3-9 POISSON DISTRIBUTION

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sharpness. If any dull blade is found, the assembly is replaced with a newly sharpened set of blades. (a) If 10 of the blades in an assembly are dull, what is the probability that the assembly is replaced the rst day it is evaluated (b) If 10 of the blades in an assembly are dull, what is the probability that the assembly is not replaced until the third day of evaluation [Hint: Assume the daily decisions are independent, and use the geometric distribution.] (c) Suppose on the rst day of evaluation, two of the blades are dull, on the second day of evaluation six are dull, and on the third day of evaluation, ten are dull. What is the probability that the assembly is not replaced until the third day of evaluation [Hint: Assume the daily decisions are independent. However, the probability of replacement changes every day.] 3-94. A state runs a lottery in which 6 numbers are randomly selected from 40, without replacement. A player chooses 6 numbers before the state s sample is selected. (a) What is the probability that the 6 numbers chosen by a player match all 6 numbers in the state s sample (b) What is the probability that 5 of the 6 numbers chosen by a player appear in the state s sample

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(c) What is the probability that 4 of the 6 numbers chosen by a player appear in the state s sample (d) If a player enters one lottery each week, what is the expected number of weeks until a player matches all 6 numbers in the state s sample 3-95. Continuation of Exercises 3-86 and 3-87. (a) Calculate the nite population corrections for Exercises 3-86 and 3-87. For which exercise should the binomial approximation to the distribution of X be better (b) For Exercise 3-86, calculate P1X 12 and P1X 42 assuming that X has a binomial distribution and compare these results to results derived from the hypergeometric distribution. (c) For Exercise 3-87, calculate P1X 12 and P1X 42 assuming that X has a binomial distribution and compare these results to the results derived from the hypergeometric distribution. 3-96. Use the binomial approximation to the hypergeometric distribution to approximate the probabilities in Exercise 3-92. What is the nite population correction in this exercise

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POISSON DISTRIBUTION

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We introduce the Poisson distribution with an example.

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EXAMPLE 3-30

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Consider the transmission of n bits over a digital communication channel. Let the random variable X equal the number of bits in error. When the probability that a bit is in error is constant and the transmissions are independent, X has a binomial distribution. Let p denote the probability that a bit is in error. Let and pn. Then, E1x2 pn P1X x2 n a b P x 11 x p2 n

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x x n a b a n b a1 x

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Now, suppose that the number of bits transmitted increases and the probability of an error decreases exactly enough that pn remains equal to a constant. That is, n increases and p decreases accordingly, such that E(X) remains constant. Then, with some work, it can be shown that limnS P1X x2 e x!

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0, 1, 2, p

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Also, because the number of bits transmitted tends to in nity, the number of errors can equal any nonnegative integer. Therefore, the range of X is the integers from zero to in nity. The distribution obtained as the limit in the above example is more useful than the derivation above implies. The following example illustrates the broader applicability.

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CHAPTER 3 DISCRETE RANDOM VARIABLES AND PROBABILITY DISTRIBUTIONS

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EXAMPLE 3-31

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Flaws occur at random along the length of a thin copper wire. Let X denote the random variable that counts the number of flaws in a length of L millimeters of wire and suppose that the average number of flaws in L millimeters is . The probability distribution of X can be found by reasoning in a manner similar to the previous example. Partition the length of wire into n subintervals of small length, say, 1 micrometer each. If the subinterval chosen is small enough, the probability that more than one flaw occurs in the subinterval is negligible. Furthermore, we can interpret the assumption that flaws occur at random to imply that every subinterval has the same probability of containing a flaw, say, p. Finally, if we assume that the probability that a subinterval contains a flaw is independent of other subintervals, we can model the distribution of X as approximately a binomial random variable. Because E1X2 we obtain p n np

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That is, the probability that a subinterval contains a flaw is n. With small enough subintervals, n is very large and p is very small. Therefore, the distribution of X is obtained as in the previous example. Example 3-31 can be generalized to include a broad array of random experiments. The interval that was partitioned was a length of wire. However, the same reasoning can be applied to any interval, including an interval of time, an area, or a volume. For example, counts of (1) particles of contamination in semiconductor manufacturing, (2) flaws in rolls of textiles, (3) calls to a telephone exchange, (4) power outages, and (5) atomic particles emitted from a specimen have all been successfully modeled by the probability mass function in the following definition.

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De nition Given an interval of real numbers, assume counts occur at random throughout the interval. If the interval can be partitioned into subintervals of small enough length such that (1) the probability of more than one count in a subinterval is zero, (2) the probability of one count in a subinterval is the same for all subintervals and proportional to the length of the subinterval, and (3) the count in each subinterval is independent of other subintervals, the random experiment is called a Poisson process. The random variable X that equals the number of counts in the interval is a Poisson random variable with parameter 0 , and the probability mass function of X is f 1x2 e x!

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