ECEN 303: Homework 2 Solutions - About people.tamu.edu

ECEN 303: Homework 2 Solutions

Problem 1:

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

B0cw.4
Cwc

Cw
0.3 0.6

Bw
Bc 0.3
0.4

Cwc
0.7
Bwc

(a) The event that the prize is never awarded is (B ∩ Bwc ∩ Cwc ) ∪ (Bc ∩ Cwc ∩ Bwc ). Thus

Pr(no prize awarded) = Pr(B ∩ Bwc ∩ Cwc ) + Pr(Bc ∩ Cwc ∩ Bwc )
= Pr(B) Pr(Bwc | B) Pr(Cwc | B ∩ Bwc ) + Pr(Bc) Pr(Cwc | Bc) Pr(Bwc | Bc ∩ Cwc )
= (0.7) · (0.7) · (0.4) + (0.3) · (0.4) · (0.7) = 0.28

(b) The event that Bo wins is given by (B ∩ Bw) ∪ (Bc ∩ Cwc ∩ Bw). Thus

Pr(Bo wins) = Pr(B ∩ Bw) + Pr(Bc ∩ Cwc ∩ Bw)
= Pr(B) Pr(Bw | B) + Pr(Bc) Pr(Cwc | Bc) Pr(Bw | B ∩ Cwc )
= (0.7) · (0.3) + (0.3) · (0.4) · (0.3) = 0.246

(c)

Pr(Ci entered first | Bo wins)) = Pr( Ci entered first and Bo wins)
Pr(Bo wins)

= Pr(Bc ∩ Cwc ∩ Bw) = (0.3) · (0.4) · (0.3) = 0.146
Pr(Bo wins) 0.246

(d) 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

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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) 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). Since E1 ⊂ E2, we have Pr(E1 ∩ E2) = Pr(E1). Thus the events
E1 and E2 are independent if and only if Pr(E1) Pr(E2) = Pr(E1). This is only possible if either
Pr(E1) = 0 or Pr(E2) = 1. We check both of these cases as follows.

• Pr(E1) = P · (0.3) + (1 − P ) · (0.6). This equals 0 only if P = 2 which is not possible. Thus
Pr(E1) = 0 for all possible values of P .

• We have Pr(E2c) = P · (0.7) · (0.6) + (1 − P ) · (0.4) · (0.3). This equals 0 only if P = −0.4
which is not possible. Thus, Pr(E2c) > 0 for all possible P , which means that Pr(E2) < 1 for
all possible P

Thus there is no value of P such that the two events are independent.

Problem 2:
(a) 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(F c | D1) = 0.6
Pr(F | D2) = 0 Pr(F c | D2) = 1
Pr(F | D3) = 0 Pr(F c | D3) = 1
Pr(F | D4) = 0 Pr(F c | D4) = 1

We want to find Pr(Di | F c) for i = 1, 2, 3, 4. Using Bayes’ Rule

Pr(Di | F c) = Pr(Di) Pr(F c | Di)

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Pr(Dk) Pr(F c | Dk)

k=1

Substituting the values, we get

Pr(D1 | F c) = 0.1666
Pr(D2 | F c) = Pr(D3 | F c) = Pr(D4 | F c) = 0.2777

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Problem 3:
Consider a similar problem of two communication links in series, with failure probabilities p1 and
p2. Let G1 denote the event that the first link is good and G2 denote the event that the second
link is good. For two communication links in series, communication is impossible if at least one of
the link fails. Thus we can replace it by an equivalent communication link with failure probability
p given by

p = Pr(Gc1 ∪ G2c ) = Pr(G1c ) + Pr(Gc2) − Pr(G1c ∩ Gc2) = p1 + p2 − p1p2

For two communication links in parallel, communication becomes impossible if both the links fail.
Therefore, we can replace it by an equivalent communication link with failure probability p given
by

p = Pr(Gc1 ∪ G2c ) = Pr(G1c ) Pr(G2c ) = p1p2

Similarly three communication links in parallel can be replaced by an equivalent communication
link with failure probability p given by

p = p1p2p3

Thus we can simplify the network (see figure on the next page) and obtain the probability that A

and B can communicate as 1 − 51 = 205 = 0.8008.
256 256

Problem 4:
(a) 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 = W1W2W3, W1W2L3, W1L2W3, W1L2L3, L1W2W3, L1W2L3, L1L2W3, L1L2L3

The probability law is

P ({W1W2W3}) = p1p2p3, P ({W1W2L3}) = p1p2(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, suppose you
first play against player 1, then player 2 and then player 3. 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 − p1p3]. Thus the probability of winning the tournament is

p2(p1 + p3 − p1p3),

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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 − p1p3) ≥ p1(p2 + p3 − p2p3),
p2(p1 + p3 − p1p3) ≥ p3(p2 + p1 − p2p1)

It can be seen that the first inequality above is equivalent to p2 ≥ p1, while the second inequality
above is equivalent to p2 ≥ p3. Thus if you use the optimal order of playing the opponents , your
probability of winning the tournament is p2(p1 + p3 − p1p3).

Problem 5: 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 P ({HH}) = p2, P ({HT }) = P ({T H}) = p(1 − p)
and P ({T T }) = (1 − p)2. 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

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or not. P (A ∩ B) p2
P (B) p(1
P (A | B) = = p2 + − p) (1)

and P (A ∩ C) p2
P (C) 2p(1
P (A | C) = = p2 + − 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) is 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:
(a) Assuming Pr(B) > 0. Q satisfies the non-negativity axiom

Q(A) = Pr(A|B) = Pr(A ∩ B) ≥ 0
Pr(B)

Q satisfies the normalization axiom:

Q(S) = Pr(S|B) = Pr(S ∩ B) = Pr(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 ∪ A2) ∩ B) = Pr ((A1 ∩ B) ∪ (A2 ∩ B))
Pr(B) Pr(B)

= Pr(A1 ∩ B) + Pr(A2 ∩ B)
Pr(B) Pr(B)

= Q(A1) + Q(A2)

(b) Using part (a), assume Q(C) > 0, we have

Q(A|C ) = Q(A ∩ C) = Pr(A ∩ C|B) = Pr(A ∩ C ∩ B) = Pr(A|B, C)
Q(C ) Pr(C |B ) Pr(C ∩ B)

and Q(A|C ) = Q(C |A)Q(A) = Pr(C|A, B) Pr(A|B)
Thus, Q(C ) Pr(C |B )

Pr(A|B, C) = Pr(C|A, B) Pr(A|B) .

Pr(C |B )

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