Let $(B_t)$ a Brownian motion. I denote $\tau_a=\inf\{t\geq 0\mid B_t=a\}$, $T_{a,b}=\tau a\wedge \tau b$ and $p_x\{A\}=p\{A\mid B_0=x\}$.
Why $$p_x\{B_{T_{a,b}}=b\}=p_x\{\tau_a<\tau_b\}\ \ \ ?$$
To me, $$p_x\{B_{T_{a,b}}=b\}=p_x\{T_{a,b}=\tau_b\}=p_x\{\tau_b<\tau_a\},$$ so what's wrong in my argument ?