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My earlier question became too long so succintly:

What is $P(T\cap \neg B)$ if $P(T)=P(A\cup B)$ (OR-gate) where $P(A)=0$ and $P(B)=0$?

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hhh
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2 Answers2

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If $P(A)=0$ and $P(B)=0$ then $P(A\cup B)=0$ so $P(T)=0$ so $P(T\cap{\rm anything})=0$.

Gerry Myerson
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  • What about this $P(T\cap \neg B)=P(\neg T \cup B)$? $P(\neg T)=1$ because of $P(T)=0$? $P(B)=0$ by definition. What is the intersection or the union? 0 or 1? I do here the step of reductio ad absurdium but I cannot see where. – hhh Feb 26 '13 at 00:51
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    $T\cap\neg B$ is the complement of $\neg T\cup B$, so their probabilities add up to $1$. – Gerry Myerson Feb 26 '13 at 01:04
  • Of course, $P(A)^C=1-P(\neg A)$ -- sorry haven't studied probability for years so have to recap, yes that is correct +1. Thank you. – hhh Feb 26 '13 at 01:07
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The realization $P(A)^C=1-P(\neg A)$ solved a problem in the original thread, namely the case when a system works with zero-probability-working components. It was not reductio ad absurdum. We need a new concept called interval probabilities to specify the phrases such as "working component A" meaning $P(A)>0$.

Let's dig into this with a few examples.

What about if $\neg A$ works and $\neg B$ works i.e. $P(\neg A\cap \neg B)=1$ when $P(A)=0$ and $P(B)=0$?

We have $P(A\cup B)=1-P(\neg A\cap \neg B)$ so $P(\neg A\cap \neg B)=1-P(A\cup B)=1-(0+0)=1.$ So the system -- the OR-port -- works i.e. $P(T)=1$ but how is this possible with a system where each component is broken? Is this Reductio ad absurdum or is the thinking correct i.e. it is possible to get the system working with zero-probability-working components?

System's working when $P(\neg A)$ working and $P(A)=0$. What does it mean that $P(\neg A)$ is working?

We know $P(\neg A)=1$ because of $P(A)=0$ for a part in the OR-system that has no-working component. We could also use the earlier formula for verification i.e. $P(\neg A)=1-P(A)=1-0=1$. Now $$P(T|\neg A)=P(T\cap \neg A)/P(\neg A)=P(T\cap \neg A)=0$$ and $$P(T|\neg B)=P(T\cap \neg B)/P(\neg B)=P(T\cap \neg B)=0$$

because the system cannot work and the part is working $\{0\}\cap \{1\}=\emptyset$. This is because the probability of the system to work is always 0 because each component has zero probability even though the assumption $\neg A$ works aka $P(\neg A)\in [1,0[$ where the term "working" means that the component has non-zero working probability.

INTERVAL PROBABILITIES

I need interval probabilities to attribute specific mathematical meaning to the phrases such as "working component A" aka $P(A)>0$.

Example

Does the system $T$ aka the OR-gate work if $P(A)\in]0,0.1]$ and $P(B)\in [0.9,1[$?

<p>Now $P(\neg A\cap \neg B)=1-P(A\cup B)$ where $P(A\cup B)\in [0.9,1[$ so $P(\neg A\cap \neg B)\in ]0,0.1].$ So the system works with the probability range $]0,0.1]$.</p>
hhh
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