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
Problem 1.5.2
Let
s
i
denote the outcome that the roll is
i
. So, for 1
i
6,
R
i
s
i
. Similarly,
G
j
s
j
1
s
6
.
(a) Since
G
1
s
2
s
3
s
4
s
5
s
6
and all outcomes have probability 1 6,
P G
1
5 6. The event
R
3
G
1
s
3
and
P R
3
G
1
1 6 so that
P R
3
G
1
P R
3
G
1
P G
1
1
5
(b) The conditional probability that 6 is rolled given that the roll is greater than 3 is
P R
6
G
3
P R
6
G
3
P G
3
P s
6
P s
4
s
5
s
6
1 6
3 6
(c) The event
E
that the roll is even is
E
s
2
s
4
s
6
and has probability 3 6. The joint proba
bility of
G
3
and
E
is
P G
3
E
P s
4
s
6
1 3
The conditional probabilities of
G
3
given
E
is
P G
3
E
P G
3
E
P E
1 3
1 2
2
3
(d) The conditional probability that the roll is even given that it’s greater than 3 is
P E G
3
P EG
3
P G
3
1 3
1 2
2
3
Problem 1.5.3
Since the 2 of clubs is an even numbered card,
C
2
E
so that
P C
2
E
P C
2
1 3. Since
P E
2 3,
P C
2
E
P C
2
E
P E
1 3
2 3
1 2
The probability that an even numbered card is picked given that the 2 is picked is
P E C
2
P C
2
E
P C
2
1 3
1 3
1
1
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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
S
234 243 324 342 423 432
and each of the six outcomes has probability 1 6. The events
E
1
E
2
E
3
O
1
O
2
O
3
are
E
1
234 243 423 432
O
1
324 342
E
2
243 324 342 423
O
2
234 432
E
3
234 324 342 432
O
3
243 423
(a) The conditional probability the second card is even given that the first card is even is
P E
2
E
1
P E
2
E
1
P E
1
P
243 423
P
234 243 423 432
2 6
4 6
1 2
(b) The conditional probability the first card is even given that the second card is even is
P E
1
E
2
P E
1
E
2
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 Spring '08
 Duman
 Conditional Probability, Probability, Probability theory

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