Answer:
(a) The probability of exactly 1 page has error is 0.271.
(b) The probability that there are at most 3 pages has error is 0.857.
Step-by-step explanation:
Let <em>X</em> = number of typos.
The probability of a typo is, P (X) = <em>p </em>= 0.005.
The number of pages in the novel is, <em>n</em> = 400.
The random variable <em>X</em> follows a Binomial distribution with parameter <em>n</em> and <em>p</em>.
But as the probability is very small and the sample size is too large we can use Poisson distribution to approximate the binomial distribution.
This distribution has parameter,
.
The probability mass function of the Poisson distribution is:

(a)
Compute the probability of exactly 1 page has error as follows:

Thus, the probability of exactly 1 page has error is 0.271.
(b)
Compute the probability that there are at most 3 pages has error as follows:
P (X ≤ 3) = P (X = 0) + P (X = 1) + P (X = 2) + P (X = 3)

Thus, the probability that there are at most 3 pages has error is 0.857.