How do I prove the following:   $\operatorname{Pr}(A \cup B \cup C) = \operatorname{Pr}(A) + \operatorname{Pr}(B) + \operatorname{Pr}(C) $  $\qquad − \operatorname{Pr}(A \cap B) − \operatorname{Pr}(A \cap C) − \operatorname{Pr}(B \cap C) + \operatorname{Pr}(A \cap B \cap C)$? $\max( \operatorname{Pr}(A), \operatorname{Pr}(B)) \leq \operatorname{Pr} (A \cup B)$
120472
Let $(\Omega,\mathcal F, P)$ be a probability space and let $A_1,A_2,...,A_n$ be events in $\mathcal F$.  Prove the following inclusion-exclusion formula  $P(\bigcup_{i=1}^nA_i)=\sum_{k=1}^n$ $\sum_{\mathcal J \subset \{1,...,n\}; |\mathcal J|=k} (-1)^{k+1}P(\bigcap_{i \in \mathcal J} A_i)$  I am trying to prove this formula by induction; for $n=2$, let $A, B$ be two events in $\mathcal F$. We can write $A=(A \setminus B) \cup (A\cap B)$, $B=(B \setminus A) \cup (A\cap B)$, since these are disjoint unions, then  $P(A)=P(A \setminus B)+P(A\cap B)$ and $P(B)=P(B \setminus A)+P(A\cap B)$.  Now, $A \cup B$ can be expressed as $A \cup B= (A\setminus B) \cup (A \cap B) \cup (B \setminus A)$, which it's also union of disjoint sets, so  $P(A \cup B)=P(A \setminus B)+P(A\cap B)+P(B \setminus A)=P(A)+P(B)-P(A \cap B)$.  This proves that for $n=2$, the formula is correct.  Now, I want to show that the formula is true for $n \implies$ the formula is satisfied for $n+1$, I got stuck at this point, maybe I could express $P(\cup_{i=1}^{n+1} A_i)=P((\cup_{i=1}^n A_i \setminus A_{n+1}) \cup A_{n+1})=P(\cup_{i=1}^n A_i \setminus A_{n+1})+P(A_{n+1})$ (1)  I can use the inductive hypothesis on the term $P(\cup_{i=1}^n A_i \setminus A_{n+1})$, so $P(\cup_{i=1}^n (A_i \setminus A_{n+1}))=\sum_{k=1}^n$ $\sum_{\mathcal J \subset \{1,...,n\}; |\mathcal J|=k} (-1)^{k+1}P(\bigcap_{i \in \mathcal J} (A_i\setminus A_{n+1}))$  Expression (1) becomes $\sum_{\mathcal J \subset \{1,...,n\}; |\mathcal J|=k} (-1)^{k+1}P(\bigcap_{i \in \mathcal J} (A_i\setminus A_{n+1})) +P(A_{n+1})$  I don't know to to take this expression to the one I want, which is  $\sum_{k=1}^{n+1}$ $\sum_{\mathcal J \subset \{1,...,n+1\}; |\mathcal J|=k} (-1)^{k+1}P(\bigcap_{i \in \mathcal J} A_i)$  I would like to know how to continue this part of the exercise.
80693
Say we have a venn diagram with A on the left, intersection C in the middle and B on the right.   I want to show that $P(A\cup B) = P(A) + P(B)  - P(A\cap B)$  I know what it can be written as A+B+C, but a lot of the proofs  go from:  $P(A\cup B) = P(A) + P(B)  + P(C)$ to  $P(A) + P(B) = P(A\cup B)  + P(C) $  and then rearrange to  get the proof and get $P(A\cup B) =P(A) + P(B)-  P(C) $ . I just don't get how the left side of the inequality get go from $P(A\cup B)$  to $P(A) + P(B)$. Doesn't that only happen if we assume that they are disjoint. In this case, do we assume that they are disjoint? I know how to write the proof but just having a hard time actually understanding it.
106469
$$ f\left(\bigcup\limits_{\lambda \in \wedge} A_{\lambda}\right) = \bigcup\limits_{\lambda \in \wedge} f(A_\lambda)$$  Let $b \in f(\bigcup\limits_{\lambda \in \wedge} A_{\lambda})$ $\rightarrow b=f(a)$ for some $a \in (\bigcup\limits_{\lambda \in \wedge} A_{\lambda})$.  Since $a \in (\bigcup\limits_{\lambda \in \wedge} A_{\lambda}) \rightarrow a \in A_{\lambda}$ for some $\lambda \in \wedge$.  Since $b=f(a)$, then $b \in f(A_{\lambda})$ for some $\lambda \in \wedge$.  $\rightarrow b \in \bigcup\limits_{\lambda \in \wedge} f(A_\lambda)$ $\rightarrow f(\bigcup\limits_{\lambda \in \wedge} A_{\lambda}) \subset \bigcup\limits_{\lambda \in \wedge} f(A_\lambda)$.  I used a similar argument to prove $\bigcup\limits_{\lambda \in \wedge} f(A_\lambda) \subset f(\bigcup\limits_{\lambda \in \wedge} A_{\lambda})$.  Which shows $ f(\bigcup\limits_{\lambda \in \wedge} A_{\lambda}) = \bigcup\limits_{\lambda \in \wedge} f(A_\lambda)$.  Please let me know if this is valid.
