Math 300 , Spring 2002, Day 5, W, Feb 6 Hit reload to get most current version

(Relative) Area is a good metaphor for probability.

Principle of inclusion and exclusion (p. 21):
This formula also works if you are counting elements in sets.  Just change  P to #, so #(A) is the number of elements in A.
There's some nice math here:
Suppose you are looking at a Discrete sample space, so individual points have probability.  Consider a point x.
Now suppose you have two sets A and B.   #(A or B)= #(A) + #(B) - #(A&B)
    Suppose x is only in one of the sets (say A).  Then it's counted once.
    Suppose x is in two of the sets.  Then it's counted twice (the simple sets) and de-counted once (the pair).

Suppose you have a whole bunch of sets, A, B, C, D, E,.... and you look at the union #(A or B or C or....).  The general pattern holds: 1-at-a-time-terms - 2-at-a-time-terms + 3-at-a-time-terms....  Here's the beginning of a general proof.
   Suppose x is in one of the sets (A).  It's counted 1 time.
    Suppose x is in two of the sets (A, B).  How often is it counted?
                2C1 = # of single sets it's in.  Minus 2C2 = # of 2-set intersections it's in.  = 2-1=1
    Suppose x is in three of the sets (A, B, C).  How often is it counted?
                3C1 = # of single sets it's in.  Minus 3C2 = # of 2-set intersections it's in.
                        Plus 3C3= # of 3-set intersections it's in = 3 -3 + 1=1
    Suppose x is in four of the sets.  How often is it counted?  4C1 - 4C2 + 4C3 - 4C4 = 4 - 6 + 4 - 1 = 1
Etc.  So whatever point you pick, it gets counted the correct number of times.
Not proved yet:   nC1-nC2+nC3-nC4+.....+nCn = 1
We'll prove this later on, using a different approach.

HW:  Show nC1-nC2+nC3-nC4+.....+nCn = 1 when n = 5.

Mutually Exclusive = Disjoint.  Doesn't = Independent (which we haven't met yet this semester)

Good list of rules on p. 27, including
DeMorgan's Laws:
Not( Aor B) = Not A And Not B.  I don't want cake or pie = I don't want cake and I don't want pie.
Not (A and B) = Not A Or Not B:  You can't have cake and pie = You can't have cake, or you can't have pie.

HW:  Show DeMorgan's laws true by shading a series of Venn diagrams.

HW: From Moore&McCabe: p. 359, 4.75, 4.77 (the reading is pp. 346-350top)
HW: Ash, p.27:  3, 4, 5, 2, 8, 9, 11 a&b.   Don't feel bad if you have to look at the answers.