Section Handout #7 Solutions: Probability
Will you add a note on the pattern for the students? Also you didn't
write anything in 3c)
3d): P(A  B=b, C=c) = 0.22, not 0.105... No idea how you got 0.105...
The distribution over A was 0.018, 0.064 unnormalized (directly from
the joint).
In Class
Question 1.
Consider the following 3sided dice with the given side values. Assume the dice are all fair (each side has probability ) and all rolls are independent.
A: 2, 2, 5
B: 1, 4, 4
C: 3, 3, 3
 What is the expected value of each die?
Answer: 3 for A,B,C
 Consider the indicator function better(X,Y) which has value 1 if X>Y and
value 1 if X<Y. What are the expected values of better(A, B), better(B, C), better(C, A)? Why are these sometimes called nontransitive dice?
Answer:
better(A,B) = (1) (5/9) + (1) (4/9) = 1/9
better(B,C) = (1) (2/3) + (1) (1/3) = 1/3
better(C,A) = (1) (2/3) + (1) (1/3) = 1/3
So, A is better than B, B is better than C, and C is better than
A.
Question 2.
On a day when an assignment
is due (A=a), the newsgroup tends to be busy (B=b), and the computer lab tends
to be full (C=c). Consider the following conditional probability tables
for the domain, where A = {a, ¬a}, B = {b, ¬b}, C = {c, ¬c}.
P(A) 
P(BA) 
P(CA) 

B 
A 
P 
b 
a 
0.90 
¬b 
a 
0.10 
b 
¬a 
0.40 
¬b 
¬a 
0.60 

C 
A 
P 
c 
a 
0.70 
¬c 
a 
0.30 
c 
¬a 
0.50 
¬c 
¬a 
0.50 

 Construct the joint distribution out of these conditional probabilities
tables assuming B and C are independent given A.
Answer: We use the chain rule, P(A,B,C) = P(A) P(B,C  A) and then use the
independence of B and C given A in this problem to derive
P(A,B,C) = P(A) P(BA) P(CA)
P(B,CA) 
P(A, B, C) 
A 
B 
C 
P 
a 
b 
c 
0.63 
a 
b 
¬c 
0.27 
a 
¬b 
c 
0.07 
a 
¬b 
¬c 
0.03 
¬a 
b 
c 
0.2 
¬a 
b 
¬c 
0.2 
¬a 
¬b 
c 
0.3 
¬a 
¬b 
¬c 
0.3 

A 
B 
C 
P 
a 
b 
c 
0.126 
a 
b 
¬c 
0.054 
a 
¬b 
c 
0.014 
a 
¬b 
¬c 
0.006 
¬a 
b 
c 
0.16 
¬a 
b 
¬c 
0.16 
¬a 
¬b 
c 
0.24 
¬a 
¬b 
¬c 
0.24 

 What is the marginal distribution P(B,C)? Are these two variables
absolutely independent in this model? Justify your answer using the
actual probabilities, not your intuitions.
Answer: We marginalize out A using the P(A,B,C) table from part a).
B 
C 
P 
b 
c 
0.286 
¬b 
c 
0.254 
b 
¬c 
0.214 
¬b 
¬c 
0.246 
We can marginalize out C to get P(B=b) = 0.5 and marginalize
out B to get P(C=c) = 0.54. We have P(B=b,C=c) = 0.286 which is not
equal to P(B=b) P(C=c) = 0.27. So B and C are not independent.
 What is the posterior distribution over A given that B=b, P(A 
B=b)? What is the posterior distribution over A given that C=c, P(A 
C=c)? What about P(A  B=b, C=c)? Explain the pattern among
these posteriors and why it holds.
Answer: We use P(AB=b) = P(A,B=b) / P(B=b). We get P(A,B) by marginalizing out
C from the full joint P(A,B,C) from part a).
P(A,B) 
P(AB=b) 
P(A,C) 
P(AC=c) 
P(AB=b, C=c) 
A 
B 
P 
a 
b 
0.18 
¬a 
b 
0.32 
a 
¬b 
0.02 
¬a 
¬b 
0.48 


A 
C 
P 
a 
c 
0.14 
¬a 
c 
0.40 
a 
¬c 
0.06 
¬a 
¬c 
0.40 


A 
B 
C 
P 
a 
b 
c 
0.44 
¬a 
b 
c 
0.56 

Homework
Question 1. Assume that a joint distribution
over two variables, X = {x, ¬x}and Y = {y, ¬y} is known to have the marginal distributions
P(x) = P(¬x) = P(y) = P(¬y). Give joint distributions satisfying these marginals for
each of these conditions:
 X and Y are independent
Answer:
X 
Y 
P 
x 
y 
0.25 
x 
¬y 
0.25 
¬x 
y 
0.25 
¬x 
¬y 
0.25 
 Observing Y=y increases the belief in X=x, i.e. P(x  y) > P(x)
Answer:
X 
Y 
P 
x 
y 
0.50 
x 
¬y 
0.0 
¬x 
y 
0.0 
¬x 
¬y 
0.50 
 Observing Y=y decreases the belief in X=x, i.e. P(x  y) < P(x)
Answer:
X 
Y 
P 
x 
y 
0.0 
x 
¬y 
0.5 
¬x 
y 
0.5 
¬x 
¬y 
0.0 
Question 2. Sometimes, there is traffic
(cars) on the freeway (C=c). This could either be because of a ball game (B=b) or
because of an accident (A=a). Consider the following joint probability
table for the domain, where A = {a, ¬a}, B = {b, ¬b}, C = {c, ¬c}.
P(A, B, C) 
A 
B 
C 
P 
a 
b 
c 
0.018 
a 
b 
¬c 
0.002 
a 
¬b 
c 
0.126 
a 
¬b 
¬c 
0.054 
¬a 
b 
c 
0.064 
¬a 
b 
¬c 
0.016 
¬a 
¬b 
c 
0.072 
¬a 
¬b 
¬c 
0.648 

 What is the distribution P(A,B)? Are A and B independent in this
model given no evidence? Justify your answer using the actual
probabilities, not your intuitions.
Answer:
A 
B 
P 
a 
b 
0.02 
¬a 
b 
0.08 
a 
¬b 
0.18 
¬a 
¬b 
0.72 
We have that P(A=a) = 0.2 and P(B=b) = 0.1. We see that
P(A,B) = P(A) P(B) for each setting of A and B.
 What is the marginal distribution over A given no evidence?
Answer: We have that P(A=a) = 0.2 and P(A=¬a) = 0.8.
 How does this change if we observe that C=c; what is the posterior
distribution P(A  C=c)? Does this change intuitively make
sense? Why or why not?
Answer:
P(A,C) 
P(AC=c) 
A 
C 
P 
a 
c 
0.144 
¬a 
c 
0.136 
a 
¬c 
0.056 
¬a 
¬c 
0.664 

A 
C 
P 
a 
c 
0.514 
¬a 
c 
0.486 

 What is the conditional distribution over A if we then learn there is a
ball game, P(A  B=b, C=c)? Does it make sense that observing B should
cause this update to A (called explainingaway)? Why or why not?
Answer:
P(A  B=b, C=c) 
A 
B 
C 
P 
a 
b 
c 
0.22 
¬a 
b 
c 
0.78 

Yes, the knowledge that there was a game should explain why there is traffic instead of an accident.