Define $Y_1 = X_1 + 2X_2$ and $Y_2 = X_2 + 3X_3$. Determine $P(Y_2> 4 | Y_1 = 2)$ Let $X_1, X_2$ and $X_3$ be independent random variables, such that $X_1$ and $X_3$ are distributed $N (1, 1)$ and $X_2$
$N (0, 1)$ is distributed. Define $Y_1 = X_1 + 2X_2$ and $Y_2 = X_2 + 3X_3$. Determine $P(Y_2> 4 | Y_1 = 2)$. Express the
result in terms of $\Phi$, the cumulative probability function of the standard normal distribution.
My try:
I know that if $U_i ∼ N (µ_i, {σ^2}_i), i = 1, 2$, independent random varaibles. Then  $U_1 + U_2 ∼ N (µ_1 + µ_2, {σ^2}_1 + {σ^2}_2).$
Then
$$Y_1 = X_1 + 2X_2 = X_1 + X_2 + X_2∼N(1,3) $$
$$Y_2 = X_2 + 3X_3 = X_2 + X_3 + X_3 ∼N(3,4)$$
$P(Y_2> 4 | Y_1 = 2) = 1 - P(Y_2 \leq 4 | Y_1 = 2) = 1 - \frac{P(Y_2 \leq 4 , Y_1 = 2)}{P(Y_1=2)}$
I don´t know If I´m on the right track. Any suggestions would be great!
 A: Note that $Y_1$ and $Y_2$ are jointly Gaussian, so we can use the following fact of jointly Gaussian variables $Y_1 \sim \mathcal{N}(\mu_1, \sigma^2_1)$ and $Y_2 \sim \mathcal{N}(\mu_2, \sigma^2_2)$:
$$Y_2 ~|~ (Y_1 = y) \sim \mathcal{N}\left(\mu_2 + \rho \cdot\frac{\sigma_2}{\sigma_1} \cdot (y - \mu_1), (1 - \rho^2) \cdot\sigma^2_2\right)$$
where $\rho$ is the correlation coefficient between $Y_1$ and $Y_2$:
$$\rho = \frac{\text{Cov}(Y_1, Y_2)}{\sigma_1 \sigma_2}$$
In this case, $Y_1 \sim \mathcal{N}(1, 5)$ and $Y_2 \sim \mathcal{N}(3, 10)$ (note this is not exactly what you've written; when multiplying a normal distribution by a constant $c$ the variance is scaled by $c^2$, not $c$), so
$$\text{Cov}(Y_1, Y_2) = \text{Cov}(X_1, X_2) + \text{Cov}(X_1, 3 X_3) + \text{Cov}(2 X_2, X_2) + \text{Cov}(2 X_2, X_3)$$
and all terms vanish except for the third, which evaluates to $2 \cdot \text{Cov}(X_2, X_2) = 2 \cdot \text{Var}(X_2) = 2$. Thus, $\rho = 2 / \sqrt{50}$. Plugging in our formula, the conditional distribution is then given by
$$Y_2 ~|~ (Y_1 = 2) \sim \mathcal{N}\left(3 + \frac{2}{\sqrt{50}} \cdot \frac{\sqrt{17}}{\sqrt{5}}, \frac{23}{25} \cdot 10 \right) = \mathcal{N}\left(\frac{17}{5}, \frac{46}{5} \right)$$
So it follows that
$$\mathbb{P}(Y_2 > 4 ~|~ Y_1 = 2) = 1 - \Phi\left(\frac{3/5}{\sqrt{46 / 5}}\right) = \boxed{1 - \Phi\left(\frac{3}{\sqrt{230}}\right)}$$
which matches up with the WA answer in the comments! Thanks to Maxim for pointing out the error.
