The number of errors in a textbook follow a Poisson distribution with a mean of 0.03 errors per page. What is the probability that there are 3 or less errors in 100 pages? Round your answer to four decimal places (e. g. 98.7654).

Therefore the required probability is 0.6472.

Step-by-step explanation:

Poisson distribution: A Poisson distribution is discrete distribution.

Let X be a discrete variable the number of event. Let λ be the mean of X.

Then the probability of k event is

Here mean of each page is =0.03

Mean of 100 pages = (0.03×100)= 3

λ = 3

The required probability

P(X≤3)  

= P(X=0)+P(X=1)+P(X=2)+P(X=3)

≈ 0.6472

Therefore the required probability is 0.6472

a = 11 : 4

b = 38

c = 1 for every 6

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