Question

Suppose that a book of 600 pages contains 40 print mistakes. Assume that these errors are randomly distributed throughout the book and the number of errors per page follows a Poisson distribution. The probability that all the 10 pages selected at random will have no print mistakes is:

• A. \(\frac{1}{3}e^{-1}\)
• B. \(2e^{-1/3}\)
• C. \(e^{-2/3}\)
• D. \(\frac{1}{3}e^{-2}\)

Hint:

For a Poisson distribution with parameter \(\lambda\), \[ P(X=0)=e^{-\lambda}. \] When several independent events are involved, multiply the corresponding probabilities.

Answer 1

The Correct Option is C

Step 1: Find Poisson parameter per page.
\(\lambda=\dfrac{40}{600}=\dfrac{1}{15}\) mistakes per page. Probability of zero mistakes on one page: \(P(X=0)=e^{-1/15}\).

Step 2: Extend to 10 independent pages.
\(P(\text{all 10 pages have 0 mistakes})=(e^{-1/15})^{10}=e^{-10/15}=e^{-2/3}\). \[ \boxed{e^{-2/3}} \]