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Unformatted text preview: Stat 211 Prof Parzen CHAPTER 3 Binomial Probability, Random Variables, Expectation A Binomial Probability problem considers independent trials whose outcome are 1 or (also called failure or success) accord ing as a specified event A does not or does occur. Let p denote the probability of observing 1 on a trial; let q=1p denote the probability of failure. Two variables are of general interest: K , the number of times 1 occurs among the n trials, called the number of successes in n trials; W, the number of trials until a success (outcome 1) is ob served. We call W a waiting time. Probability theory assumes we know the population probabil ity p and calculates probabilities of observing K to have values in a specified interval. Statistical theory considers the problem that we have observed a value of K and want to determine intervals of specified prob ability for the unknown population probability p. This is a sub tle problem because p is a constant and we are observing only one value in the sample observed. We use concept of probability to describe our uncertainty about the value of p based on evi dence (data) and statistical methods. Important examples of W: (1) the number of children a woman has if she stops after giving birth to a boy; note that in Western countries, approximately .512 p = (significantly differ ent from .5 p = ). (2) A married couple interested in conceiving a child want to predict W, the number of months to success; to do this one must assume a value for the probability p of suc cess (pregnancy is achieved this month) which experience indi cates is between .15 and .25 (take p=.2 to compute an answer). To find probability of an event we apply: (1) counting prob ability (when the sample space is finite and we assume equally likely outcomes); (2) axiomatic probability (in general)  2  Game of Odd Person Out: Consider m friends who toss fair coins to determine who will perform a certain task. If there is one player whose outcome (heads or tails) is different from all the rest, that person is declared “odd person out” and performs the task. We want to determine the probability [ ] Pr p A = of the event A that the m fair coin tosses result in an odd person out. Sample space ( 29 { } 1 , , : 1 m i S x x x or = = … has ( 29 2 m Size S = . Event A can be described as a subset of S: ( 29 { 1 , , : m A x x = … exactly one j x = and others 1 = , or exactly one 1 j x = and others } = Applying counting probability (assume equally likely out comes) we find probability of event A by formula [ ] ( 29 ( 29 Pr A size A size S = . One can show ( 29 2 Size A m = . Therefore [ ] [ ] 1 Pr Pr odd person out 2 m p A m = = = For ( 29 ( 29 { } [ ] 3, ' 1,1,1 , 0,0,0 , Pr 6 8 3 4 m A A = = = = In the game of odd person out waiting time W is how many times the players must toss coins in order to determine an odd person out....
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 Fall '07
 Parzen
 Binomial, Normal Distribution, Probability, Probability theory, 1 m, 0 K, 0 K, 1 2 k

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