Question

The mean number of errors per page made by a member of the word processing pool for a large company is thought to be 1.8 with the number of errors distributed according to a Poisson distribution. If a page is​ examined, what is the probability that more than two errors will be​ observed?

The probability that more than two errors will be observed is ___________?

Answer 1

Solution :

Given that ,

mean = $\mu$ = 1.8

Using poisson probability formula,

P(X = x) = (e^{- $\mu$} * $\mu$ ^x ) / x!

P(X > 2) = 1 - P(X $\leq$ 2)

= 1 - (P(X = 0) + P(X = 1) + P(X = 2))

= 1 - (e^-1.8 * 1.8⁰⁾ / 0! + e^-1.8 * 1.8¹⁾ / 1! + e^-1.8 * 1.8²⁾ / 2! )

= 1 - 0.73062

= 0.26938

Probability = 0.26938

The probability that more than two errors will be observed is 0.26938 .