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2015-02-18T19:38:47+05:30

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We can assume a probability distribution for this kind of problem.   We can say that the number of defective pages is a random variable X.  The probability of a page being defective is random and  = 100/1000 = 0.1

In the first 50 pages, the expected number of defective pages is 50 * 0.1 = 5.  Let X denote the number of defective pages in the first 50 pages.  

We can have 0 defective pages, 1 or 2 or 5 or 10 or 19 or 20 or up to 50 defective pages in the first 50 pages of the book.  That means,

  0 <= X <= 50         and   E(X) = λ = mean or expected value = 5

We can assume the Poission probability distribution or the Normal probability distribution to solve this.  We assume that  X follows the Possion's probability distribution  function.

P(X=k) = \frac{e^{-\lambda}\ \lambda^k}{k !}\\

P(X \ge k)= \frac{e^{-\lambda}(e\ \lambda)^k}{k^k}\\

Let us calculate the probabilities, with λ = 5.

    P(X=10) <= 0.0181 = the probability that the number of defective pages is 10.

     P(X>= 10)  = 0.145.    This is the probability that the number of defective pages is equal to or more than 10.
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for info:
we can also calculate  P(X=0) = 0.006738
 P(X=1)=0.0337                 P(X=5) = 0.175    
P(X=8) = 0.0653          P(X=9) = 0.0362

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