Mathematics Department, UCLA
T. Richthammer
winter 09, ﬁnal
Mar 02, 2009
Final: Math 170A Probability, Sec. 1
1.
(6 pts)
Consider a probability model with
S
=
{
1
,
2
,
3
,
4
}
,
P
(
{
1
,
2
}
) =
1
2
,
P
(
{
1
,
3
}
) =
1
4
.
(a) Is it possible that
P
(
{
4
}
) = 0?
(b) What are the possible values of
P
(
{
4
}
)?
Answer:
Let
p
i
:=
P
(
{
i
}
). We have
p
1
+
p
2
=
1
2
,
p
1
+
p
3
=
1
4
and
p
1
+
p
2
+
p
3
+
p
4
= 1.
(a) If
p
4
= 0, then from above we get
p
2
= 1

1
4
=
3
4
> p
1
+
p
2
, which is not possible.
(b) For given
p
4
we have :
p
3
= 1

(
p
1
+
p
2
)

p
4
=
1
2

p
4
,
p
1
= (
p
1
+
p
3
)

p
3
=
1
4

1
2
+
p
4
=
p
4

1
4
and
p
2
= (
p
1
+
p
2
)

p
1
=
1
2

p
4
+
1
4
=
3
4

p
4
. These probabilities are all in [0
,
1] for
p
4
∈
[
1
4
,
1
2
],
so these are the possible values for
p
4
.
2.
(10 pts)
Deﬁne probability models for the following situations, and give a precise description of
the event
A
. (
Don’t
calculate P(A)!)
(a) A fair die is rolled 4 times.
A
= “The sum of the values is at most 10”.
(b) A point is chosen within a cube of side length 8 completely at random.
A
= ”The distance of
the point to the surface of the cube is less than 1“.
(c) A number is chosen completely at random from
{
1
, . . . , n
}
. If this number is
k
, then a fair
coin is tossed
k
times and the number of heads is recorded.
A
= ”Head appears in more the
half of the coin tosses”.
Answer:
(a)
S
=
{
1
, . . . ,
6
}
4
,
P
(
A
) =

A

/

S

. The projections
X
1
, . . . , X
4
denote the outcomes of the 1st,
. . . 4th roll.
A
=
{
X
1
+
. . .
+
X
4
= 10
}
.
(b)
S
= [0
,
8]
3
,
P
(
A
) =
λ
3
(
A
)
/λ
3
(
S
), where
λ
3
denotes volume.
A
=
S

[1
,
7]
3
.
(c)
S
=
{
1
, . . . , n
}
2
. The ﬁrst projection
X
1
denotes the number chosen, the second projection
X
2
the number of heads.
P
((
k, l
)) =
p
k
p
l

k
, where
p
k
=
1
n
and
p
l

k
=
(
k
l
)
1
2
k
(number of successes in
k
independent trials).
A
=
{
X
2
> X
1
/
2
}
.
3.
(9 pts)
Suppose that 8 candies are distributed to 8 persons. Determine the probability that
everybody gets a candy, assuming that all possible distributions are equally likely and
(a) the candies are distinct
(b) the candies are identical.
In both cases deﬁne a model. You
don’t
have to simplify your results, but determine in which case
the probability is higher.
Answer:
(a)
S
=
{
1
, . . . ,
8
}
8
=
{
(
x
1
, . . . , x
8
) :
x
i
∈ {
1
, . . . ,
8
}}
, where
x
i
denotes the person that gets candy
i
.
P
(
A
) =

A

/

S

. Here

S

= 8
8
,

A

= 8!, so
P
(
A
) = 8!
/
8
8
.
(b)
S
=
{
(
x
1
, . . . , x
8
) :
x
i
∈ {
0
,
1
,
2
. . .
}
, x
1
+
. . . x
8
= 8
}
, where
x
i
denotes the number of candies
for person
i
.
P
(
A
) =

A

/

S

. Here

S

=
(
15
7
)
,

A

= 1, so
P
(
A
) =
7!8!