKO

Our goal is to determine the probability that aball will land in slot “D”. A win occurs when theball falls to the Right 3 times and to the Left 3times in any order.

•Let the random variable X = number oftimes the ball falls to the right.

•Our goal is to find P (X=3)

•Use the random number generator on your calculatorto generate six numbers representing the positionsA, B, C, E, F, and G.

_ A__ ___B___ ___C___ ___D__ ___E___ ___F___ ___G___

RandInt (1,2,6) where 1 = Left and 2 = Right

Lands in D __________________________

Not in D _____________________________

Lands in D _____________________________

Not in D ______________________________

CLASS TOTALS in D _________

Not in D _________

Determine the total number of possible outcomes forPlinko…list them systematically

HINT: How many positions to left and right?

Use powers of 2 to find to combinations of

R’s and L’s I’ll get you started…see the board

LLLLLL

RLLLLL

LRLLLL

RRLLLL

LLRLLL

RLRLLL

LRRLLL

RRRLLL

•Complete the probability distribution for therandom variable X = # times the ball falls tothe right (i.e.) 0 R’s, 1 R, 2 R’s, # R’s etc.

P(X)

The features of this experiment are as follows:

•There are two outcomes (L, R) or heads/tails,evens/odds (success/failure)

•6 digits are drawn for a single trial

•The flips or draws are independent

(one outcome has no influence on the

next)

•The probability of success (falling to right) is thesame each trial

A situation in which these fourconditions are satisfied is called abinomial setting

To reiterate these characteristics…

The Binomial Setting

1.each observation falls into one of twocategories (success and failure)

2.there is a fixed number n ofobservations

3.The n observations are all independent.

4. the probability of success, p, is the samefor each observation

The distribution of the count X ofsuccesses in the binomial setting is thebinomial distribution with parameters nand p. The parameters n is the number ofobservations, and p is the probability of asuccess on any one observation. Thepossible values of X are the wholenumbers from 0 to n.B ( n, p)

Continuing with PLINKO

•Suppose 5 PLINKO balls are dropped downthe board in succession. Find the probabilitythat all of them will land in slot “D”.

•Find the probability that exactly 2 of themland in slot “D”.

PLINKO

•P(lands in slot D) = .3125

•So all 5 in slot D would be

P(all in slot D) = (.3125)5= .00298

•P(2 in slot D) = 2 in slot D & 3 not in D

= (.3125)2(1-.3125)3 = .0317

PLINKO

•If a ball landing in slot “A” or “G” pays$50, a ball landing in “B” or “F” pays $25,a ball in “C” and “E” pay $10, and a balllanding in “D” pays $5, find the expectedwinnings(mean)when 5 balls are dropped.

•What is the standard deviation of the totalamount won?

PLINKO

Pay

out

$50

$25

$10

$25

$50

P(X)

.0156

.0938

.2344

.3125

.2344

.0938

.0156

Expected value =50(.0156)+25(.0938)+10(.2344)+5(.3125)+10(.2344)+25(.0938)+50(.0156)

= $12.50

PLINKO

Pay

out

$50

$25

$10

$25

$50

P(X)

.0156

.0938

.2344

.3125

.2344

.0938

.0156

VAR (X) = (50-12.50)2(.0156) + (25-12.5)2(.0938) + (10-12.50)2(.2344) + (5- 12.50)2(.3125) + (10-12.50)2(.2344) + (25-12.5)2(.0938) + (50-12.50)2(.0156)

VAR (X) = 93.7499

STD DEV(X) = $9.68

Example 1

Suppose Dolores is a 65% free throw shooter. If weassume that the repeated shots are independent“what is the probability that Dolores makes exactly7 of her next 10 free throws?” If X is the binomialrandom variable that gives us the count of successesfor the experiment, we say X has B(10,.65)

The question is to find P(X=7)

use n C r under MATH PRB where r = x

10 C 7 (.65)7(.35)3 = .252

(b) What is Dolores’ probability that she makes nomore than 5 free throws?

This time the question is asking what is

P (x ≤ 5)?

P (x ≤ 5) = P(x=0) + P(x=1) + P(x=2) + P(x=3)+ P(x=4) + P(x =5)

= 10 C 0 (.65)0(.35)10 + 10 C 1 (.65)1(.35)9 + 10 C 2(.65)2(.35)8 + 10 C 3 (.65)3(.35)7 + 10 C 4(.65)4(.35)6 + 10 C 5 (.65)5(.35)5

= .249 about 25% chance

What is the probability that Dolores will make at least 6free throws?

