# Probability of choosing a specific point on a sphere

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 $$p = p2$$?

• Unless I missed your point, the probability of choosing a single point on a surface (a continuous variable taking a single value) is 0. Jun 2, 2019 at 9:30
• @AugSB Consider for example if we want to approximate an integral onto the surface of a semi-hemisphere with Monte Carlo Integration. We would pick the value of $n$ number of randomly distributed points, and we would divide their sum by the $pdf$ of choosing those points, which would be $1/(2 \pi \ R^2)$. Jun 2, 2019 at 9:39