Re: [OT] Math

Nathan wrote:
> 
> From: Brian Quirt <baqrt@mta.ca>
> Date: Sunday, January 21, 2001 02:33
> Subject: Re: [OT] Math
> 
> > That depends on whether you want EXACTLY 4 hits,
> > or AT LEAST 4 hits.
> 
> In reply to both this and Schoon's posting, I would say LL
> wants to know the chance of exactly four out of his salvo
> of six missiles hitting. The chance of four or more of the
> six hitting would also be interesting...
> 
> > For EXACTLY 4, your probability is:
> >
> > P = (1/6)^4*(5/6)^2 (4 missiles hit, 2 miss).
> 
> Which works out at 0.0536%, or less than the 0.0772%
> chance of every missile hitting from a salvo of four. If I
> read your formula correctly, you've stated the chance of
> the first four missiles hitting and the fifth and sixth
> missing; one exact case out of our 64.

	Oops. I forgot about that. You have to add in the combinations
possible. I guess that would mean something slightly different. Note
that I will be using 6C4 as my abbreviation of "6 choose 4" (combination
notation)

	Thus 6C4*(1/6)^4*(5/6)^2  = 0.8% (which is the same as you got).
Note
that my own statistical ability is (currently) based on one class in
high school ("Finite math" - covered probability, statistics, induction,
combinations/permutations, binomial theorem, etc. etc. but all briefly)
(although in getting my degree I'm very likely to be taking more at some
point - I suspect that either Physics or Computer Science will require
it).

> > 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.
> 
> This correctly sums three specific outcomes, not
> the general "any four (or more) hit" scenario.
> Common sense dictates that if every missile has
> the same chance of hitting, six must have a better
> chance of achieving the objective than four.

Again, here we should have:

P = 6C4*(1/6)^4*(5/6)^2 + 6C5*(1/6)^5*(5/6) + 6C6*(1/6)^6 = 0.87% = 0.9%
(again as you got).

In general, when you want to know whether or not exactly n events, with
P=a will occur out of a set of m events, m>n,

P = mCn*(a)^n*(1-a)^(m-n)

Hope that helps,