In a call center the number of received calls in a day can be modeled by a Poisson random variable. We know that on average about 0.5% of the time the call center receives no calls at all. How can I find the distribution of the number of received calls in a day?
Poisson distribution phone calls
probability
Related Solutions
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).
Grinding it out: Let $A$ be the event there was at least one call on Tuesday, and $B$ be the event there were exactly two days with no call. We want $\Pr(A\mid B)$. It is easier to think about $\Pr(A^c\mid B)$.
Let $p$ be the probability of no call on a particular day. You know how to find $p$.
Then the probability of exactly two call-free days in a "week" is $\binom{6}{2}p^2(1-p)^4$.
To finish calculating $\Pr(A^c\mid B)$, we need $\Pr(A^c\cap B)$. This is the probability of no call on Tuesday, times the probability exactly one of the other $5$ days is call-free. This is $p\binom{5}{1}p(1-p)^4$.
Divide. There is cancellation and we get $\frac{2}{6}$.
Another way: If there were two call-free days, the probability one of them was Tuesday is obviously $\frac{2}{6}$. So the conditional probability of at least one call on Tuesday is $\frac{4}{6}$.
Best Answer
Let $X$ be the number of received calls in a day, i.e. $X\sim Pois(\lambda)$.
$P(X=0)=e^{-\lambda} = 1/200 \Rightarrow \lambda = \ln(200)$. Our number of calls in a day is distributed $Pois(\ln(200))$.