SOLUTION: If the inverse conditional posterior probability formulas are P(A|B) = [P(B|A)*P(A)]/[P(B|A)*P(A)+P(B|¬A)*P(¬A)], and P(¬A|¬B) = [P(¬B|¬A)*P(¬A)]/[P(¬B|¬A)*P(¬A)+P(¬B|

SOLUTION: If the inverse conditional posterior probability formulas are P(A|B) = [P(B|A)P(A)]/[P(B|A)P(A)+P(B|¬A)P(¬A)], and P(¬A|¬B) = [P(¬B|¬A)P(¬A)]/[P(¬B|¬A)*P(¬A)+P(¬B|

Question 1190835: If the inverse conditional posterior probability formulas are
P(A|B) = [P(B|A)*P(A)]/[P(B|A)*P(A)+P(B|¬A)*P(¬A)], and
P(¬A|¬B) = [P(¬B|¬A)*P(¬A)]/[P(¬B|¬A)*P(¬A)+P(¬B|A)*P(A)],
then what are the posterior probability equations for P(A|¬B), and P(¬A|B)?
P(A|¬B) = ?
P(¬A|B) = ?
Thanks.
Answer by math_tutor2020(3817) (Show Source): You can put this solution on YOUR website!

You only really need to memorize or write down on a reference sheet one formula, and it is
P(A|B) = [P(B|A)*P(A)]/[P(B|A)*P(A)+P(B|¬A)*P(¬A)]

This is because the second formula is derived from the first one.

Notice how replacing every A with ¬A gets us this:
P(¬A|B) = [P(B|¬A)*P(¬A)]/[P(B|¬A)*P(¬A)+P(B|¬¬A)*P(¬¬A)]
If you see ¬¬A, then that's the same as A
P(¬A|B) = [P(B|¬A)*P(¬A)]/[P(B|¬A)*P(¬A)+P(B|A)*P(A)]

Now if you replaced every B with ¬B, then you should end up with the second formula you mentioned.

Use this idea to compute what P(A|¬B) would be.
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.

Side note:
P(B) = P(B|A)*P(A) + P(B|¬A)*P(¬A)
due to the law of total probability

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