- #1

- 126

- 2

## 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