Stat 330
Solution to Homework 3
1
Independence, Baron p. 40
Events
A
and
B
are independent. Show, intuitively, and mathematically, that
(a) Their complements are also independent.
We know, that
P
(
A
∩
B
) =
P
(
A
)
P
(
B
)
.
We want to ﬁnd out about the probability of the intersection of the complements. Intuitively, the
complements should be independent, since the occurrence of the event or the negative of the events
should not matter in the case of independence.
Mathematically, we get:
P
(
A
∩
B
) =
P
(
A
) +
P
(
B
)

P
(
A
∪
B
) =
=
P
(
A
) +
P
(
B
)

(1

P
(
A
∩
B
)) =
=
P
(
A
) +
P
(
B
)

(1

P
(
A
)
·
P
(
B
)) =
= 1

P
(
A
) +
P
(
B
)

1 +
P
(
A
)
·
P
(
B
)) =
=
P
(
A
)
(
1

P
(
B
)) +
P
(
B
) =
P
(
A
)
·
P
(
B
)
(b) If they are disjoint, then
P
(
A
) = 0 or
P
(
B
) = 0.
Since we know, that
P
(
A
∩
B
) =
P
(
A
)
P
(
B
)
the statement follows directly, if we also assume, that the
events are disjoint, i.e.
P
(
A
∩
B
) = 0
(c) If they are exhaustive (i.e.
A
∪
B
= Ω), then
P
(
A
) = 1 or
P
(
B
) = 1.
From
1 =
P
(
A
∪
B
)
we get,
1 =
P
(
A
∪
B
) = 1

P
(
A
∩
B
) = 1

P
(
A
)
·
P
(
B
)
Therefore
P
(
A
)
·
P
(
B
) = 0
,
which implies that either
P
(
A
) = 0
or
P
(
B
) = 0
. Then either
P
(
A
) = 1
or
P
(
B
) = 1
.
(4 points)
2
Bayes’ Theorem
(a) Suppose that a barrel contains many small plastic eggs. Some eggs are painted red and some are painted
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 Spring '08
 Staff
 Probability theory, Bayesian probability, 0.999..., test pos

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