6.3c Geometric Random Variables
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Transcript of 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.