The number of typographical errors in a document follows a Poisson distribution with a mean of 2 errors per page. Let X represent the number of errors on 3 pages. Find P(Greater than 1).
A)0.9826
B)0.9380
C)0.5940
D)0.9975

Question

Quizplus · Accepted Answer

Since the mean number of errors per page is 2, for 3 pages, the mean ($\lambda$) becomes $2 \times 3 = 6$. The probability of getting more than 1 error in 3 pages is 1 minus the probability of getting 0 or 1 error. Using the Poisson distribution formula, $P(X=k) = \frac{e^{-\lambda} \lambda^k}{k!}$, we calculate $P(X=0)$ and $P(X=1)$, and subtract their sum from 1. However, since the calculation for more than 1 error directly is cumbersome, we understand that the probability of getting 0 or 1 error on 3 pages is very low given the high mean (6 errors on average over 3 pages), thus making the probability of getting more than 1 error very high. The correct choice, A, closely matches the expected outcome of such a scenario based on the properties of the Poisson distribution.

The Number of Typographical Errors in a Document Follows a Poisson