Math 300 , Spring 2008, Day 2, Wed. Jan.30 Hit
reload to get most current version
Reading: Review M&M 5th ed p. 266-70
(Independence, multiplication rule) Continue with 1-2. Look ahead
at
1-3
HW: ( Just counting
first probabilities next.) p. 14, #6, #8
Do A and B to HAND IN: Work with/get help
from anyone you can.
A) When I first moved to NY, license plates
were 3 letters followed by 3 numbers (TKU397)
a) How many different license
plates could there be back then?
b) Now there are 7
characters--there
is still some pattern, but I expect eventually there won't be.
How
many plates can there be if any of the 7 characters can be any letter
or
any digit?
c) Can you count "exactly
3 letters, together, but anywhere in the list" (ACD2355, 2ACD355,
etc.).
?
**d) I'm not sure what the
present patterns are, but there seem to be clumps of letters and
digits.
Observe in the parking lot(s) and see if you can see the likely
structures
(not counting vanity plates.) Can you count how many,
if your analysis of the structures is correct?
B) Ann, Betty, Carol, Xavier and Zane are going
to the movies. They want to sit Girl, Boy, Girl, Boy, Girl.
How many possible ways are there to do this?
How many ways if Ann and Xavier refuse to sit
next to each other? (You'll need a tree, or a list...and/or a trick.)
(Some simple probabilities)
p. 13: 1, 2, 3, 4 (a
poker hand is 5 cards here)
All of these go smoothly using "committees"
(combinations). For 1b, 3, 4a, do them also using lineups
(permutations) and get the same answers as before. Try 1a using
permutations and explain why it's harder to get it right using
permutations than it is using combinations.
"Combinatorics"--the tools of counting--is a
subject that can get amazingly hard amazingly quickly. Some
people
are gifted at it (they often end up working for the government in the
code-breaking/making
area). We'll just look at some of the most mainstream and useful
problems.
Multiplication principle: For experiments that can be thought of
as a sequence of stages.
M&M: If A and B are independent, P(A and
B) =
P(A)P(B).
Ash: Successive stages (slots): find how
many ways
each slot can be filled. Total # of outcomes = product of # of
ways at each stage (as long as # is independent of previous stage).
A
TREE is a very general
tool, useful even (especially?) when general principles don't quite
work. We'll see more examples later. If you're having
trouble figuring out a problem, cut the numbers down (e.g.10-and-3
objects to 4 or 5-and-2) and start drawing trees to look for
structures. For instance, Suppose that A is Hillary, B is Obama, and C
is Edwards, and we need a Presidential and a Vice-Presidential
candidate. The above tree structure shows the 6
possibilities. Now suppose Hillary refuses to play VP to Obama;
the BA branch is not possible, but you can still draw a tree and count
the rest.
Sampling with replacement, r things chosen from a population of
n:
To keep track, you must write each down before putting it back,
so it's an Ordered sample (you know which is first, which second, etc)
Number of samples: nn...n (n multiplied by itself r
times: nr )
(Modern physics deals with unordered,
sampled with replacement, but that's weird.)
Sampling without replacement
--Number of Permutations
(lineups) of length r
chosen from n things:
(Permutations of n things taken r at a time)
Number of Ordered samples of size r from a population of n (sampling
without replacement)
nPr =
n(n-1)(n-2)...(n-r+1) =n!/(n-r)!
--Number of Combinations , "committees" of n
things taken r at a time:
"n choose r" = Binomial coefficient.
(Properties: p. 9)
Number of Unordered samples of size r from a population of n ("always"
assume sampling
without replacement)
Probability pattern is still: Find a
sample
space of equally likely outcomes; count them. Then count the
outcomes
favorable to your event, divide by # in sample space.
Outcomes for many problems can be lineups,
or committees. (Sometimes one is better, sometimes equally
effective)
Most
common error: switching kinds between counting sample space and
counting
event.
Example: Choose 2 from {A, B, C}. Prob that A is
chosen. See from pictures 4/6 or 2/3. Generalize?
Easier to find Prob that A is Not chosen.
Combinations: 3C2 total, 2C2 with A not chosen. P = 1/3
Permutations: 3(2) total, 2(1) with A not chosen. P =
2/6.
Next: Choosing (without replacement) r from n
objects, where k are "defective". P(# of defectives chosen is x)
This page belongs to Sally Sievers who is solely
responsible
for its content. Please see our statement
of responsibility.