Uniform RV $X$ has $E[X] = 10$ and $P[X\geq 12] = 1/4$, calculate $P[X<9]$ 
$X$ is a uniform random variable. It has the expected value $E[X] = 10$ and the probability $P[X\geq 12] = \frac{1}{4}$. What is the probability $P[X\leq 9]$?

Here is how I attempted it: $P[X\geq 10] = \frac{1}{2}$ because it is uniform. $P[X\geq 12]$ is $2$ away from $P[X\geq 10]$. 
$$(1/4)/2 = 1/8.$$ 
So it is $\frac{1}{8}$ between each integer. $P[X\leq 9]$ is $1$ integer away from $P[X\geq 10]$, so 
$$P[X\leq 9] = \frac{1}{2} - \frac{1}{8} = \frac{3}{8}.$$
I know this is a messed up way to solve it and I don't even know if it's right. There must be a more coherent way?
 A: To be serious, I guess you could assume $X$ is uniform continuously in the interval $[a,b]$ with a PDF:
$$f_X(x) = \cases{\dfrac1{b-a}&for a < x < b\\0&otherwise}$$
The expected value $E[X] = 10$, which gives one linear equation:
$$\frac{a+b}{2} = 10$$
From $P[X\ge 12]$,
$$\begin{align*}
\frac14 &= P[X\ge 12]\\
&= \int _{12}^b \frac{dx}{b-a}\\
&= \left[\frac{x}{b-a}\right]_{12}^b\\
&= \frac{b-12}{b-a}\\
48 &= a + 3b
\end{align*}$$
Solve for $a$, compare it with the $9$ in $P[X<9]$, and since $a = 6 < 9$,
$$\begin{align*}
P[X<9] &= \int_a^9 \frac{dx}{b-a}\\
&= \left[\frac x{b-a}\right]_a^9\\
&= \frac{3}{8}
\end{align*}$$
A: It looks good to me. You could spin this several ways, but I'll just branch off your approach.
Assume $X\sim \operatorname{unif}[a,b]$. Well, it has constant density, call it $h$. As you noted, we know that
$$P(X> 10) = h(b-10) = \frac{1}{2}$$
(the height times the length is just the "area under the curve").
We were also given that 
$$P(X> 12) = h(b-12) = \frac{1}{4}.$$
Solving this system gives $h = \frac{1}{8}$ and $b = 14$. This implies $a = 6$ since $h(b-a) = 1.$
Now, 
$$P(X\leq 9) = \frac{1}{8}(9-6) = \frac{3}{8}.$$
Alternatively, since we know $P(X>12) = \frac{1}{4}$, then $P(X\leq 12) = \frac{3}{4}$, and by ratios
$$\frac{h(12-a)}{h(10-a)} = \frac{3/4}{1/2}.$$
This immediately gives $a = 6$, and $P(X\leq 9)$ is given as above.
A: $X\sim U(a,b)$. Note that:
$$\mathbb E(X)=\mu =\frac{a+b}2=10 \ \text{implies} \ \ P(10\le X\le b)=\frac12\\
 \ P(X\ge 12)=\frac14 \ \ \text{implies} \ \ P(12\le X\le b)=\frac14$$
Hence, $b=14$ (because $b-10=2(b-12)$) and $a=6$ (because $a+b=20$). 
Thus:
$$P(X\le 9)=\frac{9-6}{14-6}=\frac38.$$
