Question

The number of errors in a textbook follow a poisson distribution with a mean of 0.01 errors per page. what is the probability that there are 3 or less errors in 100 pages?

Answer 1

`\text{The formula of the Poisson distribution is the following:}\\P_v(n)=(v^ne^(-v))/(n!)\\\text{Wherein the parameter } v=pN=0.01*100=1\\\text{So the formula becomes:}\\P(n)=(1)/(e* n!)\\\text{3 errors means either no error, one error, two errors one three errors}\\\text{So we will compute the probability for n = 0, 1, 2 and 3:}\\P(0)=(1)/(e),P(1)=(1)/(e),P(2)=(1)/(2e),P(3)=(1)/(6e)\\\text{Adding we get the probability to have less than 3 errors:}\\0.98`