Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
We know that the classical probability density of finding the ball at a height $h$ is given by
$$
P_{\text{class}}(y) = \frac{2}{T v_{(y)}}\tag 1
$$
Where $ T $ is the period of oscillation which is the time it takes for the ball to bounce from the ground to $ h $ and back, and $ v_{(y)} $ is the velocity of the ball as a function of height $ y $.
Applying the energy conservation on the ball,
$$K_i+U_i=K_f+U_f$$
The speed of the ball at the highest point is zero, so $K_i=0$.
$$
0+\color{red}{\bf\not} mgh = \frac{1}{2} \color{red}{\bf\not} m v^2 + \color{red}{\bf\not} mg y
$$
Solving for $ v $;
$$
v_{(y)} = \sqrt{2g(h - y)}\tag 2
$$
We can find the time for the ball to reach the ground from the height $ h $ by using the kinematic formula of
$$\Delta y=v_{iy}t+\frac{1}{2}a_yt^2$$
where $v_{iy}=0$, so
$$ \Delta y = h = \frac{1}{2} g t^2 $$
Solving for $ t $:
$$
t = \sqrt{\frac{2h}{g}}
$$
The total period $ T $ is twice this time (up and down), thus
$$
T = 2 \;\;\sqrt{\frac{2h}{g}}\tag 3
$$
Substitute $ T $ from (3), and $ v_{(y)} $ from (2) into (1)
$$
P_{\text{class}}(y) = \frac{2}{2 \sqrt{\frac{2h}{g}} \sqrt{2g(h - y)}}$$
$$ P_{\text{class}}(y) = \frac{1}{ \sqrt{\dfrac{2h}{\color{red}{\bf\not} g}\cdot 2\color{red}{\bf\not} g(h - y)}}
$$
$$
\boxed{P_{\text{class}}(y) = \frac{1}{2 \sqrt{h^2 - hy}}}
$$
This is the classical probability density of finding the ball at a height $ y $.
$$\color{blue}{\bf [b]}$$
See the graph below.
$$\color{blue}{\bf [c]}$$
The graph of $ P_{\text{class}}(y) $ shows how the probability density changes with height. The graph starts low at $ y = 0 $, increases gradually, and rises sharply near $ y = h $.
This makes sense because the ball moves faster near the ground and slower near the maximum height. The probability density is higher near $ y = h $ because the ball spends more time there due to its low velocity.
And since the ball spends the least amount of time near the ground, where it is moving the fastest, the probability density is lowest at $ y = 0 $
