Proof

(related to Corollary: Probability of Laplace Experiments)

According to the axiom probability of certain event we have that \[p(\Omega)=p(\{\omega_1,\ldots,\omega_n\})=1.\] By definition of Laplace experiments, all elementary events \(\omega_i\) are mutually exclusive, we can add their probabilities, resulting in \[p(\Omega)=p(\{\omega_1\})+\ldots+p(\{\omega_n\})=1.\] Because we have a Laplace experiment by hypothesis, it follows that \[p(\Omega)=p(\{\omega_1\})+\ldots+p(\{\omega_n\})=\underbrace{p + \ldots + p}_{n\text{ times}}=1,\] and therefore \[p(\omega_i)=\frac 1n~~~~~~~~~~~~(i=1,\ldots n).\]

It follows for the finite subset \(A\subseteq\Omega\) that we have to sum the probability \(\frac 1n\) the number of times, the event \(\omega_i\) belongs to the event \(A\). Again, because all events are mutualy exclusive, we get

\[p(A)=\sum_{\omega_i\in A}\frac 1n=\frac{|A|}{|\Omega|}.\]


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References

Bibliography

  1. Bosch, Karl: "Elementare Einf├╝hrung in die Wahrscheinlichkeitsrechnung", vieweg Studium, 1995, 6th Edition