Do we have to redo all of the calculations or is thereanother way? If she makes at least 6 how many will shemiss? Hmmm…????? How is this related to the previousproblem?

P(x ≥ 6) is the same as P( 1- P(x ≤ 5)) =

= 1 - .249

= .751

Mean and Standard Deviation

Clearly we can calculate the mean andstandard deviation of a binomial randomvariable using the methods from Chapter 7but there is another way…

μx = np

Normal Approximation for BinomialDistribution

When np and n(1-p) are SUFFICIENTLYLARGE i.e. both are ≥ 10, the binomialrandom variable X has an approximatelynormal distribution.

The mean μ = np and

Community College Problem

Nationally, 15% of community college students livemore than 6 miles from campus. Data from asimple random sample of 400 students at onecommunity college is analyzed.

(a)What are the mean and standard deviation for thenumber of students in the sample trial?

X has B(400,.15)

μ = np = 400(.15) = 60

Community College Problem

Community College Problem

(b) Use a normal approximation to calculateprobability that at least 65 of the students inthe sample live more than 6 miles fromcampus. Because 400(.15) = 60 ≥10 and400(.85) = 340 ≥ 10, we can use the normalapproximation to the binomial withN(60,7.14)

Community College Problem

Find P (x ≥ 65)

P (z ≥ (65-60)/7.14 = .70

From Table A the P (z < .70) = .7580 so

P (z ≥ .70 ) = (1-.7580)= .242

Credit Card Example

Suppose 60% of adults have credit carddebt. If we survey 2500 adults, what is theprobability more than 1520 would havecredit card debt?

X = # adults who have credit card debt outof 2500

X is B(2500,.60)

We want to find P(X > 1520)

Credit Card Example

Is np ≥ 10 ? 2500(.60) = 1500 ≥ 10

Is n(1-p) ≥ 10? 2500(.40) = 1000 ≥ 10

Yes, it approximates a normal distribution.

μ = np = 1500

Credit Card Example

We want to find P (x >1520)

P (z >

Do problems 8.2, 8.8 and 8.16

Technology ToolboxExploring binomialdistributions

Chapter 8 Section 8.2 TheGeometric Distribution

Hawaiian Villager Problem

On the island of Oahu in the village of Nankuli,80% of the residents are of Hawaiian ancestry. Ifyou visit Nanakuli, what is the probability thefirst village you meet is Hawaiian?

X = # villagers you must meet

P(X = 1)

P(X = n) = (1 – p)n-1p -- probability it is (p) *probability it is not (1-p)

P (X = 1) = (1-.8)1-1(.8) = .8

RULE FOR CALCULATINGGEOMETRIC PROBABILITIES

If X has a geometric distribution with probability p ofsuccess and (1-p) of failure on each observation, thepossible values of X are 1,2,3,…If n is any one ofthese, the probability that the first success occurs onthe nth trial is

Hawaiian Villager Problem

What is the P( you don’t meet a Hawaiianuntil the 2nd villager?)

P (X =2) (1-.8)2-1(.8)= .16

Let’s extend this concept for third, fourth,fifth villagers…

Hawaiian Villager Problem

P (X =1) (1-.8)1-1(.8)= .8

P (X =2) (1-.8)2-1(.8)= .16

P (X =3) (1-.8)3-1(.8)= .032

P (X =4) (1-.8)4-1(.8)= .0064

P (X =5) (1-.8)5-1(.8)= .00128

Hawaiian Villager Problem

When this data is graphed what do younotice?

Graphs of Geometric Distributions have a‘step ladder” appearance since you aremultiplying the height of each bar by anumber less than 1. Each bar will beshorter than the previous bar. Thehistogram is ALWAYS right skewed.

Characteristics of a Geometric Distribution

Hawaiian Villager Problem

Find the probability it will take more than 4villagers to meet a native Hawaiian.

P(x > 4) = (1-p)n = (1-.8)4 = (.2) 4 = .0016

Hawaiian Villager Problem

Find the average number of villagers it willtake to meet a native Hawaiian.

