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Chapter  8  Binomial  and  Geometric  Distribu7ons  

•  8.2 Geometric Distributions

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•  Waiting for Sammy Sosa – geometric distribution Children’s cereals sometimes contain prizes. Imagine that packages of Chocolate-

Coated Sugar Bombs contain one of three baseball cards: Mark McGwire, Sammy Sosa, or Barry Bonds. Shak wanted to get a Sammy Sosa card and had to buy eight boxes until getting his desired card. Shak feels especially unlucky.

Should Shak consider himself especially unlucky? On average, how many boxes would a person have to buy to get the Sammy Sosa card?

In this activity, you will become familiar with the geometric distribution, including the shape of the distribution and how to find its mean.

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•  Waiting for Sammy Sosa – geometric distribution

1.   Roll  your  die:    If  the  side  with  1  or  2  dots  lands  up,  this  will  represent  the  event  of  buying  a  box  of  Chocolate-­‐Coated  Sugar  Bombs  and  ge@ng  a  Sammy  Sosa  card.    If  one  of  the  other  sides  lands  on  top,  roll  again.    Count  the  number  of  rolls  unBl  you  get  a  1  or  2.  •  Add  to  the  class  histogram  showing  the  number  of  rolls  it  took  to  get  

your  first  Sammy  Sosa  card.    (Do  this  10  Bmes  and  record  each  on  the  class  histogram  so  that  we  have  more  data.)  

•  Describe  the  shape  of  this  distribuBon.  •  What  was  the  average  number  of  “boxes”  purchased  to  get  a  Sammy  

Sosa  card?  •  EsBmate  the  chance  that  Shak  would  have  to  buy  eight  or  more  boxes  to  

get  his  card.  •  What  assumpBons  did  you  make  in  this  simulaBon  about  the  

distribuBon  of  prizes?    Do  you  think  they  are  reasonable  ones?  

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•  Waiting for Sammy Sosa – geometric distribution

1.  Roll  your  die:  2.   Doubles:    In  some  games,  a  player  must  roll  doubles  before  conBnuing,  such  

as  when  in  jail  in  Monopoly.    Use  a  pair  of  dice  or  random  numbers  to  simulate  rolling  a  pair  of  dice.    Count  the  number  of  rolls  unBl  you  get  doubles.  •  Add  your  informaBon  to  the  class  histogram  of  the  number  of  rolls  our  

class  required  to  roll  doubles.    (Do  this  10  Bmes,  too.)  •  Describe  the  shape  of  the  distribuBon.  •  What  was  the  average  number  of  rolls  required?  

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•  Waiting for Sammy Sosa – geometric distribution

1.  Roll  your  die:  2.  Doubles:  3.   Wai5ng  Time  Distribu5on:    In  steps  1  &  2,  you  constructed  a  waiBng-­‐Bme  

distribuBon  using  simulaBon.    Now  construct  a  theoreBcal  waiBng-­‐Bme  distribuBon  for  ge@ng  a  different  cereal  prize.    Boxes  of  Post’s  Coca  Pebbles  recently  contained  one  of  four  endangered  animal  sBckers:    a  parrot,  an  African  elephant,  a  Bger,  or  a  crocodile.    Suppose  4096  children  want  a  sBcker  of  a  parrot.  •  How  many  of  them  would  you  expect  to  get  a  parrot  in  the  first  box  of  

Cocoa  Pebbles  they  buy?    What  assumpBons  are  you  making?  •  How  many  children  do  you  expect  will  have  to  buy  a  second  box?  •  How  many  of  them  do  you  expect  will  get  a  parrot  in  the  second  box?  

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•  Waiting for Sammy Sosa – geometric distribution

1.  Roll  your  die:  2.  Doubles:  3.  WaiBng  Time  DistribuBon:  4.   Fill  in  the  following  table:  

               •  Make  a  histogram  of  the  theoreBcal  waiBng-­‐Bme  distribuBon.  •  The  height  of  each  bar  of  the  histogram  is  what  proporBon  of  the  height  

of  the  bar  to  its  lea?  •  What  is  the  average  number  of  boxes  purchased?      

