Problem
My question is about the probability of choosing a specific point on a sphere.
Let's suppose I divide my sphere into 4 equally-sized sectors (so 4 semi-hemispheres), and I choose a point randomly distributed on the semi-hemisphere surface, while the selection of the semi-hemisphere depends on a probability distribution.
My question is: what is the probability of choosing that specific point?
Example
I divide my sphere into 4 semi-hemispheres. The probabilities of choosing a specific semi-hemisphere are $q = [0.2, 0.3, 0.4, 0.1]$, meaning the probability of choosing the first semi-hemisphere is 0.2, the second 0.3, etc..
After choosing a semi-hemisphere, I choose a random point on the surface of it.
My solution
My solution consists of two steps:
1) First of all I choose a semi-hemisphere. The probability is $$p1 = \frac{prob(S1)}{Total Probability}$$ where $prob(S1)$ is a specific probability sampled from $q$, choosing a random number in range {0,1} and checking when this number is lower then the cumulative probability (sorry if this is not clear, I'll do the actual calculus below). $TotalProbability$ is the normalized sum (which, in this case, is already normalized).
2) In the second step I need to choose a random point in this semi-hemisphere. In this case, the probability would be:
$$p2 = \frac{1}{\frac{Area}{NumberPartitions}} = \frac{1}{\frac{4\pi R^2}{4}} = \frac{1}{\pi R^2}$$
For this reason, the combined probability should be:
$$ p = p1 * p2$$
Let's suppose the random number sampled is 0.43. Checking the cumulative probability of the probability bins in $q$, this number belongs to the second bin, 0.3. The probability would be:
$$ p = p1 * p2 = \frac{0.3}{1} * \frac{1}{\pi R^2} $$
Is this correct, or I am missing something? Thanks
EDIT:
Bonus question 1
What if I can only choose a specific point in each semi-hemisphere (so I don't have the option to randomly choose a point on the semi-hemisphere surface)? In this case, my intuition suggests that the solution would be only the first part of the previous equation, so $p = p1$. Is this correct?
Bonus question 2
What if I choose only a random point on the semi-hemisphere with the highest relative probability? This case really confuses me. Indeed, there is no probability involved in choosing a semi-hemisphere, since I just take the highest one. But this choice depends on the probability distribution. Should I only consider the second part of the previous equation, so $ p = p2$?
