6.3c Geometric Random Variables

7/26/2019 6.3c Geometric Random Variables

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GeometricRandom Variables

Target Goal:

I can find probabilities involving geometric randomvariables

6.3c

h.w: pg 405: 93 99 odd, 101 - 103


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Review Binomial:

The # of trials n is fixed.

Xcounts the number of successes.

Possible values of X are 0, 1, 2, n Probability for success same for all n

Independence


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Consider:

Flip a coin until you get a head.

Roll a die until you get a 3.

Shoot three pointers until you make 1.

What is the main difference?


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Geometric Distributions

Countsthe number of trials until an event

happens.1. Success or failures

2. The probability of success p is the same for

all events.3. Observations are independent.

4. The variable of interest is (X= 1, 2, 3, , );

the number of trials required to obtain thefirst success.


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What does represent?

You will never get a success loser.


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Which is a Geometric Distribution? Check the conditions.

Roll Die until 3 Draw an Ace

Success or failures

The prob. samefor all events

Observations areindependent

Execute untilevent occurs?

Y Y

Y

Y Y

Y:1/6N: First draw: 4/522nd draw: 4/51

N: previous pick effects

the next.


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Rules for Calculating GeometricProbabilities

The probability of the first success on thenthtrialis:

P(X=n) = (1-p)n-1p for X = 1, 2, 3, ..

{s/a qn-1p}

(1-p) s/a q: Probability of failure withp being the probability of success


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Note:

The longer it takesto get the first success,the closer the probability gets to 0.

The table of probabilities could have noend.


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Example: Roll a Die

Construct the probability distribution table

for X= the number of rolls of a die until athree occurs.

P(X=1) = (5/6)0(1/6)1= 0.1667

P(X=2) = (5/6)1(1/6)1=

P(X=3) =

P(X=4) =

Complete and fill in table.

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


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The probability histogram for ageometric distributionis always skewedto the right.


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Exercise: Hard Drive

Suppose we have data that suggest that3%of a companys hard drives are

defective.You have been asked to determine theprobability that the first defective hard driveis the fifth unit tested.


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a) Verify that this is a geometric setting.

Success or failures?

The prob. same for all events?

Observations are independent? Execute until event occurs?


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Identify the random variable:

X = number of drives tested in order to findthe first defective

What constitutes success in this situation?

Success is a defective hard drive.


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b) What is the probability that the first defectivehard drive is the fifth unit tested?

P(X=5) = (1-0.03)5-1(.03)= (0.97)4(.03)

= .0266

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


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c) Find the first four entries in the table of thepdffor the random variable X.

X 1 2 3 4

P(X) .03 .0291 .0282 .0274

P(X=1), P(X=2), etc. (2min)


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Mean or Expected Value

The Mean or Expected Valueof a geometricvariable is:

The Varianceof X is:

2= (1-p)/p2

=

x

1

= p

2/ pq


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The probability that it takes more than n

trails to see the first success is:P(X>n) = (1-p)n or qn

Ex. Roll a die until a 3 is observed.

The probability that it takes more than 6rolls to observe a 3 is:

P(X>6)= (1-p)n

= (5/6)6

0.335


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Exploring Geometric Distributions:Calculator

Verify our previous results.

Enter the list of the # of trials, 1 to 7 in L1.


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Highlight L2and enter geometric pdfs;Select 2nd VARS: geometpdf (1/6,L1):

Nspire: name after you enter formula


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Menu:data,summary plot


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Simulating Geometric Experiments

Called wait time because you continue toconduct trails until a success is observed.

Example : Show me the Money! Cheerios claims a free $1 bill every 20th

box.

Lets simulate to determine how manyboxes you need to buyto get the money.


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Simulation with Table D

Let 2 digit numbers 00 to 99 represent abox of Cheerios.

Let 01 to 05 represent a box with $1.

Let 00, 06 to 99 represent a box w/o $1 Read Table, line 127:

Form pairs and organize into 5 rows, ten

across until a 01 to 05 is found.Ex. 23 33 06


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How many boxes did it take?

Why?

55!

Check the variation.


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Calculate the Variance and StandardDeviationto better understand the largenumber of trails.

p= 1/20 = 0.05

E(X) = 1/p= 20So why did we get 55?

2= (1-p)/p2= .95/.0025 = 380

(X) = 19.49


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How many standard deviations is our resultfrom the mean?

55 is about 1.8 s to the right of the mean20. (35 away from 20)

So it is reasonable.

Recall:

is not an appropriate measure of spread

for strongly skewed distributions.

Our geometric distribution is stronglyskewed right.

6.3c Geometric Random Variables

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