Say we have a sample of DNA from population A and population B as shown below. (Both samples taken from today).
Find the following
$$ \mathbb{P}(M>0 | T_2,...,Tn) $$ where $T_i$ is the length of branch $i$, and $M$ is the total number of mutations happening on the branch with a solid line in the above picture. Hence, find $ \mathbb{P}(M>0) $.
I know that:
$T_i$ are independent exponential random variables with parameter $i \choose{2}$
$l$ is the total tree length: $l = 2T_2 + 3T_3 + \ldots nT_n$.
The mutations occur as a Poisson process along the edges of the tree at rate $\frac{\theta}{2}$ (meaning in total it is of rate $\frac{\theta l}{2}$)
My guess: $$ \mathbb{P}(M>0 | T_2,...,Tn) = 1- \mathbb{P}(M=0 | T_2,...,Tn) = 1-e^{-\theta l/2} \sim exp\left(\frac{\theta l }{2}\right)$$ Is this right? If so how would we find $P(M>0)$?