Given Information

• The average number of e-mails received per hour is 5.
• The number of e-mails received follows a Poisson distribution.

Poisson Distribution Formula

The Poisson probability of observing $k$ events in a given time period is:

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

Where:

• $\lambda$ is the average number of events (e-mails in this case) in the given time period.
• $k$ is the actual number of events observed.
• $e$ is approximately 2.71828.

a. Probability of Receiving No E-mails During an Hour

Here, $\lambda = 5$ e-mails per hour, and we need to find the probability of receiving no e-mails ($k = 0$):

$P(X = 0) = \frac{e^{-5} \cdot 5^0}{0!} = e^{-5} \cdot 1 = e^{-5}$

Let's calculate this:

$P(X = 0) = e^{-5} \approx 0.0067$

So, the probability of receiving no e-mails during an hour is 0.0067.

b. Probability of Receiving At Least Three E-mails During an Hour

To find this, we first calculate the probabilities of receiving 0, 1, and 2 e-mails and subtract their sum from 1:

$P(X \geq 3) = 1 - P(X < 3) = 1 - (P(X = 0) + P(X = 1) + P(X = 2))$

We already calculated $P(X = 0) = 0.0067$.

Now, let's calculate $P(X = 1)$ and $P(X = 2)$:

$P(X = 1) = \frac{e^{-5} \cdot 5^1}{1!} = 5e^{-5} \approx 0.0337$

$P(X = 2) = \frac{e^{-5} \cdot 5^2}{2!} = \frac{25e^{-5}}{2} \approx 0.0842$

Now, sum these probabilities:

$P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) \approx 0.0067 + 0.0337 + 0.0842 = 0.1246$

So,

$P(X \geq 3) = 1 - 0.1246 \approx 0.8754$

Thus, the probability of receiving at least three e-mails during an hour is 0.8754.

c. Expected Number of E-mails Received During 15 Minutes

The rate of e-mails per hour is $\lambda = 5$.

To find the expected number of e-mails in 15 minutes, convert 15 minutes to hours:

$\lambda_{15 \text{ min}} = \lambda_{1 \text{ hour}} \times \frac{15}{60} = 5 \times \frac{1}{4} = 1.25$

Thus, the expected number of e-mails received during 15 minutes is 1.25.

d. Probability that No E-mails are Received During 15 Minutes

Here, $\lambda = 1.25$ e-mails in 15 minutes. The probability of receiving no e-mails ($k = 0$):

$P(X = 0) = \frac{e^{-1.25} \cdot 1.25^0}{0!} = e^{-1.25}$

Let's calculate this:

$P(X = 0) = e^{-1.25} \approx 0.2865$

So, the probability that no e-mails are received during 15 minutes is 0.2865.

Do you want more details, or do you have any questions?

Here are 5 related questions:

1. What is the probability of receiving exactly 5 e-mails in an hour?
2. How does the expected number of e-mails change if the average increases to 10 e-mails per hour?
3. What is the probability of receiving fewer than 2 e-mails in a 15-minute period?
4. What is the variance of the number of e-mails received during an hour?
5. How would the probabilities change if the time period was reduced to 10 minutes?

Tip: When dealing with Poisson distributions, remember that the time period and the average rate are crucial to determining probabilities.