Probability question

[Steven G]

Let's assume that when we throw a dart at a dart board we will hit the board and where we hit will be random.

Let p_i be a point on the board for i in the Naturals

Now P(hit p₁) = 0 or P(hit p₁) = a where 0 < a < 1

Case 1: P(hit p₁) = 0

\(\displaystyle \lim_{n\to \infty}(\sum_{i=0}^n P(p_i)) \ \lim_{n\to \infty}(\sum_{i=0}^n 0) = \lim_{n\to \infty}(n*0) = \lim_{n\to \infty}(0) = 0 \neq1\)

Case 2: P(hit p₁) = a where 0 < a < 1

\(\displaystyle \lim_{n\to \infty}(\sum_{i=0}^n P(p_i)) \ \lim_{n\to \infty}(\sum_{i=0}^n a) = \lim_{n\to \infty}(n*a) = \infty\neq1\)

I know that \(\displaystyle P(hit \ p_1) = 0\) but I am not clear where my errors are above.

 

[Romsek]

Your underlying distribution is 2D continuous uniform \(\displaystyle f_X(x) = \dfrac{1}{\pi r^2},~|x|\leq r\)

I'm not entirely sure what you're trying to show. It looks like you want to show the probability of hitting a set with measure 0,
i.e. a point, is 0. Is that correct?

 

[Steven G]

I was hoping to show that if you add up the prob of hitting each point on the dart board that the sum would be 1.

 

[Romsek]

\(\displaystyle A=\pi r^2\)
The probability of hitting any point is uniform on this area with probability \(\displaystyle \dfrac{1}{\pi r^2}\)

Total probability will be 1.

This is a continuous distribution. You can't use sums.