Question

Messages reach to a computer server according to a Poisson distribution with a mean rate of 20 per hour. Find the length of an interval of time such that the probability that some messages reach during this interval is 0.10.

Answer 1

First we need to find the value of lambda that gives a probability of 1.00 - 0.1 = 0.90 that no messages will reach the server.
\[P(X=0)=0.9=\frac{e ^{- \lambda} \times \lambda ^{0}}{0!}=e ^{- \lambda}\]
Simplifying we get
\[- \lambda=\ln 0.9=-0.10536\]
Therefore we have the result that for a probability of 0.9 that no messages reach the server and a probability of 0.1 that one or more messages reach the server
\[\lambda=0.10536\]
Let t be the length of time in seconds that corresponds to the above value of lambda. Then we can write the following equation:
\[\frac{20}{0.10536}=\frac{60^{2}}{t}\]
Solving gives the value of t as 18.965 seconds.

Answer 2

I arrived at the same answer but rather than starting with the Poisson Random Variable, \(X\), I started with the Poisson Random Process, \(X(t)\)
$$
\large
P\left (X(t)=0\right )=0.9=\frac{e ^{- {\lambda}t} \times \left ({\lambda t }\right )^{0}}{0!}=e ^{- {\lambda}t}
$$
Where \(\lambda=20\).
Solving for \(t\):
$$
\large{
e ^{- {\lambda}t}=0.9\\
- {\lambda}t=\ln {0.9}\\
t=\cfrac{\ln {0.9}}{-\lambda}=\cfrac{\ln {0.9}}{-20}~hours
}
$$
Converting this to seconds:
$$
\large{
t=\cfrac{\ln {0.9}}{-20}\times \cfrac{3600 ~seconds}{1~hour}=18.965~seconds
}
$$
Exactly as @kropot72 found.