Chapter 1
Problems
(I) (a) By the generalized basic principle of counting there are
26- 26 10 10- 1(3- 10- 10:67,600,000
(b) 2625-10-9'8-7-6219,656,{}09
xx,
/
There are 4! possible arrangements. By assigning instruments to Jay, Jack, John and Jim, in
that
Chapter 3
Problems
G H6 1 different} = P{6, different}/P{diff6rentl
- P{Ist = 6,2nd i 6} + P{lst ¢ 6,2nd = 6}
5/6
= 21/6 5/6 :18
could also have been solved by using reduced sample spacefor given that outcomes differ it
is the same as asking for the prob
Chapter 2
Problems
(9
13.
(a) S = {(F, r), (r, 3), (Kb), (8, r). (g, 8). (8,17), (5. r), b, g), (b, 45)}
(b) S = {(r, 8), (I: b). (g, r), (8, b), (b, 90,07,191
S = {(n,x1, .
(nan, .
event (uleEnY is the event that 6 never appears.
that she or he is a ciga
Name:
Section:
Multiple Choice
1.(2 pts.) The area of the region enclosed by the curves y = x2 and y = (x 1)2 + 1 is
given by the integral:
1
1
(x 1)2 + 1) dx (b)
(a)
1
0
(2x 2x2 ) dx
0
(2x 2x2 ) dx
(c)
1
1
1
(d)
1
(x2 ) dx
(2x2 2x) dx
(e)
0
Solution:
Fir
3 2 1
2. The rank of A is
1. Let A = 1 3
4 5
1
(a) 2
(b) 3
(c) 0
(d) 4
(e) 1
2. Let P2 = cfw_a0 +a1 t+a2 t2 where cfw_a0 , a1 , a2 range over all real numbers, and let T : P2 P2
be a linear transformation dedined by
T (a0 + a1 t + a2 t2 ) = a1 + 9a2 t