Probability and Stochastic Processes:
A Friendly Introduction for Electrical and Computer Engineers
Roy D. Yates and David J. Goodman
Problem Solutions
: Yates and Goodman,1.5.1 1.5.3 1.5.4 1.5.6 1.6.6 1.7.1 1.7.3 1.7.6 1.7.9
1.8.2 1.8.4 and 1.8.5
Problem 1.5.1
probability.
(a) Note that the probability a call is brief is
P
[
B
] =
P
[
H
0
B
]+
P
[
H
1
B
]+
P
[
H
2
B
] =
0
.
6
.
The probability a brief call will have no handoffs is
P
[
H
0

B
] =
P
[
H
0
B
]
P
[
B
]
=
0
.
4
0
.
6
=
2
3
(b) The probability of one handoff is
P
[
H
1
] =
P
[
H
1
B
]+
P
[
H
1
L
] =
0
.
2. The probability that a call
with one handoff will be long is
P
[
L

H
1
] =
P
[
H
1
L
]
P
[
H
1
]
=
0
.
1
0
.
2
=
1
2
(c) The probability a call is long is
P
[
L
] =
1

P
[
B
] =
0
.
4. The probability that a long call will
have one or more handoffs is
P
[
H
1
∪
H
2

L
] =
P
[
H
1
L
∪
H
2
L
]
P
[
L
]
=
P
[
H
1
L
]+
P
[
H
2
L
]
P
[
L
]
=
0
.
1
+
0
.
2
0
.
4
=
3
4
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
Problem 1.5.4
Define
D
as the event that a pea plant has two dominant
y
genes. To find the conditional proba
bility of
D
given the event
Y
, corresponding to a plant having yellow seeds, we look to evaluate
P
[
D

Y
] =
P
[
DY
]
P
[
Y
]
1
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Note that
P
[
DY
]
is just the probability of the genotype
yy
. From Problem 1.4.3, we found that with
respect to the color of the peas, the genotypes
yy
,
yg
,
gy
, and
gg
were all equally likely. This implies
P
[
DY
] =
P
[
yy
] =
1
/
4
P
[
Y
] =
P
[
yy
,
gy
,
yg
] =
3
/
4
Thus, the conditional probability can be expressed as
P
[
D

Y
] =
P
[
DY
]
P
[
Y
]
=
1
/
4
3
/
4
=
1
/
3
Problem 1.5.6
The problem statement yields the obvious facts that
P
[
L
] =
0
.
16 and
P
[
H
] =
0
.
10. The words
“10% of the ticks that had either Lyme disease or HGE carried both diseases” can be written as
P
[
LH

L
∪
H
] =
0
.
10
(a) Since
LH
⊂
L
∪
H
,
P
[
LH

L
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 Spring '09
 MarshallSylvan
 Conditional Probability, Probability, Probability theory, Designated hitter, Field Players

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