The following question is taken from Mark Joshi's Quant Job Inteview.
Question: What is the probability that a Brownian motion will cross level $1$ before time $t$?
(A follow up question): What is the probability that a Brownian motion will eventually cross the level $1$ (that is, the limit as $t\to\infty)?$
Let $$M(t) = \max_{0\leq s\leq t}W(t)$$ where $(W(t))_{t\geq 0}$ is Brownian motion.
Note that \begin{align*} \mathbb{P}(M(t) \geq 1) & = \mathbb{P}(M(t) \geq 1, W(t) \geq 1) + \mathbb{P}(M_t \geq 1, W(t) \leq 1) \\ & = \mathbb{P}(M(t) \geq 1, W(t) \geq 1). \end{align*}
If the first passage time of Brownian motion at level $1$ is denoted by $$\tau_1 = \inf\{t\geq 0: W(t) = 1\},$$ then $$M(t) \geq 1 \quad \Leftrightarrow \tau_1 \leq t.$$ So, \begin{align*} \mathbb{P}(M(t) \geq 1, W(t) \geq 1) & = \mathbb{P}(\tau_1 \leq t, W(t) \geq 1) \\ & = \mathbb{P}(W(t) \geq 1). \end{align*} So, I got $$\mathbb{P}(W(t) \geq 1).$$ However, answer given in the book is $$2\mathbb{P}(W(t) \geq 1).$$
I do not see where is my mistake in my calculations above.
As pointed out by @NateEldredge in comment below, the probability $$\mathbb{P}(M(t) \geq 1, W(t) \leq 1) \neq 0.$$ In fact, $$\mathbb{P}(M(t) \geq 1, W(t) \leq 1) = \mathbb{P}(M(t) \geq 1, W(t) \geq 1).$$ This justifies the factor $2$ of the final answer.