Conditional probabilities summing to one

Question

Let's assume that we have two Bernoulli random variables: $A$ (can be true or false) and $B$ (can be true or false), and further assume we have been given $P(A=\text{true}\mid B=\text{true})$ and $P(A=\text{true}\mid B=\text{false})$.

Is it possible to calculate $P(A=\text{false}\mid B=\text{true})$ and $P(A=\text{false}\mid B=\text{false})$ from this? I think what it must hold is that these four terms must sum to one, i.e. $$P(A=\text{true}\mid B=\text{true}) + P(A=\text{true}\mid B=\text{false}) \\+ P(A=\text{false}\mid B=\text{true}) + P(A=\text{false}\mid B=\text{false}) = 1.$$

Answer 1

No it's incorrect. $$P(A=true|B=true)+P(A=false|B=true)$$ $$=\frac{P(A=true,B=true)}{P(B=true)}+\frac{P(A=false,B=true)}{P(B=true)}$$ $$=\frac{P(B=true)}{P(B=true)}=1$$ Similarly, $$P(A=true|B=false)+P(A=false|B=false)=1$$ Thus, yes you can calculate the variables required, but the equation you wrote was incorrect

Answer 2

Not quite. You have across all joint possibilities $$P(A=\text{true} \text{ and } B=\text{true}) + P(A=\text{true}\text{ and }B=\text{false}) \\\;\,\,+ P(A=\text{false}\text{ and } B=\text{true}) + P(A=\text{false}\text{ and } B=\text{false}) = 1$$

but for conditional probabilities you have two equations $$P(A=\text{true}\mid B=\text{true}) + P(A=\text{false}\mid B=\text{true}) =1\\P(A=\text{true}\mid B=\text{false}) + P(A=\text{false}\mid B=\text{false}) = 1$$

Answer 3

The summation you mention gives sum $2$ (not $1$).

This because: $$P(A=\text{true}\mid B=\text{true})+P(A=\text{false}\mid B=\text{true})=1$$ and: $$P(A=\text{true}\mid B=\text{false})+P(A=\text{false}\mid B=\text{false})=1$$

In both cases if you know one of terms then you also know the other one.