If the phone calls from a central are made in a Poisson process ( N(t), t≥ 0 ), in average every 10 minutes they have one phone call.
calculate the probability that no call is received in the range ( 0 to 10 minutes ), and exacly one call between 10 and 15 minutes.
Best Answer
If you have a Poisson process with parameter $\lambda$ ($\lambda$ is the average number of events occurring in an interval of given unit length), and if $X$ is the number of events occurring in an interval of length $t$, then $X$ has Poisson distribution with parameter $\lambda t$. That is $$ P[X=k] =(\lambda t)^k {e^{-\lambda t}\over k!},\quad k=0,1,2,\ldots $$
Now on to your specific problem.
You are free to take the "unit time" to be whatever you like. Let's take it to be one hour. Then, since the average number of calls every 10 minutes is 1, the parameter for the Poisson process is $\lambda=6$ (events per hour). For the problem we will need to express the length $t$ of a time interval in hours.
For the first part:
If $X$ is the number of events occurring in the first 10 minutes, then $t=1/6$ and $\lambda t=6\cdot{1\over 6}=1$; so $$ P[X=k] =(1)^k {e^{-1}\over k!}, \quad k=0,1,2,\ldots $$
You need to find $P[X=0]$.
For the second part of the problem, using the independence assumption for disjoint time intervals in a Poisson process, you can find the probability that exactly one call is made in the first 5 minutes. If $Y$ is the number of calls made in the first 5 minutes, then $t=1/12$ and $\lambda t=6\cdot{1\over12}=1/2$; so $$ P[Y=k] =(1/2)^k {e^{-1/2}\over k!}, \quad k=0,1,2,\ldots. $$
You need to find $P[Y=1]$.
If you want the probability that both parts of your problem hold (it's not clear to me if this is what you want), by independence, you may multiply the two probabilities found above (remember, in a Poisson process, events that happen in one time period do not influence events that happen in another, disjoint time period).