Re: [OT] Math

Laserlight wrote:
> 
> A long time ago, in a university far, far away....I took statistics.
> And have since forgotten most of it.
> 
> If I fire six missiles, each of which hits with a 1 in 6 chance, I
> know the chance of getting six hits is (1/6)^6 and the chance of
> getting six misses is (5/6)^6.  Let's say, though, that I want to know
> what the probability is for getting four hits?  I could work out all
> the 46000+ possible results and track them....

	That depends on whether you want EXACTLY 4 hits, or AT LEAST 4
hits.
For EXACTLY 4, your probability is:

P = (1/6)^4*(5/6)^2 (4 missiles hit, 2 miss). For AT LEAST 4, your
probability is

P = (1/6)^4*(5/6)^2 + (1/6)^5*(5/6) + (1/6)^6 (4, 5, or 6 hits). That
may simplify down to (1/6)^4, but I'm not sure.