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Proof is as follows:
Pr(A) = Pr(B)*Pr(A|B) + Pr(Bcomplement)*Pr(A|Bcomplement) Pr(Bcomplement) = 1-Pr(B) =>Pr(A) = Pr(B)*Pr(A|B) + (1-Pr(B))*Pr(A|Bcomplement) Pr(A) = Pr(A|B) (defn of independence) =>Pr(A) = Pr(B)*Pr(A) + (1-Pr(B))*Pr(A|Bcomplement) =>(1-Pr(B))*Pr(A) = (1-Pr(B))*Pr(A|Bcomplement) If Pr(B) = 1 (or 0) then statistical independence is meaningless, so: =>Pr(A) = Pr(A|Bcomplement) QED The second part is obvious once this is proven |
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Ozzy -
(Assuming you mean very basic algebra...) A lot of the time, the biggest stumbling block to learning algebra is the introduction of letters. Try using it without "x" but with "__" instead, so give him: __ + 5 = 8 Have him solve a bunch of those. Then give him x + 5 = 8 etc. for the same exact problems When he has trouble with them, show him that it's the same exact problems he had before. It's also possible that he can solve __ + 5 = 8 by mentally subtracting 5 from 8, but doesn't get the idea of subtracting from both sides -- that's a bit harder to get across, usually, because it's the "unconscious" response - it's so obvious to do it that he doesn't realize what he's doing. That can be tough, but usually using more concrete examples - something like (if he's a football fan): If the Bears score 6 field goals and the Redskins score 5 field goals, how many points do the Redskins need to tie the game? 18 = 15 + x (or 6 = 5 + x) and then explain that it's one more field goal; you can tell because if you take away the number of field goals the Redskins scored, from how many the Bears scored, that leaves remaining how many more field goals they need. Often adding in something the person is very familiar with helps, because they get more interested in it and/or have some additional 'unconscoius help' from their mind being in a more comfortable area. |
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