Conditional expectation based on uniform RV 
A motorist just had an accident. The accident is minor
  with probability 0.75 and is otherwise major. Let b be a positive
  constant. If the accident is minor, then the loss amount follows a
  uniform distribution on the interval [0, b]. If the accident is major,
  then the loss amount follows a uniform distribution on the interval
  [b, 3b]. The median loss amount due to this accident is 672. Calculate
  the mean loss amount due to this accident.

Correct Answer: 882
My atttempt: 
Let X = event that accident is minor, P(X) = .75, Let Y = event that accident is major, P(Y) = .25
Let L = loss. Then we are given conditional distributions $f_{L|X}$~ Unif(0, b) and $f_{L|Y}$ ~  Unif(b, 3b). We are also given that median loss is 672, or $F_L(672) = 0.5$ . We want to find $E[L]$. 
By law of total expectation/double expectation, $E[L]= E[L|X]P(X) + E[L|Y]P(Y) = (\frac{b}{2})(0.75) + (2b)(0.25) = \frac{7b}{8}. $
If we find b, we are done. We can find using the information about median, in terms of the pdf of L, the median info can be written as
$(0.75)\int_0^{672}\frac{1}{b}dl  + (0.25)\int_0^{672}\frac{1}{2b}dl = 0.5$, We see that b > 672 from this equation, otherwise the first term is simply 0.75 > 0.5 (because the left hand integral would be over the entire range of Unif(0, b)). Thus calculating this integral under this assumption, we get that b = 1176. Then $E[L] = \frac{7*1176}{8} = 1029$
 A: Your equation for the median is wrong here, this is why you don't get the correct answer.

You know that the mean is $\frac{7}{8}b$. Let's try to find $b$ from the median, which will give us the expectation.
We know that $0.75$ of the probability lies in $[0,b]$, i.e., $\mathbb{P}\{ \textrm{loss} \leq b \} = 0.75$. Therefore, by definition of the median $m$ (as the value $m$ s.t. $\mathbb{P}\{ L \leq m \} \geq \frac{1}{2}$ and $\mathbb{P}\{ L \geq m \} \geq \frac{1}{2}$), we have that $m \leq b$ (if you don't see why, draw a picture about where $m$ could be among the two intervals $[0,b]$ and $[b, 3b]$ and see how much probability that gives on each side of $m$). Which is great: everything being continuous (uniform distribution on $[0,b]$), we know that $m$ is the value such that
$$\mathbb{P}\{ L \leq m \} = \frac{1}{2}
$$
and since the 0.75 probability mass is uniform on $[0,b]$, this implies $m = b\cdot  \frac{1/2}{0.75} = \frac{2}{3}b$. Since $m=672$, we get $\boxed{b = 1008}$, and therefore $\mathbb{E}[L] = \frac{7}{8}\cdot 1008 = \boxed{882}$.
