MATH 1780 Lecture Notes Chapter 3 Section 5 and 6

MATH 1780 Lecture Notes Chapter 3 Section 5 and 6 - As...

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As before, in the Binomial Distribution, we will be  tossing a coin repeatedly.  However this time, we will  not be tossing it only n times.  We are going to be  tossing it an undetermined number of times.  This  time, we have an infinite sequence of Bernoulli  Random Variables, X 1 , X 2,  ... Suppose X i  is 1 if the ith trial is a success and 0 if it is  a failure. Let X=i if and only if X i =1 and for all j<i, X j =0.  In other  words X is the trial on which the first success is  observed.
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Suppose for each trial the probability of success is  p, and the probability of failure is (1 – p). What are the possible outcomes of X? The possible outcomes are 1, 2, 3, 4, . .. What is P(X=1)? In order for X to be 1, you must see a success on  the first trial.  Thus P(X=1)=p. What is P(X=2)? In order for X to be 2, you must see a failure on  the first trial, and a success on the second trial.   Thus P(X=2)=(1-p) * p.
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In order to try to see a pattern, what is P(X=3)? Thus the first two trials must be failures and the  third trial must be a success.  Thus P(X=3)=(1-p) *  (1-p) * p = (1-p) 2  * p. Can you see the pattern? What is P(X=r)? Thus the first r-1 trials must be failures and the rth  trial must be a success.  Therefore P(X=r)=(1-p) r-1   * p. This distribution is called the  Geometric  Distribution .
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A recruiting firms finds that 20% of all job  applicants are “qualified” for a certain job.   Applicants are interviewed sequentially until a  qualified applicant is found.  That applicant is then  hired.  What is the probability that the first  qualified applicant is the fourth applicant  interviewed? Let X denote the interview on which the first  qualified applicant is interviewed.
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This note was uploaded on 09/30/2008 for the course MATH 1780 taught by Professor Snyder during the Fall '08 term at North Texas.

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MATH 1780 Lecture Notes Chapter 3 Section 5 and 6 - As...

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