Finite Math 
Final Exam

Dec. 13, 2000

A Track 
R. Hammack


Name: ________________________ 
Score: _________

(1) Solve the following systems of equations.
(a)
3x_{1}  +  6x_{2}  +  9x_{3}  +  6x_{4}  =  9 
2x_{1}  +  4x_{2}  +  6x_{3}  +  2x_{4}  =  4 
Setting up the augmented matrix and row reducing:
1/3R_{1} > R_{1} 1/2R_{2} > R_{2} 
R_{1} +R_{2} > R_{2} 
R_{2} > R_{2}  2R_{2} + R_{1} > R_{1} 
This last (reduced) matrix corresponds to the system
x_{1}  +  2x_{2}  +  3x_{3}  =  1  
x_{4}  =  1 
Thus the solutions are:
x_{1} = 1  2m 3n
x_{2} = m
x_{3} = n
x_{4} = 1
(b)
x

+

y



z

=

0

x

+

y

+

z

=

2

x



y

+

z

=

2

R_{1} +R_{2} > R_{2}
R_{1} + R_{3} > R_{3} 
R_{2} <> R_{3}  1/2R_{2} > R_{2} 1/2R_{3} > R_{3} 
R_{2} + R_{1} > R_{1}  R_{3} + R_{2} > R_{2} 
From the reduced matrix, you can see that there is just one solution, namely x = 1, y = 0, z = 1.
(2) Find the maximum of the objective function P = x  10y subject to
the following inequalities
x + y

≤ 4 
x  y

≤ 2 
x

≥ 0 
y

≥ 1 
Notice that this linear programming problem is not in standard form. Thus the simplex method will not work. Solve it by graphing.
When you graph the feasible region, it looks as follows. There are 4 corner points, tabulated on the right.
As you can see from the table, the objective function is maximized at 11 when x = 1 and y = 1 
(3) Use the simplex method to maximize P = 2x + 3y + z, subject to the following constraints:
2x

+

y

+

z

≤ 2

4x

+

4y

≤ 4


3x

+

3y

+

z

≤ 9

x, y, z

≥ 0

There are three problem contraints, so there will be 3 slack variables. The system becomes:
2x

_{+ y }

+ z

+ s_{1}

= 2  
4x

+ 4y

+ s_{2}

= 4  
3x

+ 3y

+ z

+ s_{3}

= 9  
2x

 3y

z

+ P

= 0 
Now we set up the simplex tableau. The pivot column and row are indicated.
x

y

z

s_{1}

s_{2}

s_{3}

P


s_{1}

2

1

1

1

0

0

0



2

2/1 = 2  
s_{2}

4

4

0

0

1

0

0



4

4/4 = 1 (smallest)  
s_{3}

3

3

1

0

0

1

0



9

9/3 = 3  










P

2

3

1

0

0

0

1



0

1/4 R_{2} > R_{2}
x

y

z

s_{1}

s_{2}

s_{3}

P


s_{1}

2

1

1

1

0

0

0



2


s_{2}

1

1

0

0

1/4

0

0



1


s_{3}

3

3

1

0

0

1

0



9












P

2

3

1

0

0

0

1



0

R_{2} + R_{1} > R_{1}
3R_{2} + R_{3} > R_{3}
3R_{2} + R_{4} > R_{4}
x

y

z

s_{1}

s_{2}

s_{3}

P


s_{1}

1

0

1

1

1/4

0

0



1

1/1 = 1 (smallest)  
y

1

1

0

0

1/4

0

0



1

(can't divide by 0)  
s_{3}

0

0

1

0

3/4

1

0



6

6/1 = 6  










P

1

0

1

0

3/4

0

1



3

There's still a negative indicator above, so it's time for another pivot operation.
The pivot row and column are indicated above.
The pivot entry is already 1, so that's good.
 R_{1} + R_{3} > R_{3}
R_{1} + R_{4} > R_{4}
x

