Prepare the following exercises for discussion in class on
Thursday, November 17.
Exercises
Given the full joint distribution in Figure 13.3 of the course
book, calculate the following:
P(toothache)
P(Cavity)
P(Toothache|cavity)
P(Cavity|toothache ∨ catch)
Show that the statement
P(A,B|C)=P(A|C)P(B|C)
is equivalent to either of the statements
P(A|B,C)=P(A|C) and P(B|A,C)=P(B|C)
Often we need to carry out reasoning over some pair of variables X, Y
conditioned on the value of other variable E.
[a]Using the definitions of conditional probabilities, prove the
conditionalized version of the product rule:
P(x, y | e) = P(x | y, e)
P(y | e)
Prove the conditionalized version of Bayes’ rule:
P(y | x, e) = P(x |
y, e) P(y | e) / P(x | e)
Following the example on medical diagnosis of Section 13.5.1 of
the text book, give the probability P(M|s) by completing the
normalization calculation. Assume 0.05 as a reasonable value for
P(s|¬ m) and ignore the value for P(s).
Three prisoners A, B, C are in their cells. They are told
that one of them will be executed the next day and the others will be
pardoned. Only the gorvenor knows who will be executed. Prisoner A
asks the guard a favor. “Please ask the governor who will be
executed, and then tell either prisoner B or C that they will be
pardoned.” The guard does as was asked and then comes back and tells
prisoner A that he has told prisoner B that he (B) will be
pardoned. What are prisoner A’s chances of being executed, given
this message? Is there more information than before his request to the
guard? At least two approaches are possible to answer this questions:
classical probability and via Bayes’ rule. Develop them both. [This
problem adapted from Pearl 1988, is also a variant of another rather
famous puzzle:
you are given the choice of three doors. Behind one door is a car;
behind the others, goats. They are placed randomly. After you have
chosen a door, the door remains closed. Another person, who knows what
is behind the doors, now has to open one of the two remaining doors,
and the door he opens must have a goat behind it. If both remaining
doors have goats behind them, he chooses one randomly. You are now
posed with the question, whether you want to stay with your first
choice or to switch to the last remaining door. ]
Suppose that you go out to purchase an automobile. The probability
that you will go to dealer 1, d1 is 0.2. The probability of going
to dealer 2, d2 is 0.4. There are only three dealers you are
considering and the probability that you go to the third, d3 is
0.4. At d1 the probability of purchasing a particular automobile,
a1, is 0.2.; at dealer d2 the probability of purchasing a
automobile a1 is 0.4. Finally, at dealer d3, the probability of
purchasing a1 is 0.3. Suppose you purchase automobile a1. What
is the probability that you purchased it at dealer d2?