Number of Boxes Purchased to get first Parrot Sticker

Number of Children

1 … 20

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•  Waiting for Sammy Sosa – geometric distribution •  Wrap-Up

1.  Describe the shape of a waiting-time (geometric) distribution for a given probability p of success on each trial. Will the first bar in a waiting-time distribution always be the highest? Why or why not? The height of each bar is what proportion of the height of the bar to its left?

2.  What are some other situations that can be modeled by a waiting-time distribution?

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•  Geometric Settings In a binomial setting, the number of trials n is fixed and the binomial random variable

X counts the number of successes. In other situations, the goal is to repeat a chance behavior until a success occurs. These situations are called geometric settings.

Defini7on:  A  geometric  se;ng  arises  when  we  perform  independent  trials  of  the  same  chance  process  and  record  the  number  of  trials  un;l  a  par;cular  outcome  occurs.  The  four  condi;ons  for  a  geometric  se?ng  are  

• Binary? The possible outcomes of each trial can be classified as “success” or “failure.”

•  Independent?  Trials  must  be  independent;  that  is,  knowing  the  result  of  one  trial  must  not  have  any  effect  on  the  result  of  any  other  trial.  

•  Trials?  The  goal  is  to  count  the  number  of  trials  un;l  the  first  success  occurs.  

• Success? On each trial, the probability p of success must be the same.

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•  Geometric Random Variable In a geometric setting, if we define the random variable Y to be the

number of trials needed to get the first success, then Y is called a geometric random variable. The probability distribution of Y is called a geometric distribution.

Definition:

The number of trials Y that it takes to get a success in a geometric setting is a geometric random variable. The probability distribution of Y is a geometric distribution with parameter p, the probability of a success on any trial. The possible values of Y are 1, 2, 3, ….

Note: Like binomial random variables, it is important to be able to distinguish situations in which the geometric distribution does and doesn’t apply!

•  Example: The Birthday Game Ms. Raskin is planning to give you 10 problems for homework. As an alternative, you can agree to play the Birthday

Game. Here’s how it works. A student will be selected at random from the class and asked to guess the day of the week on which Ms. Raskin’s best friend was born. If the student guesses correctly, the class will have only one homework problem. If the student guesses the wrong day of the week, Ms. Raskin will once again select a student from the class at random. The chosen student will try to guess the day of the week on which a different one of Ms. Raskin’s many friends was born. If this student gets it right, the class will have two homework problems. The game continues until a student correctly guesses the day on which one of Ms. Raskin’s many friends was born. Ms. Raskin will assign a number of homework problems that is equal to the total number of guesses made by members of the class. Are you ready to play the Birthday Game?

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•  Example: The Birthday Game The random variable of interest in this game is Y = the number of guesses it takes to

correctly identify the birth day of one of your teacher’s friends. What is the probability the first student guesses correctly? The second? Third? What is the probability the kth student guesses correctly?

Verify that Y is a geometric random variable.

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P(Y =1) =1/7

B: Success = correct guess, Failure = incorrect guess I: The result of one student’s guess has no effect on the result of any other guess. T: We’re counting the number of guesses up to and including the first correct guess. S: On each trial, the probability of a correct guess is 1/7.

Calculate P(Y = 1), P(Y = 2), P(Y = 3), and P(Y = k)

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P(Y = 2) = (6 /7)(1/7) = 0.1224

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P(Y = 3) = (6 /7)(6 /7)(1/7) = 0.1050NoBce  the  pabern?  

If Y has the geometric distribution with probability p of success on each trial, the possible values of Y are 1, 2, 3, … . If k is any one of these values,

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P(Y = k) = (1− p)k−1 p

Geometric  Probability  

•  Example: Monopoly In the board game Monopoly, one way to get out of jail is to roll doubles. How likely is it that someone

in jail would roll doubles on his first, second, or third attempt? If this was the only way to get out of jail, how many turns would it take, on average? The random variable of interest in this game is Y = the number of attempts it takes to roll doubles once. What is the probability of rolling doubles the first attempt? The second? Third? What is the probability of rolling doubles on the kth attempt?

