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

probabilityprobability distributionsprobability theory

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)

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

Best Answer

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