I just installed a home security system that is supposed to alert me on my smartphone if someone tries to break into my house. Unfortunately, the smartphone app doesn't seem to be compatible with iOS 10, so it doesn't work; fortunately the security system also sounds an audible alarm, and my upstairs neighbor John and my downstairs neighbor Mary will probably hear the alarm and call me if it goes off. They're a bit hard of hearing, though, so they might call me even if it doesn't go off or they may not hear it go off if it does. To make matters worse, the alarm may incorrectly go off for other reasons, like if there's a big storm outside. Suppose that on any given day, the probability of a burglary is 0.001 and the probability of a big storm is 0.002. If the alarm goes off, there's a 90% chance John will call me. There's a 5% chance he'll call me even if the alarm doesn't go off. Meanwhile, there's a 70% chance Mary will call me if the alarm goes off and a 1% chance she'll call if it doesn't. There's a 95% chance the alarm will go off if there's both a storm and a burglary, a 94% chance if there's a burglary but no storm, a 29% chance if there's a storm but no burglary, and a 0.1% chance if there's neither a burglary nor a storm. Question: If both Mary and John call me, what's the probability that there's a storm? In other words, what is P(S=t | J=t, M=t)? You can again answer this using the Variable Elimination (VE) algorithm. This time it should be a little less work than in module 9.6 because we already have some evidence (both John and Mary call me). Recall the following slide on how to handle evidence: In other words, once you computed P ( S , J = t , M = t ) using VE, you can compute the conditional P ( S | J = t , M = t ) as:
I just installed a home security system that is supposed to alert me on my smartphone if someone tries to break into my house. Unfortunately, the smartphone app doesn't seem to be compatible with iOS 10, so it doesn't work; fortunately the security system also sounds an audible alarm, and my upstairs neighbor John and my downstairs neighbor Mary will probably hear the alarm and call me if it goes off. They're a bit hard of hearing, though, so they might call me even if it doesn't go off or they may not hear it go off if it does. To make matters worse, the alarm may incorrectly go off for other reasons, like if there's a big storm outside.
Suppose that on any given day, the probability of a burglary is 0.001 and the probability of a big storm is 0.002. If the alarm goes off, there's a 90% chance John will call me. There's a 5% chance he'll call me even if the alarm doesn't go off. Meanwhile, there's a 70% chance Mary will call me if the alarm goes off and a 1% chance she'll call if it doesn't. There's a 95% chance the alarm will go off if there's both a storm and a burglary, a 94% chance if there's a burglary but no storm, a 29% chance if there's a storm but no burglary, and a 0.1% chance if there's neither a burglary nor a storm.
Question: If both Mary and John call me, what's the probability that there's a storm? In other words, what is P(S=t | J=t, M=t)?
You can again answer this using the Variable Elimination (VE) algorithm. This time it should be a little less work than in module 9.6 because we already have some evidence (both John and Mary call me). Recall the following slide on how to handle evidence:
In other words, once you computed P ( S , J = t , M = t ) using VE, you can compute the conditional P ( S | J = t , M = t ) as:
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