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Suppose we are picking points uniformly at random from the surface of the Earth. I want to compute the probability that I pick a point in the Western hemisphere, given that I pick a point on the equator. The answer should clearly be $1/2$.

From the definition, we have

$$P(A|B)=\frac{P(A\cap B)}{P(B)},$$

where event $A$ is choosing a point in the Western hemisphere and $B$ is choosing a point on the equator. As the equator is a $1$-dimensional smooth line on a $2$-dimensional surface, it has measure $0$. So I compute $P(B)=0$. But using the conditional probability formula requires $P(B)>0$. In fact, this is the definition of conditional probability! So how do I make sense of $P(A|B)$? Clearly, it should work out to be $1/2$, but what is the rigorous way to compute it?

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,sorry i forget territory how equator is located,but why is probability $1/2$? –  dato datuashvili Nov 14 '12 at 6:08
See en.wikipedia.org/wiki/Equator. I reason that the answer is $1/2$ because precisely $1/2$ of the equator lies in the Western hemisphere. –  Potato Nov 14 '12 at 6:09
then yes,if first is equator,then it would $1/2$ –  dato datuashvili Nov 14 '12 at 6:13
you can count probability of $B$ as $1$,because if you take point from equator,it does not matter from which part you take,probability is just $1$ –  dato datuashvili Nov 14 '12 at 6:15
But clearly the probability of selection a point on the equator from a uniform distribution on the Earth is 0. –  Potato Nov 14 '12 at 6:17

1 Answer 1

up vote 3 down vote accepted

This is a surprisingly philosophical question, and as such, here is a link to a philosophical paper about it: http://philrsss.anu.edu.au/people-defaults/alanh/papers/what_cp_couldnt_be.pdf

Practically speaking though, you're absolutely correct - this probability is 1/2. However it is difficult to describe this fact using conditional probability the way it is usually understood.

The way I would "rigorously" approach this problem the following: let's say you have a probability space $(X,\Sigma,P)$ and a subspace $Y\subset X$ such that $P(Y)=0$. How do we 'condition' on this space? Well, the same way we consider a "line" integral in $\mathbb{R}^2$: $Y$ becomes your new universe, so you have to define a new probability space $(Y,\Sigma_2,P_2)$ where you can answer questions such as this. The statement $P(A\cap B)/P(B)$ is somewhat like trying to measure the length of a line segment using a bathroom scale - the scale ignores the line segment, so you have to get a ruler instead!

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thank you very much for link,$+1$ –  dato datuashvili Nov 14 '12 at 6:34

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