To solve this problem, we will use the Poisson distribution formula, which is:

$P(X = k) = \frac{{\lambda^k e^{-\lambda}}}{k!}$

Where:

• $\lambda$ is the average number of errors per page (which is 2.7 in this case),
• $k$ is the number of errors,
• $e$ is Euler's number (approximately 2.71828),
• $k!$ is the factorial of $k$.

We are asked to find the probability that more than 2 errors are observed, i.e., $P(X > 2)$. This is equal to:

$P(X > 2) = 1 - P(X \leq 2)$

So first, we need to calculate $P(X \leq 2)$, which is the sum of the probabilities for 0, 1, and 2 errors:

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)$

Now, let's calculate each of these:

1. $P(X = 0) = \frac{{2.7^0 e^{-2.7}}}{0!} = e^{-2.7}$
2. $P(X = 1) = \frac{{2.7^1 e^{-2.7}}}{1!} = 2.7 e^{-2.7}$
3. $P(X = 2) = \frac{{2.7^2 e^{-2.7}}}{2!} = \frac{{2.7^2}}{2} e^{-2.7}$

Now, we can sum these probabilities and subtract them from 1 to find $P(X > 2)$.

Let me calculate that for you.The probability that more than two errors will be observed on a page is approximately 0.5064 (rounded to four decimal places).

Would you like further details or have any questions about this? Here are 5 related questions to explore further:

1. How would the probability change if the mean number of errors was different?
2. What is the probability of observing exactly three errors on a page?
3. How do we compute cumulative probabilities for other values of $k$ in a Poisson distribution?
4. Can the Poisson distribution be used to model other types of events?
5. How does the Poisson distribution compare to the binomial distribution?

Tip: When solving Poisson distribution problems, it helps to remember that it models rare events over a fixed interval or space, like errors per page or arrivals per hour.