Probability Question

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Homework Statement



Suppose that for two events A and B. A [itex]\subseteq[/itex] B. Show that [itex]B^{c} \subseteq A^{c}[/itex].

Homework Equations





The Attempt at a Solution



[itex]1- B \subseteq 1- A[/itex]

I'm not sure where to go from here. Thanks for any help you can provide.
 

Answers and Replies

  • #2
HallsofIvy
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That's really a "set" problem rather than a probability problem. I presume your "1" represents the "universal set" rather than a number?

The "standard" way to prove "[itex]X\subset Y[/itex]" is to start "let [itex]x\in X[/itex]" and, using the definitions and information you are given about X and Y, end with "therefore [itex]x\in Y[/itex]". Here, "X" is the set [itex]B^c[/itex], the complement of B, the set of all objects that are not in B.

So, if [itex]x\in B^c[/itex], then x is not in B. Since A is a subset of B (every member of A is also in B), x is not in A and so [itex]x\in A^c[/itex].
 
  • #3
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Hi HallsofIvy,

Is universal set and null set the same?

thanks,

green
 
  • #4
rollingstein
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Hi HallsofIvy,

Is universal set and null set the same?


No. Not the same.
 
  • #5
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You mean by universal set Ω?

Also how does showing that
x[itex]\in B^{c}[/itex] and x[itex]\in A^{c}[/itex]

show that

[itex]B^{c} \subseteq A^{c}[/itex]?

I'm not seeing this
 
  • #6
HallsofIvy
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Okay, it sounds like you are using [itex]\Omega[/itex] to mean the "universal set", the set of all things we are allowing in our sets.

It is NOT true that "[itex]x\in B^c[/itex] and [itex]x\in A^c[/itex]" implies [itex]B^c\subseteq A^c[/itex]" and I did not claim it was. What I showed was that if[itex]x\in B^c[/itex] implies [itex]x\in A^c[/itex], then [itex]B^c\subseteq A^c[/itex].

That follows directly from the definition of "[itex]\subset[/itex]". If you are doing problems like that you surely should know that definition.
 
  • #7
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[itex]B^{c} \subseteq A^{c}[/itex]

This means that B complement is a subset or equal to A complement. If x is an element of B complement and A complement than why must B complement be a subset or equal to A complement? How do I know that A complement is not a subset of B complement?

Since x is an element of A complement and B complement then...

B complement must be a subset or equal to A complement
or
A complement must be a subset or equal to B complement

how do I know which one is true?
 
  • #8
haruspex
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If x is an element of B complement and A complement than why must B complement be a subset or equal to A complement? How do I know that A complement is not a subset of B complement?

Since x is an element of A complement and B complement then...

B complement must be a subset or equal to A complement
or
A complement must be a subset or equal to B complement
None of that makes sense. If some x is an element of B complement and A complement then all that proves is that their intersection is non-empty. You need to think in terms "if x is an element of ... then x is an element of ..." You're trying to prove that Bc ⊆ Ac. Can you fill in the "..." ?
 
  • #9
HallsofIvy
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[itex]B^{c} \subseteq A^{c}[/itex]

This means that B complement is a subset or equal to A complement. If x is an element of B complement and A complement than why must B complement be a subset or equal to A complement?
I didn't say that! A single "x" tells us nothing. The point was that this is true for all members of B complement. That is the definition of "[itex]X\subseteq Y[/itex]": for all [itex]x\in X[/itex], [itex]x\in Y[/itex].

How do I know that A complement is not a subset of B complement?

Since x is an element of A complement and B complement then...

B complement must be a subset or equal to A complement
or
A complement must be a subset or equal to B complement

how do I know which one is true?
By knowing what "subset" and "member" mean! Go back and look at my argument, showing that any element of B complement must be a member of A complement.

And that is the definition of "B complement is a subset of A complement".

If your are asking, as you appear to be, "How can I prove this without knowing any of the definitions", you can't!
 

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