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I'm currently solving the same problem that was posted in this Stack Overflow question but had a question regarding another aspect of the problem. The source of this problem is the course regarding Probabilistic Graphical Models on Coursera. To first draw out the problem:

Consider the following model for traffic jams in a small town, which we assume can be caused by a car accident, or by a visit from the president.

enter image description here

Calculate $P(\text{Accident} = 1\ \vert\ \text{Traffic} = 1)$ and $P(\text{Accident} = 1\ \vert\ \text{Traffic} = 1, \text{President} = 1)$.

I'll only look at the first probability, since it directly shows my question. The solution offered on the question that I linked is:

$$ \begin{align} P(A = 1\ \vert\ T = 1) & = \frac{P(A = 1, T = 1)}{P(T = 1)} \\ \\ P(A = 1, T = 1) & = P(T = 1, A = 1, P = 0) + P(T = 1, A = 1, P = 1) \\\\ P(T = 1, A = 1, P = 0) & = P(T = 1\ \vert\ A = 1, P = 0) P(A = 1) P(P = 0) \\ & = 0.5 \times 0.1 \times 0.99 = 0.0495 \end{align} $$

My question is, why are we assuming that $A$ and $P$ are independent? The particular graph that is given in the problem represents a $v$-structure in Bayesian networks, and as far as I'm aware if the middle node is observed, then the two connected nodes are no longer independent.

Since we observed that there is a traffic jam (i.e., $T = 1$), how can we assume that $P(A = 1, P = 0) = P(A = 1) P(P = 0)$?

I'm obviously misunderstanding something since I got the answer wrong, but I'm just wondering where. Any tips are appreciated. Thanks.

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  • $\begingroup$ independence of $P$ and $A$ seems to be an model assumption that is given in the problem statement. So you can assume that $P$ and $A$ are not independent, but that would be solving another problem. $\endgroup$ – hakanc Dec 12 '20 at 14:29
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Since we observed that there is a traffic jam (i.e. $T=1$), how can we assume that $P(A=1,P=0)=P(A=1)P(P=0)$?

You should decide observance based on asked probability. Although it's given in the question statement, $T$ is not observed when calculating $P(A,P)$. $T$ is observed when calculating $P(A,P|T)$.

Bayesian networks, by their definition, enforce dependence relations. In this one, $A$ and $P$ are dependent if $T$ is given. If not, they're independent.

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  • $\begingroup$ "You should decide observance based on asked probability." This line will probably save me a lot of trouble in the future. I don't know why it's not explicitly mentioned in other lecture videos I've watched. Maybe it's perceived as relatively obvious. $\endgroup$ – Seankala Dec 12 '20 at 21:56

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