A or (B and C) = (A and B) or (A and C) (not in texts but sometimes useful)
I didn't demo the Siegrist virtual
laboratories http://www.math.uah.edu/stat/)
(direct
to
applets) Ball and Urn = Hypergeometric.
Other HW questions?
"New" material: Sets--OR (union), AND
(intersection),
NOT (complement), and probabilities thereof
(Relative) Area is a good metaphor for probability. Venn
Diagrams
help with counting too.
how
to shade venn diagrams, 2-set
probabilities,(same)
3-set
probabilities
diagrams
for more than 3 sets (math research)
Probability RULES: S is sample
space,
A event. (M&M numbering)
1) 0< P(A) < 1,
any A in S
2) P(S) = 1
3) P(Ac) = 1-
P(A)
(Therefore P(empty set, impossible event) = 0)
4) If A and B are disjoint ("Mutually
exclusive") ((A and B) is
empty), then P(Aor B) = P(A) + P(B)
4b) P(AorB) = P(A) + P(B) - P(A and
B)
(4 is a special case of 4 b, but more "basic")
? P (A and Not B) = P(A) - P(A and B).
fig 4.18(ed.4))
Remember OR is always "inclusive"--"and/or" in math, tho not in
English.
Exclusive or (XOR) sometimes "happens" when A and B
are defined in a problem; can use rule 4.
Warning: Mutually Exclusive = Disjoint. Doesn't =
Independent (which
we haven't met yet this semester)
- - - - - - -
Some Useful set rules: (Ash's
examples,
p. 27)
"At least 3"
List numbers and circle your set, (red bolded here)
0 1 2 3
4 5 6... See what's left if you
want the complement.
Why think if you can draw?
Visible
(same)
at least (at most, etc.) in context of Binomial
"At least one" has complement "none"
DeMorgan's Laws:
(A or B)c =
Ac and Bc
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.
(A and B)c = Ac
or Bc
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.
Distributive: A and (B or C) = (A
or B) and (A or C)
A or (B and C) = (A and B) or (A and C) (not in texts but
sometimes
useful)
3 or more sets: Look at 3 sets: P(AorBorC)
=
P(A) + P(B) + P(C)
- P(A and
B) - P(A and C) - P(B and C) (areas added twice above)
+ P(A and B and C) (area added 3
times and subtracted 3 times above) Ash p. 21 (2)
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.
(Numerator for equally likely sample space.)
There's some nice math here:
Suppose you are looking at a Discrete sample space, so individual
points
have probability. Consider a point x.
Look at 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.
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