The problem is to find:
$P_{x}(\{T_{B_{1}}<\infty\}\cap \{T_{B_{2}}<\infty\})$ where $B_{1},B_{2}$ are the two hemispheres of sphere S shown below. There are two possible paths for first hitting time: a)paths that enter S through one cap and exit via the opposite one b)paths that hit both caps for the first time without crossing the blue disk.
For a)
One idea is to first use the hitting probability for sphere S and then Poissons formula for the probability that BM will escape the S from ,say, the $B_{1}$ cap.
For $x\in B_{2}$, Poissons formula is $P_{x}(T_{S}=T_{B_{1}})=C\int_{B_{1}}\frac{1}{|x-y|^{n-2}}dS_{y}$. So as we see it depends on the entry point of BM into $B_{2}$. Thus, we don't have independence.
But we do have a lower bound, by taking the furthest point x from $B_{1}$ i.e. the pole of $B_{2}$ call it $x_{0}$
So we get the lower bound $P_{x_{0}}(T_{B_{1}}=T_{S})P_{x}(T_{S}<\infty)=C\frac{R^{n-2}}{|x|^{n-2}}\int_{B_{1}}\frac{1}{|x_{0}-y|^{n-2}}dS_{y}$.
For b)
I am not sure
Thanks