Why naive probability is calculated the way its calculated?

So why question is very naive and also its a disclaimer that I will be asking a lot of questions on probability since it is one of my favourite subject that I want to know the thought process of person who solves the problems of probability.

Coming on to question, why do you think

P(\text{soccer}) = \frac{\text{# socker player}}{\text{# total students}}

Also, this could mean I have problem understanding fractions (like why to put this number in denominator or numerator).

Hi @tbhaxor!

Thanks for your post, this is not a naive question, in fact. In the case you mentioned, it can be simple for some people to understand, but ultimately we define probability in a more general way using what is called measure spaces, and there are some imposed definitions (or axioms, but it is subject to discussion the right term here).

In the soccer example, you can think in the frequentist way: You can define the probability of a student playing soccer as the percentage of times you get a soccer player when you randomly draw one student from all students. So it is expected that as you run more and more trials, you get a number close to the value

\dfrac{\# \text{ soccer player}}{\# \text{ total students}}
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Its like if we circumscribe a circle in the square. Where diameter of circle is same as size of square, then what is the probability point lands in side the circle.

Still this is not the exact answer I am looking for. I will do my research and write a post on it and share with you :smile:

I think I see your point.

Note that the examples pointed here are quite different. The first example is an example with a finite number of outcomes whereas the latter is related to an infinite number of points (points inside a circle). The general idea of probability remains the same - we should look at how much of the square’s area, the circle is inscribed.

So one should just divide the area of the circle by the area of the square. This should tell us the proportion of points that will land in the circle.

In the case you mentioned: Area square = 6^2 = 36 and Area Circle = \pi 3^2 (note that the diameter is 6, then the radius is 3) so the probability that a point lands inside the circle is just

\frac{9 \pi }{36} = \frac{\pi}{4}

However, note that this argument supposes that the point is choosen uniformly in the square. This is why we could make that simplification. If, in the other hand, we haven’t chosen the point uniformly, e.g., the point is “more likely” to be choosen closer to the center of the circle, then the result would be different and we would need to know more about the probability distribution of the point.

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Yes, now your answer is close to what I think should be the motivation of using this formula. The first example is of discrete probability and second example is of continuous probability (with bounds).