Answer to Question #170298 in Statistics and Probability for Bikram Chaudhary

Question

The probability of getting no misprint in a page of a book e^-4. What is the probability that a page contains more than 2 misprints? (State the assumptions you make in solving this problems).

Answer 1

Assumption: the Poisson distribution

The probability of finding k errors on a page of a book is

$P(X=k) = e^{-λ} \frac{λ^k}{k!}$

Given

$P(X=0) = e^{-4}$

Thus λ = 4

The probability that a page contains more than 2 misprints:

$P(X>2) = 1 -P(X≤2) = 1 -[P(X=0) + P(X=1)+P(X=2)] \\ P(X=0) = e^{-4} \times \frac{4^0}{0!} = e^{-4} \\ P(X=1) = e^{-4} \times \frac{4^1}{1!} = 4e^{-4} \\ P(X=2) = e^{-4} \times \frac{4^2}{2!} = 8e^{-4} \\ P(X>2) = 1 -[e^{-4} + 4e^{-4} + 8e^{-4}] \\ = 1 -13e^{-4} \\ = 1 -0.2381 \\ = 0.7619$