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Jonas

Joined: 08 Sep 2005
Posts: 77 Posted: Fri May 26, 2006 12:52 am    Post subject: Simple Proof I am doing a bit of independent study for the first actuarial exam.
The textbook that I have asks that I prove that the complement of A and
B are independent if A and B are independent. I started out with the
following:

I want to show that the probability of the complement of A times the
complement of B is equal to the probability of the intercept of the
complement of A and B. Pr(\A)*Pr(\B)=Pr(\(AB))

[1-Pr(A)]*[1-Pr(B)]

1-Pr(A)-Pr(B)+Pr(A)Pr(B)

1-Pr(A)-Pr(B)+Pr(AB)

I am stuck at this point. If I were to have 1-Pr(AB), in the last
step, I would be done. I can't figuere out where I am making my
mistake.

I also tried working with the notation for conditional probability to
prove that the complements are independent but I had no success.

Any good hints or suggestions that you can give would be much
appreciated. Biba
science forum beginner

Joined: 14 Feb 2006
Posts: 4 Posted: Fri May 26, 2006 7:46 am    Post subject: Re: Simple Proof It goes like this: A^c - complement of A B^c - complement of B A+B - union of A and B (not necesserly disjoint) P(A+B)=P(A)+P(B)-P(AB) Then you have: P(A^cB^c)=1-P(A+B) =1-P(A)-P(B)+P(A)P(B) =1-(1-P(A^c))-(1-P(B^c))+(1-P(A^c))(-1P(B^c)) =1-1+P(A^c)-1+P(B^c)+1-P(B^c)-P(A^c)+P(A^c)P(B^c) =P(A^c)P(B^c) which proves the claim. I hope it helps Jonas

Joined: 08 Sep 2005
Posts: 77 Posted: Fri May 26, 2006 1:11 pm    Post subject: Re: Simple Proof This helps very much so. Thanks for your help. I see where I made my
mistake. I think that the proof you give below can be shortened by
saying that the step after the second could be 1-[P(A+B)] then it can
be said that P[(AB)^c].

Biba wrote:
 Quote: It goes like this: A^c - complement of A B^c - complement of B A+B - union of A and B (not necesserly disjoint) P(A+B)=P(A)+P(B)-P(AB) Then you have: P(A^cB^c)=1-P(A+B) =1-P(A)-P(B)+P(A)P(B) =1-(1-P(A^c))-(1-P(B^c))+(1-P(A^c))(-1P(B^c)) =1-1+P(A^c)-1+P(B^c)+1-P(B^c)-P(A^c)+P(A^c)P(B^c) =P(A^c)P(B^c) which proves the claim. I hope it helps Jonas

Joined: 08 Sep 2005
Posts: 77 Posted: Fri May 26, 2006 1:46 pm    Post subject: Re: Simple Proof What I previously wrote was wrong. Please disregard it. VijaKhara@gmail.com
science forum beginner

Joined: 30 Sep 2005
Posts: 26 Posted: Sat May 27, 2006 10:20 pm    Post subject: Re: Simple Proof Well, I know I am wrong but I cannot find out what is my wrong point.
To me , in Biba's proof, he doesn't use anything relating to the
assumtion that A and B are independent.
In BIBA's proof:
P(A+B)=P(A)+P(B)-P(AB) this identity holds for every events.
P(A^cB^c)=1-P(A+B) This holds because of De Morgan's law:
A^cB^c=(A+B)^c for all events
Plz Figure out where Biba used the assumtion that A and B are
independent to come to the result!

Thanks Mike Terry
science forum Guru Wannabe

Joined: 02 May 2005
Posts: 137 Posted: Sun May 28, 2006 12:13 am    Post subject: Re: Simple Proof <VijaKhara@gmail.com> wrote in message
 Quote: Well, I know I am wrong but I cannot find out what is my wrong point. To me , in Biba's proof, he doesn't use anything relating to the assumtion that A and B are independent. In BIBA's proof: P(A+B)=P(A)+P(B)-P(AB) this identity holds for every events. P(A^cB^c)=1-P(A+B) This holds because of De Morgan's law: A^cB^c=(A+B)^c for all events Plz Figure out where Biba used the assumtion that A and B are independent to come to the result! Thanks

Please quote Biba's proof and we might be able to help you!

