Let $x\in (0,h)$ and $(B_t)$ a Brownian motion. I would like to compute $$\mathbb P^x\left\{\sup_{0\leq s\leq t}B_s\geq h, \inf_{0\leq s\leq t}B_s>0\right\}.\tag{*}$$ In otherword, I would like to compute the probability that a Brownian starting at $x\in (0,h)$ reach $h$ before $0$.
Q1) Since $t\mapsto \inf_{0\leq s\leq t}B_s$ is continuous, we should have $$\mathbb P^x\left\{\sup_{0\leq s\leq t}B_s\geq h, \inf_{0\leq s\leq t}B_s>0\right\}=\mathbb P^x\left\{\sup_{0\leq s\leq t}B_s\geq h, \inf_{0\leq s\leq t}B_s\geq 0\right\},$$ no ? But in an other way, this doesn't make sense since the event $\left\{\sup_{0\leq s\leq t}B_s\geq h, \inf_{0\leq s\leq t}B_s\geq 0\right\}$ describe the fact that $(B_s^x)$ (i.e. the BM starting at $x$) can reach $0$ before $h$, and thus, logically, we should rather have $$\mathbb P^x\left\{\sup_{0\leq s\leq t}B_s\geq h, \inf_{0\leq s\leq t}B_s>0\right\}<\mathbb P^x\left\{\sup_{0\leq s\leq t}B_s\geq h, \inf_{0\leq s\leq t}B_s\geq 0\right\},$$ no ? Something looks strange to me here.
Q2) How can I compute $(*)$ ? If $\tau=\inf \{t>0\mid B_t^x\notin (0,h)\}$, then $$(*)=\mathbb P^x\{B_\tau=h, \tau\leq t\},$$ but this doesn't really help, since I don't know the join distribution between $B_\tau^x$ and $\tau$.