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Probability - using DeMorgan's laws - please help!

Screenshot below!

We have to apply DeMorgan'd laws twice.. And I have no clue how to do it!
Screen Shot 2016-04-27 at 15.02.47.png18.5KB
Reply 1
Avatar for Zacken
Zacken
22
Original post by JKITFC
Screenshot below!

We have to apply DeMorgan'd laws twice.. And I have no clue how to do it!


Why don't you consider it pairwise?

Ac∩Bc=(A∪B)cA^c \cap B^c = (A \cup B)^c and Cc∩Dc=(C∪D)cC^c \cap D^c = (C \cup D)^c

Now you have (A∪B)c∩(C∪D)c=⋯(A \cup B)^c \cap (C\cup D)^c = \cdots
Original post by JKITFC
Screenshot below!

We have to apply DeMorgan'd laws twice.. And I have no clue how to do it!


Have a think about what de Morgan's laws say about (A∪B)c(A \cup B)^c. Now apply the fact that A,B A, B are mutually exclusive, in considering P((A∪B)c)=1−P(A∪B) \mathbb{P}((A \cup B)^c) = 1 - \mathbb{P}(A \cup B)
Original post by Zacken
Why don't you consider it pairwise?

Ac∩Bc=(A∪B)cA^c \cap B^c = (A \cup B)^c and Cc∩Dc=(C∪D)cC^c \cap D^c = (C \cup D)^c

Now you have (A∪B)c∩(C∪D)c=⋯(A \cup B)^c \cap (C\cup D)^c = \cdots


Looks good, I got that far, didn't know the second application
Reply 4
Avatar for Zacken
Zacken
22
Original post by JKITFC
Looks good, I got that far, didn't know the second application


Awesome! Couple that with Greg's hint above and all should be good.
Original post by Zacken
Awesome! Couple that with Greg's hint above and all should be good.


What does.. (A \cup B)^c \cap (C\cup D)^c = \cdots equal?
Reply 6
Avatar for Zacken
Zacken
22
Original post by JKITFC
What does.. (A \cup B)^c \cap (C\cup D)^c = \cdots equal?


Perhaps if we write A∪B=EA \cup B = E and F=C∪DF = C \cup D that'll clarify matters: Ec∩Fc=⋯E^c \cap F^c = \cdots? Hint: De Morgan.
Original post by JKITFC
Screenshot below!

We have to apply DeMorgan'd laws twice.. And I have no clue how to do it!


Why does the thread say probability? DeMorgan's laws are about formal logic. :redface:
Reply 8
Avatar for Zacken
Zacken
22
Original post by Juichiro
Why does the thread say probability? DeMorgan's laws are about formal logic. :redface:


Because you can apply De Morgan laws to probabilistic events.
Original post by Zacken
Because you can apply De Morgan laws to probabilistic events.


Ah, that's interesting. Technically, true statements can be interpreted as statements of 100% probability so I guess it shouldn't be so surprising. Still cool though. :smile:
Reply 10
Avatar for Zacken
Zacken
22
Original post by Juichiro
Ah, that's interesting. Technically, true statements can be interpreted as statements of 100% probability.


I've never really thought about it that way before! That does sound very intriguing. :biggrin:
Well you know that the LHS is saying that something other than A or B or C or D happened. and the P(something happened) = 1
Original post by Juichiro
Ah, that's interesting. Technically, true statements can be interpreted as statements of 100% probability so I guess it shouldn't be so surprising. Still cool though. :smile:


How do you deal with sets of measure zero?
Original post by Zacken
I've never really thought about it that way before! That does sound very intriguing. :biggrin:


Yeah, I found in an intro to logic book.

Original post by Gregorius
How do you deal with sets of measure zero?


What does that mean in terms of propositional logic?
Original post by Juichiro


What does that mean in terms of propositional logic?


I am querying your assertion that true statements can be interpreted in terms of 100% probability. As probability is done in the setting of a measure defined on a sigma algebra of sets, I am wondering what you do with sets of measure zero.
Original post by Gregorius
I am querying your assertion that true statements can be interpreted in terms of 100% probability. As probability is done in the setting of a measure defined on a sigma algebra of sets, I am wondering what you do with sets of measure zero.


No idea. :smile: I just read it in a textbook. I know what it means in propositional logic but not in set theory.

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