In random variable language, let X = number of successes
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, k = 0, 1,...,n
Note if n = 1 (a Bernoulli trial), P(X = k) = pk
qn-k for k = 0 or 1 defines this simple
distribution.
Often I is used for an "indicator" random variable, which is
1 when something happens (Success), 0 when it doesn't.
(Xi is used too)
We can think of the results of a Binomial experiment as the sum of
the individual indicators for each trial:
X = I1 + I2 +...+ In
Applet--indicators,
Applet--Binomial
as coinflip, Applet--Binomial
in time sequence
Places to go from here: Multinomial
distribution.
Urn problems (cf. Hypergeometric). Binomial Coefficients.
Geometric
and Negative Binomial Distributions. Poisson Distribution.
- - - - - - - - - -
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: "Multivariate Hypergeometric" 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.
Applet:FiniteSamplingModels:Ball
& Urn experiment (Binomial/Hypergeometric
only.
Red = Success=Defective)
Model: compare
Sampling with replacement and Sampling without replacement.
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 1)
# 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!)
Multinomial coefficient (Derivation 2)
calculated using the same pattern as the binomial:
# of rearrangements of x A's, y B's, z C's, w D's, where
x+y+z+w
= n
Choose the x places to put A's in: nCx ways. Now there
are n-x free places.
Choose the y places to put B's in: (n-x)Cy ways. There are n-x-y free
places.
Choose the z places to put C's in: (n-x-y)Cz ways There
are n-x-y-z = w free places.
Choose the w places to put D's in: (n-x-y-z)Cw ways = wCw = 1
way.
Multiply these for the value of the multinomial coefficient.
Write each as a ratio of factorials, and cancel:
n!
(n-x)!
(n-x-y)!
(n-x-y-z)!
=
n!
x!(n-x)! y!(n-x-y)!
z!(n-x-y-z)!
w!(n-x-y-z-w!)
x!y!z!w!0!
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