How do I build a probability distribution from a $P(A_k| \cap_{i=1}^{k-1}A_i)=\theta^k$? A car windshield gets progressively weaker as it suffers from repeated debris strikes. Let $A_i$ be the event that the windshield survives the $i$th strike. For $0 < \theta < 1,$ let $P(A_1)=\theta$ and $P(A_k| \cap_{i=1}^{k-1}A_i)=\theta^k,$ where $k=2,3,\dots$. Define $X$ to be the number of debris strikes needed to break the windshield. Find the probability distribution of $X$.
I tried to obtain the probability distribution by going through the first few examples.
My work:
$P(X=1)=(1-\theta), P(X=2)=\theta(1-\theta), P(X=3)=\theta^2(1-\theta),\dots$
So, $P(X=x)=\theta^x(1-\theta)$. However, I do not know if this work is correct, since the probability of the $i$th debris strike breaking the windshield may not necessarily be $(1-\theta)$.
Alternative work:
$P(A_1 \cap A_2 \cap \dots \cap A_{x-1} \cap A_x^c) = P(A_x^c | \cap_{i=1}^{x-1}A_i)P(\cap_{i=1}^{x-1}A_i)$
$=[1-P(A_x | \cap_{i=1}^{x-1}A_i)]P(A_{x-1}| \cap_{i=1}^{x-2}A_i)\dots P(A_2| A_1)P(A_1)$
$=(1-\theta^x)\theta^{x-1}\theta^{x-2}\dots \theta^2\theta =(1 - \theta^x)\theta^{(x-1)x/2}.$ 
So, $P(X=x)=p(x)=(1-\theta^x)\theta^{(x-1)x/2}$.
These two different approaches seem to yield different answers. Where am I messing up?
 A: First I thought that the second way seems to be the flawed one but right now I am thinking the opposite: Of course, for general sets $A,B$ we do not necessarily have $P(A^C|B) = 1-P(A|B)$ however, in this case, this actually seems to be true. Reason:
$$\Omega = (A \cap B) ~\dot{\cup}~ (A^C \cap B) ~\dot{\cup}~ (A \cap B^C) ~\dot{\cup}~ (A^C \cap B^C)$$
so that
$$1 = P(\Omega) = P(A \cap B) + P(A^C \cap B) + P(A \cap B^C) + P(A^C \cap B^C)$$
In the case described above, $A = A_n$ and $B= \bigcap_{i=1}^{n-1} A_i$ so that $B^C \subset A^C$ (if the windshield was broken in one of the first debris strikes then it cannot survive the $n$-th strike either!). That has two important consequences:


*

*$A^C \cap B^C = B^C$

*$A \cap B^C = \emptyset$
i.e. the equation above simplifies to
$$1 = P(\Omega) = P(A \cap B) + P(A^C \cap B) + P(B^C)$$
now we divide by $P(B)$ to obtain
$$\frac{1}{P(B)} = \frac{P(A \cap B)}{P(B)} + \frac{P(A^C \cap B)}{P(B)} + \frac{P(B^C)}{P(B)}$$
The last summand is nothing else but
$$\frac{P(B^C)}{P(B)} = \frac{1-P(B)}{P(B)} = \frac{1}{P(B)} - 1$$
so that the equation further simplifies to
$$\frac{1}{P(B)} = \frac{P(A \cap B)}{P(B)} + \frac{P(A^C \cap B)}{P(B)} + \frac{1}{P(B)} - 1$$
and that indeed means nothing else but
$$1 = P(A|B) + P(A^C|B)$$
i.e. your second approach seems to be the right onw. However we should also do a quick sanity check in here and see that $\sum_{k=1}^\infty P[X=k] = 1$ however, I am too silly to prove that :-( All I got is
\begin{align*}
\sum_{k=1}^\infty P[X=k] &= \sum_{k=1}^\infty (1-\theta^k)\theta^{k(k-1)/2} \\
  &\leq \sum_{k=1}^\infty (1-0)\theta^{k(k-1)/2} = \sum_{k=1}^\infty \theta^{k(k-1)/2} \\
  &\leq \sum_{k=1}^\infty \theta^{k(k-1)/2} + \sum \text{missing exponents} \\
  &= \sum_{k=1}^\infty \theta^k = \frac{1}{1-\theta}
\end{align*}
However, a little program that I wrote tells me that this sum converges absolutely quickly against 1 (after 10 summands already for theta=0.5)... so this seems to be the right answer I guess...
EDIT: ah... sometimes it is so simple...
\begin{align*}
\sum_{k=1}^\infty (1-\theta^k)\theta^{k(k-1)/2} 
  &= \sum_{k=1}^\infty \theta^{k(k-1)/2} - \sum_{k=1}^\infty \theta^{k + k(k-1)/2} 
\end{align*}
and surprisingly (as I did not see :-))
$$k + \frac{k(k-1)}{2} = \frac{k(k+1)}{2}$$
so the right hand side kills all the summands except for the first one which is 1... mystery solved, eh?
