Find the probability that $Z_1$ is less than $Z_2$

Question

Let $Y=(Y_1,Y_2,Y_3)'\sim N_3(\mu,\sum)$ where $\mu=(1,-1,3)'$ and $$\sum= \begin{pmatrix} 1 & 1 & 0\\ 1 & 2 & 3\\ 0 & 3 & 10\\ \end{pmatrix} \quad $$

and define $Z=(Z_1,Z_2)'$, where $Z_1=Y_1+Y_2+Y_3$, $Z_2=3Y_1+Y_2-2Y_3$.

Find the probability that $Z_1$ is less than $Z_2$

What I have so far: We have to find $P(Z_1<Z_2)$

$Z$ has a normal distribution with mean $\begin{pmatrix} 3\\-4\\\end{pmatrix} \quad $ and variance $\begin{pmatrix} 21 & -14\\-14 & 45\\\end{pmatrix}. \quad $ Any comments are appreciated.

Answer 1

$$\mathbb P(Z_1<Z_2)=\mathbb P(Z_1-Z_2<0)$$ where $$Z_1-Z_2=Y_1+Y_2+Y_3-(3Y_1+Y_2-2Y_3)=-2Y_1+3Y_3\sim N(7,\,94)$$ since $$\mathbb E[-2Y_1+3Y_3]=-2\cdot 1+3\cdot 3 = 7$$ and $$\text{Var}[-2Y_1+3Y_3]=(-2)^2\text{Var}Y_1+(3^2)\text{Var}Y_3+2\text{cov}(-2Y_1,3Y_3) = 4\cdot 1+9\cdot 10-12\cdot 0=94.$$ Then $$\mathbb P(Z_1-Z_2<0)=\Phi\left(\frac{0-4}{\sqrt{94}}\right)\approx 0.34.$$