# Thread: Probability- Geometric Prob. Distribution

1. ## Probability- Geometric Prob. Distribution

If Y has a geometric distribution with success probability p, show that

P(Y= an odd integer)= p/(1-q^2)

Thanks.

2. Originally Posted by fulltwist8
If Y has a geometric distribution with success probability p, show that

P(Y= an odd integer)= p/(1-q^2)

Thanks.
What's q? What event constitutes a success?

3. That IS exactly as it's stated in the book. q=1-p. it doesn't say what constitutes a success.

4. Originally Posted by fulltwist8
That IS exactly as it's stated in the book. q=1-p. it doesn't say what constitutes a success.
Without knowing what event constitutes a success, the question makes no sense - to me, anyway. It's very .... odd (ha ha) - to me, anyway - that the question would not define it. And it doesn't even define the random variable Y .....

By the way, if you make what would seem the obvious assumption that success is getting an odd number, then the probability of getting an odd number eventually is going to be 1, I'd have thought (unless p = 0) ......

Are you sure there's not some lead in, some introduction, some broader context ...... to the question. Because as is, the question is

5. Originally Posted by fulltwist8
If Y has a geometric distribution with success probability p, show that

P(Y= an odd integer)= p/(1-q^2)

Thanks.
I am trying to do the same question, have you had any success?

6. Originally Posted by fulltwist8
If Y has a geometric distribution with success probability p, show that

P(Y= an odd integer)= p/(1-q^2)
In a geometric probability distribution, the probability of needing k attempts to get a success is $q^{k-1}p$. If you add these probabilities for all odd values of k then you get $p+q^2p+q^4p+q^6p+\ldots$. This is a simple geometric series which you ought to be able to sum.