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:
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:
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For a Poisson distribution with parameter \(\lambda\),
\[
P(X=0)=e^{-\lambda}.
\]
When several independent events are involved, multiply the corresponding probabilities.
Updated On: Jun 26, 2026
- \(\frac{1}{3}e^{-1}\)
- \(2e^{-1/3}\)
- \(e^{-2/3}\)
- \(\frac{1}{3}e^{-2}\)
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The Correct Option is C
Solution and Explanation
Step 1: Determine the Poisson parameter per page.
The book has \[ 600 \] pages and \[ 40 \] printing mistakes. Hence the average number of mistakes per page is \[ \lambda=\frac{40}{600} =\frac{1}{15}. \] Thus, the number of mistakes on a page follows a Poisson distribution with parameter \[ \lambda=\frac{1}{15}. \]
Step 2: Find the probability that a page contains no mistakes.
For a Poisson random variable, \[ P(X=0)=e^{-\lambda}. \] Therefore, \[ P(\text{no mistake on a page}) = e^{-1/15}. \]
Step 3: Find the probability for 10 pages.
Assuming independence of pages, the probability that each of the \(10\) selected pages contains no mistakes is \[ \left(e^{-1/15}\right)^{10} = e^{-10/15} = e^{-2/3}. \]
Step 4: Final conclusion.
Hence, the required probability is \[ \boxed{e^{-2/3}}. \]
The book has \[ 600 \] pages and \[ 40 \] printing mistakes. Hence the average number of mistakes per page is \[ \lambda=\frac{40}{600} =\frac{1}{15}. \] Thus, the number of mistakes on a page follows a Poisson distribution with parameter \[ \lambda=\frac{1}{15}. \]
Step 2: Find the probability that a page contains no mistakes.
For a Poisson random variable, \[ P(X=0)=e^{-\lambda}. \] Therefore, \[ P(\text{no mistake on a page}) = e^{-1/15}. \]
Step 3: Find the probability for 10 pages.
Assuming independence of pages, the probability that each of the \(10\) selected pages contains no mistakes is \[ \left(e^{-1/15}\right)^{10} = e^{-10/15} = e^{-2/3}. \]
Step 4: Final conclusion.
Hence, the required probability is \[ \boxed{e^{-2/3}}. \]
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