U~$(0,i)$ where $i$ is a fair dice roll. 
If I throw a fair dice and the result is $i$ then I choose a point $X$~$(0,i)$
1st) What is the expected value and standard deviation of $X$?
2nd) if $X > 3$ then what is the probability that the result of the dice was $6$?

I think I got the 1st part correct but the second part I am confused on what method to use.
My attempt:
1st)
expected value of a dice roll = $\frac{1}{6} *(1+2+3+4+5+6) = 3.5 $
so, $i = 3.5$ and $X$~Uni$(0,3.5)$
By definition of the Uniform distribution :
$$E(X)=\frac{3.5}{2}=1.75  $$
$$D(X)= \frac{3.5}{\sqrt{12}}   $$
2nd)
This is the part am not sure of:
if $X$~Uni$(0,3.5)$ and $P(X>3)= 1-P(X \leq3) = 1- \frac{3}{3.5}= \frac{1}{7}$
we want $$P(i=6|X>3) = \frac{P(i=6,X>3)}{P(X>3)} $$but idk how to get the  numerator here, also I may be completely wrong about both parts so please point that out if it happens to be the case!
ALso, this  might be right:
$$ P(i=6,X>3) =  P(X>3|i=6)* P(i = 6)  =  \frac{3}{6}* \frac{1}{6}$$
This would imply $$ P(i=6|X>3) = \frac{1/6 * 1/2}{1/7} = \frac{7}{12}$$
Any help is appreciated!
 A: You're confusing the expected value of a dice roll with the expected value of the random variable. For example, if you throw the dice and the result is $1$, then $X \sim U[0, 1]$ and $E(X) = 1/2$. However, for a general $i$, then, as already stated in the problem, $X \sim U[0, i] \Rightarrow E(X) = i/2$.
Alternatively, you could suppose $i = 1$, then $i = 2$, then $i = 3$, ..., and repeat the same calculations until you exaust all $6$ possibilities.
A: a)
Note that, for any $\sigma$-field $\mathcal G$, we have $E[X] = E[E[X|\mathcal G]]$
Let us consider $\mathcal G$ = $\sigma(Y)$, where Y is rv. that characterises the number we've rolled. Due to Y being discrete, we only need to find $E[X|Y=k]$ to know $E[X|Y]$.
So take $k\in\{1,2,3,4,5,6\}$
$E[X|Y=k]$ = $\frac{E[X\cdot \chi _{\{Y=k\}}]}{P(Y=k)}$ = $\frac{E[X_k*\chi _{\{Y=k\}}]}{\frac{1}{6}}$ = $E[X_k]$ = $\frac{k}{2}$, where $\chi$ is the characteristic function, $X_k$ is the uniform rv on $[0,k]$ (third equality is due to independence), so
$E[E[X|Y]]$ = $E[\frac{Y}{2}]$ = $\frac{7}{4}$
To find $D(X)$ note that, $D(X) = \sqrt{ E[X^2] - (E[X])^2 } $, and that we already have $E[X]$, to find $E[X^2]$, we proceed similarly, considering $E[X^2|Y=k]$ we find it is equal to $E[X_k^2]$ = $\frac{1}{k} \cdot \frac{k^3}{3}$ = $\frac{k^2}{3}$, so again
$E[X^2] = E[E[X^2|Y]] = E[\frac{Y^2}{3}] = \frac{1^2+2^2+3^2+4^2+5^2+6^2}{18} = \frac{91}{18}$, so plugging that: $ D(X) = \sqrt{ \frac{91}{18} - \frac{49}{16} } 
$=$ \sqrt{\frac{91\cdot 8 - 49 \cdot 9}{16\cdot9}} = \frac{\sqrt{287}}{12}$
b)
As you noted: $P(Y = 6 | X>3) = \frac{P(X>3)|Y=6)P(Y=6)}{P(X>3)}$
$P(X>3|Y=6) = P(X_6 >3) = \frac{1}{2}$
$P(Y=6) = \frac{1}{6}$
The problematic one is $P(X>3) = \sum_{k=1}^6 P(X>3|Y=k)P(Y=k) = \frac{1}{6}\sum_{k=4}^6 P(X_k>3) = \frac{1}{6}\sum_{k=4}^6\frac{k-3}{k}$ =$ \frac{1}{6}(\frac{1}{4} + \frac{2}{5} + \frac{1}{2}) = \frac{1}{6}(\frac{5+4+10}{20}) = \frac{19}{120}$
So, $P(X>3) = \frac{\frac{1}{6}\cdot \frac{1}{2}}{\frac{19}{120}} = \frac{10}{19}$
