Math 302.102 Fall 2010
Solutions to Assignment #1
1.
(a)
If
S
=
{
a,b,c,d
}
is the sample space consisting of 4 outcomes, then there are 2
4
= 16 pos-
sible events. They can be enumerated by listing all events containing 4 elements, namely
{
a,b,c,d
}
,
all events containing 3 elements, namely
{
a,b,c
}
,
{
a,b,d
}
,
{
a,c,d
}
,
{
b,c,d
}
, all events containing 2
elements, namely
{
a,b
}
,
{
a,c
}
,
{
a,d
}
,
{
b,c
}
,
{
b,d
}
,
{
c,d
}
, all events containing 1 element, namely
{
a
}
,
{
b
}
,
{
c
}
,
{
d
}
, and all events containing 0 elements, namely
∅
. That is,
F
=
±
∅
,
{
a
}
,
{
b
}
,
{
c
}
,
{
d
}
,
{
a,b
}
,
{
a,c
}
,
{
a,d
}
,
{
b,c
}
,
{
b,d
}
,
{
c,d
}
,
{
a,b,c
}
,
{
a,b,d
}
,
{
a,c,d
}
,
{
b,c,d
}
,
{
a,b,c,d
}
²
.
(b)
If all events are equally likely, then the probability of any event is simply the number of
outcomes in that event divided by 4. That is,
P
{
A
}
=
#(
A
)
4
for any
A
∈ F
where #(
A
) denotes the number of elements in
A
.
2.
In order to solve this problem, we need to consider all possible scenarios where player 1 (your
opponent) chooses a coin and then player 2 (you) chooses a coin. There are six separate scenarios
that we will consider in turn.
(i) Suppose that player 1 chooses A and player 2 chooses B. Then
P
{
player 2 wins
}
=
P
{
coin
A
shows tails
}
=
2
5
.
(ii) Suppose that player 1 chooses A and player 2 chooses C. Then
P
{
player 2 wins
}
=
P
{
(coin
C
shows tails) or (coin
A
shows tails and
C
shows heads)
}
=
P
{
coin
C
shows tails
}
+
P
{
coin
A
shows tails and
C
shows heads
}
=
2
5
+
2
5
·
3
5
=
16
25
.
(iii) Suppose that player 1 chooses B and player 2 chooses A. Then
P
{
player 2 wins
}
=
P
{
coin
A
shows heads
}
=
3
5
.
(iv) Suppose that player 1 chooses B and player 2 chooses C. Then
P
{
player 2 wins
}
=
P
{
coin
C
shows tails
}
=
2
5
.