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1. Introduction and Examples

   What does it MEAN to say that

    a) The probability of Point Up for a thumbtack is P[U]=1/2?
    b) The probability of Heads for a 100Esc coin  is P[H]=1/2?
    c) The probability that IBM stock rises $1 today, P[+]=1/2?

Three interpretations of probability:

    *) Symmetry:  If there are k equally-likely outcomes, each has
                  P[E] = 1/k

    *) Frequency: If you can repeat an experiment indefinitely,
                  P[E] =  lim   [# of occur in n tries] / [n]
                         n->oo

    *) Belief:    If you are indifferent between winning $1 if E
                  occurs or winning $1 if you draw a blue chip
                  from a box with $100p$ blue chips, rest red,
                  P[E] = p

Limitations and shortcomings of each... nested... agree where possible.

Math Background:  Should know:

   \sum_3^{100} (4/5)^k :: \int_0^10 x e^{-2x} dx :: \lim_{x->oo} x^3/e^x

   \lim_{n->oo} (1-2/n)^n      ::    (a+b)^n = \sum_0^n {n:k} a^k b^{n-k}

   \int\int (x+y) dx dy over triangle with vertices (0,0), (1,0), (0,1)

------------------------------

Example:  If Duke, UNC, and NCSU have 800, 1000, 600 accounts on NCSC
supercomputer, and if accounts are six letters starting with D, U, or S
respectively, what is the probability that  "UABCXY" is a valid account?


Answ:  1000/26^5

------------------------------

2. Counting

When we choose "at random" one from a set of some number N of objects,
the intention is usually that each object has the same probability---
and, since the sum of the probabilities is one, that probability must be 1/N.

The hard part is figuring out what N is!

The business of counting objects is called "combinatorics"; we don't
need to know much of it, which is good because for most of us it's a
hard business.  Here's an example:

Q1:  In how many different ways can 10 people be ordered 1,2,...,10?

A1:  We can choose any of 10 for the 1st spot; once we've done that,
     we can choose any of  9 for the 2nd spot; ...
     ==> Answ = 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1 = 10! = 3628800

     In general, N objects can be placed in any of N! = N*(N-1)*..*2*1
     orders.  This number grows VERY fast... 70! > 10^100.  Try to avoid
     calculating high factorials... you can find them by logs if needed,
       N! = 10^[sum log(k): 1<k<=N]  <-- FACTORIAL

Q2:  How many different subcommittees of 3 people may be chosen from
     a group of 10 people?

A2:  Hmmmm.  One solution: let's pick the subcommittee in order:
     We can choose any of 10 for the 1st spot; once we've done that,
     we can choose any of  9 for the 2nd spot; once we've done that,
     we can choose any of  8 for the 3rd spot; altogether we've got
     ==> Answ = 10 * 9 * 8 = 720 subsets.

     Is that right?

     Well, no...  if the group members are A B C D E F G H I J, we
     have counted the subcommittee "A C E" more than once.  How many
     times?

     ACE AEC CAE CEA EAC ECA  <-- 6 times, so answer is 720/6=120.

                  /N\    N*(N-1)*...*(N-k+1)      N! / (N-k)!
     In general: (   ) = -------------------  =  ------------
                  \k/    k * (k-1) * ... * 1      k!

                                          /10\
     "Binomial Coefficients".  Example:  (    ) = 10*9*8/3*2*1 = 120.
                                          \ 3/

     Special cases: k=1; k=0; k=n; k <-> n-k


Q3:  The Duke R/C Airplane Club has ten members: 6 men and 4 women.
     All three officers (Pres, VP, Treas) are female.  What is the
     probability that would happen by chance alone?

A3:  There are (10:3)=120 possible subsets of 3 people from the club.
     There are (4:3)=4 possible subsets of 3 women from the club.  If
     all 120 subsets were equally likely (for example, if we pulled
     names out of a hat), then the probability of an all-female exec
     committee would be

               (4:3)/(10:3) = 4/120 = 1/30 = 0.0333

     Statistical note:  The line of reasoning often used to explore
     questions of discrimination goes as follows:

     EITHER  officers were selected in a nondiscriminary way and we
             saw a relatively rare event--- 1/30, about the same as
             rolling double six's with dice, or tossing 5 Heads in
             a row---

     OR      officers were not selected in a way that makes every
             member equally likely to be chosen.

Note that:

     1)  1/30 isn't all that rare--- if there were, say, 4 women in
         a club of 100 R/C members the prob would drop to
               (4:3)/(100:3) = 4/161700 = 2.473717/10^5,
         a minor miracle, way too much of a coincidence to believe.

     2)  Rejecting "selected at random" isn't the same as saying
         "biased and unfair"--- for example, the women might be
         the only really experienced fliers, etc.

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Mathematical Bit:

     (a+b)^N = (a+b)*(a+b)*(a+b)*(a+b)*(a+b)*(a+b)*...*(a+b)

             =    b *   b *   b *   b *   b *   b *...*   b
              + a   *   b *   b *   b *   b *   b *...*   b
              +   b * a   *   b *   b *   b *   b *...*   b
        +...+ +   b *   b *   b *   b *   b *   b *...* a
        +...
             = (n:0) a^0 b^n + (n:1) a^1*b^(n-1) + (n:2) a^2 * b^(n-2) + ...
             = sum (n:k) a^k b^(n-k)

     BINOMIAL THEOREM

Example:

     (a+b)^10 = b^10 + 10 a b^9 + 45 a^2 b^8 + 120 a^3 b^7 + ...
                                   +45 a^8 b^2 + 10 a^9 b + a^10

Properties:
                     n!   
      1. (n:k) = --------- = (n:(n-k)); so if k>n/2, switch to n-k.
                 k! (n-k)!

                  n*(n-1)*...*(n-k+1)
      2. (n:k)  = ------------------- ==> n doesn't have to be an integer!
                           k!


      In fact,

      (a+b)^z = sum     (z:k) a^k b^(z-k)  for any real or even complex z...
                k=0:oo

       ___
      Va+1  = (a+1)^(1/2) = sum (.5:k) a^k = 1 + a/2 - a^2/4 + 3a^3/16 - ...