Suppose that $B_{t}$ is a standard Brownian Motion. What is the probability that both $B_{3}$ and $B_{6}$ take positive values?
This is what I've tried but then I get stuck and I'm not sure how to proceed next:
\begin{align*} P(B_{3} > 0, B_{6})&= P(B_{3} - B_{2} > B_{2}, B_{6}-B_{5} > -B_{5})\\ &=\int_{-b_{5}}^{\infty}\int_{-b_{2}}^{\infty} \phi(b_{3})\phi(b_{6}) db_{3}db_{6}\\ &=\int_{-b_{5}}^{\infty} [\Phi(b_{3})]_{-b_{2}}^{\infty} \phi(b_{6}) db_{6}\\ &=\int_{-b_{5}}^{\infty} \Phi(b_{2})\phi(b_{6}) db_{6}\\ \end{align*}
This is where I get stuck. I'm not sure how to integrate it at this point. Is there an alternative approach to this problem?