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GATE 2024 DA Q.64: solution and answer

GATE 2024 · DAQ.64 · 2 marks · NAT

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

The value of (rounded off to three decimal places).

GATE 2024 DA question 64: Directed acyclic graph with four nodes U, V, W, Z.Tap to open full size
Figure from the GATE 2024 DA paper. It is part of the question.
Figure description

Directed acyclic graph with four nodes U, V, W, Z. Edges: U→V, U→W, V→Z, W→Z (a diamond with U at the top, V and W in the middle, Z at the bottom). Tables: P(U=0)=0.5, P(U=1)=0.5. P(V=0|U) and P(V=1|U): for U=0 it is 0.5 and 0.5; for U=1 it is 0.5 and 0.5. P(W=0|U) and P(W=1|U): for U=0 it is 1 and 0; for U=1 it is 0 and 1. P(Z=0|V,W) and P(Z=1|V,W): (V=0,W=0) is 0.5 and 0.5; (V=0,W=1) is 1 and 0; (V=1,W=0) is 1 and 0; (V=1,W=1) is 0.5 and 0.5.

Answer and solution

ANSWER

.

Official answer: 0.125Source: Official answer key, GATE 2024 DA (IISc Bengaluru)

  1. The network has edges , , , , so the joint distribution factorises as

  2. Prior: from the first table, .

  3. From the table for , row : .

  4. From the table for , row : .

  5. From the table for , row : .

  6. Multiply the four factors:

KEY IDEA

A Bayesian network's joint probability is the product of each node's conditional probability given its parents. Read the one row of each table that matches the assignment asked for, then multiply.

WATCH OUT FOR

Reading the wrong row of a table, for example using the row of the table. Another slip is to treat and as independent of and drop the conditioning. Here is fully determined by , so , not .

From the GATE 2024 DA paper. Every question of this paper · Download the full paper · All 3 Uncertainty questions, with notes

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