# Conditional probabilties in the OR-gate $T=A\cdot B$ with zero-probabilities in $A$ and $B$?

My earlier question became too long so succintly:

What are $P(T|A)=P(T\cap A)/P(A)$ and $P(T|B)=P(T\cap B)/P(B)$ if $P(A)=0$ and $P(B)=0$?

I think they are undefined because of the division by zero. How can I specify the conditional probabilities now? Please, note that the basic events $A$ and $B$ depend on $T$ because $T$ consists of them, namely $T=A \cup B$.

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Yes, it is undefined in general. It is generally pointless to ask for the conditional probability of $T$ when $A$ occurs when it is known that $A$ almost surely never happens.

But a meaningful specification in your particular case that $T = A\cup B$ is by some intuitive notion of continuity. For $P(A) \neq 0$, if $C\supseteq A$ we must have $P(C|A) = 1$. Hence one can argue from a subjective interpretation of probability that $P(T|A) = 1$ since $T\supseteq A$. And similarly for $P(T|B)$.

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I cannot understand the point "by some intuitive notion of continuity". –  hhh Feb 26 '13 at 20:22
If we make the "intuitive" assumption that $P(T|A)$ for $T = A\cup B$ is a continuous function of $P(A)$, then by taking the limit $P(A)\to 0$ we have that $P(T|A) = 1$. It is in analogy to the fact that the function $x/x$ is not defined at $x = 0$. But there exists a unique continuous function on $\mathbb{R}$ that equals $x/x$ away from $x=0$. –  Willie Wong Feb 26 '13 at 23:31
The situation is perhaps better illuminated by the even specialer case of $B = \emptyset\implies T = A$. I think it is more than reasonable to assume $P(A|A) = 1$ even for probability zero events $A$. –  Willie Wong Feb 26 '13 at 23:34

I discussed with my teacher and he said that the conditional probability formula is a tautology. If you assume zero probability in some input of the OR, then you have different topology and hence you cannot use the formula like that -- you need reform the problem -- because OR port can be simplified with certain zero-probabilities.

The key is to differentiate the different layers: logic, probability and real life. You can claim whatever you want in the logical layer before you think about probabilities -- even though you may get undefined in probabilistic layer. The real-world can be then anything between the logic and probability.

Example with interval probability

Suppose $T=A\cup B$. Now if $P(B)=0$, then $T=A$. Now if $P(B)\in [0,0.2]$, then $T=A$ if $P(B)=0$ and $T=A\cup B$ if $P(B)\not = 0$. So interval probabilities become pretty messy if you assume extreme probabilities such as $0$.

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