Probability with covariance matrix

Question

Let $X_1$, $X_2$ be normal with covariance matrix:

$$\Sigma = \begin{pmatrix}6&-3\\ -3&7\end{pmatrix}$$ Find the probability that: $P(X_1 - X_2 \ge 1)$

MY ATTEMPT:

Let: $Y = X_1 - X_2$ $$\begin{aligned} Var(Y) = Var(X_1) + Var(X_2)-2Cov(X_1 - X_2) & = 6 + 7 - 2\cdot0 \\& = \color{red}{13} \end{aligned}$$ So: $$ \begin{aligned} P(X_1 - X_2 \ge 1) & = P(Y \ge 1) \\ & = 1 - P(Y \lt 1) \\ & = 1 - \Phi(\frac{1}{\sqrt{13}}) \\ & \approx 1 - \Phi(0.2773) \\& \approx \color{red}{0.3938} \end{aligned} $$ I don't know if the result is right, i can't check it, but i don't think that this is the method to solve this.

Answer 1

$$Y = X_1-X_2$$

$$Var(Y) = Var(X_1) + Var(X_2) -2Cov(X_1,X_2) = 6+7-2(-3) = 19$$

$$P(Y \geq 1) = 1-P(Y < 1) = 1-\Phi(\frac{1}{\sqrt{19}})$$

Remember that,

$Y = X_1 + X_2 + \ldots + X_n$

$$Var(Y) = \sum_{i=1}^{n}\sum_{j=1}^{n}Cov(X_i,X_j)$$

Using $Cov(X_i,X_i) = Var(X_i)$ and $Cov(X_i, X_j) = Cov(X_j, X_i)$,

$$Var(Y) = \sum_{i=1}^{n}Var(X_i) + 2\sum_{i=1}^{n}\sum_{j=i+1}^{n}Cov(X_i,X_j)$$