HOMEWORK-1 SOLUTIONS
STA-130A, FALL 2016
1. (Problem 2, Page 27) Two six-sided dice are thrown sequentially, and
the face values that come up are recorded.
(a) The sample space is:
= cfw_(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2,
Stat 130A Homework 2, Solutions
1. Problem 46, page 30
Solution. Urn A has three red balls and two white balls, and urn B has two red balls and five white
balls. A fair coin is tossed. If it lands heads up, a ball is drawn from urn A; otherwise, a ball is
HOMEWORK-3 SOLUTIONS
STA-130A, FALL 2016
4. (Problem 24, Page 66) (Banach Match Problem) A pipe smoker carries
one box of matches in his left pocket and one box in his right. Initially, each
box contains n matches. If he needs a match, the smoker is equal
Homework #4 Solutions
Textbook Problems:
Exercise 3.3.1
n
(i) X B(n, p) implies that P (X = x) = n px q nx = nx q nx pn(nx) where q =
x
n
n
n
1 p, since x = nx . However, nx q nx pn(nx) is equal to P (Y = n x), where
Y B(n, q). Therefore, P (X = x) = P (Y
Homework #1 Solutions
Textbook Problems:
Exercise 1.2.9
(i) If A B C = A, then A (B C) = A. This implies that B C A. Since
B B C and C B C, we see that B A and C A.
(ii) If A B C = A, then A (B C) = A. This implies that A B C. Since
B C B and B C C, we se
Homework #2 Solutions
Textbook Problems:
Exercise 2.3.3.
1. P (blonde) =
20
50
=
2
5
= 0.4.
P (blonde and blue eyes)
P (blue eyes)
2. P (blonde|blue eyes) =
=
15/50
21/50
=
15
21
=
5
7
0.714.
Exercise 2.3.11.
Denote Ri , Bi to be the events that ith ball
STA 130A Homework #6 Solutions
Textbook Problems:
Exercise 4.1.3
Since P (X < Y ) + P (Y < X) = 1, and the p.d.f. is symmetric with respect to x and y, one
would expect that P (X < Y ) = 0.5. Indeed,
1
1
(x + y)dxdy =
fx,y (X, Y )dxdy =
2
1
0
0
0
y
x2
2
y
Homework #3 Solutions
Textbook Problems:
Exercise 2.2.1
(i) The random variable X may take on the values 0, 1, 2, 3. Therefore, the set is cfw_0, 1, 2, 3.
(ii) Let f (x) be the p.d.f of X when X = x. Then, f (0) =
3
f (2) = 8 = 0.375, and f (3) = 1 = 0.12
STA 130A Homework #7 Solutions
Textbook Problems:
Exercise 6.2.3.
1
(i) From u = 2 (x + y) and v = 1 (x y), we get u + v = 1 (2x) = x and u v = 1 (2y) = y.
2
2
2
Hence, x = u + v and y = u v. Since x > 0 and y > 0, we have v > 2u and v < u,
whereas u > 0
Handout, STA131C-S07, W. Polonik
On some applications of linear transformations, in particular projections and rotations, in statistics
Some notation: Vectors in Rd are denoted by bold symbols, such as a, b, e, x, y . . . etc. For a vector a we deno