Quae infra videbimus est dux exercitationum (resolutarum) quarum propositum est demonstrare nonnulla theoremata utilia ad theoriam probabilitatum. Experire eas solvere et postea compara tua responsa >:D

1. Demonstra P(A^c) = 1 -P(A) et, ex hoc fac deductionem quae sinat arguere P(\emptyset) = 0
   MOSTRAR SOLUCIÓN

   Ex definitione mensurae probabilitatis habetur quod, si A et B sunt eventus mensurabiles quicumque, tum valebit

   [a]	0\leq P(A) \leq 1
   [b]	A\cap B = \emptyset \rightarrow P(A\cup B) = P(A) + P(B)
   	P(\Omega) = 1

   Nunc, cum A\cap A^c = \emptyset, ex parte [b], habebitur:

   [d]	A\cap A^c = \emptyset \rightarrow P(A\cup A^c) = P(A) + P(A^c)

   Cum A\cap A^c = \emptyset semper verum sit et A\cup A^c = \Omega, tum habebitur

   1=P(\Omega) = P(A\cup A^c) = P(A) + P(A^c)

   Et, proinde

   P(A^c) = 1-P(A)

   Quod erat demonstrandum.

   Ad demonstrandum P(\emptyset)=0, sufficit sumere A=\Omega in relatione nuper probata et habebitur

   P(\emptyset) = P(\Omega^c) =1 - P(\Omega) = 1-1 = 0

2. a) Demonstra A\subseteq B \rightarrow P(B\setminus A) = P(B) - P(A),num aequalitas in universum vera est?b) Demonstra A\subseteq B \rightarrow P(B)\leq P(A)MOSTRAR SOLUCIÓN a)

   (1)	A\subseteq B; Praemissa
   	\equiv A\cap B = A
   (2)	B\setminus A = B\cap A^c; Definitio Theor. Coniunctorum
   (3)	B= (B\cap A) \cup (B\cap A^c); Proprietas coniunctorum
   (4)	(B\cap A)\cap (B\cap A^c)=\emptyset; Proprietas coniunctorum
   (5)	P(B)= P[(B\cap A) \cup (B\cap A^c)]; Ex (3)
   	P(B)= P (B\cap A) + P(B\cap A^c); Ex (4) + Def. Mensurae Probabilitatis
   	P(B)= P (B\cap A) + P(B\setminus A); Ex (2)
   	P(B\setminus A) =P(B) - P (B\cap A)
   	{P(B\setminus A) =P(B) - P (A) }; Ex (1)

   Itaque {\{A\subseteq B\}\vdash P(B\setminus A) = P(B) - P(A)}.

   Animadvertamus hanc aequalitatem non universaliter veram esse, quoniam pendet ex eo quod A\subseteq B sit ad obtinendum. Ex his deductionibus videri potest quod, si hoc non impletur, tum habebitur P(B\setminus A) = P(B) - P(A\cap B).

   MOSTRAR SOLUCIÓN b)

   Ex iis quae in a) rationata sunt habetur:

   \{A\subseteq B\}\vdash P(B\setminus A) = P(B) - P(A)

   \equiv \{A\subseteq B\}\vdash P(A) + P(B\setminus A) = P(B)

   Denique, cum P sit mensura probabilitatis, habetur \forall X (P(X)\geq 0), ita ut, proinde:

   {\{A\subseteq B\}\vdash P(A) \leq P(B)}.

    

3. a) Demonstra P(A\cup B) = P(A) + P(B) - P(A\cup B). Comitetur demonstrationem cum diagrammate.b) Utendo priori resultado demonstra P(A\cup B) \leq P(A) + P(B)MOSTRAR SOLUCIÓN PARTE a)

   (1)	A\cup B = (A\triangle B) \cup (A\cap B); Proprietas coniunctorum
   (2)	A\triangle B := (A\setminus B) \cup (B\setminus A); Definitio differentiae symmetricae
   (3)	(A\triangle B)\cap ( A \cap B) = \emptyset; proprietas coniunctorum
   (4)	(A\setminus B)\cap (B\setminus A) = \emptyset; proprietas coniunctorum
   (5)	P(A\cup B) = P[(A\triangle B) \cup (A\cap B)]; Ex (1)
   	P(A\cup B) = P(A\triangle B) + P(A\cap B)]; Ex (3)
   	P(A\cup B) = P(A\setminus B) + P(B\setminus A) + P(A\cap B)]; Ex (2,4)
   (6)	P(B\setminus A) = P(B) - P(A\cap B); Applicando eventus exercitationis 2
   (7)	P(A\setminus B) = P(A) - P(A\cap B); Idem quod (6)
   (8)	P(A\cup B) = P(A) - P(A\cap B) + P(B) - P(A\cap B) + P(A\cap B); ex (5,6,7)
   	{P(A\cup B) = P(A) + P(B) - P(A\cap B)}.

   MOSTRAR SOLUCIÓN PARTE b)

   \vdash P(A\cup B) = P(A) + P(B) - P(A\cap B); eventus partis a)

   \equiv\; \vdash P(A\cup B) + P(A\cap B) = P(A) + P(B)

   \equiv\; \vdash {P(A\cup B) \leq P(A) + P(B)}

    

4. Si \{E_n\} est familia infinita eventuum talis ut E_1\supseteq E_2 \supseteq E_2 \supseteq \cdots \supseteq E_n \supseteq E_{n+1}\supseteq \cdots. Demonstra verum esse

   \displaystyle P\left( \bigcap_{n=1}^\infty E_n \right) = \lim_{n\to \infty}P(E_n) MOSTRAR SOLUCIÓN

   (1)	E_n \supseteq E_{n+1}; Hypothesis
   (2)	E_n^c \subseteq E_{n+1}^c; Per complementum ex (1)
   (3)	\displaystyle(E_n^c \subseteq E_{n+1}^c) \rightarrow P\left( \bigcup_{n=1}^\infty E_n^c \right)= \lim_{n\to\infty}P(E_n^c); Proprietas Continuitatis
   (4)	\displaystyle \bigcup_{n=1}^\infty E_n^c = \left( \bigcap_{n=1}^\infty E_n \right)^c; Leges DeMorgan
   (5)	P\left(E_n^c\right) = 1 - P(E_n); Demonstratum in exercitio 1
   (6)	\displaystyle P\left(\left[ \bigcap_{n=1}^\infty E_n\right]^c \right) = \lim_{n\to\infty}[1-P(E_n)] = 1 - \lim_{n\to\infty}P(E_n); ex (2,4,5) applicatum super (3)
   	\displaystyle 1 - P\left( \bigcap_{n=1}^\infty E_n \right) = 1 - \lim_{n\to\infty}P(E_n)
   	{\displaystyle P\left( \bigcap_{n=1}^\infty E_n \right) = \lim_{n\to\infty}P(E_n)}

   \displaystyle\therefore\{E_n \supseteq E_{n+1}\}\vdash P\left( \bigcap_{n=1}^\infty E_n \right) = \lim_{n\to\infty}P(E_n).