Since it took me a considerable amount of time to come up with how to answer this question to Stanford University’s Probabilistic Graphical Models 1: Representation course, I decided I would be posting it online. Not because I want people to copy and paste it to get a full grade, but because, for a couple of hours, I wanted the course to give me a more thorough explanation on how to approach 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 (and the accompanying security motorcade).

image Created by Stanford University.

Calculate P(Accident = 1 | Traffic = 1) and P(Accident = 1 | Traffic = 1, President = 1). Separate your answers with a space, e.g., an answer of

0.15 0.25

means that P(Accident = 1 | Traffic = 1) = 0.15 and P(Accident = 1 | Traffic = 1, President = 1) = 0.25. Round your answers to two decimal places and write a leading zero, like in the example above.

What is given

  • President visiting town

P(P=1)=0.01P(P=0)=1−P(P=1)=0.99 P(P=1) = 0.01 \\ P(P=0) = 1 - P(P=1) = 0.99

  • Car accident on highway

P(A=1)=0.1P(A=0)=1−P(A=1)=0.9 P(A=1) = 0.1 \\ P(A=0) = 1 - P(A=1) = 0.9

  • Traffic jam

P(T=1∣P=0,A=0)=0.1P(T=1∣P=0,A=1)=0.5P(T=1∣P=1,A=0)=0.6P(T=1∣P=1,A=1)=0.9 P(T=1 \mid P=0, A=0) = 0.1 \\ P(T=1 \mid P=0, A=1) = 0.5 \\ P(T=1 \mid P=1, A=0) = 0.6 \\ P(T=1 \mid P=1, A=1) = 0.9

What is asked

  • Given it was observed a traffic jam, what are the odds of having a car accident on highway

P(A=1∣T=1)=P(A=1,T=1)P(T=1)=0.05040.1449=0.3478261≈0.35 \begin{aligned} P(A=1 \mid T=1) &= \frac{P(A=1 , T=1)}{P(T=1)} \\ &= \frac{0.0504}{0.1449} \\ &= 0.3478261 \\ &\approx 0.35 \end{aligned}

  • Given it was observed a traffic jam and the president visit to town, what are the odds of having a car accident on highway

P(A=1∣T=1,P=1)=P(A=1,T=1,P=1)P(T=1,P=1)=0.00090.0063=0.1428571≈0.14 \begin{aligned} P(A=1 \mid T=1, P=1) &= \frac{P(A=1 , T=1, P=1)}{P(T=1, P=1)} \\ &= \frac{0.0009}{0.0063} \\ &= 0.1428571 \\ &\approx 0.14 \end{aligned}

Development

P(A=1,T=1)=P(P=0,A=1,T=1)+P(P=1,A=1,T=1)=0.0495+0.0009=0.0504 \begin{aligned} P(A=1 , T=1) &= P(P=0, A=1 , T=1) + P(P=1, A=1 , T=1) \\ &= 0.0495 + 0.0009 \\ &= 0.0504 \end{aligned}

P(P=0,A=1,T=1)=P(P=0)×P(A=1)×P(T=1∣P=0,A=1)=0.99×0.1×0.5=0.0495 \begin{aligned} P(P=0, A=1 , T=1) &= P(P=0) \times P(A=1) \times P(T=1 \mid P=0, A=1) \\ &= 0.99 \times 0.1 \times 0.5 \\ &= 0.0495 \end{aligned}

P(P=1,A=1,T=1)=P(P=1)×P(A=1)×P(T=1∣P=1,A=1)=0.01×0.1×0.9=0.0009 \begin{aligned} P(P=1, A=1 , T=1) &= P(P=1) \times P(A=1) \times P(T=1 \mid P=1, A=1) \\ &= 0.01 \times 0.1 \times 0.9 \\ &= 0.0009 \end{aligned}

P(T=1)=P(T=1,P=0,A=0)+P(T=1,P=0,A=1)+P(T=1,P=1,A=0)+P(T=1,P=1,A=1)=0.0891+0.0495+0.0054+0.0009=0.1449 \begin{aligned} P(T=1) &= P(T=1, P=0, A=0) + P(T=1, P=0, A=1) + P(T=1, P=1, A=0) + P(T=1, P=1, A=1) \\ &= 0.0891 + 0.0495 + 0.0054 + 0.0009 \\ &= 0.1449 \end{aligned}

P(T=1,P=0,A=0)=P(P=0)×P(A=0)×P(T=1∣P=0,A=0)=0.99×0.9×0.1=0.0891 \begin{aligned} P(T=1, P=0, A=0) &= P(P=0) \times P(A=0) \times P(T=1 \mid P=0, A=0) \\ &= 0.99 \times 0.9 \times 0.1 \\ &= 0.0891 \end{aligned}

P(T=1,P=0,A=1)=P(P=0)×P(A=1)×P(T=1∣P=0,A=1)=0.99×0.1×0.5=0.0495 \begin{aligned} P(T=1, P=0, A=1) &= P(P=0) \times P(A=1) \times P(T=1 \mid P=0, A=1) \\ &= 0.99 \times 0.1 \times 0.5 \\ &= 0.0495 \end{aligned}

P(T=1,P=1,A=0)=P(P=1)×P(A=0)×P(T=1∣P=1,A=0)=0.01×0.9×0.6=0.0054 \begin{aligned} P(T=1, P=1, A=0) &= P(P=1) \times P(A=0) \times P(T=1 \mid P=1, A=0) \\ &= 0.01 \times 0.9 \times 0.6 \\ &= 0.0054 \end{aligned}

P(T=1,P=1)=P(A=0,T=1,P=1)+P(A=1,T=1,P=1)=0.0054+0.0009=0.0063 \begin{aligned} P(T=1, P=1) &= P(A=0, T=1, P=1) + P(A=1, T=1, P=1) \\ &= 0.0054 + 0.0009 \\ &= 0.0063 \end{aligned}