Question

In a book of 500 pages , there are 500 misprinting errors, find the probability of at most three misprinting 3 pages selected at random from the book.

Answer 1

On an average, the misprints occur at a rate of 500500=1500500=1 misprint per page.
This problem can be solved my assuming that the distribution of misprints throughout 500 pages is Poisson distribution.
Let X be the random variable representing the number of misprints.
X ∼Po(λ)∼Po(λ)
Since λλ represents the rate of occurrence of events in a time interval. λ=1λ=1
Probability that a page contains at most 3 misprints,
P(X ≤3)=P(X=0)+P(X=1)+P(X=2)+P(X=3)≤3)=P(X=0)+P(X=1)+P(X=2)+P(X=3)
P(X ≤3)=e−1100!+e−1111!+e−1122!+e−1133!≤3)=e−1100!+e−1111!+e−1122!+e−1133!

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Since P(X=x)=e−λ(λ)xx!P(X=x)=e−λ(λ)xx!

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P(X ≤3)=e−1+e−1+e−12!+e−13!≤3)=e−1+e−1+e−12!+e−13!
P(X ≤3)=2e+12e+16e≤3)=2e+12e+16e
P(X ≤3)=166e=0.98101≤3)=166e=0.98101

Hence, this is the answer.