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check they are right way round in survey, its very late

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Robin Clark 2011-01-09 00:44:16 +00:00
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survey/survey.tex
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@ -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
$$
P(S_k|B_n) = \frac{5.10^{-8} \; 0.1 }{ 10^{-7}} = 0.05 = 5\%
$$
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