Originally Posted by

**thecontractor** Sorry for bumping this thread but im still stuck. I've given a go at it in finding the pdf.

This is what i have so far:

Basically, if we draw the tree diagram we get a left-weighted binary-tree

We know that we need more girls then boys so we cannot have X = some even number.

Let $\displaystyle f_x(x)$ be our pdf.

So $\displaystyle f_x(x) = 0$, if x is even.

We just need to find the case where X = odd.

I've considered some cases for x = 1,3,5,7,9.....

$\displaystyle P(X = 1) = f_x(1) = \frac{2}{3}$, this from case {F}

$\displaystyle P(X = 3) = f_x(1) = (\frac{1}{3})(\frac{2}{3})^2$, this is from case {MFF}

$\displaystyle P(X = 5) = f_x(5) = 2(\frac{1}{3})^2(\frac{2}{3})^3$, this is for case {MMFFF, MFMFF}

$\displaystyle P(X = 7) = f_x(7) = 4(\frac{1}{3})^3(\frac{2}{3})^4$, for cases {MMMFFFF,MFMMFFF,MMFMFFF,MMFFMFF}

**Now**, this is where i get stuck:

$\displaystyle P(X = 9) = f_x(9) = \bold{c_9}(\frac{1}{3})^4(\frac{2}{3})^5$

for general case:

$\displaystyle P(X = n) = f_x(n) = \bold{c_n}(\frac{1}{3})^{\frac{n-1}{2}}(\frac{2}{3})^{\frac{n+1}{2}}$, for n odd

I'm not sure how to find $\displaystyle \bold{c_n}$. I guess it will be some kind of counting problem and may involve combinations.

Any ideas will be appreciated