Probability and Information: An Integrated Approach (2nd Ed., 2008) – Solutions Manual – Applebaum

ALL 10 CHAPTERS COVERED




SOLUTIONS MANUAL

,Probability and information
solutions manual

Chapter 2
1. 12.

2. (a) 7! = 5040, (b) 6! = 720.

3. There are n ways of choosing the first, (n − 1) ways of choosing the
second, . . ., two ways of choosing the (n − 1)th and one way of choosing
the last; hence by Theorem 2.2, the total number of arrangements is
n(n − 1) . . . 2.1 = n!

4. 15, 20, 35.
µ ¶
52
5. = 2 598 960.
5

6. Put r = 4 and (n − r) = 8, then n = 12.
µ ¶ µ ¶
n n n! n!
7. + = +
r r+1 (n − r)!r! (n − r − 1)!(r + 1)!
n!(r + 1 + n − r) (n + 1)!
= =
µ(n − r)!(r
¶ + 1)! (n − r)!(r + 1)!
n+1
=
r+1
Hence each term in Pascal’s triangle is the sum of the two above it.

8. Sampling without replacement, order relevant . · . number of winning
15!
combinations is (15−3)! = 2730.

, 9. At each turn there are seven names to be allocated to seven people and
therefore 72 = 49 different possible guesses of which seven are correct.
The player may guess correctly on the first, second or third go, or not
at all. Hence, the number of different possibilities is

7 + (42 × 7) + (42 × 42 × 7) + (42 × 42 × 42) = 86 737.

10. If the man buys r grapefruit, the woman must buy (4 − r), where
r = 0, 1, 2, 3 or 4. Hence the total number of ways is
µ ¶ µ ¶ µ ¶ µ ¶ µ ¶ µ ¶
10 8 10 8 10 8
· + · + ·
4 0 3 1 2 2
µ ¶ µ ¶ µ ¶ µ ¶
10 8 10 8
+ · + · = 3060.
1 3 0 4
¡10¢
11. (a) 7
= 120.
¡8¢
(b) Either
¡8¢ neither goes or one goes, so total number of ways is 7
+
2 × 6 = 64.
¡9¢
(c) 6 = 84.

12. 2n = (1 + 1)n = nr=0 nr 1r 1n−r = nr=0 nr .
P ¡ ¢ P ¡ ¢

13. The first n1 can be chosen in nN1 different ways, the next n2 in N n−n
¡ ¢ ¡ 1
¢
2
different ways, . . ., and the final nr in N −n1 −nn2 −···−n
¡ ¢
r
r−1
different ways.
Hence, the total number of ways is

N! (N − n1 )!
× × ···
(N − n1 )!n1 ! (N − n1 − n2 )!n2 !
(N − n1 − n2 − · · · − nr−1 )!
×
nr !0!
µ ¶
N
= .
n1 , n 2 , . . . , n r

N! N
¡ ¢
14. n1 !n2 !...nk−1 !0!
= n1 ,n2 ,...,nk−1
.
µ ¶
12
15. = 27 720.
3, 5, 4

,16. Add the (n − 1) barriers to the r objects from which we must choose
r, so the number of ways is
µ ¶
n+r−1
.
r

17. If we first put m objects into each container we have (r − nm) left, so,
by the above argument, the number of ways is
µ ¶
r − nm − n − 1
r − nm
When m = 1 this reduces to
µ ¶ µ ¶
r−1 r−1
= .
r−n n−1

18. (i) If x = cos2 (θ) then dx
dθ
= −2 sin(θ) cos(θ), hence
Z 0
β(m, n) = − cos2m−2 (θ)(1 − cos2 (θ))n−1 × 2 cos(θ) sin(θ) dθ
π/2

and the required result follows from the identity 1 − cos2 (θ) =
sin2 (θ).
dy 1
R∞ 2
(ii) If y 2 = x then dx = 2y , hence Γ(m) = 0 y m−2 e−y × 2y dy and
the required result follows.
R∞R∞ 2 2
(iii) Γ(m)Γ(n) = 4 0 0 e−(x +y ) x2m−1 y 2n−1 dx dy
Now substitute x = r cos(θ), y = r sin(θ) to obtain
Z ∞ Z π/2
2
Γ(m)Γ(n) = 4 e−r r2m−1 cos2m−1 (θ)r2n−1
0 0
× sin2n−1 (θ)r dr dθ
Z ∞
2
= 2 e−r r2(m+n)−1 dr
0
Z π/2
×2 cos2m−1 (θ) sin2n−1 (θ) dθ
0

Now use the result of (ii) to write
Z ∞
2
Γ(m + n) = 2 e−r r2(m+n)−1 dr
0

and the required result follows from (i).

, (iv) Using the result of (iii), we have

m!(n − 1)!
B(m + 1, n) =
(m + n)!

and the result follows upon multiplying top and bottom by n.


