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September 14th, 2017, 05:55 AM   #1
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Problem with Bernoulli trials

I have the following problem:

We conduct a series of indipendent Bernoulli trials with probability of success p for each trial and let N be the number of trials until the first success. Next we conduct N independent Bernoulli trials with probability of success q for each trial and let X be the number of successful trials. Calculate P(X=1)

Any ideas on how to solve this?
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September 14th, 2017, 12:38 PM   #2
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Looking at the 2nd set of trials first.

$P[X=1|N] = \dbinom{N}{1}q^1 (1-q)^{N-1} = N q (1-q)^{N-1}$

so

$P[X=1] = \displaystyle \sum_{N=1}^\infty~(N q (1-q)^{N-1})P[N]$

now looking at the first set of trials, $N$ has a geometric distribution

$P[N] = p^{N-1}p$

so

$P[X=1] = \displaystyle \sum_{N=1}^\infty~(N q (1-q)^{N-1})p^{N-1}p =

\dfrac{p q (1-q)}{(1-q)(1- p + pq)^2}$

Naturally you'll have to show the last summation result.

I wouldn't mind a second set of eyes on this one.
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September 15th, 2017, 03:11 AM   #3
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Sorry but I'm still confused... Why isn't $\displaystyle P[X=1|N] = \dbinom{N}{1}q^1 (1-q)^{N-1} = N q (1-q)^{N-1}$ the correct and final answer? Also how did you come up with that: $\displaystyle P[X=1] = \displaystyle \sum_{N=1}^\infty~(N q (1-q)^{N-1})P[N]$ did you use the law of total probability?

Sorry if my questions are stupid but I'm fairly new to probabilities. Could you be more detailed about the solution please?
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September 15th, 2017, 06:07 AM   #4
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Quote:
Originally Posted by Vaki View Post
Sorry but I'm still confused... Why isn't $\displaystyle P[X=1|N] = \dbinom{N}{1}q^1 (1-q)^{N-1} = N q (1-q)^{N-1}$ the correct and final answer?
Because $N$ can vary according to the first set of trials. It too is a random variable.


Quote:
Also how did you come up with that: $\displaystyle P[X=1] = \displaystyle \sum_{N=1}^\infty~(N q (1-q)^{N-1})P[N]$ did you use the law of total probability?
yes.

In general for discrete random variables $V$, and mutually exclusive random variables $U_k$

$P[V] = \displaystyle \sum_k P[V|U_k]P[U_k]$

Quote:
Sorry if my questions are stupid but I'm fairly new to probabilities. Could you be more detailed about the solution please?
This is a bit more difficult than the garden variety probability problem. Have you mastered easier problems yet?
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