Calculate this probability: $$ P(B_1 <x, B_2 < y) , $$ where $B_t$ is Brownian motion.
If $B_1 $ and $B_2$ were independent, it is easy, because this probability would be product of two probabilities, but in this case $B_1$ is not independent with $B_2$ and I don't know what to do.
It seem to me I found the answer. As I said in the comment, we use the independence, and we will have. $$P(B_{2}<y, B_{1}=t) = P(B_{1}=t)P(B_{2}-B_{1}<y-t|B_{1}=t)=P(B_{1}=t)P(B_{2}-B_{1}<y-t)$$
It remains to integrate up to $x$.