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Unformatted text preview: Probability and Stochastic Processes:
A Friendly Introduction for Electrical and Computer Engineers
Roy D. Yates and David J. Goodman
Problem Solutions : Yates and Goodman,1.5.2 1.5.3 1.5.5 1.6.1 1.6.4 1.6.7 1.7.4 1.8.3 1.8.4 and
1.9.1 6, Ri si . Similarly, G j ¤ £¡
¢ (a) Since G1
s2 s3 s4 s5 s6 and all outcomes have probability 1 6, P G1
s3 and P R3 G1
1 6 so that
R 3 G1 sj 1 s6 . 5 6. The event ¡ P R 3 G1
P G1 ¤ ¡
¢
¡
¤ ¦ ¦ ¦ ¦ ¡
¢ 1
5 ¡ ¡ P R 3 G1 ¡ i ¦ §§§
©¨¨¨¦ ¥ Problem 1.5.2
Let si denote the outcome that the roll is i. So, for 1 (b) The conditional probability that 6 is rolled given that the roll is greater than 3 is
P R 6 G3
P G3 P s6
P s4 s5 s6 16
36 ¡ ¦
¤¦¦
¦¦ ¡
¡ ¡
¢
¡
¡ P R 6 G3 (c) The event E that the roll is even is E
s2 s4 s6 and has probability 3 6. The joint probability of G3 and E is
P s4 s6
13
P G3 E The conditional probabilities of G3 given E is ¡ ¡ ¡ 2
3 P E G3
P G3 2
3 P G3 E P G3 E
PE 13
12 (d) The conditional probability that the roll is even given that it’s greater than 3 is Problem 1.5.3
Since the 2 of clubs is an even numbered card, C2 E so that P C2 E
2 3,
P C2 E
13
12
P C2 E
PE
23 P C2 1 3. Since P E ¡ ¡ ¡ ¡ ¡ The probability that an even numbered card is picked given that the 2 is picked is
P C2 E
P C2 13
13 ¡ ¡ ¡ P E C2 1 1 ¡ 13
12 ¡ ¡ ¡ P E G3 Problem 1.5.5
written i jk where the ﬁrst card drawn is i, the second is j and the third is k. The sample space is
234 243 324 342 423 432 ¤ ¦ ¤¦¦
¤¦¦
¤¦¦
¦¦¦¦ ¢
¡ S and each of the six outcomes has probability 1 6. The events E1 E2 E3 O1 O2 O3 are O3 243 423 ¦ 234 324 342 432 234 432 ¦ O2 ¢¡
¢¡
¢
¡ 243 324 342 423 324 342 ¦ ¦
¦
¦ ¢¡
¢¡
¢
¡ E3 O1 ¦¦ E2 234 243 423 432 ¤
¤
¤
¦¦¦ E1 (a) The conditional probability the second card is even given that the ﬁrst card is even is
P E2 E1
P E1 P 243 423
P 234 243 423 432 26
46 ¡ ¡ ¦ ¦¦ ¦ ¡ P E2 E1 12 ¡ (b) The conditional probability the ﬁrst card is even given that the second card is even is
P E1 E2
P E2 P 243 423
P 243 324 342 423 26
46 ¡ ¡ ¦ ¦¦ ¦ ¡ ¡ P E1 E2 12 (c) the probability the ﬁrst two cards are even given the third card is even is
P E1 E2 E3
P E3 ¡ ¡ P E1 E2 E3 0 (d) The conditional probabilities the second card is even given that the ﬁrst card is odd is
P O1 E 2
P O1 P O1
P O1 ¡ ¡ ¡ P E 2 O1 1 (e) The conditional probability the second card is odd given that the ﬁrst card is odd is
P O1 O2
P O1 ¡ ¡ P O2 O1 0 Problem 1.6.1
This problems asks whether A and B can be independent events yet satisfy A B? By deﬁnition,
P A P B . We can see that if A B, that is
events A and B are independent if and only if P AB
they are the same set, then ¡ ¡ P AA PA PB § ¡ ¡ ¡ Thus, for A and B to be the same set and also independent,
PA P AB PAPB PA ! ¡ ¡ ¡
There are two ways that this requirement can be satisﬁed: PA 0 implying A B S. B φ. ¡¡ ¡ "
¡ " 1 implying A ¡¡ PA 2 2 ¡ P AB Problem 1.6.4
/
0, P A B 0. ¡ # (a) Since A B
¡# (b) To ﬁnd P B , we can write
PA PB PA B 58 38 PB 0 '
( % ¡
# ( & ¡ $
'
% Thus, P B PA B 1 4. ¡ (c) Since A is a subset of Bc , P A Bc PA (d) Since A is a subset of Bc , P A Bc P Bc 3 8. ¡ ¡ $
¡ ¡ # 3 4. (e) The events A and B are dependent because
0 ¡
¡
)¡ ¡ (f) Since C and D are independent P CD P CD
PC PD PC D %
& ¡ $ (h) Since C and D are independent events, P C D
(i) P C Dc PC PC D (j) P C Dc PC PD ¡ #
¡
#
¡ 13 P C Dc # '& 0 ¡ $
%
'
# 0 ¡ # (k) By De Morgan’s Law, Cc 12 Dc P Cc Dc 23 12 23 13 ¡ $ ' ¡ $
$
% ¡
¡'
'
0 PC 13
12 56 '
¡ 1 '
¡ ¡
¡ ' % ¡ # PC D PAPB P C P D . So PD
(g) This permits us to write 3 32 ¡ ¡
§ ¡ P AB PC 1 2. 16 12 1 23 16 23 C D c . This implies
PC D c 1 PC D 16 (l) By Deﬁnition 1.6, events C and Dc are independent because
P C Dc 16 P C P Dc 12 13 ¡ ¡ ¡ #
Problem 1.6.7
(a) For any events A and B, we can write the law of total probability in the form of
P AB '
¡ ( ¡
¡
¡ Since A and B are independent, P AB
P ABc PA P ABc ¡ ! '
0
% PA P A P B . This implies PAPB Thus A and Bc are independent. 3 PA 1 PB P A P Bc (b) Proving that Ac and B are independent is not really necessary. Since A and B are arbitrary
labels, it is really the same claim as in part (a). That is, simply reversing the labels of A and
B proves the claim. Alternatively, one can construct exactly the same proof as in part (a) with
the labels A and B reversed.
