Suppose the robot is randomly placed into the world, so that it can be in any location with equal probability (that is, we use a uniform prior). If we now observe a door, what is the probability that we really are at a location with a door? Try solving this both by hand and using the applet, then compare the results.
- The posterior probability is proportional to the prior times the conditional probability:
P(at door | sense door) = P(at door)*P(sense door | at door)/P(sense door) = (1/4)*(4/5)/((1/4)*(4/5) + (3/4)*(1/10)) = (1/5)/(11/40) = 8/11
There are multiple ways to arrive at the same answer. The applet gives a probability of 2/11 = 0.182 for each location containing a door. Since there are four such locations, their total probability is 4*(2/11) = 8/11, which is in agreement with our previous answer.
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