Math 300 , Spring 2004, Day 3, Fri, Feb 6Hit reload to get most current version

Reading: Review 1-2. Read 1-3.  Don't get wrapped up in "Double counting"--
"Combinatorics", continued.
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.

A standard "committee" pattern:  9 balls, 5 green and 4 red.  Draw 3 balls.  P(2 are red)=?.  "Hypergeometric" Hypergeometric urn
   The 3 balls are a committee.  Number of Possible committees (sample space) = 9C3.
      How do we count the committees with exactly 2 red members?  Must have 2 red and 1 other (=green).  Think of them as the red subcommittee and the green subcommittee.  All-red subcommittee of 2 members, is drawn from 4 reds.  4C2.  All-green subcommittee of 1 member, is drawn from 5 greens.  5C1.  Any red subcommittee could be paired with any green subcommittee, so multiplication rule applies, there are (4C2 times 5C1) committees of 3 with exactly 2 reds.

Note nC1 = n.  Ash often writes "n" when "nC1" would illuminate the structure better.  Likewise, nC0 = 1, there is 1 way to choose an empty committee (= nCn, 1 committee of the whole)
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"Matching" problems: Can usually be made into a "slot" problem.  If you want to match two sets of individuals, make one set into a labeled "row" of slots (fixed order) and then distribute the other set.   
Sometimes it will make a problem easier to "see" if you put labels on the things.

E.g. Musical chairs.  5 kids and 3 chairs. How many ways can they sit down? (Oh, in the game the kids are in a fixed row,  harder.  Assume they're running across the room toward the chairs.)  Chairs 1__, 2__, 3__ .  Kids A, B, C, D, E.  First chair can get any of the 5, then next gets any of the remaining 4, 3rd gets any of the remaining 3.  5x4x3 .  3 kids and 5 chairs?  Let the kids be the slots.
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HW:
Start a page of "Named Probability Distributions."
A) The first is the Hypergeometric (not in Ash):  If you have n objects, and k are "defective", and you take a sample of r of the objects:  What is the probability of getting 0, 1, 2,...defectives in your sample?
Write down the general formula for x defectives.  (People who did 251 this fall saw this.)
    What limitations are there on x  if k < r?  Is there any case where it is impossible for x to be 0 (it is impossible to get no defectives?)  Explain.
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p. 13:    1, 2, 3, 4
5 (There's an easy way to do part b.  You get to choose which woman gets handed her coat first, second, etc.  The probabilities will be the same whichever woman you list first.  You can do it other ways but it's harder.),
7 (this is the most complicated one.  But still straightforward.  I had another sample space model, same answer: )
10, 11
Ash, p. 19: 1, 2

A) Work separately, HAND this one IN WEDNESDAY.  Using Example 4, pp. 15-16
a)  Show that the two probabilities of Method 1 and Method 2 (p.16)  are equal (set them equal, expand the "choose"'s, cancel numerators and denominators till the two sides match up.  You shouldn't have to actually multiply anything.) 
b)  Ash chose to treat the sample space as "lineups" (Method 1:36P7 in the denominator; finding the committee for the numerator and then multiplying it by 7! to get the possible lineups of the committee.  Method 2 starts with lineups in the top.)  You can also do the problem with a sample space of committees. Find the number of committees with 2 digits, 4 consonants, 1 vowel.  Find the count for the whole sample space. Show that the resulting probability equals the probability of Method 1.


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