TEST BANK FOR Probability and Random Processes for Electrical and Computer Engine By John A. Gubner

1. W = {1,2,3,4,5,6}.
2. W = {0,1,2, . . . ,24,25}.
3. W = [0,¥). RTT > 10 ms is given by the event (10,¥).
4. (a) W = {(x,y) 2 IR2 : x2+y2 100}.
(b) {(x,y) 2 IR2 : 4 x2+y2 25}.
5. (a) [2,3]c = (−¥,2)[(3,¥).
(b) (1,3)[(2,4) = (1,4).
(c) (1,3)\[2,4) = [2,3).
(d) (3,6] \(5,7) = (3,5].
6. Sketches:
x
y
1
1
1 B
0
B
x
y
−1
B
−1
−1
x
y
x
y
x
y
3
x
y
3
C1 H3 J3
1
2 Chapter 1 Problem Solutions
x
y
x
y
3
3
3
3
H3 J3
U = M3 H U 3 J3 N3
=
x
y
2
M2 N3
U = M2
2
x
y
4
3
M4UN3
4
3
7. (a) [1,4]\

[0,2][[3,5]

=

[1,4]\[0,2]

[

[1,4]\[3,5]

= [1,2][[3,4].
(b)

[0,1][[2,3]
c
= [0,1]c \[2,3]c
=
h
(−¥,0)[(1,¥)
i
\
h
(−¥,2)[(3,¥)
i
=

(−¥,0)\
h
(−¥,2)[(3,¥)
i
[

(1,¥)\
h
(−¥,2)[(3,¥)
i
= (−¥,0)[(1,2)[(3,¥).
(c)
¥\
n=1
(−1
n , 1
n ) = {0}.
(d)
¥\
n=1
[0,3+ 1
2n ) = [0,3].
(e)
¥[
n=1
[5,7− 1
3n ] = [5,7).
(f)
¥[
n=1
[0,n] = [0,¥).
Chapter 1 Problem Solutions 3
8. We first let C A and show that for all B, (A\B)[C = A\(B[C). Write
A\(B[C) = (A\B)[(A\C), by the distributive law,
= (A\B)[C, since C A)A\C =C.
For the second part of the problem, suppose (A\B)[C = A\(B[C). We must show
that C A. Let w 2 C. Then w 2 (A \ B) [C. But then w 2 A \ (B [C), which
implies w 2 A.
9. Let I := {w 2 W : w 2 A)w 2 B}. We must show that A\I = A\B.
: Let w 2 A\I. Then w 2 A and w 2 I. Therefore, w 2 B, and then w 2 A\B.
: Let w 2 A\B. Then w 2 A and w 2 B. We must show that w 2 I too. In other
words, we must show that w 2 A)w 2 B. But we already have w 2 B.
10. The function f : (−¥,¥)![0,¥) with f (x) = x3 is not well defined because not all
values of f (x) lie in the claimed co-domain [0,¥).
11. (a) The function will be invertible if Y = [−1,1].
(b) {x : f (x) 1/2} = [−p/2,p/6].
(c) {x : f (x) < 0} = [−p/2,0).
12. (a) Since f is not one-to-one, no choice of co-domain Y can make f : [0,p] !Y
invertible.
(b) {x : f (x) 1/2} = [0,p/6][[5p/6,p].
(c) {x : f (x) < 0} =?.
13. For B IR,
f −1(B) =
8>>< >>:
X, 0 2 B and 1 2 B,
A, 1 2 B but 0 /2 B,
Ac, 0 2 B but 1 /2 B,
?, 0 /2 B and 1 /2 B.
14. Let f :X !Y be a function such that f takes only n distinct values, say y1, . . . ,yn.
Let B Y be such that f −1(B) is nonempty. By definition, each x 2 f −1(B) has the
property that f (x) 2 B. But f (x) must be one of the values y1, . . . ,yn, say yi. Now
f (x) = yi if and only if x 2 Ai := f −1({yi}). Hence,
f −1(B) =
[
i:yi2B
Ai.
15. (a) f (x) 2 Bc , f (x) /2 B,x /2 f −1(B),x 2 f −1(B)c.
(b) f (x) 2
¥[
n=1
Bn if and only if f (x) 2 Bn for some n; i.e., if and only if x 2 f −1(Bn)
for some n. But this says that x 2
¥[
n=1
f −1(Bn).
4