Math 300 , Spring 2008, Day 12, F, Feb 22 .after
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Reading: Ash pp. 51-2. Also Finite Geom. / Geometric Series, p.
36. Ahead, Poisson dist, sec 2-5, pp. 63-67
HW:
Multinomial: Ash: pp. 52ff. For these, write out the formulas
for the results. Calculate if it's not too hard; or, if suitable, stick
it into an appropriate applet. (Repeated from last day)
1(answer to 1b is wrong; should be 4 fair not 3, also in multinomial
coeff.), 4, 7 (7 is Multinomial, with 5 outcomes. See if you can
figure out how to see it that way, before looking at the answer.)
Binomial coefficients,Pascal's triangle, and (x+y)n:
postpone A.
Remember Day 5: We can now prove ( nC1-nC2+nC3-nC4+.....+
nCn = 1). Hint: Let x = 1 and y = 1 and look at the expansion of (x-y)n
. The inclusion/exclusion theorem, for P(A or B or C....), is used in problem
7d, p. 52, assigned today.
add AA: Find (1 + x +
x2 +...+ xn)(1 - x). Do it for n = 2, 3, 4, and generalize.
B. Pascal's triangle: a) Complete the row for n = 8, using the
handout and the row for 7. answers
b) Use algebra/arithmetic to show nCk = (n-1)C(k-1) + (n-1)Ck. Try
for smallish numbers, n =6, k=4, etc. See if you can show it in
the general case, using algebra.
c) Starting with (x+y)3 = x3y0
+ 3x2y1 + 3x1y2 + x0y3,
multiply by (x+y) and collect terms to get (x+y)4. Notice which
3Ck terms you sum to get each 4Ck coefficient. Check with triangle. This
should give some insight into how/why the triangle works as it does.
Postpone the rest
C. Geometric/Neg. Binomial: a) Give closed
(no "...")
formula for " Success happened on or before y'th" = p
+ qp + q2p + q3p + ...+ qy-1p =
? Use the Finite Geometric Series . b) Check that this plus
P(no successes in first y) =1.
D. a) Make a tree to model
the Negative binomial with k=2 (wait for 2nd success), through 4 trials
(it gets fat quickly!)
Write out the sequences that lead to the 2nd success on
the 4th trial: SFFS is one.
Convince yourself that (y-1)C(r-1) pr-1qy-r
p is the correct probability in this case.
b) Use Applet:NegativeBinomialExperiment
to get the histogram for the Geometric distribution (k=1). Copy,
roughly, the distributions when p = .2 and p = .8, on the same
scale.
Note each probability is q times the previous one.
c) Use the applet as in b. With k = 2, (wait for 2nd success)
run the value for p back and forth (between .2 and 1) to see the
distribution.
Repeat with k= 3, 4, 5. Write what the shape is like at the
extremes
and how it changes.
Freund problem sheet:
13, 15, 17
Ash p. 53, 16 (a,b,c are straightforward. There are
several
ways of getting d. The shortest is the trickiest. For e,
write
down the favorable sequences and their probabilities. Then you
can
use the Geometric Series.)
17. Make a tree of the fav's.
= = = = = = = = = = = = = = = = = = = = = = =
Comment on proof : if P(A and B) = P(A)P( B) then P(Ac
and Bc) = P(Ac)P( Bc), another method.
Ac and Bc = (A or B)c ,
by DeMorgan's Law, so P(Ac and Bc) = 1 - P(A or
B).
But P(A or B) = P(A) + P( B) - P(A and B) =
P(A) + P( B) - P(A )P( B) by the given.
So P(Ac and Bc) = 1 - P(A) -
P( B) + P(A )P( B) = [1 - P(A)][1 - P(B)] = P(Ac)P( Bc)
Handout on HIV; note differing
probs of having HIV depending on the risk pool.
Multinomial Distribution: See details Day 11
4 outcomes
A,
B, C, D. Prob's P(A)....P(D) , n
trials. Each possible
result is an n-string of A, B, C, D.
Let x, y, z, w be the number of each kind of outcome in the
result.
x + y +z + w = n
P(x, y, z, w) = [n! / (x!y!z!w!)] P(A)xP(B)yP(C)zP(D)w
+ + + + + + + + + + + + pick up here Monday
More about Binomial Coefficients: (handout)
A) (x+y)n has binomial coefficients in the expansion.
nCk xkyn-k.
Use this to show that Binomial distribution satisfies the law 1
= P(0) +P(1) +.....+P(n) (Prob. of everything = 1)
How?
Let x = p, y = q. Then 1 = (p + q)n = P(0) +P(1) +.....+P(n)
B) Pascal's triangle: Interactive Probability: Bernoulli Trials. Galton
Board Experiment
Each pair sums to the one below. nCk = (n-1)C(k-1) + (n-1)Ck
Sum of paths argument: number of paths to position (n,k) = sum of number
of paths to 2 positions just
above . (For Discrete alums--a setup to an argument by induction)
..
Geometric distribution: How many
flips to the first head? 
Applet:
BinomialTimelineExperiment
Applet:NegativeBinomialExperiment
Geometric: k=1
Model. Identical Bernoulli trials: Continue till first
Success. (Note, not exactly independent because you quit
at
the first head, but probability of success on any single trial stays
the
same--is independent of number of trials.)
Y = # of trials. Sample space: (1,2,3,4,..........) A discrete
but not finite sample space. Do tree.
P(Y = y) = (1-p)y-1p = qy-1p (the sequence
FF....FS, y-1 failures followed by 1 success)
Some books use W = # of failures before the first
success.
Y = W+1.
Does it work? Do the probabilities sum to
1?
p + qp + q2p + q3p +.......= 1.
See p. 36, Geometric Series, Finite Geometric
Series
.
"It takes at least y to get a success" Prob = No success in first
y-1. = qy-1
" Success happened on or before y'th" =
p + qp + q2p + q3p + ...+ qy-1p
= Finite Geometric Series .
Negative binomial (p. 52) Same as Geometric, only
number
of trials till the r'th success (k'th in applet). Formula for y
trials:
The last in the string is S; before that are y-1 letters, of which r-1
are S's (and y-r are F's). So that part is binomial.
P(Y = y) = (y-1)C(r-1) pr-1qy-r p = (y-1)C(r-1)
prqy-r What are the possible
values for y?
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