Added a partition c diagram to describe unitary

state using set theory and an Euler diagram.

component_failure_modes_definition/component_failure_modes_definition.tex

@@ -720,6 +720,22 @@ of interest when analysing systems from a statistical perspective.
 This is of interest for the application of conditional probability calculations
 such as Bayes theorem~\cite{probstat}.
 
+Another way to view this is to consider the failure modes of
+component, with the $OK$ state, as a universal set $\Omega$, where
+all sets within $\Omega$ are partitioned.
+Figure \ref{fig:partitioncfm} shows a partitioned set representing 
+component failure modes $\{ B_1 ... B_8, OK \}$ obeying unitary state conditions.
+
+\begin{figure}[h]
+ \centering
+ \includegraphics[width=350pt,keepaspectratio=true]{./component_failure_modes_definition/partitioncfm.jpg}
+ % partition.jpg: 510x264 pixel, 72dpi, 17.99x9.31 cm, bb=0 0 510 264
+ \caption{Base Component Failure Modes with OK mode partitioned}
+ \label{fig:partitioncfm}
+\end{figure}
+
+
+
 
 %%-
 %%- Need a complete and more complicated UML diagram here

survey/survey.tex

@@ -665,15 +665,19 @@ the probability that $S_k$ is caused by $B_n$ thus
 
 
 $$
-P(S_k|B_n) = \frac{5.10^{-8} \; 0.1  }{ 10^{-7}} = 0.05 = 5\%
+P(S_k|B_n) = \frac{5.10^{-8} .\; 0.1  }{ 10^{-7}} = 0.05 = 5\%
 $$
 
 
-To take an example from the diagram (see figure \ref{fig:partitionbcfm2}), where the base component fault cannot 
+Taking an example from the diagram (figure \ref{fig:partitionbcfm2}), where the base component fault cannot 
-lead to the system failure $S_k$. Taking say $B_9$ which does not overlap with $S_k$
+lead to the system failure $S_k$. 
+Taking say $B_9$ which does not overlap with $S_k$ (i.e. $B_9 \cap S_k = \emptyset  $),
 we can see that $P(S_k | B_9) = 0$. 
-Bayes theorem applied to $B_9$ becomes $P(S_k|B_9) = \frac{P(B_9) \; 0  }{ 10^{-7}}$
+Bayes theorem applied to $B_9$ becomes 
-As this is a factor in the numerator,
+
+$$P(S_k|B_9) = \frac{P(B_9) .\; 0  }{ 10^{-7}} = 0 = 0\%$$
+
+As $ P(S_k | B_n)$  is a factor in the numerator,
 the application of bayes theorem to $B_9$ being a cause for $S_k$ has a probability
 of zero, as we would expect.