The number of Bernoulli trials required to produce exactly 1 success and at least 1 failure.

Let $$X$$ be the number of Bernoulli $$(p)$$ trials required to produce exactly $$1$$ success and at least $$1$$ failure. Find the distribution of $$X$$.

How can I answer this question if I don't know how many trials have taken place?

For context, the section this comes from was about Poisson distribution.

$$\binom{n}{k}p^k(1-p)^{(n-k)}\rightarrow \frac{e^{-\mu}\mu^{k}}{k!}$$ $$\text{as }n\rightarrow \infty \text{ and } p\rightarrow 0 \text{ with } np=\mu$$

However I read in a different book that you can use the Geometric probability model for Bernoulli trials: Geom(p)

$$P(X=x)=q^{X-1}P$$ $$\text{ where p is prob. of success, X is number of trials, and q is prob. of failure }$$

• Observe that you can condition on the first trial to be success or failure and with this, obtain the probability. For the "experiment" can be loosely described by sequences SF, SSF, SSSF, ... and FS, FFS, FFFS,... – Will M. Feb 25 '19 at 18:49
• If the distribution would be something like SF, FS, SSF, SFS, FSS, FFS, FSF, SFF, .... Would there ever be an end? Would that be the distribution? – Project 2501 Feb 25 '19 at 21:03
• If $N$ is the number of trial until the experiment is concluded then $N = 3$ is the same as the occurrence of one of the following two sequences $SSF$ or $FFS.$ Observe these two sequences are mutually exclusive. Now think of the general $N = p.$ – Will M. Feb 25 '19 at 21:11
• ok, N=2 {FS, Sf}=2!, N=3 {SSF, ...SFF}= 6!, N=4 {SSSF, ......, FFFS}=4!=24 but I don't understand how to find a finite number? – Project 2501 Feb 25 '19 at 21:24
• You seem to not be understanding the problem, the experiment finishes as soon as two different letters appear. Hence, if you start with an S, you can only have S until the first F (this is geometric); but you could also start with an F until the first S appears (another geometric). This two sequences are mutually exclusive: the first cannot be both success and failure, it has to be either or; therefore, the probabilities add up. – Will M. Feb 25 '19 at 21:26

the only posibilities for X ( since exactly one sucess is allowed)

x=2 --> {SF,FS}

x=3 --> {FFS}

x=4 ---> {FFFS}

and so on...

So, if $$p = P(S)$$

$$P(X=2) = \frac{ 2.p.(1-p) } {1-p^{2}}$$

$$P(X=x) = \frac{(1-p)^2 .(1-p)^{x-2}.p}{1-p^2} = \frac{(1-p)^{x}.p}{1-p^2} ~~ \forall x \geq 3$$

the denominator $$1-p^2$$ is because you're not allowing the secuence SS

• If you sum these probabilities then you get $1-p^2$, though... – Math1000 Feb 26 '19 at 3:24
• Yes, fixed now, i forgot to compensate for the sequences starting with SS, thanks again – DrFran Feb 26 '19 at 3:44