What is the probability that a $B$ Brownian motion reaches the $y=10$ level on time interval $\left[0,100\right]$ and $B_{100}<0$?
My idea is the following...
If we denote the time of reaching the given level as $\tau$, then it is known that $\mathbf{P}\left(\tau\leq t\right)=\mathbf{P}\left(\frac{y^{2}}{Z^{2}}\leq t\right)$, where $Z$ is a standard normal distributed random variable. Because the definition of $\tau$, $B_{\tau}=10$ and $$\left\{ \tau\leq 100\right\} \cap\left\{ B_{\tau}-B_{100}>10\right\} \Longleftrightarrow\left\{ \tau\leq 100\right\} \cap\left\{ 10-B_{100}>10\right\}$$
So in this case, the question is $$\mathbf{P}\left(\left\{ \tau\leq 100\right\} \cap\left\{ B_{\tau}-B_{100}>10\right\} \right) =\mathbf{P}\left(\left\{ \tau\leq 100\right\} \cap\left\{ 10-B_{100}>10\right\} \right)=...$$
Because $B_{\tau}-B_{100}$ is independent from $B_{\tau}$, I think we can continue as $$\ldots\mathbf{P}\left(\tau\leq 100\right)\cdot\mathbf{P}\left(10-B_{100}>10\right)={\mathbf{P}\left(\frac{10^{2}}{Z^{2}}<100\right)\cdot\mathbf{P}\left(B_{100}<0\right)=\mathbf{P}\left(1<Z^{2}\right)\cdot\frac{1}{2}=}$$
$${\mathbf{P}\left(\left\{ Z<-1\right\} \cup\left\{ Z>1\right\} \right)}\cdot\frac{1}{2}=2\left(1-\varPhi\left(1\right)\right)\cdot\frac{1}{2}=1-\varPhi\left(1\right).$$
Here $\varPhi$ denotes the cumulative distribution function of the standard normal distribution. Is it a good reasoning? Do I miss something? I'm not sure that we can factor the probability into product in the middle, because I'm not sure if $\left\{ \tau\leq t\right\}$ and $\left\{ B_{\tau}-B_{100}>10\right\}$ are independent.