Solutions to Homework 7
April 7, 2007
6.1: 2
There are 5 choices here, so guessing right on one of them will increase the
score. Hence, the probability of increasing the score from one guess is
1
5
. The probability
of guessing a question wrong is
4
5
. The probability of guessing two questions wrong is
4
5
·
4
5
=
16
25
. So, the probability of increasing the score, i.e. not getting questions wrong,
is 1

16
25
=
9
25
.
6.2: 5
(b) (
A
∩
B
))
∪
(
B
∩
C
)
∪
(
A
∩
C
). (Note that it is redundant to include
A
∩
B
∩
C
Do you see why?)
(d) If at most one happens, then at least two do not happen. As in (b), you could
say (
A
∩
B
)
∪
(
B
∩
C
)
∪
(
A
∩
C
) Another way is to say (
A

B

C
)
∪
(
B

A

C
)
∪
(
C

A

B
)
∪
A
∪
B
∪
C
. There are many possible answers. For example, if
S
is deﬁned as
in the problem, then
A
∪
B
∪
C
could be replaced by
S

(
A
∪
B
∪
C
).
(e) The best way is to say (
A
∩
B
∩
C
)
∪
(
B
∩
A
∩
C
)
∪
(
C
∩
A
∩
B
). Is it clear why
this is not the same as the answer to (d)?
Comments:
1. Remember that events are themselves sets. That means that answers like
{
(
A,B
)
,
(
B,C
)
,
(
A,C
)
}
and
{
x

x
⊆ {
A,B,C
}
and

x
 ≤
2
}
, while close,
are incorrect. It is a matter of distinguishing
set
from
set of sets
—two
very diﬀerent things.
2. You have to be careful with set notation. You can’t write things like
A
+
B

2
C
. It makes sense to talk

A

+

B
  
A
∩
B

(as we do
in the InclusionExclusion Principle), since we can add and subtract
cardinalities
of sets, which are numbers. But we can’t add and subtract
sets; they’re not numbers.
6.2: 9
Let
A
be any set. Then
A
∪ ∅
=
A
and
A
and
∅
are disjoint (since
A
∩ ∅
=
∅
).
We can invoke the third rule of the deﬁnition of probability measure, which says that
Pr(
A
∪ ∅
) = Pr(
A
) + Pr(
∅
). But since
A
∪ ∅
=
A
, we have Pr(
A
) = Pr(
A
∪ ∅
). Thus,
1