From 925d55089021f28a5456506a686a5553133cad55 Mon Sep 17 00:00:00 2001
From: Robin Clark <robin48gx@gmail.com>
Date: Sun, 9 Jan 2011 00:44:16 +0000
Subject: [PATCH] check they are right way round in survey, its very late

---
 survey/survey.tex | 62 +++++++++++++++++++++++++++++------------------
 1 file changed, 39 insertions(+), 23 deletions(-)

diff --git a/survey/survey.tex b/survey/survey.tex
index 8a596d2..2e3d3db 100644
--- a/survey/survey.tex
+++ b/survey/survey.tex
@@ -53,16 +53,35 @@ abd let $S$ be a system level failure mode.
 
 We can say that the conditional probability of $S$ given $B$ is denoted as 
 \begin{equation}
-\label{eqn:bayes1}
- P(S|B) = P(S \cap B) / P(S)
+\label{eqn:condprob}
+ P(S|B) = \frac{P(S \cap B)}{P(S)}
 \end{equation}
 
-Or in other words we can say that the probability of $B$ and $S$ occurring
-divided by the probability of $S$ occurring due to any cause, is the probability
-the $B$ caused $S$. We can call this the {\em conditional probability} of $S$ given $B$.
+%Or in other words we can say that the probability of $B$ and $S$ occurring
+%divided by the probability of $S$ occurring due to any cause, is the probability
+%the $B$ caused $S$. 
+We can call this the {\em conditional probability} of $S$ given $B$.
 Re-arranging \ref{eqn:bayes1}
 
-$$ P(S|B) P(S)  = P(S \cap B)  $$
+$$ P(S) P(S|B)   = P(S \cap B)  $$
+
+The inverse condition, $B$ given $S$ is
+
+$$ P(B) P(B|S)   = P(S \cap B)  $$
+
+As for one being the cause of the other, both equations must be equal,
+we can state,
+
+$$ P(B) P(B|S)   = P(S \cap B) = P(S) P(S|B)  $$
+
+we can now re-arrange the equation to remove the intersection $P(S \cap B)$ term
+thus
+
+\begin{equation}
+\label{eqn:bayes1}
+ P(S|B) =  \frac{P(S) P(B|S)}{P(B)} .
+\end{equation}
+
 
 \paragraph{Multiple Events and  conditional Probability}
 
@@ -81,27 +100,24 @@ Thus if B is any event, we can apply bayes theorem
 to determine the statistical likelihood that a given failure mode $B_n$
 will cause the system level error $S_k$
 
-IN ENGLEEEESH Inverse causality.....
-Prob $B_n$ caused $S_k$ is the prob $S_k$ caused by $B_n$ divided by prob of $B_n$
+%IN ENGLEEEESH Inverse causality.....
+%Prob $B_n$ caused $S_k$ is the prob $S_k$ caused by $B_n$ divided by prob of $B_n$
 
 $$
 P(S_k|B_n) = \frac{P(S_k) \; P(B_n | S_k) }{P(B_n)}
-%P(B|S_k) = \frac{P(S_k | B)\, P(B)}{P(S_k)}
 $$
-%%% because  the probability of $B_n$ in the sample space SS
-%%%is the sum of all probabilities off all failure modes in the indexed set $SS$
-%%%multiplied by the probability of each failure mode causing
-%%%the system failure mode $S_k$.
-%%%
-%%%$$ 
-%%%P(B_n) = {\sum_j^n P(B_j \cap S_k)} = {\sum_j^n P(B_n|A_i) P(A_i)}
-%%%$$
-%%%
-%%%we can express this as
-%%%\begin{equation}
-%%%\label{eqn:bayes2}
-%%% P(S_k|B) = \frac{P(S_k) \; P(B_n|S_k)}{ \sum__{j=1}^{n}  P(B_j)P(S_k | B_j). } 
-%%%\end{equation}
+
+For example were we to have a component that has a failure mode $B_n$ with an MTTF of $10^{-7}$ hours
+and its associated system failure mode $S_k$ has a MTTF of $5.10^{-8}$ hours, and given that
+when the system error $S_k$ occurs, there is a 10\% probability that $B_n$ had occured, we can determine
+the probability that $S_k$ is caused by $B_n$ thus
+
+
+$$
+P(S_k|B_n) = \frac{5.10^{-8} \; 0.1  }{ 10^{-7}} = 0.05 = 5\%
+$$
+
+
 
 RESTRICTIONS: