1. CSE 468/568 Deadlines
Lab1 grades out tomorrow (5/1) HW2 grades out by weekend Lab 3 grades out next weekend HW3 – Probability homework out Due 5/7 FINALS: Monday (5/12) from 11:45 – 1:45pm NSC 201 Closed book, bring calculators

2. Joint and Conditional Probability
P(X=x and Y=y) = P(x,y) If X and Y are independent, then P(x,y) = P(x).P(y) P(x|y) is the probability that x occurs given y has occurred, or x given y P(x|y) = P(x,y)/P(y) P(x,y) = P(x|y).P(y) P(x|y) = P(x)

3. Example
The director of an insurance company’s computing center estimates that the company’s computer has a 20% chance of catching a computer virus. However, she feels that there is only a 6% chance of the computer’s catching a virus that will completely disable its operating system. If the company’s computer should catch a virus, what is the probability that the operating system will be completely disabled? P(A) = 0.2 P(B)=0.06 P(B|A)=P(A,B)/P(A) = 0.06/0.2 = .3

4. Law of Total Probability
DISCRETE CONTINUOUS

5. Example
There are three boxes each containing light bulbs. First box has ten of which four are dead. Second has six of which one is dead. Third has eight bulbs of which three are dead. What is the probability of choosing a dead light bulb if one is chosen at random from one of the boxes? P(dead) = P(dead|B1).P(B1) + P(dead|B2).P(B2) + P(dead|B3).P(B3) = 4/10*1/3 + 1/6*1/3 + 3/8*1/3 = 113/360

6. Example 2
Let 60% of be spam. Let 90% of spam have a forged header. Let 20% of non-spam have forged header. What is the probability that an with forged header is spam? Let A = message contains forged header Partition sample space: = {spam, not spam} B = message is spam Bc = message is not spam P(B|A) = [P(A|B).P(B)]/[P(A|B).P(B) + P(A|Bc).P(Bc)] = 0.9*0.6/(0.9* *0.4) = 0.871

7. Example 3
Consider a well shuffled card deck. What is the probability that the second card in the deck is an ace? Partition sample space A = {First card is an ace} Ac = {First card is not an ace} B = {second card is an ace} P(B) = P(B|A).P(A) + P(B|Ac)|P(Ac) = 3/51*4/52 + 4/51*48/52 = 4/52

8. State Estimation Example
Let the robot obtain measurement z What is P(open|z=open)?

9. Probabilistic Sensing
P(open|z) is our estimate
P(z|open) is sensor model
Sensor model computed apriori
Bayes rule allows us to use it to estimate state of the world:

10. Example P(z=open|open) = 0.6 P(z=open|open) = 0.3
P(open) = P(open) = 0.5
Let z = open
P(open|z=open) = P(z=open|open).P(open)/(P(z=open|open).P(open) + P(z=open|open).p(open))
P(open|z=open) = 0.6 * 0.5 / (0.6* *0.5) = 0.67
Z raises the probability that the door is open

11. Combining Evidence Let robot obtain a second measurement z2
How can we combine this with the previous information?
P(z2=open|open) = P(z2=open|open) = 0.7
P(z2=open|open) = P(z2=open|open) = 0.3
P(open|z1) = 2/3
Let z2 = open
P(open|z2, z1) = P(z2|open).P(open|z1)/(P(z2|open).P(open|z1) + P(z2|open).P(open|z1)
= 6/10.2/3/(6/10.2/3 + 4/10.1/3) = 0.75
z2 lowers the probability that the door is open !

12. Actions World is inherently dynamic because of
Actions taken by the robot
Actions taken by other agents in the environment
Passage of time changes the world
How can we incorporate actions into our model?

13. Typical Actions and their Effect
Robot turns its wheels to move
Robot uses its manipulator to grasp an object
Actions cannot be carried out in absolute certainty i.e., they are inherently uncertain
Actions typically increase the uncertainty of robot state

14. Modeling Actions
Incorporating outcome of an action u into the current belief is done by using the conditional pdf P(x|u,x’) It is the probability density function that executing u changes the robot state from x’ to x

15. Example: Door closing

16. Action Model P(x|u,x’) for u = “close door”:
If the door is open, the action “close door” succeeds in 90% of all cases.

17. Integrating Action Outcome
Continuous case: Discrete case:

18. Belief after Door Closing