Here's the question (Sorry if the translation is a little weird, if anything isn't clear tell me):
In a phone company there's two types of callers:
First type (A) calls to buy a cellphone.
Second type (B) calls to fix a bug in their cellphone.
We know that the number of times a caller calls is independent and is distributed Poisson distribution with parameters $\lambda_A=1$ and $\lambda_B=\frac{1}{2}$.
So the time between the calls are also independent and are distributed exponentially with parameters $\lambda_A,\lambda_B$.
What is the probability that the first caller from type A, called after the third caller from type B?
My attempts:
I tried to find a way to apply a known formula:
Let's say the time between the calls of caller A is $Q^i_A$, and times between calls of caller $B$ is $Q^i_B$ (where $i$ is $1$ between the first two calls, $2$ between second and third call etc...
And I know that:
$T_B^3=Q_B^1+Q_B^2+Q_B^3 \Longrightarrow T_B^n\sim Gamma(3,\lambda_B)$
$T_A^1= Q_A^1$ (Just to be consistent).
And Here I tried to connect the question to this formula:
$P(T_n\le t)=P(N_t \ge n) =1-\sum^{n-1}_{k=0}\frac{e^{-\lambda t (\lambda t)^k}}{k!}$. Where $N_t\sim Pois(\lambda t)$
But now I'm struggling to link it to my question here, how do I write "first caller of type A has called after the third caller from type B" the formula doesn't give me some type of connection, it just tells me that "the probability that the time is less than $t$ is the probability that the caller has called more or equal to $n$ times".Using this fact: Given two independent random variables, $X_1\sim exp(\lambda_1) , X_2\sim exp(\lambda_2)$,
We know that: $P(X_1<X_2)=\frac{\lambda_1}{\lambda_1 + \lambda_2}$.
But again I'm struggling to link it to my question here, if I want the time that the third caller of type B called, I can say that $T^3_B\sim Gamma(3,\lambda_B)$ and so $P(T_A^1 < T_B^3)$ but here I have lost the exponential variable, and I don't think I can use the formula any more.
I think I have to somehow use the second approach for each of the $Q_B^i$, but that doesn't make any sense for me, it's like saying "Time between calls of caller type B is greater than time between calls of caller A", which isn't the case.
The last idea I had about this formula, I was a little confused, if I have to reconfigure the definition of $Q_A^0$ of time between the start and the first call, I could say that $P(Q_B^0 < Q_A^0)$, and also $P(Q_B^1 < Q_A^0)$, and $P(Q_B^2 < Q_A^0)$ and since they're independent, I would multiply them.
But what's bothering me is for example lets say $Q_B^0=5$, $Q_B^1$ could be $4$ too (since $Q$ is the time between calls). which messes up my last attempt, since if $Q_A^0=12$ and $Q_B^0=5, Q_B^1=5, Q_B^2=5$, that satisfies my formula, but it shouldn't since when $Q_B^2$ happens, $Q_A^0$ has already occurred.
Sorry if I made this long, I've tried to include everything I did and every thought of mine, in order to improve my understanding of this subject as I've felt lost in this question.
Any help or feedback is really appreciated, Thanks in advance!
Your first approach almost got you there already. Reusing that notation, the probability you are looking for is $$P(T_A^1 > T_B^3)$$ where the random variables $T_A^1 \sim \text{Gamma}(1,\lambda_A)$ and $T_B^3 \sim \text{Gamma}(3,\lambda_B)$ are independent. According to wikipedia (https://en.wikipedia.org/wiki/Gamma_distribution#Related_distributions) the fraction $T_B^3/T_A^1$ has a generalized beta prime distribution, i.e. $$\frac{T_B^3}{T_A^1} \sim \beta'(3,1,1,\lambda_A/\lambda_B).$$ With this you can calculate $P(T_A^1 > T_B^3)$.