based on the https://djsaunde.github.io/read/books/pdfs/probabilistic%20graphical%20models.pdf at page 91

It claims that $P(l^1 | i^0)$ is 0.389. However, I am unable to get this value or something close to it.

For this i am summing up $i^0$ rows.
$P(g^1, i^0) = (0.3 + 0.05)/2 = 0.175$
$P(g^2, i^0) = (0.4 + 0.25)/2 = 0.375$
$P(g^3, i^0) = (0.3 + 0.7)/2 = 0.5$

$P(L^1 | i^0) = 0.9 * 0.175 + 0.6 * 0.375 + 0.01 * 0.5 = 0.3425$

Subsequently it also claims that $P(i^1 | g^3)$ ~ 0.079 which i am unable to derive this answer also.

I am new to probability so can someone explain why is this the answer to me?


1 Answer 1


First expression to calculate is: $$p(l^1|i^0)=\sum_{g\in\{g^1,g^2,g^3\}} p(l^1|g)p(g|i^0)$$

You've $p(l^1|g)$ as a table, so $p(g|i^0)$ needs to be plugged in for each $g$.

For $g=g^1$:

$$p(g^1|i^0)=\sum_{d\in\{d_0,d_1\}} p(g^1|d,i^0)p(d)=0.3\times0.6+0.05\times0.4=0.2$$

Following this way, I could verify the result, and I believe you can, too.

Beware that the tables are conditional probabilities and you need to account for priors as well (not just mean the two probabilities as you did in your calculations).

  • $\begingroup$ Can you explain your first expression @gunes $\endgroup$
    – aceminer
    Commented Jan 7, 2022 at 19:00
  • $\begingroup$ It's the total probability law. We'll have $p(l|g,i)=p(l|g)$ because of the conditional independence imposed by the network itself (based on the graph) $\endgroup$
    – gunes
    Commented Jan 7, 2022 at 19:06

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