GATE DS&AI 2024 | Question: 54

Question

Given the following Bayesian Network consisting of four Bernoulli random variables and the associated conditional probability tables:

\begin{array}{|c|c|}
\hline & P(\cdot) \\
\hline U=0 & 0.5 \\
\hline U=1 & 0.5 \\
\hline
\end{array}
\begin{array}{|c|c|c|}
\hline & P(V=0 \mid \cdot) & P(V=1 \mid \cdot) \\
\hline U=0 & 0.5 & 0.5 \\
\hline U=1 & 0.5 & 0.5 \\
\hline
\end{array}
\begin{array}{|c|c|c|}
\hline & P(W=0 \mid \cdot) & P(W=1 \mid \cdot) \\
\hline U=0 & 1 & 0 \\
\hline U=1 & 0 & 1 \\
\hline
\end{array}
\begin{array}{|c|c|c|c|}
\hline & & P(Z=0 \mid \cdot) & P(Z=1 \mid \cdot) \\
\hline V=0 & W=0 & 0.5 & 0.5 \\
\hline V=0 & W=1 & 1 & 0 \\
\hline V=1 & W=0 & 1 & 0 \\
\hline V=1 & W=1 & 0.5 & 0.5 \\
\hline
\end{array}

The value of $P(U=1, V=1, W=1, Z=1)= \_\_\_\_\_\_\_$ (rounded off to three decimal places).

Answer 1

0.5*0.5*0.5= 0.125

Answer 2

The given directed acyclic graph represents the Bayesian Belief Network (BBN) which represents the conditional dependencies between the random variables.

Here, $Z$ depends on $V$ and $W$ and both $V$ and $W$ depends on $U.$

So the joint probability:

$\mathbb{P}(Z,V,W,U)=\mathbb{P}(Z|V \cap W)*\mathbb{P}(V \cap W)$       $(1)$

Now,

$V$ and $W$ are conditionally independent given $U.$

So,

$\mathbb{P}(V \cap W) = \mathbb{P}(V \cap W | U)\mathbb{P}(U)+\mathbb{P}(V \cap W | \neg U)\mathbb{P}(\neg U)$ 

Since, $V$ and $W$ are conditionally independent so $\mathbb{P}(V \cap W | U) =\mathbb{P}(V| U)*\mathbb{P}(W | U)$ and $\mathbb{P}(V \cap W | \neg U) =\mathbb{P}(V| \neg U)*\mathbb{P}(W | \neg U)$

Hence,

$\mathbb{P}(V \cap W) =\mathbb{P}(V| U)*\mathbb{P}(W | U) \mathbb{P}(U)+\mathbb{P}(V| \neg U)*\mathbb{P}(W | \neg U)\mathbb{P}(\neg U)$ 

So, from $(1)$

$\mathbb{P}(Z,V,W,U)=\mathbb{P}(Z|V \cap W)*[\mathbb{P}(V| U)*\mathbb{P}(W | U) \mathbb{P}(U)+\mathbb{P}(V| \neg U)*\mathbb{P}(W | \neg U)\mathbb{P}(\neg U)]$

  

Therefore,

  

$\mathbb{P}(Z=1,V=1,W=1,U=1)=\mathbb{P}(Z=1|V=1, W=1)*[\mathbb{P}(V=1| U=1)*\mathbb{P}(W=1 | U=1) \mathbb{P}(U=1)+$

$\mathbb{P}(V=1| U=0)*\mathbb{P}(W=1 | U=0)\mathbb{P}(U=0)]$

   

$\mathbb{P}(Z=1,V=1,W=1,U=1)=0.5[0.5*1*0.5+0.5*0]=0.125$