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

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SUMMARY

The discussion centers on the probability function P defined for a sequence of Bernoulli trials, specifically when playing until the rth success is achieved. The probability of an elementary event represented as a k-tuple is given by the formula q^{k-r}p^r, where p is the probability of success and q is the probability of failure. The key question raised is whether P(\Omega) equals 1, necessitating the proof that the sum of the probabilities from k=r to infinity, represented by the equation ∑_{k=r}^{\infty}binom{k-1}{r-1}q^{k-r}p^r, equals 1. This indicates that the function P is indeed a valid probability measure on the fundamental set Ω.

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quasar987
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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 occurrence of

q^{k-r}p^r

because the trials are independent 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?
 
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that looks like a negative binomial distribution... not sure. but you should probably prove it sums to 1.
 

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