It was given that $p_{2,3} = 2$ and $p_{0,1} = 12$
$p_{k, k+1} = \frac{P(X=k+1)}{P(X=k)}, k=0,1,2,...,n = (\frac{n-k}{k+1})(\frac{1-\theta}{\theta})$
The question was: Find $P(X \geq 2)$. Answer: $n=4, \theta = \frac{1}{4}$
My problem is I'm not getting $n=4$. What am I doing wrong. I just solved using simultaneous equations.
$p_{2,3} = (\frac{n-2}{3})(\frac{1-\theta}{\theta}) = 2$ ... (1)
$p_{0,1} = (\frac{n}{2})(\frac{1-\theta}{\theta}) = 12$ ... (2)
(2) / (1)
$(\frac{n}{2})(\frac{3}{n-2}) = 6$
$\frac{3n}{2n-4} = 6$
$3n = 6(2n-4)$
$3n = 12n - 24$
$9n = 24$
$n = \frac{8}{3}$?
Answer:
$$P_{0,1} = n\frac{1-\theta}{\theta} = 12$$ $$\frac{1-\theta}{\theta} = \frac{12}{n}$$ $$P_{2,3} = \frac{n-2}{3}\frac{1-\theta}{\theta} = 2$$ $$12n - 24 = 6n$$ $$n = 4$$ $$\frac{1-\theta}{\theta} = 3$$ $$\theta = \frac{1}{4}$$
Now that you have theta and n, you can easily find P(X>=2). If you need help on that let me know
Thanks
Satish