(c) To prove that Ac and Bc are independent, we apply the result of part (a) to the sets A and Bc .
Since we know from part (a) that A and Bc are independent, part (b) says that Ac and Bc are
independent.
Problem 1.7.4
The tree for this experiment is 5 BH H 5 T 38 38 BT 18 3 14 AT 3 34 18 3 4 434 4 4
4
3
3
2
2 232 2 2 4
434 4 4 4
2
2 232 2 2 B T AH 5 12 34 5 A H 3 14 12 The probability that you guess correctly is
PC P BH 38 38 34 %
¡ % ¡ ( ¡ Problem 1.8.3 P AT (a) The experiment of picking two cards and recording them in the order in which they were selected can be modeled by two subexperiments. The ﬁrst is to pick the ﬁrst card and record it,
the second subexperiment is to pick the second card without replacing the ﬁrst and recording
it. For the ﬁrst subexperiment we can have any one of the possible 52 cards for a total of 52
possibilities. The second experiment consists of all the cards minus the one that was picked
ﬁrst(because we are sampling without replacement) for a total of 51 possible outcomes. So the
total number of outcomes is the product of the number of outcomes for each subexperiment.
52 51 2652 outcomes ¡6 (b) To have the same card but different suit we can make the following subexperiments. First we
need to pick one of the 52 cards. Then we need to pick one of the 3 remaining cards that are of
the same type but different suit out of the remaining 51 cards. So the total number outcomes
is
52 3 156 outcomes ¡6 (c) The probability that the two cards are of the same type but different suit is the number of outcomes that are of the same type but different suit divided by the total number of outcomes
involved in picking two cards at random from a deck of 52 cards. ¡ 4 156
2652 ¡ P same type, different suit 1
17 (d) Now we are not concerned with the ordering of the cards. So before, the outcomes K 8
and 8 K were distinct. Now, those two outcomes are not distinct and are only considered
to be the single outcome that a King of hearts and 8 of diamonds were selected. So every pair of
outcomes before collapses to a single outcome when we disregard ordering. So we can redo
parts (a) and (b) above by halving the corresponding values found in parts (a) and (b). The
probability however, does not change because both the numerator and the denominator have
been reduced by an equal factor of 2, which does not change their ratio. 8¦7 7 ¦8 Problem 1.8.4
In this case, the designated hitter must be chosen from the 15 ﬁeld players. We can break down the
experiment of choosing a starting lineup into two subexperiments. The ﬁrst is to choose 1 of the 10
pitchers, the second is to choose the remaining 9 batting positions out of the 15 ﬁeld players. Here
we are concerned about the ordering of our selections because a new ordering speciﬁes a new starting
lineup. So the total number of starting lineups when the DH is selected among the ﬁeld players is
10
1 9 is read as 15 ”permute” 9 and is equal to 15! 9! 1 816 214 400 18 162 144 000 different lineups ¦¦¦ ¡ Problem 1.9.1 9 ¡ This gives a total of N1 15 ¦¦¦¡ 9 Where 15 15 (a) Since the probability of a zero is 0.8, we can express the probability of the code word 00111
as 2 occurrences of a 0 and three occurrences of a 1. Therefore
P 00111 08 2 02 3 0 00512 § ¡ § § ¡
(b) The probability that a code word has exactly three 1’s is
5
3 5 08 2 02 3 0 0512 § ¡ § § P three 1’s ¡ ...
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This note was uploaded on 02/11/2012 for the course EEE 350 taught by Professor Duman during the Spring '08 term at ASU.
 Spring '08
 Duman

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