Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
    
,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!

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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.