Math 300 , Spring 2004, Day 5, W, Feb 11 Hit reload to get most current version

Principle of inclusion and exclusion (p. 21): P( A or B or C.....):  Subtract P's of intersections of all even-set intersections, add P's of intersections of all odd-set intersections.
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: the needed general fact,    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.

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.
Warning:  Mutually Exclusive = Disjoint.  Doesn't = Independent (which we haven't met yet this semester)
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Ash 1-5: Continuous uniform distributions:
When any point in an interval or in an area  is as likely as any other ("uniform" distribution), then probabilities can be calculated directly from lengths, or from areas.
Bulletin board: I throw a dart at it. Suppose I am bad enough at throwing  darts that every point on the bulletin board is as likely to be hit as any other point.  What is the probability that my dart lands in the circle?  P = (Area of circle)/(area of whole square).

(Note--from review, p. 29: to tell which side of a boundary line satisfies an inequality, it is enough to test one point.)
I'll discuss Buffon's needle Monday.

HW Ash p. 34, 1, 2, 6, 7.  Read #3 and its answer.