Using probability of camera flash interval to get the probability density equation in Griffith's Quantum Mechanics book In Griffith's QM, example 1 chapter 1, what is the intuition behind using the probability of camera flash interval to get the probability density equation in terms of "dx".
Griffith says that evidently 
$$
\rho(x) = \frac{1}{2\sqrt{hx}}
$$
which is not so evident for me, how does the probability of camera flashing get the probability density in terms of "dx"
$$
\frac{dt}{T}=\frac{1}{2\sqrt{hx}}dx
$$
I am having a tough time digesting the logic 
 A: I think that Griffiths' explanation is indeed a bit lacking, but it isn't hard to see what he's doing.
First, let's consider an alternative way to solve the problem. Let $T$ be the amount of time the ball spends in the air. Then, 
$$x(t) = \frac 1 2 g t^2 \rightarrow h = \frac 1 2 g T^2 \rightarrow T = \sqrt{\frac{2h}{g}}$$
Since the probability of a sample being taken in any time between $t$ and  $t + dt$ is constant and since the probability that a given sample occurs sometime between $t = 0$ and $t = T$ must be $1$, we have
$$\rho (t)dt = \frac 1 T dt.$$
Now, to find the average of $x$ over this time interval, we take (Equation 1.9 from Section 1.1 in Introduction to Quantum Mechanics, 2nd ed.)
$$\langle x\rangle=\int_0^T x(t)\rho(t)dt = \int_0^T \frac{1}{2}gt^2\frac 1 T dt = \frac{gT^2}{6} = \frac h 3.$$
Griffiths, instead of doing this, chose to make a change of the integration variable from $t$ to $x$. Again, 
$$x = \frac 1 2 g t^2 \rightarrow dx = g t dt \rightarrow \frac {dx} {gtT} = \frac{dt}{T}.$$
Substitiuing in $t$ and $T$ gives, 
$$\frac{dt}{T} = \frac {dx} {g} \sqrt{\frac{g}{2x}\frac{g}{2h}} = \frac{dx}{2\sqrt{hx}} \rightarrow \rho(x) = \frac{1}{2\sqrt{hx}},$$
as needed. This gives the same answer. I don't know why Griffiths wanted to do it like this, though. Perhaps he wanted to keep things consistent with working in $x$.
