Second Problem Assignment
EECS 401
Due on January 19, 2007
PROBLEM 1 (30 points) Bo and Ci are the only two people who will enter the Rover Dog
Fod jingle contest. Only one entry is allowed per contestant, and the judge (Rover) will
delare the one winner as soon is he receivers a suitable inane entry, whih may be never.
Bo writes inane jingles rapidly but poorly. He has probabilitly 0.7 of submitting his entry
first. If Ci has not already won the contest, Bo’s entry will be declared the winner with
probability 0.3. Ci writes slowly, but he has a gift for this sort of thing. If Bo has not
already won the cntest by the time of Ci’s entry, Ci will be declared the winner with
probability 0.6.
Solution Let B be the event that Bo enters firsts. Then Bc is the event that Ci enters
first. Let Bw be the event that Bo wins the contest and Cw be the event that Ci wins the
contest. We have the following probability tree
Bw
0.3
B
Cw
0.6
0.7
0.7
Bcw
0.4
Ccw
Cw
0.3
0.6
Bw
Bc
0.3
0.4
Ccw
0.7
Bcw
1
Due on January 19, 2007
Solutions
Second Problem Assignment
(a) What is the probability that the prize never will be awarded?
Solution The event that the prize is never awarded is (B ∩ Bcw ∩ Ccw ) ∪ (Bc ∩ Ccw ∩ Bcw ).
Thus
Pr (no prize awarded) = Pr(B ∩ Bcw ∩ Ccw ) + Pr(Bc ∩ Ccw ∩ Bcw )
= Pr(B) Pr(Bcw | B) Pr(Ccw | B ∩ Bcw ) + Pr(Bc ) Pr(Ccw | Bc ) Pr(Bcw | Bc ∩ Ccw )
= (0.7) · (0.7) · (0.4) + (0.3) · (0.4) · (0.7) =
0.28
(b) What is the probability that Bo will win?
Solution The event that Bo wins is given by (B ∩ Bw ) ∪ (Bc ∩ Ccw ∩ Bw ). Thus
Pr(Bo wins) = Pr(B ∩ Bw ) + Pr(Bc ∩ Ccw ∩ Bw )
= Pr(B) Pr(Bw | B) + Pr(Bc ) Pr(Ccw | Bc ) Pr(Bw | B ∩ Ccw )
= (0.7) · (0.3) + (0.3) · (0.4) · (0.3) =
0.246
(c) Given that Bo wins, what is the probability that Ci’s entry arrived fisrt?
Solution
Pr(Ci entered first | Bo wins)) =
=
Pr( Ci entered first and Bo wins)
Pr(Bo wins)
Pr(Bc ∩ Ccw ∩ Bw )
(0.3) · (0.4) · (0.3)
=
= 0.146
Pr(Bo wins)
0.246
(d) What is the probability that the first entry wins the contest?
Solution Let B be the event that Bo enters firsts. Then Bc is the event that Ci enters
first. Let Bw be the event that Bo wins the contest and Cw be the event that Ci wins the
contest. We have the following probability tree
Pr(first entry wins) = Pr(B ∩ Bw ) + Pr(Bc ∩ Cw ) = Pr(B) Pr(Bw | B) + Pr(Bc ) Pr(Cw | Bc )
= (0.7) · (0.3) + (0.3) · (0.6) =
0.39
(e) Sppose that the probability that Bo’s entry arrived first were P instead of 0.7. Can
you find a value of P for which “First entry wins” and “Second entry does not win”
are independent events?
2
Due on January 19, 2007
Solutions
Second Problem Assignment
Solution Let E1 be the event that “First entry wins” and E2 be the event that “Second
entry does not win”. Note that E1 ⊂ E2 . By definition, events E1 and E2 are independent
if and only if Pr(E1 ∩ E2 ) = Pr(E1 ) Pr(E2 ). However, note that E1 ⊂ E2 implies that
Pr(E1 ∩ E2 ) = Pr(E1 ). Thus, events E1 and E2 are independent if and only if Pr(E1 ) =
Pr(E1 ) Pr(E2 ). This is only possible if either Pr(E1 ) = 0 or Pr(E2 ) = 1. We check both of
these cases as follows:
− Pr(E1 ) = 0 ≡ P · (0.3) + (1 − P) · (0.6) = 0 ≡ P = 2 which is not possible.
− Pr(Ec2 ) = 0 ≡ P · (0.7) · (0.6) + (1 − P) · (0.4) · (0.3)) = 0 ≡ P = −0.4 which is not
possible.