Mike. VijaKhara@gmail.com
science forum beginner

Joined: 30 Sep 2005
Posts: 26 Posted: Sun May 28, 2006 2:59 am    Post subject: Re: Simple Proof Hi there, the following is his proof:
It goes like this:

A^c - complement of A
B^c - complement of B
A+B - union of A and B (not necesserly disjoint)
P(A+B)=P(A)+P(B)-P(AB)

Then you have:

P(A^cB^c)=1-P(A+B)
=1-P(A)-P(B)+P(A)P(B)
=1-(1-P(A^c))-(1-P(B^c))+(1-P(A^c))(-1P(B^c))
=1-1+P(A^c)-1+P(B^c)+1-P(B^c)-P(A^c)+P(A^c)P(B^c)
=P(A^c)P(B^c)

which proves the claim.

Thanks
Mike Terry wrote:
 Quote: VijaKhara@gmail.com> wrote in message news:1148768450.895960.84020@y43g2000cwc.googlegroups.com... Well, I know I am wrong but I cannot find out what is my wrong point. To me , in Biba's proof, he doesn't use anything relating to the assumtion that A and B are independent. In BIBA's proof: P(A+B)=P(A)+P(B)-P(AB) this identity holds for every events. P(A^cB^c)=1-P(A+B) This holds because of De Morgan's law: A^cB^c=(A+B)^c for all events Plz Figure out where Biba used the assumtion that A and B are independent to come to the result! Thanks Please quote Biba's proof and we might be able to help you! Mike. Mike Terry
science forum Guru Wannabe

Joined: 02 May 2005
Posts: 137 Posted: Sun May 28, 2006 2:07 pm    Post subject: Re: Simple Proof <VijaKhara@gmail.com> wrote in message
 Quote: Hi there, the following is his proof: It goes like this: A^c - complement of A B^c - complement of B A+B - union of A and B (not necesserly disjoint) P(A+B)=P(A)+P(B)-P(AB) Then you have: P(A^cB^c)=1-P(A+B)

=1-P(A)-P(B)+P(AB) # as you note below
=1-P(A)-P(B)+P(A)P(B) # INDEPENDENT EVENTS

 Quote: =1-P(A)-P(B)+P(A)P(B) =1-(1-P(A^c))-(1-P(B^c))+(1-P(A^c))(-1P(B^c)) =1-1+P(A^c)-1+P(B^c)+1-P(B^c)-P(A^c)+P(A^c)P(B^c) =P(A^c)P(B^c)

The above is correct but maybe over complex? How about just:

=1-P(A)-P(B)+P(A)P(B) # INDEPENDENT EVENTS
=(1-P(A))(1-P(B))
=P(A^c)P(B^c)

Mike.

 Quote: which proves the claim. Thanks Mike Terry wrote: VijaKhara@gmail.com> wrote in message news:1148768450.895960.84020@y43g2000cwc.googlegroups.com... Well, I know I am wrong but I cannot find out what is my wrong point. To me , in Biba's proof, he doesn't use anything relating to the assumtion that A and B are independent. In BIBA's proof: P(A+B)=P(A)+P(B)-P(AB) this identity holds for every events. P(A^cB^c)=1-P(A+B) This holds because of De Morgan's law: A^cB^c=(A+B)^c for all events Plz Figure out where Biba used the assumtion that A and B are independent to come to the result! Thanks Please quote Biba's proof and we might be able to help you! Mike. Biba
science forum beginner

Joined: 14 Feb 2006
Posts: 4 Posted: Mon May 29, 2006 9:09 am    Post subject: Re: Simple Proof Mike Terry wrote:
 Quote: =1-P(A)-P(B)+P(A)P(B) # INDEPENDENT EVENTS =(1-P(A))(1-P(B)) =P(A^c)P(B^c)

Sure, I just wanted to write down every step of the proof, in details.  Display posts from previous: All Posts1 Day7 Days2 Weeks1 Month3 Months6 Months1 Year Oldest FirstNewest First
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