y

z

s_{1}

s_{2}

s_{3}

P


z

1

0

1

1

1/4

0

0



1


y

1

1

0

0

1/4

0

0



1


s_{3}

1

0

0

1

1/2

1

0



5












P

2

0

0

1

1/2

0

1



4

There are no more negative indicators, so we are done.
P is maximized at 4 when x = 0, y = 1, and z = 1.
(4) A hand of 4 cards is dealt off of a standard 52card deck. How many such hands are possible if:
(a) Order is considered? P_{52, 4} = (52)(51)(50)(49) = 6,497,400.
(b) Order is not considered? C_{52, 4} = P_{52, 4} / 4! = 6,497,400 / 24 = 270,725.
(5) Four cards are dealt off a wellshuffled deck.
(a) You win $1 if the first card is a club, the second is a diamond, the third is a heart, and the fourth is a spade. Find the probability of your winning.
P(E) = n(E)/n(S) = 13^{4} / P_{52, 4} = 28561 / 6,497,400 ~ 0.439%
Alternate solution: (13/52)(13/51)(13/50)(13/49) ~ 0.439%
(b) You win $1 if the four cards are of different suits. Find the probability of your winning.
Let E = "cards are of different suits." Here is how you could make
an outcome in E:
1. Choose suit for 1st card (4 ways)
2. choose 1st card (13 ways)
3. choose suit for 2nd card (3 ways)
4. choose 2nd card (13 ways)
5. choose suit for 3rd card (2 ways)
6. choose 3rd card (13 ways)
7. choose suit for 4th card (1 way)
8. choose 4th card (13 ways)
By multiplication principle, n(E) = (4)(13)(3)(13)(2)(13)(1)(13) = (4!)(13^{4)}
P(E) = n(E)/n(S) = (4!)(13^{4)}/ P_{52, 4} ~ 10.549%
(6) A fair die is rolled 5 times. Find the probabilities of the following events.
(a) All rolls have an even number of spots.
(1/2)(1/2)(1/2)(1/2)(1/2) = 1/32 = 3.125%
(b) It is not the case that all rolls have an even number of spots.
1  1/32 = 31/32 = 96.875%
(7) Suppose A and B are events. Suppose that P(A) = 0.5, P(B) = 0.6,
and P(A ∩ B) = 0.3.
(a) Find P(A ∪ B).
P(A ∪ B) = P(A) + P(B)  P(A ∩ B) = 0.5 + 0.6  0.3 = 0.8
(b) Are A and B independent? Explain.
Notice that P(A)P(B) = (0.5)(0.6) = 0.3 = P(A ∩ B).
Thus, since P(A)P(B) = P(A ∩ B), it follows that A and B are independent.
(c) Are A and B mutually exclusive? Explain.
If A and B were mutually exclusive, we would have P(A ∪ B) = P(A) + P(B).
However, as was computed above, P(A ∪ B) = 0.8, while P(A) + P(B) = 0.5
+ 0.6 = 1.1
Therefore A and B are not mutually exclusive.
(8) Two cards are dealt off a wellshuffled deck.
(a) What is the probability that both cards are black? (26/52)(25/51) = 650 / 2652 ~ 24.509%
(b) What is the probability that either both cards are black or both are aces?
A = "both black"
B = "both aces"
P(A or B) = P(A) + P(B)  P(A and B) =
(26/52)(25/51) + (4/52)(3/51)  (2/52)(1/51) = 660 / 2652 = 155 / 221 ~ 24.886%
(c) What is the probability that both cards are black and both are aces?
As was computed above, P(A and B) = (2/52)(1/51) = 1/ 1326 ~ 0.000754%
Here's another way to look at it.
There is only one way that "A and B" can happen, and that is if the
hand consists of the ace of clubs and the ace of spades.
The probability of this is thus 1/C_{52, 2} = 1/1326
(9) Find the optimal stratigies for each player in the following matrix
games. Determine if each game is fair, and if not, which player has the advantage.
(a)
[ 
3

4

] 
1

2

1 is a saddle value. The game is strictly determined.
R should always play second row.
C should always play first column.
Game is not fair, for even when these optimal strategies are used, C always
wins a point and R loses.
(b)
[ 
4

2

3

] 
2

1

1

C should never play column 3, for column 2 is always better for C than column 3 no matter what choice R makes.
Thus this game reduces the the non strictly determined game
[ 
4

2

] 
2

1

Using the formula:
D = (4  1)  ( 2 + 2) = 9.
P ^{*} = [ 1/3 2/3 ]
Q^{*} = 
[

1/3

] 
2/3


0

Game value = ((4)(1)  (2)(2)) / D = 0, so the game is fair
(10) Imagine that you own a movie rental business with two locations, Location A and Location B. A movie rented at one location may be returned to the other. You notice that each week, 90% of the movies rented at Location A are returned to Location A, while 70% of those rented at Location B are returned to Location B. An inventory taken today suggests that 40% of the movies are at Location A, and the remaining are at Location B.
(a) What percentages of the movies will be at the two locations next week?
S_{0} = [ .4 .6 ]
P = [

.9

.1

] 
.3

.7

S_{1} = S_{0} P = [ (.4)(.9)+(.6)(.3) (.4)(.1)+(.6)(.7) ] = [.54 .46]
Thus after one week 54% are at location A and 46% are at location B.
(b) What percentages will be at the two locations in the long run?
The answer is given by the stationary matrix S = [x y], which has the property SP = S, or
[x y][

.9

.1

]= [x y] 
.3

.7

Multiplying, this gives [.9x+.3y .1x+.7y] = [x y].
This gives the following system which must be solved in order to find S=[x
y]:
.9x + .3y = x
.1x + .7y = y
x + y = 1
This system simplifies to:
.1x + .3y = 0
.1x  .3y = 0
x + y = 1
The last equation gives y = 1  x. Plugging this into the first equation,
.1x + .3(1  x) = 0
.1x + .3 .3x = 0
.4x = .3
x = .3/.4 = 3/4 = .75
So y = 1  x = 1  .75 = .25
Therefore the stationary matrix is S = [ .75 .25 ].
In the long run, 75% of the movies will be at Location A and 25% will be at
Location B.