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 September 14th, 2017, 04:55 AM #1 Member   Joined: Apr 2014 From: Greece Posts: 58 Thanks: 0 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?
 September 14th, 2017, 11:38 AM #2 Senior Member     Joined: Sep 2015 From: USA Posts: 2,009 Thanks: 1041 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.
 September 15th, 2017, 02:11 AM #3 Member   Joined: Apr 2014 From: Greece Posts: 58 Thanks: 0 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?
September 15th, 2017, 05:07 AM   #4
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Quote:
 Originally Posted by Vaki 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?

 Tags bernoulli, problem, trials

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