Mr S...I STILL have a problem with the solution..

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Posted by Subhotosh Khan on September 19, 2002 at 12:27:39:

In Reply to: Re: Mr S...I have a problem with the solution.. posted by Soroban on September 19, 2002 at 09:44:40:

: : After maximizing, we get
: : u = v = 1/sqrt(2)

: : so now u + v = sqrt(2) > 1

: : Arn't you implicitly assuming u + v = 1 when you wrote P = (1-u)(1-v)!

: No, I assumed that: u + (not u) = 1.
**********************************

I assume 'v' is a subset of (not u)

so (not u) = v + (not (u or v))

then u + (not u) = u + v + (not (u or v))

All these are positive numbers.

thus I am back to original dilema. If all those are true and (u+v) > 1 then [u + (not u)] > 1

***********************************************

: : Am I missing something (these probability problems always confuse me..)??

: Understood - I've been there.
: (And can still be found there.)

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Comments:
: : : After maximizing, we get : : : u = v = 1/sqrt(2) : : : so now u + v = sqrt(2) > 1 : : : Arn't you implicitly assuming u + v = 1 when you wrote P = (1-u)(1-v)! : : No, I assumed that: u + (not u) = 1. : ********************************** : <b> : I assume 'v' is a subset of (not u) : so (not u) = v + (not (u or v)) : then u + (not u) = u + v + (not (u or v)) : All these are positive numbers. : thus I am back to original dilema. If all those are true and (u+v) > 1 then [u + (not u)] > 1 : </b> : *********************************************** : : : Am I missing something (these probability problems always confuse me..)?? : : Understood - I've been there. : : (And can still be found there.)

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