Proof

(related to Proposition: Probability of Included Event)

We first note that

\[B=\Omega\cap B=(A\cup \overline A)\cap B=(A\cap B)\cup (\overline A\cap B).\]

By assumption of the inclusion \(A\subset B\), we have that \(A\cap B=A\). Therefore

\[B=A\cup (\overline A\cap B).\]

Because \(A \cap (\overline A\cap B)=\emptyset\), the events \(A\) and \((\overline A\cap B)\) are mutually exclusive. Therefore, it follows from the definition of probability that

\[p(B)=p(A)+p(\overline A\cap B)\]

Because \(p(\overline A\cap B)\ge 0\), we have \(p(A)\le p(B)\), as required.


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References

Bibliography

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