Math 141 WIR, Spring 2007, c Benjamin Aurispa
Math 141 Exam 2 Review Answer Key
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 = { (1 , 6) , (2 , 3) , (3 , 2) , (6 , 1) , (2 , 4) , (4 , 2) }
P ( E ) =
1
6
(c) F = { (1 , 4) , (4 , 1) , (2 , 2) , (1 , 3) , (3 , 1) }
P ( F ) =
5
36
3.
(a)
Simple Event Y G O R P
Probability
13
44
11
44
9
44
6
44
5
44
(b) P ( G ∪ O ) =
11
44
+
9
44
=
20
44
(c) P ( P c ) = 1 − P ( P ) = 1 − or
5
11
5
44
=
39
44
4. Note: I had changed some of the numbers for this problem to P ( E ) = .
4 , P ( F ) = .
5, and
P ( E ∪ F ) = .
6 so that it would actually make sense. These changes have been made on the problem sets and are also reflected in the full solutions.
(a) 0 .
3
(b) 0 .
5
(c) 0 .
7
(d) 0 .
2
5.
(a)
(b)
(c)
(d)
89
500
175
500
350
500
325
500
= 0 .
178
= 0 .
35
= 0 .
7
= 0 .
65
6.
(a)
(b)
(c)
(d)
4
13
9
13
3
4
1
2
7.
(a) P (20 , 20) = 20!
(b) 3!
· 9!
· 6!
· 5!
8.
P (200 , 2) · C (198 , 18)
1

Math 141 WIR, Spring 2007, c Benjamin Aurispa
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!
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 = 1200
14. See full solutions from the live review for graph.
Corner Points C = 2 x + 4 y
(2 , 8) 36
(2 , 6)
(6 , 4)
28
28
C is minimized at infinitely many points which all lie on the line segment between (2 , 6) and
(6 , 4). The minimum value of C is 28.
15. Note: I changed the problem to 35 ounces of jelly available instead of 24. This change is reflected on the problem sets and on the full solutions from the live review.
x = number of small sandwiches sold each hour y = the number of large sandwiches sold each hour
Maximize P = 2 x + 5 y
Subject to:
3 x + 5 y ≤ 30
4 x + 5 y ≤ 35 x ≥ 0 y ≥ 0
2

Math 141 WIR, Spring 2007, c Benjamin Aurispa
Corner Points P = 2 x + 5 y
(0 , 0) 0
(0 , 6)
(8 .
75 , 0)
(5 , 3)
30
17.5
25
Profit is maximized when 0 small and 6 large sandwiches are sold. The maximum profit is $30.
There are no ounces of peanut butter left over, but there are 5 ounces of jelly left over.
16.
x = number of children tickets sold y = number of student tickets sold z = number of adult tickets sold
Maximize P = 10 x + 20 y + 25 z
Subject to: x + y + z ≤ 15000 z ≥ 4 x
1
20 y ≤ z ≥ 1
3
(25 z )
2 x ≥ 0
( x + y + z ) y ≥ 0 z ≥ 0
3