1!Discrete r.v. (III) Juana Sanchez!UCLA Dept of Statistics 13.7.10! Stat 100A-Introduction to Probability! 1!Other discrete random variables • There are many discrete random variables. We see a few more in these lecture notes: geometric, negative binomial and hypergeometric. !• In each case, changing the parameters gives us a different member of the same family. !13.7.10! Stat 100A/Sanchez-Introduction to Probability!2!13.7.10! Stat 100A-Introduction to Probability! 3! For example, the binomial is really a family. The shape of the pmf changes with n and p.If we fix n at n=4 and change p from "(left to right) p=0.1 to 0.5 (top row) to 0.9 to 0.99 (bottom row!13.7.10! Stat 100A-Introduction to Probability! 4!If we keep p fixed at p=0.1 and change n from n=4 to n=20 to n=100 to n=1000"13.7.10! Stat 100A-Introduction to Probability! 5! The Geometric Random Variable!• Let Xi=1 if the trial of an experiment results in a success and Xi=0 otherwise. Let P(Xi=1)=p be constant across trials. The trials are independent of each other. What is the number of the trial in which the first success occurs? Let’s call this number Y. !• Y is a new r.v. and its probability distribution is ……!13.7.10! Stat 100A-Introduction to Probability! 6!P(Y=y)=P(x1=0,x2=0,xy-1=0,xy=1)=(1-p)y-1 p "for y=1,2,…..countably infinite!• Example of variables with this distribution:!(a) The year that a dam is in service before it overflows; (b) the number of automobiles going through a radar check before the first speeder is detected; (c ) the number of mistakes before getting it right; (d) and many others. !2!13.7.10! Stat 100A-Introduction to Probability! 7! Example !• If one third of the persons donating blood at a clinic have O+ blood, find the probability that the first O+ donor is the fourth donor of the day. !P(Y=4) = (1-1/3)3(1/3) = 0.0987 !13.7.10! Stat 100A-Introduction to Probability! 8!The pmf for this problem is….!13.7.10! Stat 100A-Introduction to Probability! 9!Expected Value and Variance of a geometric random variable!E(X) = 1/p !Var(X) = (1-p)/p2!How many times should you expect to roll a die before you get a 6? E(X) = 6 times = 1/(1/6) give or take the standard deviation (i.e., give or take 5.4 times) 13.7.10! Stat 100A-Introduction to Probability! 10!Example !• A recruiting firm finds that 30 percent of the applicants for a certain industrial job have received advanced training in computer programming. Applicants are interviewed sequentially and selected at random from the pool. !(a) Find the probability that the first applicant who has received advanced training in programming is found on the fifth interview. !Y=number of interview on which 1st applicant having advanced training is found. ! P(Y=5) = (1-0.3)4 0.3 = 0.072 !13.7.10! Stat 100A-Introduction to Probability! 11!If the parameter p changes, the distribution changes. !13.7.10! Stat 100A-Introduction to Probability! 12!The negative binomial random variable!The geometric distribution models the probabilistic behavior of the number of the trial on which the first success occurs in a sequence of independent Bernoulli trials. But what if we were interested in the number of the trial for the second success, or the third success, or (in general) the rth success? The distribution governing probabilistic behavior in these cases is called the negative binomial distribution. !3!13.7.10! Stat 100A-Introduction to Probability! 13!Neg Binomial (cont’d)!Let Y denote the number of the trial on which the rth success occurs in a sequence of independent Bernoulli trials…."P(Y=y) =P[(1st (y-1) trials contain (r-1) successes and the yth trial is a success)]!=P[1st (y-1) trials contain r-1 successes] x! P[yth trial is a success] !The first probability statement is identical to the one that results in a binomial model and "13.7.10! Stat 100A-Introduction to Probability! 14!hence....!,.....1,)1(11)1(11)(1+=−⎟⎟⎠⎞⎜⎜⎝⎛−−=−⎟⎟⎠⎞⎜⎜⎝⎛−−==−−−rrypprypppryyYPryrryr13.7.10! Stat 100A-Introduction to Probability! 15!Example !• 30% of the applicants for a certain position have received advanced training in computer programming. Suppose that three jobs that require advanced programming training are open. Find the probability that the third qualified applicant is found on the fifth interview, if the applicants are interviewed sequentially and at random. !13.7.10! Stat 100A-Introduction to Probability! 16!The answer is…. !Assume independent trials, with 0.3 being the probability of finding a qualified candidate on any one trial. Let Y denote the number of the trial on which the third qualified candidate is found. Y can be reasonably assumed to have a negative binomial distribution!079.0)7.0()3.0(24)5(23=⎟⎟⎠⎞⎜⎜⎝⎛==YP13.7.10! Stat 100A-Introduction to Probability! 17!Expected Value and Variance of the negative binomial! E(X) =rpVar(X)=r(1− p)p2In the example of the candidates for the jobs, how many interviews should we expect to conduct to find the three experienced candidates? 3/0.3 =10, the expected value, but we could need 4,88 more or less Interviews than that. (4.830) is the standard deviation. 13.7.10! Stat 100A-Introduction to Probability! 18!The Hypergeometric Random Variable!• A sample of size n is to be chosen randomly (without replacement) from a box containing N balls, of which k are white and N-k are black. If we let X denote the number of white balls selected, then X is a hypergeometric random variably with parameters N, k, n. The pmf of X is !kxnNxnkNxkxXP,.....,1))(=⎟⎟⎠⎞⎜⎜⎝⎛⎟⎟⎠⎞⎜⎜⎝⎛−−⎟⎟⎠⎞⎜⎜⎝⎛==4!13.7.10! Stat 100A-Introduction to Probability! 19!Example !• A warehouse contains ten printing machines, four of which are defective. A company randomly selects five of the machines for purchase. What is the probability that all of the machines are Not defective? !Y=# of Non defectives, N=10, n=5!k=6 =number of non defective machines!0238.05100456)5( =⎟⎟⎠⎞⎜⎜⎝⎛⎟⎟⎠⎞⎜⎜⎝⎛⎟⎟⎠⎞⎜⎜⎝⎛==YP13.7.10! Stat 100A-Introduction to Probability! 20!Example !• A purchaser of electrical components buys them in lots of size 10. It is his policy to inspect 3 components randomly from a lot and to accept the lot only if all 3 are nondefective. If 30 percent of the lots have 4 defective components and 70 percent have only 1,
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