Math 300 , Spring 2002, Day 12, F, Feb 22 Hit reload to get most current version

Binomial Distribution B(n, p):  n independent, identical "Bernoulli" trials, each one Success or Failure.
    p = P(S) = prob of success on a single trial.
    q = P(F) = prob. of failure.
In random variable language, let X = number of successes  Interactive Probability:Bernoulli trials
Each possible result is an n-string of S's and F's.  Any particular string with k S's (and n-k F's) has probability pk qn-k
P(X=k) = sum of pk qn-k terms, one for each possible string with k S's.
There are nCk different strings.  nCk = "Binomial Coefficient" = n choose k
             (Choose the k places to put the S's.  The F's go in the remaining places)
P(X=k) = (nCk) pk qn-k

Places to go from here:  Multinomial distribution.  Urn problems (cf. Hypergeometric).  Binomial Coefficients.  Geometric Distribution.  Poisson Distribution.

Multinomial Distribution: Still  n independent, identical trials,
Generalize from Two outcomes to 3, (4, etc.) outcomes.  4 outcomes A, B, C,  D.   Prob's P(A)....P(D)
     Suppose 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
Any particular string has prob P(A)xP(B)yP(C)zP(D)w
 How many different strings? (Ash p. 45-6)
Multinomial coefficient (Derivation like Binomial)  # of rearrangements of x A's, y B's, z C's, w D's.
Take the n letters, label them so all distinguishable.  A1  A2  A3  B1 B2 C1 C2 D.   n! rearrangements.
How many look alike without the labels?  The A's can be rearranged among themselves 3! = x! different ways , The B's 2! = y! different ways, etc.
     Each rearrangement of A's can go with any rearrangement of B's, with any rearrangement of C's, etc.
So for any given list, e.g. ABBACADC, there are x!y!z!w! different versions if you can distinguish the A's, the B's etc.
                   (One is   A3B1B2 A2C1A1DC2 )
Multinomial coefficient = n! / (x!y!z!w!)

Urn problems (Ash pp. 48-50): N balls.  Labeled S, F. (or A, B, C, D)
   Drawing n with replacement: = Binomial/Multinomial because the urn is the same each time.
   Drawing n without replacement:
            Removing each ball changes the contents of the urn by 1.
            Tree model has subsequent probabilities changing correspondingly.
    Easier to calculate using committees/poker&bridge hand techniques.
            2 kinds of balls = Hypergeometric Dist.  (D defectives)
            More kinds of balls:  Like poker/bridge hands.
   Limit-result:  If N is large in proportion to n, and the number of balls of each kind drawn is small in proportion to the number in the urn,  then drawing without replacement may be approximated by drawing with replacement.
   Interactive probability:  "Urn" (Binomial/Multinomial only.  Red = Success=Defective)
          Model: compare Sampling with replacement and Sampling without replacement.
- - - - - - - - - - - - - - - - -  - -
Reading:  Ash pp. 46-51, M&M 375-9, 387-9 (Ash postpones means and s.d.'s.  Do a little review in M&M to remember what the words mean, for a distribution.)

HW:
A) Urn problem:  N = 12 balls, 4 Red.  Draw n =3.
    1) Without replacement:  Calculate the distribution values, for x = 0, 1, 2, 3 Reds.
             Do it two ways:  Hypergeometric formula, and tree with 3 branchings.
    2) With replacement: Calculate the distribution values, for x = 0, 1, 2, 3 Reds.
            Do it two ways: Binomial formula, and tree with 3 branchings.
    3) Go to the Interactive Probability: Urn program and check your answers.  Also record the mean and s.d. for both distributions. On the same graph, graph the two bar graphs showing the distribution. Which is more spread out, the with or without replacement version?
    4) Still in Interactive Probability, increase the number of balls, but keep the proportion.  N = 99, R = 33. Draw n = 3.
Record the distributions and mean and s.d., with and without replacement.
Make the two trees, and label the branches.  Note similarity/difference.

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 Interactive Probability.
1,  2,  4,  5,  8,  9, 12
7 (This is Multinomial, with 5 outcomes.  See if you can figure out how to see it that way, before looking at the answer.)
6
Freund Handout  7, 11
 
 


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