So, I'm kinda stuck here at this question. Any help would be appreciated. (this is question 50 of chapter 3 of Introduction to probability models by Ross)
Suppose that N, the number of flips made of a coin that comes up heads with probability p, is a geometric random variable with parameter α, independent of the results of the flips. Let A be the event that all flips land heads.
(a) Find P(A) by conditioning on N. (b) Find P(A) by conditioning on the result of the first flip.
For part a) what I have is: Let X be r.v. No. of flips land heads. Let A be all flips heads. Let N be No. of flips.
$P(A) = P(X = n) = \sum_0^{\infty} P(A|N=n) P(N=n) = \sum_0^{\infty} ({n \choose n} p^n (1-p)^{n-n})((1-\alpha)^{n-1} . \alpha) = \sum_0^{\infty} P^n.\alpha.(1-\alpha)^{n-1}$
Is it in the right direction? Thanks.
For part a,
Let $N$ be a geometric r.v. with parameter $\alpha$, and let $X|N=n$ be binomial with parameters $n$ and $p$.
The probability of $A$ can be expressed as
$ P(A) = \sum_{n=1}^{\infty} \alpha(1-\alpha)^{n-1} p^n = \frac{\alpha}{1-\alpha} \sum_{n=1}^{\infty} [(1-\alpha)p]^{n}$
Since $0 < (1-\alpha)p < 1$,
$\sum_{n=1}^{\infty} [(1-\alpha)p]^{n} = \frac{(1-\alpha)p}{1-(1-\alpha)p}$, so
$P(A) = \frac{\alpha p}{1 - (1-\alpha)p}$