document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1190835>Question 1190835</A>: <I><SPAN STYLE=\"word-break: break-all;\"> If the inverse conditional posterior probability formulas are<BR>\n" );
document.write( "P(A|B) = [P(B|A)*P(A)]/[P(B|A)*P(A)+P(B|¬A)*P(¬A)], and<BR>\n" );
document.write( "P(¬A|¬B) = [P(¬B|¬A)*P(¬A)]/[P(¬B|¬A)*P(¬A)+P(¬B|A)*P(A)],<BR>\n" );
document.write( "then what are the posterior probability equations for P(A|¬B), and P(¬A|B)?<BR>\n" );
document.write( "P(A|¬B) = ?<BR>\n" );
document.write( "P(¬A|B) = ?\r<BR>\n" );
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document.write( "Thanks. </SPAN>\n" );
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document.write( "You only really need to memorize or write down on a reference sheet one formula, and it is<BR>\n" );
document.write( "P(A|B) = [P(B|A)*P(A)]/[P(B|A)*P(A)+P(B|¬A)*P(¬A)]\r<BR>\n" );
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document.write( "This is because the second formula is derived from the first one. \r<BR>\n" );
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document.write( "Notice how replacing every A with ¬A gets us this:<BR>\n" );
document.write( "P(¬A|B) = [P(B|¬A)*P(¬A)]/[P(B|¬A)*P(¬A)+P(B|¬¬A)*P(¬¬A)]<BR>\n" );
document.write( "If you see ¬¬A, then that's the same as A<BR>\n" );
document.write( "P(¬A|B) = [P(B|¬A)*P(¬A)]/[P(B|¬A)*P(¬A)+P(B|A)*P(A)]\r<BR>\n" );
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document.write( "Now if you replaced every B with ¬B, then you should end up with the second formula you mentioned.\r<BR>\n" );
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document.write( "Use this idea to compute what P(A|¬B) would be. <BR>\n" );
document.write( "You would of course need to start back over with the P(A|B) formula. Then replace every B with ¬B and simplify any ¬¬B into B.\r<BR>\n" );
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document.write( "Side note:<BR>\n" );
document.write( "P(B) = P(B|A)*P(A) + P(B|¬A)*P(¬A)<BR>\n" );
document.write( "due to the law of total probability<BR>\n" );
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