Jonah
Math 426: Probability, Homework 2 Selected Solutions
Exercises from Ross, §1, §2:
•
•
•
•
§1,
§1,
§2,
§2,
p.
p.
p.
p.
15
17
47
47
‘Problems’ (26, 27)
‘Theoretical exercises’ (14)
‘Problems’ (2, 3, 15, 19)
‘Theoretical exercises’ (1, 2, 3, 4, 12, 14)
Problem 26 §1. Expand (x1 + 2x2 + 3x3 )4 .
Solution. By the multinomial theorem on page 10, we have
X
4
4
xn1 1 (2x2 )n2 (3x3 )n3
(x1 + 2x2 + 3x3 ) =
n
,
n
,
n
1
2
3
n1 +n2 +n3 =4
4
4
4
4
4
3
1
3
1
=
x1 +
x1 (2x2 ) +
x1 (3x3 ) +
x21 (2x2 )2
4, 0, 0
3, 1, 0
3, 0, 1
2, 2, 0
4
4
4
4
2
1
1
2
2
1
3
+
x (2x2 ) (3x3 ) +
x (3x3 ) +
x (2x2 ) +
x1 (2x2 )2 (3x3 )1
2, 1, 1 1
2, 0, 2 1
1, 3, 0 1
1, 2, 1 1
4
4
4
4
1
1
2
1
3
4
+
x (2x2 ) (3x3 ) +
x (3x3 ) +
(2x2 ) +
(2x2 )3 (3x3 )1
1, 1, 2 1
1, 0, 3 1
0, 4, 0
0, 3, 1
4
4
4
2
2
3
+
(2x2 ) (3x3 ) +
(2x2 )(3x3 ) +
(3x3 )4
0, 2, 2
0, 1, 3
0, 0, 4
= x41 + 16x42 + 81x43 + 8x31 x2 + 12x31 x3 + 32x1 x32 + 96x32 x3 + 108x1 x33 + 216x2 x33
+ 24x21 x22 + 54x21 x23 + 216x22 x23 + 72x21 x2 x3 + 144x1 x22 x3 + 216x1 x2 x23 .
Problem 27 §1. If 12 people are are to be divided into 3 committees of respective sizes 3, 4, and 5,
how many divisions are possible?
Solution. The answer is given by
12
12!
=
= 27720.
3!4!5!
3, 4, 5
Theoretical Exercise 14 §2. From a set of n people, a committee of size j is to be chosen, and
from this committee, a subcommittee of size i, i ≤ j, is also to be chosen.
(a) Derive a combinatorial identity by computing, in two ways, the number of possible choices of the
committee and subcommittee.
n X
n j
n n−i
(b) Use part (a) to prove
=
2
for i ≤ n.
j
i
i
j=1
n X
n j
(c) Use part (a) and Theoretical Exercise 13 to show that
(−1)n−j = 0 for i < n.
j
i
j=1
1
n
Solution (a). There are
ways of choosing the committee of size j, and for each of these choices
j
j
there are
choices of a subcommittee of size i, so the total number of possible choices of committee
i
n j
and subcommittee are
. On the other hand, we can also count this number by observing that
j
i
n
n−i
there are
choices of the subcommittee and, for each of these,
ways to complete this to
i
j−i
the committee. Thus we find
n j
n n−i
=
.
j
i
i
j−i
Solution (b). We have
n X
n j
j=i
j
i
=
n X
n n−i
j−i
X
n n
n−i
=
i j=i j − i
X
n−i n
n−i
=
i k=0
k
n n−i
=
2 ,
i
j=i
i
where the first line uses the identity shown in part (a), the third line uses the change of index k = j − i,
and the last line uses the identity shown in Example 4e.
Solution (c). Performing the same simplifications as in part (b), we have
n n X
X
n j
n n−i
n−j
(−1)
=
(−1)n−j
j
i
i
j
−
i
j=i
j=i
X
n n
n−i
=
(−1)n−j
i j=i j − i
X
n−i n
n−i
=
(−1)n−k−i
i k=0
k
X
n−i n−i
n−2k−i n
= (−1)
(−1)k
i k=0
k
= 0,
where the last line holds by Theoretical Exercise 13.
Problem 2 §2. A die is rolled continually until a 6 appears, at which point the experiment stops.
What is the sample space of this experiment? Let En denote the event that n rolls are necessary to
2
S
c
complete the experiment. What points of the sample space are contained in En ? What is ( ∞
1 En ) ?
Solution. The sample space S can be represented by the set of sequences, finite or infinite, on the
numbers 1, 2, 3, 4, 5, 6 satisfying: an element of the sequence is equal to 6 if and only if it is the last
element of the sequence.
S∞ The cevent En consists of the elements of the sample space that are sequences
of length n. The set ( 1 En ) corresponds precisely to the set of infinite sequences of S. In particular,
this is the set of outcomes where a 6 is never rolled.
Problem 19 §2. Two symmetric dice have had two of their sides painted red, two painted black, one
painted yellow, and the other painted white. When this pair of dice is rolled, what is the probability
that both dice land with the same color face up?
Solution. The event that the dice land with the same color face up is the disjoint union of the event
that they are both red, both black, both yellow, and both white. Each of these probabilities can be
computed by counting:
2∗2
6∗6
2∗2
P (both black) =
6∗6
1∗1
P (both yellow) =
6∗6
1∗1
P (both white) =
6∗6
P (both red) =
1
= ,
9
1
= ,
9
1
= , and
36
1
= .
36
Thus the probability that they land on the same color is 2 ∗ (1/9) + 2 ∗ (1/36) = 5/18.
Theoretical Exercise 4 §2. Show that
∞
[
!
Ei
∞
[
∩ F = (Ei ∩ F ) and
1
1
∞
\
1
Solution. We have
x∈
∞
[
!
Ei
∩ F ⇐⇒ x ∈
∞
[
1
Ei and x ∈ F
1
⇐⇒ ∃i such that x ∈ Ei and x ∈ F
⇐⇒ ∃i such that x ∈ Ei ∩ F
∞
[
⇐⇒ x ∈ (Ei ∩ F ),
1
3
!
Ei ∪ F =
∞
\
1
(Ei ∪ F ).
and similarly
x∈
∞
\
!
Ei
∪ F ⇐⇒ x ∈
1
∞
\
Ei or x ∈ F
1
⇐⇒ ∀i x ∈ Ei or x ∈ F
⇐⇒ ∀i x ∈ Ei ∪ F
∞
\
⇐⇒ x ∈ (Ei ∪ F ).
1
4