Originally posted by smomofoIn stats a uniform distribution is a distribution where all possibilities are equally possible. So A is just as likely to land at a1 as a2 no matter where a1 and a2 are on Planet X.
Zero, if they are uniformly distributed. I think. Wouldn't any two make an angle of 120 from center?
Originally posted by SPMarsWould I be right in thinking that if just *two* ships were involved the probability would be 0.5?
Planet X is a ball with centre O.
Three spaceships A, B, C land at random positions on the surface of Planet X, these positions being independent and uniformly distributed.
Once landed, a pair A, B of spaceships can communicate provided that the angle AOB is less than 90 degrees. Similarly for B, C and C, A.
What is the probability that all three spaceships can stay in touch (with, for example, A communicating with C via B if necessary)?
Originally posted by howardbradleyYes that's right for two ships: without loss of generality we can suppose ship A has landed at the north pole. Then there is probability 1/2 that ship B lands in the northern hemisphere, and that's the region where the two can communicate.
Would I be right in thinking that if just *two* ships were involved the probability would be 0.5?
My reasoning being that one ship landing would define a hemisphere in which communication was possible - with the ship at the pole. The other ship would then have a 50/50 chance of landing in the same hemisphere.
I'm not sure this helps with the final answer. Intriguing problem though.
Originally posted by smomofoAlas, it is not 1/8.
Well then I change my answer to a probability of 0.125. How 'bout that?
Originally posted by XanthosNZCertainly that is a good idea, and would get you a feel for the sphere case. And the integration is easier over a circle.
In approaching this problem you may find it easier to look at the case of a circle instead of a sphere first. Think of the two cases (B lands connected to A and B lands unconnected to A) as seperate and work from there.
I hope this helps someone.
Originally posted by smomofoYep integration is one way. And spherical polar coordinates are probably helpful.
Ugh. This going to require integration isn't it? It's been waaay too long for that. Plus, I never dealt with polar coordinates well, and I suspect this might be a problem where they are appropriate. I was on to the whole 0.5 thing, at least!
My new answer is: I don't know. For me, it would be fastest to do it by experiment.