Normal Bivariate distribution and conditional expectation I've been dealing with this proability problem, any advice would be really appreciated:

Let $(X,Y)$ be a normal bivariate random vector, such that: $\mathbb{E}[X]=2$, $Var[X]=4$, and  $\mathbb{E}[Y|X=x]=2+x$ and $Var[Y|X=x]=5$. Calculate a) $\mathbb{P}(Y>5)$ and b) $\mathbb{E}[Y^2|X=7]$

My attempt:
I have no clue on the a) part, since I can't know how Y distributes, because I don't know if X,Y are independent. 
For part b) I used the following, relative to conditional variance:
$Var[Y|X=x]=\mathbb{E}[Y^2|X=x]-(\mathbb{E}[Y|X=x])^2$
So I only computed when $X=7$ and obtained:
$Var[Y|X=7]=\mathbb{E}[Y^2|X=7]-(\mathbb{E}[Y|X=7])^2$
$5=\mathbb{E}[Y^2|X=7]-(2+7)^2=\mathbb{E}[Y^2|X=7]-81$
So:
$\mathbb{E}[Y^2|X=7]=86$. But I'm not sure if I did something wrong. 
Thanks in advance.
 A: Step 1: Conditionally on $X$
First work conditionally on $X$. Here you use the property of a normal distribution, that if $(X,Y)$ are jointly normal, then $Y \mid X=x$ is also normal. Hence, since you know the conditional mean and variance you know that:
$$ Y \mid X=x \sim \mathcal{N}(2+x, 5)$$
Based on the above it is easy to calculate $ \mathbb P[ Y > 5 \mid X=x]$.
Step 2: Marginalize over $X$
Now notice that $\mathbb P[Y > 5 \mid X]$ is a function of $X$. Again by properties of multivariate normal you know that $X$ is normal and since you know its mean and variance, you know the distribution of $X$. Now just integrate over $X$, i.e.
$$ \mathbb P[Y > 5] = \mathbb E[ \mathbb P[Y >5 \mid X]] $$
I'll leave the calculations to you.
Alternative way:
Again for a normal, there is a very explicit expression for the conditional distribution, see e.g. the wiki article. From this and your information, you can quickly figure out what the variance and mean of $Y$ are, and since you know $Y$ must be normal too, you know its distribution.
