in a book, the mean number of errors per chapter is 0.8. What is the
probability that there are less than 2 errors in a particular chapter?

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Clarification of Question by mojomabris-ga on 10 Apr 2006 15:53 PDT

well, that's all I was given. to me it looks like the Poisson
distribution formula can be used, I just don't know how. is this
possible?

I don't think there is enough information provided to give a correct answer.

Do you know the standard deviation or variance... # of chapters could
even help if there aren't many.

Jack_of_few_trades is correct, you do need more information.  What
type of distribution is it?  The Poisson distribution is often used
for counting things.  If it is the Poisson distribution, the mean and
the variance are the same.

Since no additional information is given, we have to assume that the
errors are randomly distributed. I won't give the solution for this
homework problem, but I'll give a hint. This is identical to the case
of radioactive decay. Suppose the average number of decays from a lump
of uranium is 0.8 in one second, what is the probability of 0 decays
in one second? What is the probability of 1 decay? Add them up.

Here is a link to an explanation of the Poisson distribution.

http://en.wikipedia.org/wiki/Poisson_distribution

It is a discrete distribution.  You can calculate the number of
occurrences expected for a given mean.  Here is the formula for k
occurrences for a given mean ?.

The formula is f(k|?) = [e^(-?)]*(?^k)/k!

The probability of less than two occurrences (errors) is the sum of
the probabilities of zero and one errors.

f(0|0.8) = 0.449329
f(1|0.8) = 0.359463

Sum      = 0.808792