given that set B is ccontained in a set A, show that the probability of A is greater than or equal to that of B
Short proof: All the elements in B are also in A.
Long proof:
$\displaystyle \displaystyle B \subseteq A$
$\displaystyle \displaystyle n(B) \leq n(A)$
$\displaystyle \displaystyle \frac{n(B)}{n(\varepsilon)} \leq \frac{n(A)}{n(\varepsilon)}$
$\displaystyle \displaystyle Pr(B) \leq Pr(A)$.
I disagree, this is standard notation - even year 8 students should be aware of the notation of $\displaystyle \displaystyle \varepsilon$ to represent the universal set, $\displaystyle \displaystyle n(A)$ to represent the number of elements in set A, and that $\displaystyle \displaystyle Pr(A) = \frac{n(A)}{n(\varepsilon)}$. Even if they are infinite, the same logic holds if you were to draw a Venn Diagram of the situation.
Ha ha ha ... you jest
Cardinality? That is not the notation I know.$\displaystyle \displaystyle n(A)$ to represent the number of elements in set A,
Let your $\displaystyle \varepsilon$ be the unit disk, let A be some disk contained in $\displaystyle \varepsilon$, now what does your notation mean?and that $\displaystyle \displaystyle Pr(A) = \frac{n(A)}{n(\varepsilon)}$. Even if they are infinite, the same logic holds if you were to draw a Venn Diagram of the situation.
CB
The title did say “from the axioms”
Recall that $\displaystyle \left( {\forall C} \right)\left[ {P(C) \geqslant 0} \right]$ and $\displaystyle C \cap D = \emptyset \; \Rightarrow \;P(C \cup D) = P(C) + P(D)$.
Use those. $\displaystyle A = B \cup \left( {A\backslash B} \right)$, therefore $\displaystyle P(A) = P(B) + P\left( {A\backslash B} \right) \geqslant P(B)$.