ECE 190 — Fall, 2006
Assignment 8 Solutions
Section 6.9
•
1
See text.
•
2
P
(
X

Y
)=
P
(
X
∩
Y
)
P
(
Y
)
P
(
X
∩
Y
P
(
X

Y
)
P
(
Y
)
=
1
3
×
1
4
=
1
12
•
4
See text.
•
5
See text.
•
7
(a) See text.
Since both of the people chosen are treated equally (they both become Fnalists), the order of
choosing is immaterial.
±urthermore, since names are drawn at random, we are dealing with equally likely outcomes.
Thus, to determine probabilities, we can use ratios of cardinalities.

S

=
±
10
2
²
=45
(b)
P
(2 men) =
number of ways to choose 2 men

S

=
(
7
2
)
45
=
21
45
=
7
15
(c)
P
(1 woman, 1 man) =
number of ways to choose 1 woman, 1 man

S

=
(
3
1
)(
7
1
)
45
=
21
45
=
7
15
•
10
See text.
•
13
See text.
•
19
S
=
{
HHH,HHT,HTH,THH,HTT,HTH,THT,TTT
}
A
=
{
HHH,HHT,HTH,HTT
}
B
=
{
HHT,HTH,THH,TTT
}
A
∩
B
=
{
HHT,HTH
}
1
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P
(
A

B
)=
P
(
A
∩
B
)
P
(
B
)
=

A
∩
B


B

because the coin is fair
=
2
4
=
1
2
But
P
(
A

A


S

=
4
8
=
1
2
Thus
P
(
A

B
P
(
A
)
(b)
P
(
B

A
P
(
B
∩
A
)
P
(
A
)
=

A
∩
B


A

because the coin is fair
=
2
4
=
1
2
But
P
(
B

B


S

=
4
8
=
1
2
Thus
P
(
B

A
P
(
B
)
•
20
Since
A
∩
B
=
∅
,P
(
A
∩
B
)=0
Since
A
and
B
are independent,
P
(
A
∩
B
P
(
A
)
P
(
B
)
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 Fall '06
 Carter
 Probability, Harshad number, The Order, Emoticon, ASCII art, Triangular number

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