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Let = f1; 2; 3; 4; 5; 6g, A = f;; ; f1; 2g; f3; 4; 5; 6gg and P : A ! R be de ned as follows:
P (;) = 0, P ( ) = 1, P (f1; 2g) = 0:1, P (f3; 4; 5; 6g) = 0:9. Is ( ; A; P ) a probability space or not? Explain your answer (simply answering Yes/No will give you no credit.)
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 nd interesting to read the wikipedia page describing the roulette game to numer-ically assess why the game is designed to ensure the house has an advantage https://en.wikipedia.org/wiki/Roulette
Recall Bayes rule:
P (AjB) = P (BjA)P (A)
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
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 speci cally, 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 (AjB), 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.)