Math 300 , Spring 2008, Day 4, Mon, Feb 4
Hit
reload to get most current version
Repeat from
Friday's nonclass, plus some
Reading: Review 1-2. Read 1-3. Don't
get wrapped up in "Double counting".
Read ahead, Sets--OR (union), AND
(intersection),
NOT (complement), and probabilities thereof
Ash 1-4, pp 20-27 (postpone
P(AorBorC)
pp21 on first reading)
Math 251 did a little of this.4.2: M&M
5th ed:
pp259-266 (not Independence), 4.5, pp 311-15 for and, or
(notConditional)
HW:
Ash 1-2, p. 13: 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 (Which events are Not (another event)?), 11
Ash 1-3, p. 19: 1, 2
5 They mean "exactly"
3 W and 2R (how many "other" does that leave?)
4 (like example 4)
3 (harder? I think it's easier with an
ordered
list of 3. After you've chosen the first person, who CAN'T you
choose?)
6 Hint: let the people pick the seats.
10 Hints: 1) Will the probability be the
same whether i = 1, 2, or 3...?. . 2) I suggest doing it with i =
1, for 2 boys, 3 girls, then building up, 4 boys, 6 girls. Look
for
a general approach.
11 Optional--read answers. I don't want
to put bad ideas in your head.
- - - - - - - -
A) Do Handout on Hypergeometric distribution.
B) Use Siegrist's Virtual
Laboratories
( http://www.math.uah.edu/stat/)
to simulate the Hypergeometric distribution:
Bottom of main page, Applets. Finite
Sampling
Models> Ball and Urn Experiment. (Direct link to applets:
direct
to
applets (The i Applet instructions
button gives good
info
about the applets in general. In the experiment applet, hover over the
options for labels, or scroll down for
text.)
Use 12 balls, 4 red, draw a sample of size
3. Single step a few times to see how it goes. Check your computation
for
the distribution in part A above. Reset. Set Update 10, Stop
100.
Run. See how close the bar graph of data matches (doesn't).
Run again and again (without resetting) until you like the match.
How many runs did you take? (scroll down the left data panel to see the
last run)
Play with the pop. size, proportion of red, size
of sample other values, look at the shape of the distribution in
general.
C)
We find for
Hypergeometric (not
in Ash): If you have N objects, and r are "defective", (or
red) , and you
take
a sample of n of the objects: What is the probability of getting
0, 1, 2,...defectives in your sample?
Write down the general formula for x defectives.
Answer: hypergeometric, N objects, r "defective", a
sample of n,
P(x defectives) = (rCx ×(N-r)C(n-x))
÷ NCn. In class we noted that this only works for
x
< n, x < r. We also noted that when N=5,
r=3, n=3, you could not have x =0 because there were not enough "good"
objects.
Find a general inequality
involving
x, with N, r, n, to reflect the possibility of not enough
good
objects.
D) Start a page of "Named Probability
Distributions." (Keep it separate...)
The first is the Hypergeometric
N objects, r "defective", a sample of n,
P(x defectives) Give the limitations (as from #C).
- - - - - - -
Old B) Work
separately,
HAND this one IN WEDNESDAY.
FRIDAY
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.
= = = = = = = = = = = = = = = = = = = = = =
NOTES: "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.
Another Notation: n(n-1)...(n-r+1) = [n]r
- - - - - - - - - -
"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.
- - - - - - - - - -
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)
This pattern can be extended to 12 balls: 5 green, 4 red, 3 blue.
Draw 4 balls. Probability of 1Green, 2Red, (and
therefore 1 blue):
12C4 is total count for sample space.
Top of the fraction: #(1G, 2R, 1B) : 5C1 (for
the G) times 4C2 (for the R) times 3C1 (for the B)
Suppose you are in a room with 20 mosquitos
and 5 carry malaria. You get bit by 6.
What is the prob.that None are malarial?
1? 2? 3? 4? 5? 6?
(Hypergeometric)
General formula, N objects, r "defective", a sample of
n,
P(x defectives) = (rCx ×(N-r)C(n-x))
÷ NCn. Note this only works for
x
< n, x < r. (don't run out of balls, don'r
run out of "defectives.") (Other potential problem: run out of
"good"="other" balls)
Supppse now that 3 of the non-malarial
mosquitos carry dengue fever. You get bit by 6.
What is the prob. that you are bit by 3 malarial and 2 dengue
mosquitos? Only"clean" mosquitos?.
More complex combinations of these techniques
(example 4 p. 15). Again be careful: space is lineups or
committees?
This page belongs to Sally Sievers who is solely
responsible
for its content. Please see our statement
of responsibility.