Verify that Y is a geometric random variable.

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P(Y =1) =1/6

B: Success = roll doubles, Failure = not rolling doubles I: The result of one roll has no effect on other rolls. T: We’re counting the number of rolls up to and including the first doubles roll. S: On each trial, the probability of a correct guess is 1/6.

Calculate P(Y = 1), P(Y = 2), P(Y = 3), and P(Y = k)

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P(Y = 2) = (5 /6)(1/6) = 0.13889

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P(Y = 3) = (5 /6)(5 /6)(1/6) = 0.11574

NoBce  the  pabern?   If Y has the geometric distribution with probability p of success on each trial, the possible values of Y are 1, 2, 3, … . If k is any one of these values,

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P(Y = k) = (1− p)k−1 p

Geometric  Probability  

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•  Mean of a Geometric Distribution The table below shows part of the probability distribution of Y. We can’t show

the entire distribution because the number of trials it takes to get the first success could be an incredibly large number. Binom

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Shape: The heavily right-skewed shape is characteristic of any geometric distribution. That’s because the most likely value is 1.

Center: The mean of Y is µY = 7. We’d expect it to take 7 guesses to get our first success.

Spread: The standard deviation of Y is σY = 6.48. If the class played the Birth Day game many times, the number of homework problems the students receive would differ from 7 by an average of 6.48.

yi 1 2 3 4 5 6 …

pi 0.143 0.122 0.105 0.090 0.077 0.066

If Y is a geometric random variable with probability p of success on each trial, then its mean (expected value) is E(Y) = µY = 1/p.

Mean  (Expected  Value)  of  Geometric  Random  Variable  

•  Geometric Probability on the Calculator

geometpdf(n,p,k) computes P(Y = k) geometcdf(n,p,k) computes P(Y < k)

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TI-83/84: These commands are found in the distributions menu (2nd/VARS)

TI-89: These commands are found in CATALOG under Flash Apps

For the Birthday Game with probability of success p = 1/7 on each trial:

P(Y = 10) = geometpdf(1/7, 10)

= 0.0356763859

To find P(Y < 10)

= geometcdf(1/7, 10)

= 0.7502652985

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Binomial and Geometric Random Variables

In this section, we learned that…

ü  A binomial setting consists of n independent trials of the same chance process, each resulting in a success or a failure, with probability of success p on each trial. The count X of successes is a binomial random variable. Its probability distribution is a binomial distribution.

ü  The binomial coefficient counts the number of ways k successes can be arranged among n trials.

ü  If X has the binomial distribution with parameters n and p, the possible values of X are the whole numbers 0, 1, 2, . . . , n. The binomial probability of observing k successes in n trials is

Summary

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P(X = k) =nk⎛

⎝ ⎜ ⎞

⎠ ⎟ pk (1− p)n−k

Binomial and Geometric Random Variables

In this section, we learned that…

ü  The mean and standard deviation of a binomial random variable X are

ü  The Normal approximation to the binomial distribution says that if X is a count having the binomial distribution with parameters n and p, then when n is large, X is approximately Normally distributed. We will use this approximation when np ≥ 10 and n(1 - p) ≥ 10.

Summary

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µX = np

σX = np(1− p)

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Binomial and Geometric Random Variables

In this section, we learned that…

ü  A geometric setting consists of repeated trials of the same chance process in which each trial results in a success or a failure; trials are independent; each trial has the same probability p of success; and the goal is to count the number of trials until the first success occurs. If Y = the number of trials required to obtain the first success, then Y is a geometric random variable. Its probability distribution is called a geometric distribution.

ü  If Y has the geometric distribution with probability of success p, the possible values of Y are the positive integers 1, 2, 3, . . . . The geometric probability that Y takes any value is

ü  The mean (expected value) of a geometric random variable Y is 1/p.

Summary

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P(Y = k) = (1− p)k−1 p