Chapter 3
1. (a) S1 , (b) {1}, (c) {3,4,6}, (d) R3 , (e) ∅,
(f) {1,3,4,6}, (g) {2,3,4,5,6}.

2. (a) [1,4], (b) [0.5,1], (c) [−1,1].

4. These are all straightforward consequences of (B7).

5. (i) (A ∩ B) − (A ∩ C) = (A ∩ B) ∩ (A ∩ C)
= (A ∩ B) ∩ (A ∪ C) by (B5)
= (A ∩ B ∩ A) ∪ (A ∩ B ∩ C) by (B4)
= A ∩ B ∩ C) by (B7)
= A ∩ (B − C).

(ii) Use (B4).

(iii)A − (A − B) = A ∩ (A ∩ B)
= A ∩ (A ∪ B) by (B5) and (B6)
=A∩B by (B4) and (B6).
(iv) (A − B) − C = (A ∩ B) ∩ C = A ∩ (B ∩ C)
by (B2)
= A ∩ (B ∪ C) by (B5)
= A − (B ∪ C).
(v) A − (A ∩ B) = A ∩ (A ∩ B)
= A ∩ (A ∪ B) by (B5)
= (A ∩ A) ∪ (A ∩ B) by (B4)
= ∅ ∪ (A − B) = A − B by (B7).

,6. (a) (i) Use (B3). (ii) Use (B7). (iii) Use (B1).
(iv) Use (B7) noting that S − A = A.
(v) (A ∩ B) ⊙ (A ∩ C) = [(A ∩ B) ∪ (A ∩ C)] − [(A ∩ B) ∩ (A ∩ C)]



= ([(A ∩ B) ∪ A] ∩ [(A ∩ B) ∪ C]) − (A ∩ B ∩ C)
by (B4), (B2) and (B1)
= [A ∩ [(A ∩ B) ∪ C]] − (A ∩ B ∩ C) by Exercises 3.3
= ([A ∩ (A ∩ B)] ∪ (A ∩ C)) − (A ∩ B ∩ C) by (B4)
= [(A ∩ B) ∪ (A ∩ C)] − A ∩ (B ∩ C) by Exercises 3.3
= A ∩ (B ∪ C) − A ∩ (B ∩ C) by (B4)
= A ∩ (B ⊙ C) by Exercises 3.5(i).




(b) (A − B) ∪ (B − A) = (A ∩ B) ∪ (B ∩ A)
= [(A ∩ B) ∪ B] ∩ [(A ∩ B) ∪ A]
by (B4)
= (A ⊙ B)
by (B4) again, (B5),
(B6) and (B7).




7. P(S3 ) = {∅, {HH}, {HT }, {T H}, {T T }, {HH, HT },
{HH, T H}, {HH, T T }, {HT, T H}, {HT, T T },
{T H, T T }, {HH, HT, T H}, {HH, HT, T T },
{HH, T H, T T }, {HT, T H, T T }, S3 }




8. J1 ∪ J2 = (−∞, −9] ∪ [−5, 4] ∪ [6, 15]

, 9. (i) We know that A1 ∪ A2 ∈ B(S); assume as an inductive hypothesis
that B = A1 ∪A2 ∪. . .∪An−1 ∈ B(S), then we must have B∪An ∈ B(S).
(ii) and (iii) are proved similarly.

10. m(A ∪ (B ∪ C)) = m(A) + m(B ∪ C) − m(A ∩ (B ∪ C))
(by hint)
= m(A) + m(B) + m(C) − m(B ∩ C)
− m((A ∩ B) ∪ (A ∩ C))

(by (M) and (B4)). Now use (M) again on the final term and the result
follows via (B3), (B2) and (B1).

11. 0.125.

12. A similar type of consideration to that of Example 3.7.

13. (a) 0.25, (b) e−2 = 0.135 . . ., (c) 3, (d) 1.61.

14. (i) For disjoint A and B we have

(m + n)(A ∪ B) = m(A ∪ B) + n(A ∪ B)
= m(A) + m(B) + n(A) + n(B) by (M)
= (m + n)(A) + (m + n)(B).

(ii) is proved similarly.
Total masses are m(S) + n(S) and c × m(S) respectively.

15. A similar argument to Example 3.7.

16. Since PA is a partition of A we have

A = (A ∩ E1 ) ∪ (A ∩ E2 ) ∪ . . . (A ∩ En )

and the result now follows immediately from Exercises 3.9(iii).

