CE 3102
Fall 2008
Homework #2
Solutions
1. One way to make our reasoning clear is lay two columns, with claims and
justifications.
(a)
Claim
Justification
Suppose A and B are independent
Supposition
If A and B are independent, then P(AB)=P(A)
(2.13) p. 52, A&T
P(AB)
≠
P(A)
Since P(AB)>P(A)
A and B are not independent
proof by contradiction
(
a
k
a
modus tollens
)
(b)
Claim
Justification
Suppose A and B are mutually exclusive
A
∩
B=
∅
Definition of mutually
exclusive events
P(
∅
)
=
0
P
(
∅
)=1P(S)=11=0
P(A
∩
B
)
=
0
P(AB)=P(A
∩
B)/P(B)
Definition of conditional probability
P(AB)=0
Since P(A
∩
B)=0
P(AB)
≤
P(A)
Since P(A)
≥
0
P
(
A

B
)
>
P
(
A
)
G
i
v
e
n
A and B are not mutually exclusive
Proof by contraction
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Claim
Justification
P(A
∩
B)=P(AB)P(B)
Definition of conditional probability
P(AB) > P(A)
Given
P(A
∩
B)>P(A)P(B)
Substitution of P(A) for P(AB)
(d)
Claim
Justification
)
(
)
(
)
(
B
A
P
B
A
P
A
P
∩
+
∩
=
Total probability theorem
)
(
)
(
)
(
B
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 Spring '11
 Conditional Probability, Probability theory, µ, 3Days, 30 Percent

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