# Geometric distribution Problem

throneoo

## Homework Statement

a man draws balls from an infinitely large box containing either white and black balls , assume statistical independence. the man draws 1 ball each time and stops once he has at least 1 ball of each color .

if the probability of drawing a white ball is p , and and q=1-p is that of drawing a black ball , what is the:
a) average no. of balls drawn
b)average no. of white/black balls drawn
c)variance of no. of white/black balls drawn

## Homework Equations

Defining N,W,B as the total no. of balls , no. of white/black balls respectively ,

the probability generating functions (pgf) for N,W,B are respectively

θ(x)=Σpx(qx)^B from B=1 to infinity + Σqx(px)^W from W=1 to infinity
=Σ(p(qx)^B+q(px)^B)x from B=1 to infinity
=pqx^2((1-qx)^-1+(1-px)^-1)

α(x)=Σq(px)^W from W=1 to infinity
=pqx/(1-px)

β(x)=Σp(qx)^B from B=1 to infinity
=pqx/(1-qx)

## The Attempt at a Solution

Thus ,

<N>=dθ/dx at x=1

=p/q + q/p +1

<W>=dα/dx at x=1 =p/q

<B>=dβ/dx at x=1 = q/p

<W^2>=d[x*dα/dx]/dx at x=1 =p(p+1)/q^2
<B^2>=d[x*dβ/dx]/dx at x=1 =q(q+1)/p^2

Var(W)=<W^2>-<W>^2=p/q^2
Var(B)=<B^2>-<B>^2=q/p^2

4. Question

my concern is , why doesn't <N>=<W>+<B> ?

It seems to contradict N=W+B .

Also , the expression I've got for <N> has a local minimum at (p,<N>)=(0.5,3)

which is also a bit weird

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## Homework Statement

a man draws balls from an infinitely large box containing either white and black balls , assume statistical independence. the man draws 1 ball each time and stops once he has at least 1 ball of each color .

if the probability of drawing a white ball is p , and and q=1-p is that of drawing a black ball , what is the:
a) average no. of balls drawn
b)average no. of white/black balls drawn
c)variance of no. of white/black balls drawn

## Homework Equations

Defining N,W,B as the total no. of balls , no. of white/black balls respectively ,

the probability generating functions (pgf) for N,W,B are respectively

θ(x)=Σpx(qx)^B from B=1 to infinity + Σqx(px)^W from W=1 to infinity
=Σ(p(qx)^B+q(px)^B)x from B=1 to infinity
=pqx^2((1-qx)^-1+(1-px)^-1)

α(x)=Σq(px)^W from W=1 to infinity
=pqx/(1-px)

β(x)=Σp(qx)^B from B=1 to infinity
=pqx/(1-qx)

## The Attempt at a Solution

Thus ,

<N>=dθ/dx at x=1

=p/q + q/p +1

<W>=dα/dx at x=1 =p/q

<B>=dβ/dx at x=1 = q/p

<W^2>=d[x*dα/dx]/dx at x=1 =p(p+1)/q^2
<B^2>=d[x*dβ/dx]/dx at x=1 =q(q+1)/p^2

Var(W)=<W^2>-<W>^2=p/q^2
Var(B)=<B^2>-<B>^2=q/p^2

4. Question

my concern is , why doesn't <N>=<W>+<B> ?

It seems to contradict N=W+B .

Also , the expression I've got for <N> has a local minimum at (p,<N>)=(0.5,3)

which is also a bit weird

The first ball always has one of the two colors, so we need only get the probability distribution of number of draws until the opposite color. If the first is white (prob = p) we need the distribution of the number of additional draws until the first black. If the first color is black (prob = q = 1-p) we need the distribution of the number of additional draws until the first white. So, conditioned on the first color, the rest is an ordinary geometric distribution. The overall distribution of number of draws is thus a mixture of two geometric distributions. See, eg., http://en.wikipedia.org/wiki/Mixture_distribution for a brief intro to mixtures of distributions (although the article is a bit obscure and hardly introductory). Note that if ##C_1## = color of the first ball (w or b) we have
$$EN = E(N | C_1= w)\cdot P(C_1=w) + P(N | C_1 = b)\cdot P(C_1 = b)$$
and
$$EW = E(W|C_1=w) \cdot P(C_1 = w) + E(W|C_1 = b) \cdot P(C_1 = b),$$
etc.

Last edited:
throneoo
I'm too tired to read . but those equation make senes . they are just derived from bayes theorem aren't they?

throneoo
but it seems that <W>+<B> still doesn't add up to <N>...

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I'm too tired to read . but those equation make senes . they are just derived from bayes theorem aren't they?

Basically, yes.

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but it seems that <W>+<B> still doesn't add up to <N>...