Tekst toegevoegd deel 1, nog niet klaar

diff --git a/report/ass2-1.tex b/report/ass2-1.tex
index a7b576e..10b1706 100644 (file)
--- a/report/ass2-1.tex
+++ b/report/ass2-1.tex
@@ -113,16 +113,20 @@ We implemented the distributions in \textit{AILog}, see Listing~\ref{alarm.ail}
 \end{listing}
 
 \section{Queries}
-Using the following queries the probabilities are as follows:\\
+Now that we have modelled the story with the corresponding probabilities, we can have ailog calculate some other probabilities given a some observations. Down below we wrote down some probabilties and the associated ailog output. \\
+The chance that a burglary happens given that Watson calls is greater than the chance that a burglary happens without this observations, as is observerd by the difference between a and b. This makes sense as Watson calls rightly in 80\% of the time. So when Holmes receives a call by Watson, the chance that the alarm goes of increases. \\
+When we compare b to c, the same mechanisme holds. There are more observations that give evidence for a burglary as both Watson and Gibbons have called in the case of c. \\
+When you take a look at the last case, d, you see that the probability has decreased compared to c. This can be explained by an observation that is added on top of the observations of b; the radio. The variable Radio means that the newcast tells that there was an earhquake. As that is also a reason why the alarm could go of, but has nothing to do with a burglary, it decreases the probability of a burglary.
+%We kunnen misschien de kans uitrekenen dat Watson en Gibbons allebei foutief bellen? Daar mis je denk ik info over of niet?
 \begin{enumerate}[a)]
        \item $P(\text{Burglary})=
                0.002737757092501968$
        \item $P(\text{Burglary}|\text{Watson called})=
                0.005321803679438259$
-       \item $P(\text{Burglary}|\text{Watson called}\wedge\text{Gibbons called})=
+       \item $P(\text{Burglary}|\text{Watson called},\text{Gibbons called})=
                0.11180941544755249$
-       \item $P(\text{Burglary}|\text{Watson called}\wedge\text{Gibbons called}
-               \wedge\text{Radio})=0.01179672476662423$
+       \item $P(\text{Burglary}|\text{Watson called},\text{Gibbons called}
+               , \text{Radio})=0.01179672476662423$
 \end{enumerate}
 
 \begin{listing}[H]
@@ -157,7 +161,11 @@ Answer: P(burglar|Obs)=[0.01179672476662423,0.015584580594335082].
        \end{minted}
 \end{listing}
 
+ToDO write down the most probable explanation for the observed evidence
+
 \section{Comparison with manual calculation}
+ToDO: english.
+When we let ailog calculate the probability of alarm. %Wat is Obs hier??
 Querying the \textit{Alarm} variable gives the following answer:
 \begin{minted}{prolog}
        ailog: predict alarm.
@@ -166,11 +174,10 @@ Querying the \textit{Alarm} variable gives the following answer:
          [ok,more,explanations,worlds,help]: ok.
 \end{minted}
 
-Using formula: $P(i_1|C_1)+P(i_2|C_2)(1-P(i_1|C_1))$ we can calculate the
-probability of the \textit{Alarm} variable using variable elimination. This
-results in the following answer: 
-$$0.2*0.0027+0.95*0.0027*(1-0.2*0.0027)=0.00314699654673673$$
-TODOOOOOOOOOOO
+Using the formula for causal independence with a logical OR: \\ $P(Alarm|C_1, C_2) = P(i_1|C_1)+P(i_2|C_2)(1-P(i_1|C_1))$ we can calculate the probability of the \textit{Alarm} variable using variable elimination. This results in the following answer: \\
+$P(Alarm|burglar, earthquake) = P(i_1|burglar)+P(i_2|earthquake)(1-P(i_1|burglar)) =  0.2*0.0027+0.95*0.0027*(1-0.2*0.0027)=0.00314699654673673$ \\
+TODOOOOOOOOOOO %Ik weet niet of we i_1 en i_2 nog door iets anders vervangen moeten worden.
+When you compare the output of ailog and of the variable elimination, you see that they are similar.
 
 \newpage
 \section{Burglary problem with extended information}