Thus we have shown that Pr(E1 ) 6= Pr(E1 ) Pr(E2 ) for all P, which means there is no value
of P such that the two events are independent.
PROBLEM 2 (16 points) The disc containing the only copy of your term project just got
corrupted, and the disc got mixed up with three other corrupted discs that were lying
around. It is equally likely that any of the four discs holds the corrupted remains of
your term project. Your computer expert friend offers to have a lo ok, and you know
from past experience that his probability of finding your term project from any disc is
0.4 (assuming the term project is there). Given that he searches on disc 1 but cannot find
your thesis, what is the probability that your thesis is on disc i for i = 1, 2, 3, 4?
Solution Let Di be the event that the thesis in disc i and F be the event that your friend
finds the thesis in disc 1. Now, Pr(Di ) = 0.25, i = 1, 2, 3, 4 and
Pr(F | D1 ) = 0.4 Pr(Fc | D1 ) = 0.6
Pr(F | D2 ) = 0 Pr(Fc | D2 ) = 1
Pr(F | D3 ) = 0 Pr(Fc | D3 ) = 1
Pr(F | D4 ) = 0 Pr(Fc | D4 ) = 1
We want to find Pr(Di | Fc ) for i = 1, 2, 3, 4. Using Bayes’ Rule
Pr(Di | Fc ) =
Pr(Di ) Pr(Fc | Di )
4
X
Pr(Dk ) Pr(Fc | Dk )
k=1
Substituting the values, we get
Pr(D1 | Fc ) = 0.1666
Pr(D2 | Fc ) = Pr(D3 | Fc ) = Pr(D4 | Fc ) = 0.2777
3
Due on January 19, 2007
Solutions
Second Problem Assignment
PROBLEM 3 (10 points) Each −||− represents one communication link. Link failures are
independent, and each link has a probabilty of 0.5 of being out of service. Towns A and
B can communicate as long as they are connected in the communication network by a
least one path which contains only in-service links. Determine, in an eficient manner,
the probability that A and B can communicate.
Solution Consider a similar problem of two communication links in series, with failure
probabilities p1 and p2 . We can replace it by an equivalent communication link with
failure probability p given by
p = Pr(Gc1 ∪ Gc2 ) = Pr(Gc1 ) + Pr(Gc2 ) − Pr(Gc1 ∩ Gc2 ) = p1 + p2 − p1 p2
For two communication links in parallel, we can replace it by an equivalent communication link with failure probability p given by
p = Pr(Gc1 ∪ Gc2 ) = Pr(Gc1 ) Pr(Gc2 ) = p1 p2
Similarly three communication links in parallel can be replaced by an equivalent communication link with failure probability p given by
p = p1 p2 p3
Thus we can simplify the network as follows
4
Due on January 19, 2007
Solutions
Second Problem Assignment
1/2
1/2
1/2
1/2
1/2
1/2
1/2
1/4
1/8
1/2
1/2
1/2
1/2
(1)
(2)
1/2
5/8 1/8
5/16 1/8
1/2
1/2
(3)
(4)
h
51/128
a
51/256
1/2
(5)
(6)
Thus the probability that A and B can communicate is 1 −
51
=
256
205
256
= 0.8008
PROBLEM 4 (16 points) You enter a special kind of chess tournament, whereby you play
one game with each of three opponents, but you get to choose the order in which you
play your opponents. You win the tournament if you win two games in a row. You know
your probability of a win against each of the three opponents. What is your probability
of winning the tournament, assuming that you choose the optimal order of playing the
opponents? Note that the underlying probability model is not explicitly given so you
will have to define it. Give reasons why your model is a reasonable one.
Solution We assume the plays with different players are independent. Define Wi and
Li (i = 1, 2, 3) as winning and losing over the ith players respectively, and let pi denote
the probability of wining over the ith player, then
S = W1 W2 W3 , W1 W2 L3 , W1 L2 W3 , W1 L2 L3 , L1 W2 W3 , L1 W2 L3 , L1 L2 W3 , L1 L2 L3
5
Due on January 19, 2007
Solutions
Second Problem Assignment
The probability law is
P({W1 W2 W3 }) = p1 p2 p3 ,
P({W1 W2 L3 }) = p1 p2 (1 − p3 ), . . . ,
We claim that the optimal order is to play the weakest player second (the order in
which the other two opponents are played makes no difference). To see this, let pi be
the probability of winning against the opponent played in the ith turn. Then you will
win the tournament if you win against the 2nd player (prob. p2 ) and also you win against
at least one of the two other players [prob. p1 + (1 − p1 )p3 = p1 + p3 − p1 p3 ]. Thus the
probability of winning the tournament is
p2 (p1 + p3 − p1 p3 ),
The order (1, 2; 3) is optimal if and only if the above probability is no less than the
probabilities corresponding to the two alternative orders:
p2 (p1 + p3 − p1 p3 ) > p1 (p2 + p3 − p2 p3 ),
p2 (p1 + p3 − p1 p3 ) > p3 (p2 + p1 − p2 p1 )
It can be seen that the first inequality above is equivalent to p2 > p1 , while the second
inequality above is equivalent to p2 > p3 .
