PROBLEM 1 Books from a certain publisher contain on average 1 misprint per page.

Question

PROBLEM 1 Books from a certain publisher contain on average 1 misprint per page. Suppose those misprints occur according to a Poisson scatter, with a rate on 1 per page. One of the books from this publisher has 200 pages.

(a). What is the probability that there are at least 70 pages in this book with no misprints?

(b). What is the probability that there is at least one page in this book with 5 or more misprints?

(c). Suppose a proofreader goes through this book and catches each misprint with probability 0.7 (independently for different misprints). If the caught misprints are fixed, what is the probability that there are still 60 or more pages containing misprints?

(d). If this proofreader catches 3 misprints on a page, what is the probability that there are still misprints that weren't caught on that page?

Explanation / Answer

Concept Base

This problem involves application of both Poisson and Binomial distributions.

Let X = Number of misprints in a page and Y = Number of pages having a given number of misprints per page. Then,

X ~ Poisson () and Y ~ B(n, p), where = average number of misprints per page, n = number of pages in the book and p = probability that a page will have a given number of misprints per page.

We are given = 1 per page; n = 200 and p will have to be evaluated for each part of the question separately.

Back-up Theory

If X ~ B(n, p). i.e., X has Binomial Distribution with parameters n and p, where n = number of trials and p = probability of one success, then

probability mass function (pmf) of X is given by

p(x) = P(X = x) = (nCx)(px)(1 - p)n – x, x = 0, 1, 2, ……. , n ………………………………..(1)

[The above probability can also be directly obtained using Excel Function of Binomial Distribution] ……………………………………………………………………………….(1a)

If a random variable X ~ Poisson(), i.e., X has Poisson Distribution with mean then

probability mass function (pmf) of X is given by P(X = x) = e – .x/(x!) …………..(2)

where x = 0, 1, 2, ……. ,

Values of p(x) for various values of and x can be obtained by using Excel Function.

Part (a)

To calculate the probability that there are at least 70 pages in this book with no misprints, we first obtain the probability of no misprint in a page, i.e., P(X = 0)

Using Excel Function on Poisson Distribution with = 1 and x = 0, we find P(X = 0) = 0.367879.

Then, Y ~ B(200, 0.367879) and the required probability = P(Y 70) which using Excel Function on Binomial Distribution with n = 200, p = 0.367879 and Y 70, we find it as

0.32791 ANSWER

Part (b)

To calculate the probability that there is at least one page in this book with 5 or more misprints, we first obtain the probability of 5 or more misprint in a page, i.e., P(X 5)

Using Excel Function on Poisson Distribution with = 1 and x 5, we find P(X 5) = 0.00366.

Then, Y ~ B(200, 0.00366) and the required probability = P(Y 1) which using Excel Function on Binomial Distribution with n = 200, p = 0.00366 and Y 1, we find it as

0.529699 ANSWER

Part (c)

Since initially there is 1 misprint per page on an average and now given that proofreader goes through this book and catches each misprint with probability 0.7, the average number of misprints now is 0.3. To calculate the probability that there are still 60 or more pages containing misprints, we first obtain the probability of misprints in a page, i.e., P(X 1)

Using Excel Function on Poisson Distribution with = 0.3 and x 1, we find P(X 1) = 0.259182.

Then, Y ~ B(200, 0.259182) and the required probability = P(Y 60) which using Excel Function on Binomial Distribution with n = 200, p = 0.259182 and Y 60, we find it as

0.082713 ANSWER

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