The Poisson distribution is typically used to model the number of events occurring within a fixed period of time. It is defined by the following probability mass function:

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

where:

• $X$ is the random variable representing the number of events,
• $k$ is the number of occurrences,
• $\lambda$ is the average rate of occurrences (mean) over the specified interval,
• $e \approx 2.71828$ is Euler's number.

Given that the average number of views per minute is 1.5, we can proceed to solve each part:

(a) Probability of no views in a minute

We are asked to find $P(X = 0)$ when the average number of views per minute $\lambda = 1.5$.

$P(X = 0) = \frac{1.5^0 e^{-1.5}}{0!} = e^{-1.5} \approx 0.2231$

(b) Probability of two or fewer views in 10 minutes

First, determine the new rate $\lambda$ over a 10-minute interval. Since $\lambda = 1.5$ views per minute, the total rate over 10 minutes is:

$\lambda = 1.5 \times 10 = 15$

Now, we want $P(X \leq 2)$ for $\lambda = 15$.

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

Using the Poisson formula:

$P(X = k) = \frac{15^k e^{-15}}{k!}$

Calculating each term:

$P(X = 0) = \frac{15^0 e^{-15}}{0!} = e^{-15}$ $P(X = 1) = \frac{15^1 e^{-15}}{1!} = 15e^{-15}$ $P(X = 2) = \frac{15^2 e^{-15}}{2!} = \frac{225e^{-15}}{2}$

Now, summing them up:

$P(X \leq 2) = e^{-15} + 15e^{-15} + \frac{225e^{-15}}{2} = e^{-15}(1 + 15 + 112.5) = e^{-15}(128.5) \approx 3.61 \times 10^{-5}$

(c) Time interval such that the probability of no views is 0.001

Let $t$ be the length of the time interval in minutes. The probability of no views in this interval is given by:

$P(X = 0) = e^{-\lambda}$

where $\lambda$ is the average number of views in $t$ minutes. Since there are 1.5 views per minute, the rate over $t$ minutes is:

$\lambda = 1.5t$

We want $P(X = 0) = 0.001$.

$e^{-1.5t} = 0.001$

Taking the natural logarithm on both sides:

$-1.5t = \ln(0.001)$

$t = -\frac{\ln(0.001)}{1.5} \approx \frac{6.907755}{1.5} \approx 4.6052 \text{ minutes}$

So, the length of time such that the probability of no views is 0.001 is approximately 4.6052 minutes.

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Would you like more details or have any questions?

Here are 5 related questions:

1. How would the probabilities change if the average rate of views per minute increased?
2. What is the probability of getting exactly 5 views in a 10-minute interval?
3. How would you calculate the probability of getting more than 10 views in a 5-minute interval?
4. If the probability of at least one view in a minute is required to be 0.9, what should be the new average rate of views?
5. How would the results change if a different time interval, say 20 minutes, is considered?

Tip: When using the Poisson distribution, always ensure that the time interval and the average rate match. Adjust the rate based on the desired time period for accurate results.