Probability Theory and Statistics (EE/TE 3341)
Homework 1 Solutions
Problem Solutions
: Yates and Goodman,
1.1.1 1.2.2 1.3.1 1.3.2 1.4.1 1.4.2 1.4.4 1.5.1
1.5.2 and 1.5.5
Problem 1.1.1 Solution
Based on the Venn diagram
M
O
T
the answers are fairly straightforward:
(a) Since
T
∩
M
n
=
φ
,
T
and
M
are not mutually exclusive.
(b) Every pizza is either Regular (
R
), or Tuscan (
T
). Hence
R
∪
T
=
S
so that
R
and
T
are collectively exhaustive. Thus its also (trivially) true that
R
∪
T
∪
M
=
S
. That
is,
R
,
T
and
M
are also collectively exhaustive.
(c) From the Venn diagram,
T
and
O
are mutually exclusive. In words, this means that
Tuscan pizzas never have onions or pizzas with onions are never Tuscan. As an aside,
“Tuscan” is a fake pizza designation; one shouldn’t conclude that people from Tuscany
actually dislike onions.
(d) From the Venn diagram,
M
∩
T
and
O
are mutually exclusive. Thus Gerlanda’s
doesn’t make Tuscan pizza with mushrooms and onions.
(e) Yes. In terms of the Venn diagram, these pizzas are in the set (
T
∪
M
∪
O
)
c
.
Problem 1.2.2 Solution
(a) The sample space of the experiment is
S
=
{
aaa,aaf,afa,faa,ffa,faf,aff,fff
}
.
(1)
(b) The event that the circuit from
Z
fails is
Z
F
=
{
aaf,aff,faf,fff
}
.
(2)
The event that the circuit from
X
is acceptable is
X
A
=
{
aaa,aaf,afa,aff
}
.
(3)
1
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View Full Document(c) Since
Z
F
∩
X
A
=
{
aaf,aff
} n
=
φ
,
Z
F
and
X
A
are not mutually exclusive.
(d) Since
Z
F
∪
X
A
=
{
aaa,aaf,afa,aff,faf,fff
} n
=
S
,
Z
F
and
X
A
are not collectively
exhaustive.
(e) The event that more than one circuit is acceptable is
C
=
{
aaa,aaf,afa,faa
}
.
(4)
The event that at least two circuits fail is
D
=
{
ffa,faf,aff,fff
}
.
(5)
(f) Inspection shows that
C
∩
D
=
φ
so
C
and
D
are mutually exclusive.
(g) Since
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 Fall '08
 n/a
 Probability, Probability theory, Probability space, zf

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