Need a little help

perusal

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Joined
Jan 7, 2019
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21
\(\displaystyle
(e \cup f)^c = e^c \cap f^c
\)

I am struggling to prove this identity. Could someone please help me out.

Below is the kind of think i have been trying so far. But to no avail.

\(\displaystyle

e^c \cap f^c \\= e^c + f^c - e^c \cup f^c \\


= (1-e) + (1-f) - e^c \cup f^c \\

= 2-(e+f) - e^c \cup f^c \\

= (e+f)^c + 1 - e^c \cup f^c \\ \\

therefore \\ \\

e^c \cup f^c = 1 ? \\ \\
\)

pretty sure i messed up somewhere
 

pka

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Jan 29, 2005
Messages
8,527
\(\displaystyle (e \cup f)^c = e^c \cap f^c\)
I am struggling to prove this identity. Could someone please help me out.
Below is the kind of think i have been trying so far. But to no avail.
\(\displaystyle e^c \cap f^c \\= e^c + f^c - e^c \cup f^c \\
= (1-e) + (1-f) - e^c \cup f^c \\
= 2-(e+f) - e^c \cup f^c \\
= (e+f)^c + 1 - e^c \cup f^c \\ \\
therefore \\ \\
e^c \cup f^c = 1 ? \\ \\
\)
pretty sure i messed up somewhere
You certainly messed up by not telling us what any of those letters stand for.
How are they related to probability?? Are they related?
 

perusal

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Messages
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You certainly messed up by not telling us what any of those letters stand for.
How are they related to probability?? Are they related?
Ye sorry. They are each possible events in a sample space. e is short hand for p(e) and f for p(f). Neither of the events have been defined.

\(\displaystyle

e^c\)

is the complement of e
 

pka

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Messages
8,527
Ye sorry. They are each possible events in a sample space. e is short hand for p(e) and f for p(f). Neither of the events have been defined.\(\displaystyle
e^c\) is the complement of e.
I have no idea what textbook you may be using. What is your textbook?
Most all texts I have used, are based on basic set theory
What you posted seems to be a mixture of notations.
Hopefully someone else here knows how this notation works and can help you.
 

JeffM

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The notation is awful.

Let's start here:

\(\displaystyle P(E \ and \ F) = w \implies 0 \le w \le 1.\)

\(\displaystyle P(E \ and \ not \ F) = x \implies 0 \le x \le 1.\)

\(\displaystyle P(not \ E \ and \ F) = y \implies 0 \le y \le 1.\)

\(\displaystyle P(not \ E \ and \ not \ F) = z \implies 0 \le z \le 1.\)

\(\displaystyle \therefore w + x + y + z = 1.\)

\(\displaystyle P(E \ or \ F) = P(E) + P(F) - P(E \ and \ F) = (w + x) + (w + y) - w = w + x + y.\)

\(\displaystyle \therefore P(not \ \{E \ or \ F\}) = 1 - (w + x + y) = z = P(not \ E \ and \ not \ F) \text { by definition.}\)

Keep your notation for events and sets distinct from that for probabilities.

Pka is an expert in this and can probably give you an even more succinct proof.

EDIT: Your problem, I suspect, was not conceptually disaggregating into non-overlapping sets at the outset
 
Last edited:

perusal

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Thanks
 

pka

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Messages
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Suppose that \(\displaystyle \Omega\) is a set of points and \(\displaystyle \mathscr{Pr}\) is a function on the powerset \(\displaystyle \mathcal{P}(\Omega).\)
Here are Morris Marx's axioms for probability.
Axiom 1. If \(\displaystyle A\in\mathcal{P}(\Omega).\) Then \(\displaystyle A\) is an event and \(\displaystyle \mathscr{Pr}(A)\ge 0\)
Axiom 2. \(\displaystyle \mathscr{Pr}(\Omega)=1\)
Axiom 3. If \(\displaystyle \{A,B\}\subset\mathcal{P}(\Omega)~\&~A\cap B=\emptyset\) then \(\displaystyle \mathscr{Pr}(A\cup B)=\mathscr{Pr}(A)+\mathscr{Pr}( B)\)

Using those axioms one can prove the basic properties of probability.
 

JeffM

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Suppose that \(\displaystyle \Omega\) is a set of points and \(\displaystyle \mathscr{Pr}\) is a function on the powerset \(\displaystyle \mathcal{P}(\Omega).\)
Here are Morris Marx's axioms for probability.
Axiom 1. If \(\displaystyle A\in\mathcal{P}(\Omega).\) Then \(\displaystyle A\) is an event and \(\displaystyle \mathscr{Pr}(A)\ge 0\)
Axiom 2. \(\displaystyle \mathscr{Pr}(\Omega)=1\)
Axiom 3. If \(\displaystyle \{A,B\}\subset\mathcal{P}(\Omega)~\&~A\cap B=\emptyset\) then \(\displaystyle \mathscr{Pr}(A\cup B)=\mathscr{Pr}(A)+\mathscr{Pr}( B)\)

Using those axioms one can prove the basic properties of probability.
I'd have to think about those for a while to see how to deduce the laws of probability, but it is interesting that Axiom 3 involves disjoint sets, which is how I attacked the original problem.
 

pka

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Messages
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Suppose that \(\displaystyle \Omega\) is a set of points and \(\displaystyle \mathscr{Pr}\) is a function on the powerset \(\displaystyle \mathcal{P}(\Omega).\) Here are Morris Marx's axioms for probability.
Axiom 1. If \(\displaystyle A\in\mathcal{P}(\Omega).\) Then \(\displaystyle A\) is an event and \(\displaystyle \mathscr{Pr}(A)\ge 0\)
Axiom 2. \(\displaystyle \mathscr{Pr}(\Omega)=1\)
Axiom 3. If \(\displaystyle \{A,B\}\subset\mathcal{P}(\Omega)~\&~A\cap B=\emptyset\) then \(\displaystyle \mathscr{Pr}(A\cup B)=\mathscr{Pr}(A)+\mathscr{Pr}( B)\)
Note that \(\displaystyle A\cap A^c=\emptyset~\&~A\cup A^c=\Omega\)
\(\displaystyle \mathscr{Pr}(A\cup A^c)=\mathscr{Pr}(A)+\mathscr{Pr}( A^c)=\mathscr{Pr}(\Omega)=1\)
Thus \(\displaystyle \mathscr{Pr}(A^c)=1-\mathscr{Pr}(A)\)
Because \(\displaystyle \Omega^c=\emptyset\) show that \(\displaystyle \mathscr{Pr}(\emptyset)=0.\)
If \(\displaystyle A\subset B\) then show that \(\displaystyle \mathscr{Pr}(A)\le\mathscr{Pr}(B)\)

For any event \(\displaystyle A\) then show that \(\displaystyle \mathscr{Pr}(A)\le 1\)

For any events \(\displaystyle A~\&~B\) show that \(\displaystyle \mathscr{Pr}(A\cup B)=\mathscr{Pr}(A)+\mathscr{Pr}(B)-\mathscr{Pr}(A\cap B)\)
 
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