UGM June 2019 Q47

0. For a situation where 100 misprints are distributed randomly throughout the 100 pages of a book, the probability that a page selected at random will contain at most one misprint will be

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Option : D
Explanation : The number of possible misprints on a page in the book is large, but the actual misprints are only 1 per page. This is analogous to a situation when n (number of trials) is very large but p (probability of success—error in this case) is very small, and, therefore, Poisson distribution is appropriate for this situation.
Let the variable x be the number of misprints on a page. Then, the probability function of x is Poisson, given by:
where m is the mean number of misprints per page which is equal to 1, in this case. Now the probability that a page will contain at most 1 misprint is given by:
f(x  1) = f(x = 0) + f(x = 1)
Thus, P(x  1) = 0.368 + 0.368 = 0.736
Thus, the probability that the page will have at the most one misprint is 0.736.

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