The problem is a Poisson distribution problem. The Poisson distribution is a discrete probability distribution that expresses the probability of a given number of events occurring in a fixed interval of time or space if these events occur with a known constant mean rate and independently of the time since the last event.

The formula for the Poisson probability is:

P(X=k) = λ^k * e^-λ / k!

where:
- P(X=k) is the probability of k events,
- λ is the average rate (mean) of events per interval,
- e is the base of the natural logarithm (approximately equal to 2.71828),
- k! is the factorial of k.

In this case, we want to find the probability that a randomly selected page contains at least one typo, which is the same as finding the probability that a page does not contain zero typos.

So, we can find the probability of zero typos and subtract it from 1.

Let's calculate the probability of zero typos:

P(X=0) = λ^0 * e^-λ / 0!

Substituting λ with 5 (the mean number of typos per page), we get:

P(X=0) = 5^0 * e^-5 / 0!
= 1 * e^-5 / 1
= e^-5
≈ 0.0067

So, the probability that a randomly selected page contains at least one typo is:

P(X=1) = 1 - P(X=0)
= 1 - 0.0067
= 0.9933

So, there is approximately a 99.33% chance that a randomly selected page will contain at least one typo.

Question

The problem is a Poisson distribution problem. The Poisson distribution is a discrete probability distribution that expresses the probability of a given number of events occurring in a fixed interval of time or space if these events occur with a known constant mean rate and independently of the time since the last event.

The formula for the Poisson probability is:

P(X=k) = λ^k * e^-λ / k!

where:
- P(X=k) is the probability of k events,
- λ is the average rate (mean) of events per interval,
- e is the base of the natural logarithm (approximately equal to 2.71828),
- k! is the factorial of k.

In this case, we want to find the probability that a randomly selected page contains at least one typo, which is the same as finding the probability that a page does not contain zero typos.

So, we can find the probability of zero typos and subtract it from 1.

Let's calculate the probability of zero typos:

P(X=0) = λ^0 * e^-λ / 0!

Substituting λ with 5 (the mean number of typos per page), we get:

P(X=0) = 5^0 * e^-5 / 0!
       = 1 * e^-5 / 1
       = e^-5
       ≈ 0.0067

So, the probability that a randomly selected page contains at least one typo is:

P(X>=1) = 1 - P(X=0)
        = 1 - 0.0067
        = 0.9933

So, there is approximately a 99.33% chance that a randomly selected page will contain at least one typo.

Knowee AI · Accepted Answer