Probability of two times of Brownian Motion greater than 0 
For a Brownian Motion starting at $0$, I want the probability $X(2T)>0$ and $X(T)>0$. 

One idea I had was that 
\begin{align*} \mathbb{P}(X_{2T}>0,X_{T}>0) &=
\mathbb{P}(X_{2T}>0,X_{T}>0 \mid X_{2T}>X_{T})\mathbb{P}(X_{2T}>X_{T}) \\ &\quad + 
\mathbb{P}(X_{2T}>0,X_{T}>0\mid X_{T}<X_{2T})\mathbb{P}(X_{2T}<X_{T}) \end{align*}
This is $$\mathbb{P}(X_{T} > 0)\mathbb{P}(N(0,t) > 0) + \mathbb{P}(X_{2T} > 0)\mathbb{P}(N(0,t) < 0) = \frac12.$$ 
I was also thinking that every permutation of $0, X_T, X_{2T}$ are all equally likely due to symmety around $0$, and the permutations are $0, X_T, X_{2T}$ and $0, X_{2T}, X_{T}$, so it's $\frac{2}{3!} = \frac13$. 
I'm not sure if either are correct, if any. 

This is just a complement for the answer above.
$$
\mathbb{P}\left(X_{T}>0, X_{2 T}>0\right)=\frac{1}{2}-\int_{0}^{\infty} \Phi\left(-\frac{x}{\sqrt{T}}\right) \phi(x) d x
$$
The equality above in the first answer has a small mistake. The term $\phi(x)$ is the standard normal density. However, $\int_{0}^{\infty} \Phi\left(-\frac{x}{\sqrt{T}}\right) \phi(x) d x$ is derived from $\mathbb{E}\left(1_{\left\{X_{T>0\}}\right.} \Phi\left(-\frac{X_{T}}{\sqrt{T}}\right)\right)$. We can notice that $X_T$ is a normal r.v. but not standard. Therefore the density of $X_T$ should be $$\tilde\phi(x) = \dfrac{1}{\sqrt{2\pi T}}e^{-\dfrac{x^2}{2T}}$$
The integral of above equality should be $\int_{0}^{\infty} \Phi\left(-\frac{x}{\sqrt{T}}\right) \tilde\phi(x) d x$.
If we let $m = \dfrac{x}{\sqrt{T}}$, we have $\sqrt{T}dm = dx$ and
$$\sqrt{T}\tilde\phi(m\sqrt{T})  dm = \phi(m)dm$$
Now we substitute $x = \sqrt{T}m$ into $\int_{0}^{\infty} \Phi\left(-\frac{x}{\sqrt{T}}\right) \tilde\phi(x) d x$, we have
$$\begin{aligned}\int_{0}^{\infty} \Phi\left(-\frac{x}{\sqrt{T}}\right) \tilde\phi(x) d x &= 
\int_{0}^{\infty} \Phi\left(-m\right) \tilde\phi(m\sqrt{T}) d (\sqrt{T}m)\\
& =\int_{0}^{\infty} \Phi\left(-m\right) \phi(m) d m \\
&=\frac{1}{4}+\frac{1}{2 \pi} \arctan (-1)  = 0.125
\end{aligned}
$$
Hence, the answer should be $0.5-0.125 = 0.375$.
