Mathematics Department, UCLA
T. Richthammer
winter 09, sheet 2
Jan 09, 2009
Homework assignments: Math 170A Probability, Sec. 1
009. Let
S
be a sample space with probability law
P
. Let
E, F
be events with
P
(
E
) = 1
/
2,
P
(
F
c
) = 2
/
3 and
P
(
E
∪
F
) = 2
/
3. Use the properties of a probability law to calculate
P
(
F
),
P
(
E
∩
F
) and
P
(
E

F
).
Answer:
P
(
F
) = 1

P
(
F
c
) = 1
/
3.
P
(
E
∩
F
) =
P
(
E
) +
P
(
F
)

P
(
E
∪
F
) = 1
/
6.
P
(
E

F
) =
P
(
E
)

P
(
E
∩
F
) = 1
/
3.
010. Let
S
be a sample space with probability law
P
. Use the properties of a probability law
to show that for arbitrary events
E
1
, E
2
we have
(a)
P
(
E
1
∪
E
2
) =
P
(
E
1
) +
P
(
E
c
1
∩
E
2
)
(b)
P
(
E
1
∪
E
2
)
≤
P
(
E
1
) +
P
(
E
2
)
(c)
P
((
E
1

E
2
)
∪
(
E
2

E
1
)) =
P
(
E
1
) +
P
(
E
2
)

2
P
(
E
1
∩
E
2
)
Answer:
(a) This follows from the addition rule, as the set on the LHS is the disjoint union of the
two sets on the RHS.
(b) This follows from (a) as
E
c
1
∩
E
2
⊂
E
2
, and so
P
(
E
c
1
∩
E
2
)
≤
P
(
E
2
).
(c) This follows by expressing all sets as disjoint unions of
E
1

E
2
, E
2

E
1
and
E
1
∩
E
2
.
(Draw a Venn diagram!)
011. Let
S
be a sample space with probability law
P
. Use the properties of a probability law
to show that for arbitrary events
E
1
, E
2
, E
3
we have
(a)
P
(
E
1
∪
E
2
∪
E
3
) =
P
(
E
1
) +
P
(
E
c
1
∩
E
2
) +
P
(
E
c
1
∩
E
c
2
∩
E
3
)
(Hint: Venn diagram.)
(b)
P
(
E
1
∪
E
2
∪
E
3
)
≤
P
(
E
1
) +
P
(
E
2
) +
P
(
E
3
)
(Hint: Use (a))
(c) Generalize (a) and (b) to a union of
n
sets
E
1
, . . . , E
n
and prove these formulas.
Answer:
(a) This follows from the addition rule, as the set on the LHS is the disjoint union of the
two sets on the RHS.
(b) This follows from (a) as
E
2
⊃
E
c
1
∩
E
2
and
E
3
⊃
E
c
1
∩
E
c
2
∩
E
3
.
(c)
P
(
E
1
∪
. . .
∪
E
n
) =
P
(
E
1
) +
P
(
E
c
1
∩
E
2
) +
. . .
+
P
(
E
c
1
∩
. . .
∩
E
c
n

1
∩
E
n
)
≤
P
(
E
1
) +
P
(
E
2
) +
. . .
+
P
(
E
n
). Proof similarly to (a) and (b).
012. Let
S
be a sample space with probability law
P
. Use the properties of a probability law
to show that for arbitrary events
E
1
, E
2
, E
3
we have
(a)
P
(
E
1
∩
E
2
)
≥
P
(
E
1
) +
P
(
E
2
)

1
(b)
P
(
E
1
∩
E
2
∩
E
3
)
≥
P
(
E
1
) +
P
(
E
2
) +
P
(
E
3
)

2
(Hint: Use 011.(b).)
Answer:
(a)
P
(
E
1
∩
E
2
) =
P
(
E
1
) +
P
(
E
2
)

P
(
E
1
∪
E
2
)
≥
P
(
E
1
) +
P
(
E
2
)

1 as
P
(
...
)
≤
1.
(b) By 011.(b)
P
((
E
1
∩
E
2
∩
E
3
)
c
) =
P
(
E
c
1
∪
E
c
2
∪
E
c
3
)
≤
P
(
E
c
1
) +
P
(
E
c
2
) +
P
(
E
c
3
). Using
P
(
E
c
) = 1

P
(
E
) the result follows.
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013. Let
S
be a sample space with probability law
P
.
Write down the inclusionexclusion
formula explicitely (i.e. without using the summation symbol
∑
) for
n
= 2 and
n
= 3.
Answer:
n
= 2:
P
(
E
1
∪
E
2
) =
P
(
E
1
) +
P
(
E
2
)

P
(
E
1
∩
E
2
)
n
= 3:
P
(
E
1
∪
E
2
∪
E
3
) =
P
(
E
1
) +
P
(
E
2
) +
P
(
E
3
)

P
(
E
1
∩
E
2
)

P
(
E
1
∩
E
2
)

P
(
E
1
∩
E
3
)

P
(
E
2
∩
E
3
) +
P
(
E
1
∩
E
2
∩
E
3
).
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 Winter '10
 RICHTHAMMER
 Coproduct, Disjoint union, MATHEMATICS DEPARTMENT, C. Express, T. Richthammer

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