Hawaiian Villager Problem

How much variability is there in the numberof villagers required to meet a Hawaiian?

The Geometric Setting

•1. each observation falls into one of twocategories, success or failure

•2. the probability of a success is p

•3. the observations are all independent

•4. the variable of interest is the number oftrials required to obtain the first success

HANDY DANDY FORMULAS

If X is a Geometric random variable with

P(success) = p these formulas apply:

P(X=n) = (1- p)n-1(p) μx =

P(X > n) = (1- p)n

How Can You Tell Geometric fromBinomial?

•Both of these models must meet the 3conditions often called the Bernoulli trials.

(1) there are two possible outcomes

(2) the probability of a success is constant

(3) the trials are independent

•The distinguishing characteristic is:

A binomial probability model is appropriatefor a random variable that counts the # ofsuccesses in a fixed number of trials.

How Can You Tell Geometric fromBinomial?

•While a geometric probability model isappropriate for a random variable thatcounts the # of trials until the first success.(there could be an unlimited number oftrials.)

Which are these?

(1) The Los Angeles Times reported that 80%of airline passengers prefer to sleep on longflights rather than watch movies, read, etc.Consider randomly selecting 25 passengersfrom a particular long flight. Defind arandom variable X , calculate P( X=12).

Is this binomial or geometric?

Which are these?

(2) Sophie is a dog who loves to play catch.Unfortunately, she isn’t very good, and theP(catches a ball) = 0.1. Define X=# tossesrequired until Sophie to catches the ball.

Is this binomial or geometric?

Which are these?

(3) You are to take a multiple choice exam of100 questions with five possible responses(A,B,C,D,E). Suppose you have not studiedand decide to guess randomly on eachquestion. Let X = # correct responses.

Is this binomial or geometric?

Which are these?

(4) Suppose 5% of cereal boxes contain aprize. You are determined to buy cerealboxes until you win a prize.

Is this binomial or geometric?

Let’s Explore the Sophie problemand the cereal problem in moredepth.

The Sophie problem

(2) Sophie is a dog who loves to play catch.Unfortunately, she isn’t very good, and theP(catches a ball) = 0.1. Define X=# tosses requireduntil Sophie to catches the ball.

(a) calculate and interpret P(X=2)

P (X = n) = (1-p)n-1(p)

P (X =2) (1-.1)2-1(.1)= .09

(b) calculate and interpret P(X ≥ 3)

P(X > n) = (1- p)n

P(X ≥ 3) = (1- .1)3 = .729

Sophie

(2c) calculate and interpret the mean andstandard deviation of X

μx =

Cereal Problem

•Suppose 5% of cereal boxes contain a prize.You are determined to buy cereal boxes untilyou win a prize.

•(a) What is the probability you will have to buyat most 2 boxes? (X ≤ 2)

•(b) What is the probability you will have tobuy exactly 4 boxes? ( X = 4)

•(c) What is the probability you will have tobuy more than 4 boxes? (X ≥ 4)

Cereal Problem

•(a) What is the probability you will have tobuy at most 2 boxes?

X = # boxes you will buy until you win aprize.

Find P (X ≤ 2) is the same as P (1-complement) P( X > 2)

P(X > n) = (1- p)n

P (X ≤ 2) = (1-.5)2 = .25

Cereal Problem

•(b) What is the probability you willhave to buy exactly 4 boxes? ( X = 4)

P(X = n) = (1- p)n-1p

P(X = 4) = (1- .5)4-1(.5) = .0625

•(c) What is the probability you willhave to buy more than 4 boxes? (X ≥ 4)

P(X > n) = (1- p)n

P(X ≥ 4) = (1- .5)4 = .0625

#37 in the textbookWhich are binomial orgeometric?

# 37 in the book.

(a)yes, geometric X = success (tail)

failure (head)

a trial is one flip of the coin

P(tail) = .5

(b) Not independent

a trial is drawing a card with replacement

P(J) =

Dolores the Basketball Player

•Remember Dolores the basketball playerwhose free throw shooting percentage was.65? What is the probability that the firstfree throw she hits is on her 4th attempt?

•P(X = 4) (1-.65)4-1(.65)= (.35)3 (.65)= .028

•Using the TI 83/84 geometpdf (p,n)

geometpdf (.65,4) = .028

Technology ToolboxExploring geometricdistributions