17. d(A, B) + d(B, C) − d(A, C)

= m(A ⊙ B) + m(B ⊙ C) − m(A ⊙ C)
= m((A ∪ B) − (A ∩ B)) + m((B ∪ C) − (B ∩ C))
− m((A ∪ C) − (A ∩ C))

, = m(A ∪ B) − m(A ∩ B) + m(B ∪ C) − m(B ∩ C)
− m(A ∪ C) + m(A ∩ C) by Theorem 3.2 (i)
= m(A) + m(B) + m(B) + m(C) − m(A) − m(C)
by Theorem 3.2 (iv)
= 2m(B) ≥ 0.
18. (i) Note that if A and C are disjoint then A − B and C − B are also
disjoint: hence.
X
(m ∗ n)(A ∪ C) = m((A ∪ C)\B)n(B)
B⊆S
X
= m((A\B) ∪ (C\B))n(B)
B⊆S
by Exercise 3.5 (ii)
X X
= m((A\B))n(B) + m((C B))n(B)
B⊆S B⊆S
by (M)
= (m ∗ n)(A) + (m ∗ n)(C).
(ii) In the definition of (m ∗ n), write A − B instead of B and use
Exercises 3.5(iii) to obtain
X
(m ∗ n)(A) = m((A ∩ B))n(A − B)
B⊆S

But m(A ∩ B) = m(B) when B ⊆ A and 0 otherwise; hence the
result follows.
P
(iii) [(m ∗ n) ∗ p](A) = B⊆S (m ∗ n)(A\B)p(B)
XX
= m(((A\B)\C))n(C)p(B)
B⊆S C⊆S
XX
= m((A\(B ∪ C)))n(C)p(B)
B⊆S C⊆S
by Exercises 3.5 (iv)
Now write D = B ∪ C so that C = D\B to find that
XX
[(m ∗ n) ∗ p](A) = m((A\D))n(D\B)p(B)
B⊆S D⊆S
= [m ∗ (n ∗ p)](A) as required.

,Chapter 4
1. (a) principle of symmetry,
(b) subjective probability using relative frequency as evidence,
(c) as in (b) but note that historical and personal experience of many
mornings, together with the application of Newton’s laws of mo-
tion to celestial mechanics, will lead most of us to assign a prob-
ability of 1 to this event.
(d) as in (b).

3. (a) {(H, H, H), (H, H, T ), (H, T, H), (H, T, T ), (T, H, H), (T, H, T ),
(T, T, H), (T, T, T )}.
(b) i. E = {(H, H, H), (H, H, T ), (H, T, H), (H, T, T )},
ii. F = {(H, T, H), (H, T, T ), (T, T, H), (T, T, T )}.
3 1 1 1
(c) , , , .
4 4 2 2

4. Let E be the event in question, then

E = {(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)}

As #(S) = 36, the principle of symmetry yields

1
P (E) = .
6

5. By (P7), P (A ∩ B) = 0.45 + 0.22 − 0.53 = 0.14
By (B5) and (P5),

P (A ∪ B) = P (A ∩ B) = 1 − 0.14 = 0.86.

6. (a) 0.147, (b) 0.853.

7. (a) 0.119, (b) 0.179, (c) 0.032, (d) 0.328.

8. δa (IR) = 1.
Pn 1
Pn
9. i=1 P (xi ) = Z i=1 f (xi ) = 1.

, 10. By the result of Theorem 4.1 (a), we have only to show that

P (A ∩ A)
PA (A) = = 1.
P (A)

P (A ∩ B ∩ A)
11. PA (A ∩ B) = = PA (B) by (B1).
P (A)
P ((A∪B)∩C)
12. PA∪B (C) = P (A∪B)

P (A ∩ C) P (B ∩ C)
= + by (B4) and (P3)
P (A ∪ B) P (A ∪ B)
P (A) P (A ∩ C) P (B) P (B ∩ C)
= × + ×
P (A ∪ B) P (A) P (A ∪ B) P (B)
and the result follows. Note that this result shows that for fixed C ∈
B(S), P(.) (C) is not a probability measure.
7 6 7 5
13. (i) 12 × 11 = 0.318, (ii) 12 × 11
= 0.265.
5 7
(iii) 0.265 + 12 × 11 = 0.530.

14. Let D be event that the item is defective. We want
P ((X ∪ Y ) ∩ D)
PD (X ∪ Y ) =
P (D)
P (X ∩ D) + P (Y ∩ D)
=
P (D)
PX (D)P (X) + PY (D)P (Y )
=
P (D)
= 0.13.

15. Q(0) = Q(1) = 12 so that there is also maximum uncertainty of the
output:
Q0 (0) = Q1 (1) = 1 − ε, Q0 (1) = Q1 (0) = ε
So the Qi (j)’s are the same as the Pi (j)’s and the channel is completely
symmetric between input and output.

16. Q(1) = p(1 − ε), Q(E) = ε, Q(0) = (1 − p)(1 − ε).
Q0 (0) = Q1 (1) = 1, QE (0) = 1 − p, QE (1) = p.