TU Wien:Einführung in wissensbasierte Systeme VU (Egly)/Übungen SS12/Blatt 4 - Beispiel 4

Suppose we have a bag of three biased coins, $a$, $b$, and $c$, with probabilities of coming up heads of 25%, 50% and 70%, respectively. One coin is drawn randomly from the bag (with equal likelihood of drawing each of the three coins), and then the coin is flipped three times to generate the outcomes $X_{1}$, $X_{2}$, and $X_{3}$

• Create a Bayesian network corresponding to this setup and define the necessary conditional probability tables.
• Calculate which coin was most likely to have been drawn from the bag if the observed flips come out heads twice and tails once.

Lösungsvorschlag

${\begin{aligned}X_{i}&={\text{flip i turns out heads}},i\in \{1,2,3\}\\C&={\text{coin is drawn}},C\in \{a,b,c\}\end{aligned}}$

Bayesian network

${\begin{aligned}{\text{P}}(X_{i}|a)&=0.25\\{\text{P}}(X_{i}|b)&=0.50\\{\text{P}}(X_{i}|c)&=0.70\\{\text{P}}(C)&={\frac {1}{3}}\end{aligned}}$

Für jede Münze wollen wir die conditional probabilities haben (wie hoch ist die Wahrscheinlichkeit, dass C gezogen wurde, wenn die Würfe ...):

${\text{P}}(C|X_{1},X_{2},\neg X_{3})={\frac {{\text{P}}(X_{1},X_{2},\neg X_{3}|C)*{\text{P}}(C)}{P(X_{1},X_{2},\neg X_{3})}}$

also mal für jede Münze (wie hoch ist die Warhscheinlichkeit, dass die Würfe ..., wenn C gezogen wurde):

${\begin{aligned}{\text{P}}(X_{1},X_{2},\neg X_{3}|a)*{\text{P}}(a)&=(0.25*0.25*0.75)*{\frac {1}{3}}\approx 0.0156\\{\text{P}}(X_{1},X_{2},\neg X_{3}|b)*{\text{P}}(b)&=(0.50*0.50*0.50)*{\frac {1}{3}}\approx 0.0417\\{\text{P}}(X_{1},X_{2},\neg X_{3}|c)*{\text{P}}(b)&=(0.70*0.70*0.30)*{\frac {1}{3}}=0.049\\\end{aligned}}$

jetzt für die Würfe allgemein, ohne es auf eine bestimmte Münze zu beziehen:

${\begin{aligned}{\text{P}}(X_{1},X_{2},\neg X_{3})&=\sum _{C\in \{a,b,c\}}{\text{P}}(X_{1},X_{2},\neg X_{3}|C)*{\text{P}}(C)\\&=0.0156+0.0417+0.049\\&=0.1063\end{aligned}}$

und jetzt einsetzen:

${\begin{aligned}{\text{P}}(a|X_{1},X_{2},\neg X_{3})&={\frac {0.0156}{0.1063}}\approx 0.1468\\{\text{P}}(b|X_{1},X_{2},\neg X_{3})&={\frac {0.0417}{0.1063}}\approx 0.3923\\{\text{P}}(c|X_{1},X_{2},\neg X_{3})&={\frac {0.049}{0.1063}}\approx 0.461\\\end{aligned}}$

die Antwort ist also offensichtlich $c$