Question

Let A and B be events with probabilities not equal to 0 or 1. Show that if P(B|A) = 1, then P(A0 |B0 ) = 1. You may use the axioms of probability, all the theorems from the notes, and anything that was proven on the homework or in the notes. Hint: Consider slide 9 in chapter 4 for event A and show that P (A|B0 ) = 0.

Answer 1

Here, P(B|A)=1

$\Rightarrow {P(A\cap B)}={P(A)}$

So,

$P(AO|BO)=\frac{P(AO\cap BO)}{P(BO)}$

= $\frac{P(A\cup B)^{O}}{1-P(B)}$ [ from de Morgan law of set $AO\cap BO=(A\cup B)^{O}}$

= $\frac{1-P(A\cup B)}{1-P(B)}$

= $\frac{1-(P(A)+P(B)-P(A\cap B))}{1-P(B)}$ [ Applying formula for P(AUB)]

= $\frac{1-(P(A\cap B)+P(B)-P(A\cap B))}{1-P(B)}$ [ PUTIING $P(A)=P(A\cap B)$ ]

= $\frac{1-P(B)}{1-P(B)}$

=1

now for   $P(A|BO)=\frac{P(A\cap BO)}{P(BO)}$

= $\frac{P(A-B)}{P(BO)}$ [   $A\cap BO=A-B$ ]

= $\frac{P(A)-P(A\cap B)}{P(BO)}$

= $\frac{0}{P(BO)}$ [ As $P(A)=P(A\cap B)$ ]

=0

proved