Final Exam Worksheet Answers
Sections 5.1-5.3, 6.1-6.5, 7.1-7.2, 1.2-1.6, 2.1-2.4, 3.1-3.3
Fall Semester, 2007
Problem 1. Timmy has a drawer with 7 socks in it. How many ways can
Timmy pull out 5 socks if he selects socks:
(a) with replacement? [You can leave this as a mathematical expression.]
(b) without replacement? [This answer needs to be a number.]
Solution. (a) We have a Theorem that says taking k elements from a set of
n elements with replacement, we will have nk total outcomes. Here, we have
k = 5 and n = 7. So Timmy has 75 different ways of pulling out 5 socks from a
drawer with 7 socks with replacement.
(b) With replacement, we are trying to find out how many 5-permutations of a
set of 7 elements there are. So, we want to find P (7, 5):
7!
(7 − 5)!
7!
=
2!
=7·6·5·4·3
= 7 · 6 · 5 · 12
= 7 · 6 · 60
= 7 · 360
= 2520
P (7, 5) =
So Timmy has 2520 different ways of pulling out 5 socks from a drawer with 7
socks without replacement.
Problem 2. An experiment is performed 27 times, in which each each trial has
probability of success .35. What kind of process is this? What is the probability
of getting exactly 12 successes? [Do not attempt to give a decimal, but give
numbers instead of symbols.]
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Solution. This is an example of a Bernoulli Process, so we can apply Bernoulli’s
Formula with n = 27, k = 12, p = .35 and q = .65. So,
27!
Pr(12 successes)= C(27, 12)(.35)12 (.65)27−12 = 12!15!
(.35)12 (.65).15
Problem 3. Find the inverse of C =
4
7
1
2
.
Solution. Notice that
det(C) = (4)(2) − (7)(1) = 1 6= 0, so C is invertible.
2
−1
C −1 = 11
.
−7 4
Problem 4. There are two jars containing colored balls. Jar One contains 4
orange balls and 6 blue balls. Jar Two contains 4 orange balls and 8 blue balls.
First an Jar is chosen at random, then one ball is taken from the jar. What is
the probability that Jar Two was chosen given that the ball is orange.
Solution. We want to find Pr(J2 |O), where J2 is the event of choosing the
second jar and O is the event of choosing an orange ball.
Since the jars are chosen at random, there is probability 21 that either jar is
chosen. Since there are 12 balls in Jar Two, four of which are orange. We can
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conclude that Pr(O|J2 ) = 12
= 13 . To get the total probability for the event O,
we also need Pr(O|J1 ). There are 10 balls in Jar One, four of which are orange,
4
so Pr(O|J1 ) = 10
= 25 .
So, the conditional probability is given by:
Pr(J2 )Pr(O|J2 )
Pr(O)
1 1
·
= 1 22 31 1
25 + 23
P r(J2 |O) =
=
1
6
11
30
=
5
11
1 3 −1
2
,B=
2 4 0
−2
Compute, AT , 3A − 2B, AC and CB.
Problem 5. A =
2
5 −3
2 −7
1 −1
,C=
.
0 2


