Description
1 Probability spaces
Let Ω = {1, 2, 3, 4, 5, 6}, A = {∅, Ω, {1, 2}, {3, 4, 5, 6}} and P : A → R be defined as follows:
P(∅) = 0, P(Ω) = 1, P({1, 2}) = 0.1, P({3, 4, 5, 6}) = 0.9. Is (Ω, A, P) a probability space or not?
Explain your answer (simply answering Yes/No will give you no credit.)
2 Roulette Playing
Consider the game of roulette, where there are 37 pockets on the wheel (numbered from 0 to 36) and
the wheel is fair, i.e., all pockets have the same probability. The 0 pocket is green, even numbers are
black, and odd numbers are red (this is not true in reality, but it makes things simpler.) Compute
the following probabilities:
• probability that the outcome is divisible by three, given that it is black.
• probability that the outcome is divisible by three or four, given that a number is in the range
1 to 12.
Trivia: you may find interesting to read the wikipedia page describing the roulette game to numerically assess why the game is designed to ensure the house has an advantage
https://en.wikipedia.org/wiki/Roulette
3 Bayes Rule
Recall Bayes rule:
P(A|B) = P(B|A)P(A)
P(B)
Let us now apply this rule to the scenario where a robot moves along a corridor where there is
a door and is equipped with a sensor to determine if the robot is in front of the door or not (for
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example, a computer vision algorithm processing images coming from a camera on top of the robot.)
To simplify things, assume that there are only two locations for the robot, i.e., the robot is either
in front of the door or not. These two possibilities are modeled by these two events:
• A: the robot is in front of the door;
• A¯: the robot is not in front of the door.
The image processing algorithm returns a binary quantity, i.e., it either determines that the
robot is in front of the door or it determines that the robot is not in front of the door. This is
modeled by these two events:
• B: the sensor determines that the robot is in front of the door;
• B¯: the sensor determines that the robot is not in front of the door.
Finally, the sensor is not perfect but it rather makes errors. More specifically, it is possible that
the robot is in front of the door, but the algorithm fails to recognize it. But it is also possible that
the robot is not in front of the door, and the algorithm instead believes the robot is in front of the
door. These errors are modeled by the following probabilities:
P(B¯|A) = 0.1 P(B|A¯) = 0.2
Assume that initially we do not know where the robot is, and we assign equal probability to both
events. The robot then queries the sensor and the sensor returns B, i.e., the robot is in front of
the door. Based on this unique sensor reading, what is P(A|B), i.e., the probability that the robot
is indeed in front of the door?
Hint: you may think that you miss the value of P(B), but you can compute that (recall what
we saw in class and carefully study the lecture notes.)
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