40005
I'm self-studying a textbook () (not for course credit) and wanted to post my solution to a problem related to conditional probability.   Given:  $$ \begin{align} &amp;P((A \cup B)^C) = 0.6 \\ &amp;P(A \cap B) = 0.1 \\ &amp;E = A\ xor\ B\ \text{(A or B but not both)} \end{align} $$  Find $P(E\ |\ A \cup B)$  Below is my solution. Is this correct or else where did I go wrong?   $$ \begin{align} P(E\ |\ A \cup B) &amp;= \frac{P(E\ \cap\ (A \cup B))}{P(A \cup B)} \tag 1 \\  \\ P(E\ \cap (A \cup B)) &amp;= P((E \cap A)\ \cup\ (E \cap B)) \tag 2 \\ \\ &amp;= P((A \cup B) \cup (A \cup B)) \\ \\ &amp;= P(A \cup B) \\ \\ \therefore \\ \\ P(E\ |\ A \cup B) &amp;= \frac{P(A \cup B)}{P(A \cup B)} = 1 \tag 3 \\ \\ \end{align} $$  Noting that $E \cap B$ = [(A xor B) or B] = $A \cup B$ and similarly $E \cap A$ = [(A xor B) or A] = $A \cup B$.
81975
Let $a,b \in [0,1], a+b=1$ and let $\{u,v\}\subset\mathbb{R}$.   How can we prove that $$(au+bv)^2\leq au^2+bv^2?$$   Opening the brackets give me in the left side $(au+v-av)^2$, but I'm not think that's the right way to do that.  I don't find the triangle inequality very helpfull, too.  Thanks.
81310
Suppose $Pr$ is non-atomic that is $Pr(A)$ implies the existence of $B$ with $B\subset A$ such that $0&lt;Pr(B)&lt;Pr(A) $ and $A\in\mathcal{F}$ with $Pr(A)&gt;0$. (a)show that for every $\epsilon&gt;0$, we have $B\subset A$ such that $0&lt;Pr(B)&lt;\epsilon$. (b)Prove that if $0&lt;a&lt;Pr(A)$ then there exists $B\subset A$ such that $Pr(B)=a$.  Here is a hint: fix $\epsilon_n \to 0$ and define inductively numbers $x_n$ and sets $G\in \mathcal{F}$ with $H_0=\emptyset$, $H_n=\cup_{k&lt;n} G_k$, $x_n=\sup\{Pr(G): G\subset A\setminus H_n, Pr(H_n\cup G)\leq a\}$ and $G_n\subset A\setminus H_n$ such that $Pr(H_n\cup G_n)\leq a$ and $Pr(G_n)\geq (1-\epsilon_n)x_n$. Let $B=\cup_{k} G_k$.
39825
If $p,q$ are positive quantities and $0 \leq m\leq 1$ then Prove that $$(p+q)^m \leq p^m+q^m$$    Trial: For $m=0$, $(p+q)^0=1 &lt; 2= p^0+q^0$  and for $m=1$, $(p+q)^1=p+q =p^1+q^1$.   So, For $m=0,1$ the inequality is true.How I show that the inequality is also true for $0 &lt; m &lt; 1$.   Please help.
34162
I know that $\operatorname{Int}(A) \cup \operatorname{Int}(B) \subset \operatorname{Int}(A \cup B)$, but that the other direction does not hold, so can anybody please tell me whats wrong with the following proof?   Suppose $x \in \operatorname{Int}(A \cup B) \Rightarrow \exists \epsilon &gt; 0 : K(x, \epsilon) \subset A \cup B$. Assume without loss of generatlity that $K(x,\epsilon) \subset A-B \Rightarrow x \in \operatorname{Int}(A) \Rightarrow x \in \operatorname{Int}(A) \cup \operatorname{Int}(B)$.
74826
Show that $\max{\{|a|+|b|,|c|+|d|\}} \leq \max{\{|a|,|c|\}}+\max{\{|b|,|d|\}}.$  I wanted to show that $d(p,q)=\max{\{|x_1-x_2|,|y_1-y_2|\}}$ where $p=(x_1,y_1),q=(x_2,y_2)$ is a metric on $\mathbb{R^2}.$ In proving the triangle inequality let $p=(x_1,y_1),q=(x_2,y_2),r=(x_3,y_3).$ Then  \begin{align*} d(p,r)&amp;=\max{\{|x_1-x_3|,|y_1-y_3|\}}\\ &amp;\le \max{\{|x_1-x_2|+|x_2-x_3|,|y_1-y_2|+|y_2-y_3|\}}\\ &amp;\le \max{\{|x_1-x_2|,|y_1-y_2|\}}+\max{\{|x_2-x_3|,|y_2-y_3|\}}=d(p,q)+d(q,r) \end{align*}  I don't understand how this inequality is true.