PROBLEM 5 (10 points) A coin is tossed twice. Nick claims that the event of two heads is
at least as likely if we know that the first toss is a head than if we know that at least one
of the tosses is a head. Is he right? Does it make a difference if the coin is fair or unfair?
How can we generalize Nick’s reasoning? Note that the underlying probability model
is not explicitly given so you will have to define it. Give reasons why your model is a
reasonable one.
Solution The sample space is S = HH, HT, T H, T T . A reasonable model to assume
is that the two tosses are independent of each other and the probability of head in the
first toss is the same as the probability of head in the second toss. Let us assume that
P(Head) = p. Thus the probability of each event of the sample space is given by
2
P({HH}) = p2 , P({HT }) = P({T H}) = p(1 −
p) and
P({T T }) = (1 − p) . Let A be the event
that the outcome
is two heads, i.e. A = HH . Let B be the event that the first toss is
head, i.e. B = HH,
HT . And let C be the event at least one of the tosses is head, i.e.
C = HH, HT, T H . As stated in the problem, Nick claims that P(A | B) > P(A | C). Now
we can calculate both conditional probabilities to see if the claim is true or not.
P(A | B) =
P(A ∩ B)
p2
= 2
P(B)
p + p(1 − p)
(1)
and
6
Due on January 19, 2007
Solutions
Second Problem Assignment
P(A | C) =
P(A ∩ C)
p2
= 2
P(C)
p + 2p(1 − p)
(2)
Now comparing the R.H.S of (1) with the R.H.S of (2), we observe that the numerators
are the same, while the denominator of (1) is less than the denominator of (2). Thus, (1)
greater than (2), that is,
P(A | B) > P(A | C)
Generalization: The above holds for any sets A, B, C such that A ⊆ B ⊆ C.
PROBLEM 6 (12 points)
(a) We are given a sample space S and a probability law P. Let B be an event with
nonzero probability. Define the real-valued function Q on the set of events in S by
.
Q(A) = P(A | B). Show that Q satisfies the three axioms of a probability law and
hence is a valid probability law.
Solution Asuming Pr(B) > 0. Q satisfies the non-negativity axiom
Q(A) = Pr(A|B) =
Pr(A ∩ B)
>0
Pr(B)
Q satisfies the nomarlization axiom:
Q(S) = Pr(S|B) =
Pr(B)
Pr(S ∩ B)
=
=1
Pr(B)
Pr(B)
Q satisfies the additivity axiom. Let A1 and A2 be two disjoint events.
Q(A1 ∪ A2 ) = Pr(A1 ∪ A2 |B) =
=
Pr ((A1 ∩ B) ∪ (A2 ∩ B))
Pr ((A1 ∪ A2 ) ∩ B)
=
Pr(B)
Pr(B)
Pr(A1 ∩ B) Pr(A2 ∩ B)
+
= Q(A1 ) + Q(A2 )
Pr(B)
Pr(B)
(b) Using the previous part or by direct calculations, show that
P(A | B, C) =
P(C | A, B)P(A | B)
.
P(C | B)
(1)
Solution Using part (a), assume Q(C) > 0, we have
Q(A|C) =
Pr(A ∩ C|B)
Pr(A ∩ C ∩ B)
Q(A ∩ C)
=
=
= Pr(A|B, C)
Q(C)
Pr(C|B)
Pr(C ∩ B)
7
Due on January 19, 2007
Solutions
Second Problem Assignment
and
Q(A|C) =
Q(C|A)Q(A)
Pr(C|A, B) Pr(A|B)
=
Q(C)
Pr(C|B)
Thus,
Pr(A|B, C) =
8
Pr(C|A, B) Pr(A|B)
Pr(C|B)
Due on January 19, 2007