I was wondering how to derive the probability that a Brownian motion with drift lies within two thresholds, then $\mathbb{P}(b<X_{\tau}<a, \tau\in[0,t])$, where $X_t=\mu t+\sigma W_t$. It is known that $\mathbb{P}(X_{\tau}>a, \tau\in[0,t])=\mathbb{P}(T_a<t)$ follows an inverse Gaussian distribution, (where $T_a=\inf\{ t\geq 0|X_t>a \}$ is the stopping time), and it is immediate to see that $\mathbb{P}(X_{\tau}>a, \tau\in[0,t])$ and $\mathbb{P}(X_{\tau}<b, \tau\in[0,t])$ are not independent, so that no shortcut can be applied to solve the problem. Thus, I need help on how to solve my problem and derive the probability measure $\mathbb{P}(b<X_{\tau}<a, \tau\in[0,t])$.
Many thanks to all.