A standard "committee" pattern: 9 balls,
5 green and 4 red. Draw 3 balls. P(2 are red)=?. "Hypergeometric"

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.
- - - - - - - - - - - - - - -
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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