Hey meeksoup.
In terms of your answer, the A's are independent with P(An and An-1) = p^2 but P(Bn and Bn-1) = p^2 * p^2 != p^2.
Just out of curiosity, what attributes made you think of the geometric distribution?
Hi,
New to this forum. I have attached below a question that I think I have identified correctly as a geometric distribution. There are 3 parts. After many days, I have managed to complete part a. Answer attached. I wish to confirm this solution with someone. As for part b, I understand the hint but unable to apply this in reverse (i.e theta < 1). For part c, given that iam stuck in part b, I need a hint in how to solve it.
Thanks.
Hi Chiro,
Thanks for the reply.
This is how I figured this is a geometric distribution:
1) Every A has the same probability of happening. Like tossing a coin (equally likely outcome of H or T)
2) Let theta be the amount of times I toss the coin. i.e. number of times (n) >=1
3) Our success here is a bit unusual. Usually we have only one event of interest. Here we have An-1 and An.
So I was thinking that this is a geometric distribution where we are interested in the kth trial where this success occur. Theta here represents k where k starts from 1.
As a result I have been using (1-p)^k-1 * p formula. However, I think I have missed the overall picture and your answer seems more logical. Reason being if we let n =1, then:
B1=A0 intersects A1.
but A0 does not exist. So for B, n must start from 2. The formula I used accounts for A0, which as stated, does not make sense.
So I think for part a) P(Bn intersects Bn+1) = p^3. Which does not equal to p^4 (I think this is what you mean for P(Bn and Bn-1)?
Just thinking about part b and with the hints: Is it possible to prove independence by using (theta=<1) + (theta >1) = 1? Aren't complements independent of each other?
Still stuck at c.
Thanks.