1 2
Solution. AT =  3 4 .
−1 0
1 3 −1
2
3A − 2B = 3
−2
2 4 0
−2
5 −3
2 −7
=
−1 −1
10
8
3
14
.
AC is not
match up. defined
as the dimensions
don’t
1 −1
2 5 −3
4 3
4
CB =
=
0 2
−2 2 −7
−4 4 −14
Problem 6. How many distinct words can be formed from the letters in:
URLACHER?
Solution. There are 2 repeated letters, the R’s. There are C(8, 2) ways to
arrange the R’s. We have 6 distinct letters remaining. There are P(6, 6) ways
to arrange these letters. To get the total number of words, we multiply these
two numbers together to get:
8!
· 6!
(8 − 2)!2!
8!
· 6!
=
6!2!
8!
=
2!
=8·7·6·5·4·3
= 8 · 7 · 6 · 5 · 12
= 8 · 7 · 6 · 60
= 8 · 7 · 360
= 8 · 2, 520
= 20, 160
C(8, 2) · P(6, 6) =
So, there are 20,160 distinct words that can be formed from these letters.
Problem 7. A ball is chosen at random from Urn I and transferred to Urn II,
then a ball is selected from Urn II. Given the following data.
Urn I
Urn II
Orange
3
4
Blue
6
5
Given that the ball taken out of Urn II was blue, find the probably that the
transferred ball was blue.
3
Solution. Let E be the event of the transfered ball was blue, and B the event
that the ball selected from Urn II was blue. We want to find Pr(E|B). We need
to find the total probability that the ball selected was blue.
Pr(B) =Pr(E)Pr(B|E)+Pr(E 0 )Pr(B|E 0 ) =
6
9
·
6
10
+
3
9
·
5
10
=
36
90
+
15
90
=
51
90 .
Now,
Pr(E|B) =
=
=
=
Pr(E)Pr(B|E)
Pr(B)
6
6
·
9 10
51
90
36
90
51
90
36
51
Problem 8. A fair coin is tossed 3 times.
(a) What is the probability of getting tails twice in a row?
(b) Draw a probability tree illustrating this experiment, and the outcomes we
are looking for in part (a).
Solution. (a) Notice that for each coin toss, Pr(T ) = 21 . Our experiment consists of tossing a coin three times. Let E be the event of getting two tails in
a row. We need to determine how many ways there are to get tails twice in a
row. We could have the outcomes (T, T, T ), (T, T, H) or (H, T, T ) that satisfy
getting tails twice. Since the probability of getting heads or tails on any given
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toss is 12 , the probability of each outcome listed is 12 = 18 . Since these events
are mutually exclusive, that is disjoint as sets, we have that:
Pr(E) = 3 · 18 = 38 .
Problem 9. Given Pr(E ∪ F ) = .4 and Pr(E) = .2, find Pr(F ) so that E and
F will be independent.
Solution. We need to recall the inclusion exclusion formula and what it means
to be independent. Requiring independence gives us that
Pr(E ∩ F ) =Pr(E)Pr(F ) = .2Pr(F ). Plugging this into the inclusion exclusion
formula yields. Pr(E ∪ F ) =Pr(E)+Pr(F )−Pr(E ∩ F ), so
.4 = .2+Pr(F ) − .2Pr(F ). So, .8Pr(F ) = .2, so Pr(F ) = .25.
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Problem 10. Given the following cumulative distribution function for an
experiment performed 7 times with probability of success .40.
x
0
1
2
3
4
5
6
B(x)
.0280
.1586
.4199
.7102
.9037
.9812
.9984
Find the probability of getting between 3 and 5 successes. [Justify your answer.]
Solution. To get the probability of between 3 and 5 successes, we simply look
at B(5) − B(2) = .9812 − .4199 = .4713. We can do this because the probability
of getting between 3 and 6 successes is:
b(3) + b(4) + b(5) = [B(3) − B(2)] + [B(4) − B(3)] + [B(5) − B(4)] = B(5) − B(2).
Problem 11. Given two points (1, 3) and (4, 5), determine the equation of the
line they both lie on. Write the equation of this line in General Linear form and
draw the graph of this line.
5−3
Solution. Notice m = 4−1
= 23 , so the point-slope formula gives us y − 3 =
2
3 (x − 1). Multiply both sides by 3 to get 3y − 9 = 2x − 2, so the General Linear
form is −2x + 3y = 7.
Problem 12. An experiment is performed 17 times with probability of success
.40. Find the probability of getting 13 successes.
Solution. This is a Bernoulli process if we define getting Tails to be a success.
Here we have p = .4, q = 1 − .4 = .6 so Bernoulli’s Formula tells us that Pr(k
successes)=C(n, k)pk q n−k . So,
Pr(13 successes)=C(17, 13)(.4)13 (.6)17−13 =C(17, 13)(.4)13 (.6)2 .
Problem 13. A jar contains 2 pennies and 2 nickels, 2 coins are chosen at
random without replacement. If X is the value of the coins drawn, find the
probability distribution of X.
Solution.
Draw
(p,p)
(p,n)
(n,p)
(n,n)
Value of X
2
6
6
10
Probability
(2/4)(1/3) = 1/6
(2/4)(2/3) = 1/3
(2/4)(2/3) = 1/3
(2/4)(1/3) = 1/6
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Problem 14. A factory produces 1000 units in a given day, they inspect a
sample of 20 units in which 3 are found to be defective. How many defective
units should they expect to have produced in that day?
3
Solution. We expect to have µ = np = 1000( 20
) = 150 defective units.
Problem 15. Four people are running for a Senate seat. Ms. A. is twice as
likely to win as Mr. B. Mrs. C. and Mr. D. are equally likely to win. Mrs. C.
is three times as likely to win as Mr. B. No one else is running.
(a) What is the probability that each person wins?
(b) Who is most likely to win?
Solution. (a) Define the event A to be Ms. A. winning, B the event of Mr. B.
winning and so on. The given information can be formulated as:
Pr(A) = 2·Pr(B).
Pr(C) =Pr(D).
Pr(C) = 3·Pr(B).
Now, since no one else is running, we have:
Pr(A)+Pr(B)+Pr(C)+Pr(D) = 1.
Now let us get the probabilities all in terms of one event.
Pr(A) = 2·Pr(B).
Pr(C) = 3·Pr(B).
Pr(D) = 3·Pr(B).
Putting this back into our previous equation, we have:
2·Pr(B)+Pr(B) + 3·Pr(B) + 3·Pr(B) = 1, that is:
9·Pr(B) = 1. So,
Pr(B) = 19 .
Pr(A) = 2·Pr(A) = 29 .
Pr(C) = 3·Pr(B) = 39 .
Pr(D) =Pr(C) = 39 .
(b) Both Mrs. C. and Mr. D. have the highest probability of winning, so they
are most likely to win.
Problem 16. Set-up the following word problem as a linear program but do
NOT solve, but do tell me what actions you would take to solve it.
Maximize the revenue from the available resources. Stullon Steel Company
manufactures 2 types of truck bumpers from iron and nickel. Bumper A uses
10 lb iron and 2 lb of nickel, while bumper B uses 12 lb iron and 1 lb nickel.
Bumper A sells for $50 while bumper B sells for $60. They have 2400 lbs of
iron and 305 lbs of nickel on hand.
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Solution. We first need to define appropriate variables. We want to find out
how many of each type of bumper we should make, so it would make sense to
define x to be the number of Bumper A to make and y to be the number of
Bumper B.
Next, we need to define an objective function, we want to maximize revenue, so
define z = 50x + 60y to be our revenue function, which we want to maximize.
The last thing we need to do is to find our constraints. Our info can be summarized by:
Iron Nickel
Bumper A
10
2
Bumper B
12
1
Total on hand 2400 305
So, the total on hand is the most we can use. So, we have the constraints:
10x + 12y ≤ 2400
2x + y ≤ 305
x ≥ 0,
y≥0
We then find the corner points, where the boundary lines meet. Then we test
our revenue function z at these points and pick the smallest one.
Problem 17. A game is played in which a die is rolled. If an odd number is
rolled, you lose the amount on the die. If an even number is rolled, you gain
the amount on the die.
(a) Find the expected value?
(b) Is this a fair game?
Solution. (a) We first define the random variable X to be the amount gained
or lost on a roll. X takes values in {−5, −3, −1, 2, 4, 6}, and each roll is equally
likely.
1
1
1
1
1
1
E(X) = (−5)( ) + (−3)( ) + (−1)( ) + (2)( ) + (4)( ) + (6)( )
6
6
6
6
6
6
−5 − 3 − 1 + 2 + 4 + 6
=
6
3
=
6
(b) As E(X) 6= 0, the game is not fair.
Problem 18. In a jar there are 7 red balls and 14 white balls. If 3 balls are
drawn, find the expected number of red balls if the balls are drawn:
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(a) with replacement.
(b) without replacement.
(c) If we wish to maximize the number of red balls chosen, should we select with
or without replacement?
Solution. (a) If the balls are drawn with replacement, this is a Bernoulli Pro7
) = 1.
cess, so the mean is µ = np = (3)( 21
(b) If the balls are drawn without replacement, then we have the formula
a
µ = n a+b
where a is he number of red balls and b is the number of white balls.
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So, µ = (3)( 7+14
) = 1.
(c) The expected values are the same, so it does not matter how we select
the balls.
Problem 19. Solve the following multisystem.
−4x + 7y − 2z = −4
2x − y − 2z = 8
x − 2y + z = 2
− 4x + 7y − 2z = −4
2x − y − 2z = 4
x − 2y + z = 1
Solution.
Consider the augmented
matrix:



