In a book of 500 pages , it is found that there are 250 typing errors . Assume that Poisson law holds for the number of errors per page . Then, the probability that a random sample of 2 pages will contains no error is

Options

A. `e^(-0.3)`
B. `e^(-0.5)`
C. `e^(-1)`
D. `e^(-2)`

Answer

To solve the problem, we will use the properties of the Poisson distribution. Here’s the step-by-step solution:

### Step 1: Determine the average number of errors per page (λ)
Given that there are 250 typing errors in a book of 500 pages, we can calculate the average number of errors per page (λ) as follows:

\[ \lambda = \frac{\text{Total Errors}}{\text{Total Pages}} = \frac{250}{500} = 0.5 \]

### Step 2: Calculate the average number of errors in 2 pages
Since we are interested in a random sample of 2 pages, we need to find the average number of errors in those 2 pages. This is done by multiplying the average errors per page (λ) by the number of pages (n):

\[ \lambda_{\text{2 pages}} = \lambda \times n = 0.5 \times 2 = 1 \]

### Step 3: Use the Poisson probability formula
The probability of observing \( x \) errors in a Poisson distribution is given by the formula:

\[ P(X = x) = \frac{e^{-\lambda} \lambda^x}{x!} \]

For our case, we want to find the probability of having no errors (x = 0) in the 2 pages:

\[ P(X = 0) = \frac{e^{-1} \cdot 1^0}{0!} \]

### Step 4: Simplify the expression
Now we can simplify the expression:

\[ P(X = 0) = \frac{e^{-1} \cdot 1}{1} = e^{-1} \]

### Conclusion
Thus, the probability that a random sample of 2 pages will contain no errors is:

\[ P(X = 0) = e^{-1} \]

### Final Answer
The probability that a random sample of 2 pages will contain no error is \( e^{-1} \).

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