Math 141 Week in Review
Week 6 Problem Set Answers
1.
A. S = {(H, R), (H, B), (H, G), (T, R), (T, B), (T, G)}
B. E = {(H, R), (T, R)}
C. F = {(T, R), (T, B), (T, G), (H, G)}
D. No, E ∩ F = {(T, R)}
E. No, all outcomes are not equally likely. There are not an equal number of
marbles of the different colors.
2.
A. Yes, all outcomes are equally likely.
B. E = {(6, 1), (1, 6), (3, 2), (2, 3), (2, 4), (4, 2)}; P ( E ) =
C. F = {(4, 1), (1, 4), (2, 2), (1, 3), (1, 3)}; P ( F ) =
3.
1
6
5
36
A.
Simple Event
Probability
11 9
5
+
=
44 44 11
6 19
= 1! P ( R) = 1!
=
44 22
B. P (G ! O) =
( )
C. P RC
Y
13
44
4.
A. P(E ∩ F) = 0.3
B. P(FC) = 0.5
C. P(EC ∪ FC) = 0.7
D. P(EC ∩ F) = 0.2
5.
A.
89
= 0.178
500
175
B.
= 0.35
500
350
C.
= 0.7
500
325
D.
= 0.65
500
G
11
44
O
9
44
R
6
44
P
5
44
26
= 0.5
52
39
B.
= 0.75
52
16
C.
! 0.3077
52
36
D.
! 0.6923
52
6.
A.
7.
A. P(20, 20) = 20!
B. 3!⋅9!⋅6! ⋅5!
8.
P(200, 2)⋅C(198, 18)
9.
A. C(7, 4)⋅C(19, 2) = 5,985
B. Method 1: C(9, 3)C(17, 3) + C(9, 4)C(17, 2) + C(9, 5)C(17, 1) + C(9, 6) = 76,482
Method 2: C(26, 6) − [C(17, 6) + C(9, 1)C(17, 5) + C(9, 2)C(17, 4)] = 76,482
C. C(17, 6) = 12,376
D. C(10, 3)C(16, 3) + C(7, 2)C(19, 4) − C(10, 3)C(7, 2)C(9, 1) = 125,916
10.
18!
= 14,702,688
7!6!5!
11.
A. True
B. False
C. False
D. True
E. False
F. True
G. False
H. True
12.
A. 32
B. 85
C. 69
13.
5⋅5⋅4⋅4⋅3 = 1,200
14.
Corner Points: (6, 4), (2, 6), (2, 8);
The minimum value of C is 28 and is attained at infinitely points, which lie on the
line segment joining the points (6, 4) and (2, 8).
15.
x = number of small sandwiches sold each hour
y = number of large sandwiches sold each hour
Maximize P = 2x + 3y
Subject to:
3x + 4 y ! 36
2x + 4 y ! 32
x "0
y "0
Profit is maximized when 4 small and 6 large sandwiches are sold. The maximum
profit is $26. There is no peanut butter or jelly left over.
16.
x = number of child tickets sold
y = number of student tickets sold
z = number of adult tickets sold
Maximize P = 10x + 20y + 25z
Subject to:
x + y + z ! 15,000
z " 4x
20y ! 13 (25z)
z " 12 ( x + y + z)
x "0
y "0
z"0