122077
I want to proof that:  $A \subseteq B  \Leftrightarrow  A \cup B = B$  At the moment I have no idea on how to start. Please give me a hint.  Thx in advance!
110858
How do I prove the following:  Show that if $A$, $B$ and $C$ are three events such that $P(A \cap B \cap C) \neq 0$ and $P(C | A \cap B)$ = $P(C|B)$, then $P(A| B \cap C) = P(A | B)$.  Here is my attempt  $P(C|A \cap B) = { {P(A \cap B \cap C)} \over {P(A \cap B)}} = P(C|B) = {{P(C \cap B)} \over {P(B)}}$  Now, I do not know what to do. I am hoping somebody can help me. This is not a homework problem.  Bob
36112
I am trying to show that for $A$, $B \; \subset \mathbb{R}$ $A \cup B$ is Lebesgue measurable such that \begin{align} \infty &gt; \mu(A \cup B) = \mu^{*}(A) + \mu^{*}(B), \end{align} Then $A$ and $B$ are Lebesgue measurable.  Thus far, I have tried reversing the directions in the proof of Caratheodory.  I seem to want to do the following:  If $A \cup B$ is Lebesgue measurable, then, for any $E \subset \mathbb{R}$, \begin{align} \mu^{*}(E) \geq \mu^{*}(E \cap (A \cup B)) + \mu^{*}(E \cap (A \cup B)^{c}). \end{align}  I want to use this and work backwards from here adding and subtracting sets to get  \begin{align} \mu^{*}(E) \geq \mu^{*}(E \cap A) + \mu^{*}(E \cap A^{c}). \end{align} yet I am having trouble.
94535
Give short arguments for each part below to complete the reasoning. Please do not use Venn diagrams.     a. Prove that $A = (A\setminus B) \cup (A \cap B)$.   (Hint: You should prove that $(A\setminus B)\cup (A\cap B) \subseteq A$ and vice-versa)      b. Prove that $(A \setminus B) \cap (A \cap B) = \emptyset$   (Hint: Try a proof by contradiction for this statement.).    I'm not too sure exactly where to start on this. Both statements make sense logically, but I don't know how exactly to prove them. Can anyone point me in the right direction?
122574
I'm trying to prove that (without using logical equivalencies):  $\overline{A\cap B} = \bar A \cup \bar B$  by proving both sides:  (1) $ x \in \overline{A\cap B} \to x \in \bar A\cup\bar B$  (2) $ x \in \bar A\cup\bar B \to x \in \overline{A\cap B}$  I figured out the 2nd part, but I'm struggling with the first. The only thing I'm confident about now is:  Let $x \in \overline{A\cap B}$. We prove that $x \in \bar A\cup \bar B$. By definition of complement, $x \not\in A\cap B$.  I'm not sure if I should use cases, or if I should prove by contradictions. With the other variation of DeMorgan's, I could assume $x \in A$ and $x \in B$ and they would lead to contradictions with the first assumption, but I can't do that here because it's a $\cap$ instead of a $\cup$.  For reference, here's the proof I was given for the other variation of DeMorgan's:  Prove: $\overline{A \cup B} = \bar A \cap \bar B$  (1) if $x \in \overline{A \cup B}$ then $x \in \bar A \cap \bar B$  (2) if $x \in \bar A \cap \bar B$ then $x \in \overline{A \cup B}$  Proof:  (1)Let $x \in \overline{A \cup B}$. We prove that $x \in \bar A \cap \bar B$. By definition of complement, $x \not\in A \cup B$. Suppose, for contradiction, $x \not\in \bar A$. By definition of complement, $x \in A$, and by definition of union, $x \in A\cup B$, a contradiction. Thus, $x \in \bar A$. Now, suppose for contradiction, $x \not\in \bar B$. By definition of complement, $x \in B$, and by definition of union, $x \in A \cup B$, a contradiction. So, $x \in \bar B$. Therefore, $x \in \bar A$ and $x \in \bar B$, so by definition of intersection, $x \in \bar A \cap \bar B$.  (I'm leaving out the 2nd part, as I've figured out the 2nd part in my problem above)  Any ideas? I'm assuming it has to be of similar complexity.
63529
I want to prove this statement:  $$(A_1 \cup A_2)^c  =  {A_1}^c \cup {A_2}^c$$  where the $c$ means the complement.  Any help would be greatly appreciated.
105742
Let $\sqsubseteq$ be the boolean ordering of $X$, so for every $x$ and $y$ applies $x \sqsubseteq y$ if $x \sqcap y = x$. Let $v, w, a, b \in X$ with $v \sqsubseteq a$ and $w \sqsubseteq b$. Show that $v \sqcup w \sqsubseteq a\sqcup b$ and $v \sqcap w \sqsubseteq a\sqcap b$.  Should this be solved algebraically, or in a different way? And if so, where would be my starting point?
8158
How to prove $( A \times B ) \cap ( C \times D ) = ( A \cap C ) \times ( B \cap D )$ ?