Number of errors on a single page has poisson distribution with average number of errors per page is one. Find the probability that there is atleast one error on a page.

Answer

To solve the problem of finding the probability that there is at least one error on a page where the number of errors follows a Poisson distribution with an average (λ) of 1, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Poisson Distribution**:
The number of errors on a page follows a Poisson distribution with a mean (λ) of 1. The probability mass function (PMF) of a Poisson distribution is given by:
\[ P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!} \]
where \( k \) is the number of occurrences (errors in this case), and \( e \) is approximately equal to 2.71828.

2. **Define the Event**:
We want to find the probability of having at least one error on a page. This can be expressed mathematically as:
\[ P(X \geq 1) = 1 - P(X = 0) \]

3. **Calculate \( P(X = 0) \)**:
Using the PMF for \( k = 0 \):
\[ P(X = 0) = \frac{e^{-\lambda} \lambda^0}{0!} \]
Substituting \( \lambda = 1 \):
\[ P(X = 0) = \frac{e^{-1} \cdot 1^0}{1} = e^{-1} \]

4. **Substitute the Value of \( e \)**:
The value of \( e^{-1} \) can be calculated as:
\[ e^{-1} \approx \frac{1}{2.71828} \approx 0.367879 \]

5. **Calculate \( P(X \geq 1) \)**:
Now, substituting \( P(X = 0) \) back into our equation for \( P(X \geq 1) \):
\[ P(X \geq 1) = 1 - P(X = 0) = 1 - e^{-1} \approx 1 - 0.367879 \]
\[ P(X \geq 1) \approx 0.632121 \]

6. **Final Result**:
Therefore, the probability that there is at least one error on a page is approximately:
\[ P(X \geq 1) \approx 0.632 \]

### Summary:
The probability that there is at least one error on a page is approximately **0.632**.