Suppose there are 500 misprints in a book of 750 pages. Assuming that the number of misprints per page is a poisson random variate. The probability that a page selected at random has more than one mis print is

Options

A. `e^(-2//3)`
B. `(5)/(3)e^(-2//3)`
C. `1-e^(-2//3)4)`
D. `1-(5)/(3)e^(-2//3)`

Answer

To solve the problem, we need to find the probability that a randomly selected page from a book has more than one misprint, given that the number of misprints per page follows a Poisson distribution.

### Step-by-Step Solution:

1. **Determine the average number of misprints per page (λ)**:
- We are given that there are 500 misprints in a book of 750 pages.
- The average number of misprints per page (λ) can be calculated as:
\[ \lambda = \frac{\text{Total Misprints}}{\text{Total Pages}} = \frac{500}{750} = \frac{2}{3} \]

**Hint**: To find λ in a Poisson distribution, divide the total number of occurrences by the total number of trials (or pages, in this case).

2. **Understand the Poisson probability formula**:
- The probability of observing \( k \) events (misprints) in a fixed interval (page) is given by:
\[ P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!} \]
- For our case, we need to find \( P(X > 1) \).

**Hint**: Recall that for a Poisson distribution, you can find the probability of more than one event by calculating \( 1 - P(X \leq 1) \).

3. **Calculate \( P(X \leq 1) \)**:
- We can express this as:
\[ P(X \leq 1) = P(X = 0) + P(X = 1) \]
- Using the Poisson formula:
\[ P(X = 0) = \frac{e^{-\lambda} \lambda^0}{0!} = e^{-\frac{2}{3}} \cdot 1 = e^{-\frac{2}{3}} \]
\[ P(X = 1) = \frac{e^{-\lambda} \lambda^1}{1!} = e^{-\frac{2}{3}} \cdot \frac{2}{3} \]

**Hint**: Remember that \( 0! = 1 \) and \( 1! = 1 \).

4. **Combine the probabilities**:
- Now, we can combine these probabilities:
\[ P(X \leq 1) = P(X = 0) + P(X = 1) = e^{-\frac{2}{3}} + e^{-\frac{2}{3}} \cdot \frac{2}{3} \]
\[ = e^{-\frac{2}{3}} \left(1 + \frac{2}{3}\right) = e^{-\frac{2}{3}} \cdot \frac{5}{3} \]

**Hint**: Factor out common terms to simplify calculations.

5. **Calculate \( P(X > 1) \)**:
- Finally, we find the probability of more than one misprint:
\[ P(X > 1) = 1 - P(X \leq 1) = 1 - e^{-\frac{2}{3}} \cdot \frac{5}{3} \]

**Hint**: The complement rule can simplify finding probabilities of "greater than" or "less than" events.

### Final Answer:
The probability that a randomly selected page has more than one misprint is:
\[ P(X > 1) = 1 - e^{-\frac{2}{3}} \cdot \frac{5}{3} \]