A secretary makes 2 errors per page on the average. what is the probability on the next page she makes?

Question

A) 4 or more errors
B) no errors

Answer 1

This problem uses the Poisson distribution.

Let X be the number of errors on a page. X has the Poisson distribution with a mean of 2.

X ~ Poisson(2)

Where P[X = x] = exp(-2) * 2^x / x!

a) P[X > = 4] = 1 - (P[X=0] + P[X=1] + P[X=2] + P[X=3])

P[X=0] = exp(-2) = 0.135335
P[X=1] = 0.270671
P[X=2] = 0.270671
P[X=3] = 0.180447

P[X > = 4] = 0.142877

or you can solve an infinite sum, but that doesn't seem like fun.

b) see above

Answer 2

That depends on the distribution function of the number of errors on a page. The Poisson distribution is often used for this kind of thing. Assuming the Poisson distribution applies:

Let
λ = mean number of errors per page
k = actual number of errors on a page

The probability of k errors on a page given mean λ is:

P(k | λ) = {[e^(-λ)] λ^k} / k!
_______________

B) Probability of no errors.

P(0 | 2) = {[e^(-2)] 2^0} / 0! = e^(-2) ≈ 0.135335283
________________________

A) Probability of 4 or more errors.

P(4+ | 2) = 1 - P(0 | 2) - P(1 | 2) - P(2 | 2) - P(3 | 2)

P(4+ | 2) ≈ 0.14287654