Question

A website has on the average two hits per hour. Assuming a Poisson distribution for the number of hits per hour (X), calculate the probability that there are at most three hits.

Answer 1

Answer on Question #66329 – Math – Statistics and Probability

Question

A website has on the average two hits per hour. Assuming a Poisson distribution for the number of hits per hour $(X)$, calculate the probability that there are at most three hits.

Solution

Assume that $X$ has a Poisson distribution with rate $\lambda$, then we have:

$$P(X = k) = \frac{\lambda^k}{k!} e^{-\lambda}, \quad k = 0, 1, \dots$$

(see https://en.wikipedia.org/wiki/Poisson_distribution).

Since average value of $X$ (i.e. mathematical expectation of $X$) is equal to $\lambda$ (see https://en.wikipedia.org/wiki/Poisson_distribution#Mean) then we have $\lambda = 2$, and

$$P(X = k) = \frac{2^k}{k!} e^{-2}, \quad k = 0, 1, \dots$$

Then required probability is

$$\begin{array}{l} P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = e^{-2} + 2e^{-2} + 2e^{-2} + \frac{4}{3}e^{-2} = \\ = \frac{19}{3e^2} \approx 0.857. \end{array}$$

Answer: $\frac{19}{3e^2}$.

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