Is P(\Omega) Equal to 1 in Bernoulli Trials?

[quasar987]

Consider a sequence of Bernoulli trials where we play until the rth success is attained. Denote $\Omega$ the fundamental set.

We define a function P on $\Omega$ by saying say that an elementary event that is a k-tuple has a probability of occurence of

$$q^{k-r}p^r$$

because the trials are independant and the probability of success at each of them is p and the probability of failure is q.

Now I ask wheter or not with these probabilities assigned to each elementary event, we do have $P(\Omega)=1$? Did I miss something and it is implied that $P(\Omega)=1$, or we have to show that

$$\sum_{k=r}^{\infty}\binom{k-1}{r-1}q^{k-r}p^r=1$$

to show that P defined above is indeed a probability on $\Omega$?

 

[marcmtlca]

that looks like a negative binomial distribution... not sure. but you should probably prove it sums to 1.

 

[tacman]

Yes, you are correct in your understanding that P(\Omega)=1 needs to be shown in order to prove that P is a valid probability function on \Omega. This is known as the normalization condition and is a fundamental property of probability functions. In order to show this, you can use the binomial theorem to expand the summation and use the fact that \sum_{k=0}^{\infty}\binom{n}{k}x^k=(1+x)^n. This will allow you to simplify the expression and show that it is equal to 1.