Question

show that$P(A\mid B)=1\text{, then}P\left(B^c\mid A^c\right)=1\text{.}$

$P(A\mid B)=1\text{, then}P\left(B^c\mid A^c\right)=1$

Short answer

The probability of$P(A\mid B)=1\iff B\subseteq A\iff A^c\subseteq B^c\iff P\left(B^c\mid A^c\right)=1$

Step-by-step solution

Step1:P(A∣B)=1

$P(A\mid B)=1$

Logical proof: If A is certain with the knowledge of B this is equivalent to logical formulation:

$B\to A$

There is a logical rule called modus toles that states:

$B\to A\iff A^c\to B^c$

And, in the same way that we converted probability to logic:

$A^c\to B^c\Rightarrow P\left(B^c\mid A^c\right)=1$

Step2:PBc∣Ac=1

$$\begin{gathered} P(A\mid B)=1 \\ \iff \frac{P\left(AB\right)}{P\left(B\right)}=1 \\ \iff P\left(AB\right)=P\left(B\right) \end{gathered}$$

$P\left(A^c{B}^c\right)=P\left(A^c\right)-P\left(A^{c}B\right)$

$=P\left(A^c\right)-\left[P\right(B)-P(AB\left)\right]$

$=P\left(A^c\right)-0$

$=P\left(A^c\right)$

And going in a different direction than before yields:

$$\begin{gathered} P\left(A^c{B}^c\right)=P\left(B^c\right) \\ \iff \frac{P\left(A^c{B}^c\right)}{P\left(B^c\right)}=1 \\ \iff P\left(A^c\mid B^c\right)=1 \end{gathered}$$