
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