Calculating Probability based on previous Probability values So I am practising come exam questions on Normal distributions. I have this question which I can't quite get right.
$X,Y$ are normally distributed. 
We have $\overline{X} = 100, V(X)=25, \overline{Y} = 120, V(Y)=20$ where $V(X)$ is the variance of $X$.
Calculate:


*

*$p(X>92)$

*$p(Y>X)$

*$p(2X+Y < 300)$


Bits 2. and 3. are the ones that are new to me and I can't do. Any help on those is appreciated. Thank you very much.
 A: If you can handle (i), then with a bit of additional knowledge you can handle the others.
Important fact: Let $X$ and $Y$ be independent normal, with means $\mu$ and $\nu$, and variances $\sigma^2$ and $\tau^2$. Let $a$ and $b$ be constants. Then the random variable $aX+bY$ has normal distribution, mean $a\mu+b\nu$, ans variance $a^2\sigma^2+b^2\tau^2$.
For the second problem, let $W=Y-X=+(-1)X+(1)Y$.  By the "important fact,"  $W$ is normal, mean $20$, variance $45$. We want $\Pr(W\gt 0)$. You know how to calculate this sort of thing.
The third problem will feel a little simpler, since there are no minus signs.
Added: For the second problem, we are asked for the probability that $Y\gt X$. It is convenient to rewrite this inequality as $Y-X\gt 0$, or, to get things ready for the "important fact," as $(-1)X+(1)Y\gt 0$.  Let $W=(-1)X+(1)Y$. Then $W$ has the shape described above, with $a=-1$ and $b=1$. Thus $W$ is normal with mean $(-1)(100)+(1)(120)=20$. The variance of $W$ is equal to $(-1)^2(25)+(1)^2(20)=45$. Now we want to find the probability that $W\gt 0$. This is the probability that the standard normal is greater than $\frac{0-20}{\sqrt{45}}$.
For the third problem, let $W$ be the random variable $2X+Y$. Then $W$ is normally distributed, mean $(2)(100)+(1)(120)$ and variance $(2)^2(25)+(1)^2(20)=120$. So we want to find the probability that a normal with mean $320$ and variance $120$ is $\lt 300$. That's a calculation you know how to do.
Remark: The problem did not specify independence, but undoubtedly expects you to assume it. Without some assumption about the joint distributions of $X$ and $Y$, we simply cannot solve (ii) or (iii).
