Suppose that the number of typing errors per page has a Poisson distribution with an average of 4 typing errors. What is the probability that on a given page the number of typing errors will be at least 2?

Suppose that the number of
typing errors per page has a Poisson distribution with an average of 4 typing errors.
What is the probability that on a given page the number of typing errors will be at least 2?

Solution:

The probability that on a given page the number of typing errors will be at least 2 can be found by subtracting the probability of 0 or 1 error from 1. Using the Poisson probability mass function, we can find the probability of 0 errors and the probability of 1 error, and then subtract that from 1 to find the probability of at least 2 errors.

The Poisson probability mass function for x errors is:

`P(x) = (λ^x * e^-λ) / x!`

where λ is the average number of errors (4 in this case) and x is the number of errors.

So, the probability of 0 errors is:

`P(0) = (4^0 * e^-4) / 0! = e^-4 = 0.018`

And the probability of 1 error is:

`P(1) = (4^1 * e^-4) / 1! = 4e^-4 = 0.072`

To find the probability of at least 2 errors, we subtract the probability of 0 errors and 1 error from 1:

`P(≥2) = 1 - P(0) - P(1) = 1 - 0.018 - 0.072 = 0.91`

So the probability of at least 2 errors on a given page is 0.91 or 91%.