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The mean number of typing in a certain newspaper was found to be four per page. Find the probability that;
(1) Exactly three errors will be found on a page .
(2) At most three typing errors will be found on a page.
(3) More than three errors Will be found on a page.
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I just saw (and solved) several similar problems in this forum.
They are not exactly and precisely mathematical, since the condition is not completed and requires some additions.
Again, they are not exactly and precisely mathematical, but mostly serve to exercise your "common sense".
They require you to reformulate the condition in a way the problem to obtain its sense.
And they require such an addition
a) "to be in style" (which means "to be consistent"), and
b) to be likelihood.
So, again, it is not "the pure Mathematics in its pure instance" - it is the common sense hypothesizing
to make right formulation and to get the problem solved ( = to get a likelihood solution ).
After this introduction, I can now move to the solution.
An appropriate hypotheses is to assume that we have random variable X = "the number of errors in a page"
with integer random values as uniformly distributed with the minimum value of 0 and the maximum value of 8.
Then the mean average is exactly 4, as the problem states, and we are
- "in style", and
- the condition becomes "self-consistent" and "self-closed".
Then the answer to question (1) is P(X=3) = ; the answer to question (2) is P(X<=3) = ; and the answer to question (3) is P(x > 3) = .