I understand that
1: $P(A|B)=P(A∩B)/P(B)$
and
2: $P(A∣B)=1−P(A′∣B)$
But what about
3: $P(A'|B')=$?
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$\begingroup$ The desired formula might be $$P(A'\mid B')=1-\frac{P(A)-P(A\mid B)P(B)}{1-P(B)}.$$ $\endgroup$– DidOct 21, 2015 at 21:12
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1 Answer
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Applying your results, you can have
- $P(A'\mid B') = \dfrac{P(A'\cap B')}{P(B')}$
- $P(A'\mid B')=1- P(A\mid B')$
You could have something just in terms of $A$ and $B$:
- $P(A'\mid B') = \dfrac{1-P(A\cup B)}{1-P(B)}=1 - \dfrac{P(A)-P(A\cap B)}{1-P(B)}$
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$\begingroup$ P(T) = 0.7, P(F) = 0.6, also, for the question above, does it change anything if P(A) = 0.4 and P(B) = 0.3? $\endgroup$– LyrissOct 21, 2015 at 19:37
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$\begingroup$ You will also need to know the probability of the intersection or another conditional probability $\endgroup$– HenryOct 21, 2015 at 19:42
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$\begingroup$ Ok. Well they are two different scenarios. In P(A) P(B), P(A∩B) = 0. In P(T) P(F), P(T∩F) = 0.55 $\endgroup$– LyrissOct 21, 2015 at 19:46