# Conditional Probability with a double complement.

I understand that

1: $P(A|B)=P(A∩B)/P(B)$

and

2: $P(A∣B)=1−P(A′∣B)$

3: $P(A'|B')=$?

• The desired formula might be $$P(A'\mid B')=1-\frac{P(A)-P(A\mid B)P(B)}{1-P(B)}.$$
– Did
Oct 21, 2015 at 21:12

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 $B$:

• $P(A'\mid B') = \dfrac{1-P(A\cup B)}{1-P(B)}=1 - \dfrac{P(A)-P(A\cap B)}{1-P(B)}$
• Interesting, and what would P(T|F')= Oct 21, 2015 at 19:31
• What are $T$ and $F$? Is $T=F'$? Oct 21, 2015 at 19:33
• 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? Oct 21, 2015 at 19:37
• You will also need to know the probability of the intersection or another conditional probability Oct 21, 2015 at 19:42
• 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 Oct 21, 2015 at 19:46