Q. 3.26

show that $P (A ∣ B) = 1, then P (B^{c} ∣ A^{c}) = 1 .$

$P (A ∣ B) = 1, then P (B^{c} ∣ A^{c}) = 1$

Verified

Answer

The probability of $P (A ∣ B) = 1 \Leftrightarrow B \subseteq A \Leftrightarrow A^{c} \subseteq B^{c} \Leftrightarrow P (B^{c} ∣ A^{c}) = 1$

1Step1: P ( A ∣ B ) = 1

$P (A ∣ B) = 1$

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:

$B \to A \Leftrightarrow A^{c} \to B^{c}$

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

$A^{c} \to B^{c} \Rightarrow P (B^{c} ∣ A^{c}) = 1$

2Step2: P B c ∣ A c = 1

$P (A ∣ B) = 1 \Leftrightarrow \frac{P (A B)}{P (B)} = 1 \Leftrightarrow P (A B) = P (B)$

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

$= P (A^{c}) - [P (B) - P (A B)]$

$= P (A^{c}) - 0$

$= P (A^{c})$

And going in a different direction than before yields:

$P (A^{c} B^{c}) = P (B^{c}) \Leftrightarrow \frac{P (A^{c} B^{c})}{P (B^{c})} = 1 \Leftrightarrow P (A^{c} ∣ B^{c}) = 1$