How to find $\mathbb{P}\left(\int_0^1W_tdt>\dfrac{2}{\sqrt{3}}\right)$? How to find $\mathbb{P}\left(\int_0^1W_tdt>\dfrac{2}{\sqrt{3}}\right)$? Where $W_t$ is Browninan Motion.
What I have done is, for $Z\sim N(0,1)$: 
$$\begin{align}
\mathbb{P}\left(\int_0^1W_tdt>\dfrac{2}{\sqrt{3}}\right) &= \mathbb{P}\left(\int_0^1\sqrt{t}Zdt>\dfrac{2}{\sqrt{3}}\right)\\
&= \mathbb{P}\left(Z\dfrac{2}{3}>\dfrac{2}{\sqrt{3}}\right)\\
&= \mathbb{P}\left(Z>\sqrt{3}\right)\\
&= \dfrac{1}{\sqrt{2\pi}}\int_\sqrt{3}^\infty e^{-\frac{z^2}{2}}dz\\
&= -N\left(-\sqrt{3}\right)
\end{align}$$
Any help would be very much appreciated. 
 A: We first show that
$$ \int_0^t W_s \,\mathrm{d}s \sim N\left(0,\frac13 t^3 \right).$$
Indeed, the integral is defined as a limit of Riemann sums; since each of these sums are normally distributed, the limiting integral is also normally distributed (this can be proven using characteristic functions, for instance). Applying Fubini's theorem,
$$ \mathbb{E} \left[\int_0^t W_s \,\mathrm{d}s \right] = \int_0^t \mathbb{E}[W_s] \,\mathrm{d}s = \int_0^t 0 \,\mathrm{d}s = 0,$$
where the interchange of limits is permissible since w.p.1 the function $s \mapsto W_s$ is continuous and hence bounded on a compact interval. Finally,
\begin{align*}
\mathrm{Var}\left(\int_0^t W_s \,\mathrm{d}s\right)
&= \mathbb{E}\left[\left(\int_0^t W_s \,\mathrm{d}s\right)^2\right] \\
&= \mathbb{E}\left[\left(\int_0^t W_s \,\mathrm{d}s\right)\left(\int_0^t W_u \,\mathrm{d}u\right)\right] \\
&= \mathbb{E}\left[\int_0^t \int_0^t W_sW_u \,\mathrm{d}s \mathrm{d}u \right] \\
&= \int_0^t \int_0^t \mathbb{E}[W_sW_u] \,\mathrm{d}s \mathrm{d}u \\
&= \int_0^t \int_0^t \min(s,u) \,\mathrm{d}s \mathrm{d}u \\
&= 2 \int_0^t \int_0^u s \,\mathrm{d}s \mathrm{d}u  \\
&= \int_0^t u^2 \mathrm{d}u  \\
&= \frac{1}{3}t^3,
\end{align*}
where we use the fact that $\mathrm{Cov}(W_s, W_t) = \min(s, t)$ and Fubini again to pass the expectation inside. It follows that
$$ \sqrt{3} \int_0^1 W_s \,\mathrm{d}s \sim N(0,1), $$
and 
$$\mathbb{P}\left(\int_0^1 W_s \,\mathrm{d}s \geq \frac{2}{\sqrt{3}} \right) = 1 - \Phi(2),$$
where $\Phi$ is the standard normal cdf.
A: The law of $(W_t)_{0\leq t\leq 1}$ is not the same as that of $(\sqrt{t}Z)_{0\leq t \leq 1}$. So there is no reason $\int_{0}^{1} W_t \, dt$ and $\int_{0}^{1} \sqrt{t}Z \, dt$ have the same law, and as it turns out they have different laws.
Since $X = \int_{0}^{1} W_t \, dt$ is gaussian, it suffices to determine its mean and variance. And they can be computed as
$$ \mathbb{E}[X]=\int_{0}^{1} \mathbb{E}[W_t] \, dt = 0$$
and
$$ \mathbb{E}[X^2]
= \int_{0}^{1} \int_{0}^{1} \mathbb{E}[W_s W_t] \, dsdt
= \int_{0}^{1} \int_{0}^{1} s \wedge t \, dsdt = \frac{1}{3}. $$
So $X \sim \mathcal{N}(0, \frac{1}{3})$ and hence $X \stackrel{d}{=} \frac{1}{\sqrt{3}}Z$. This gives
$$ \mathbb{P}(X > \tfrac{2}{\sqrt{3}}) = \mathbb{P}(Z > 2). $$
