The problem is...
Let $B_t$ be a Brownian motion starting at point $0$, let
$T = \inf \{ t > 1; B_t = 0$ or $1 \} $
be the hitting time of $\{ 0, 1 \}$ by $B_t$ after time $1$, and let $A = [B_1 \leq 1, B_T = 1]$ be the event that at time $1$, the Brownian motion is below point $1$ and then it will reach point $1$ before reaching point $0$. Find $P_0(A)$.
At first glance, I thought this was an application of the following theorem...
"For a standard Brownian Motion, the probability that $a$ is first hit before $−b$ is given by
$p_a = \frac{b}{a + b}$ for $a > 0, b > 0$"
But this theorem is for points $a,b > 0$. Is the solution to the problem due to another version of this theorem? Or is it something completely different I should be looking at?
Take #1. Notice that $T\ge 1$. If someone came and told you $B_1=x$, then you could use the mentioned theorem from that point, right?
Additionally, $B_T - B_1$ is independent from $B_1 - B_t$ for each $t<1$.
Take #2. You may write
\begin{aligned} \mathbb P_0(A) &= \mathbb P(B_T = 1, B_1\le 1) \\ &= \int_{-\infty}^1\mathbb P(B_T = 1, B_1 = x)dx \\ &= \int_{-\infty}^1\mathbb P(B_T = 1 \mid\ B_1 = x)\mathbb P(B_1 = x)dx. \end{aligned}
Given the continuity of the Brownian trajectories and the mentioned theorem, you have $$\mathbb P(B_T=1 \mid B_1 = x) = \begin{cases} 0 & x < 0,\\ x & 0 < x < 1. \end{cases}$$
Hence, you may compute
$$\mathbb P_0(A) = \int_0^1 x\mathbb P(B_1=x)dx = \int_0^1\frac{x}{\sqrt{2\pi}}e^{-x^2/2}dx = \frac{1}{\sqrt{2\pi}}\left(1 - \frac{1}{\sqrt e}\right).$$