Re: [OT] Math

----- Original Message ----- 
From: "Brian Quirt" <baqrt@mta.ca>
To: <gzg-l@csua.berkeley.edu>
Sent: Sunday, January 21, 2001 3:32 PM
Subject: 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.
> 
> -Brian Quirt

It deosn't. (1/6)^4 is the probability of getting 4 hits with 4
missiles.

Greetings