I'm trying to find an answer to the following question:
A standard Brownian motion crosses the $t$-axis at times $t=2$ and $t=5$. Find the probability that the process exceeds level $x=1$ at time $t=4$.
If I understood correctly the exercise is asking for
$$ \Pr [B_4>1|B_2=0,B_5=0]\tag1 $$
where $B_t$ is an standard Brownian motion. It seems weird to ask for a probability conditioned to an event of probability zero, so I'm not sure if I have the correct interpretation of the exercise. Other than that what I did is the following: because $B_2=0$ then the process given by $\tilde B_{t-2}:=B_t-B_2$, for $t\geqslant 2$, is also an standard Brownian motion, and we have that (1) is equivalent to
$$ \Pr [\tilde B_2>1|\tilde B_3=0]\tag2 $$ as $\tilde B_2$ is independent of $B_2$ (however it seem weird to talk about independence of events of zero probability). Now, its also known that the process defined by $X_t:=t\tilde B_{1/t}$, with $X_0:\equiv 0$, is also an standard Brownian motion, so its easy to check that (2) is equivalent to
$$ \Pr [X_{1/6}>1/2]\approx 0.110336\tag3 $$
My questions: its this solution correct, or its just senseless? There is a different way to interpret the exercise?
UPDATE: it can be shown that the result above coincide with the result if we use a Brownian bridge formulation. It seems that the continuity of the paths of the Brownian motion give the possibility to define, consistently, conditional probabilities over events of zero probability.