1 −2 1
2
−4 7 −2 −4 −4
 2 −1 −2 8
4  →R1 ↔R3 →  −4 7 −2 −4
1 −2 1
2
1
2 −1 −2 8



1 −2 1 2 1
1 −2
0
0
R =R +4R
→R20 =R2 −2R1 →  0 −1 2 4 0  →R2 =−R2 →  0 1
3
1
3
0 3 −4 4 2
0 3



2
1
1 −2
1
−2
1
0
1
→R3 =R3 −3R2 →  0 1 −2 −4 0  →R3 = 2 R3 →  0 1
0 0
2
16 2
0 0



1
−2
0
−6
0
1 0
0
0
R =R +2R
→R20 =R2 −R 3 →  0 1 0 12 2  →R1 =R1 +2R2 →  0 1
1
3
1
0 0 1 8 1
0 0

1
−4 
4

2 1
1
−2 −4 0 
−4 4 2

1
2 1
−2 −4 0 
1
8 1

0 18 4
0 12 2 
1 8 1
This tells us that the system on the left has solution (18, 12, 8) and the system
on the right has solution (4, 2, 1).
Problem 20. Let Pr(E) = .55, Pr(F ) = .63, Pr(E ∩ F ) = .33.
(a) Find Pr(E ∪ F ).
(b) Find Pr(E ∩ F 0 ).
Solution. (a) The Inclusion-Exclusion Property says that:
Pr(E ∪ F ) =Pr(E)+Pr(F )−Pr(E ∩ F ). Plugging in what we know, we have:
8
Pr(E ∪ F ) = .55 + .63 − .33 = .85.
(b) Notice (E ∩ F 0 ) ∪ (E ∩ F ) are disjoint and their union is E.
So, Pr(E) =Pr(E∩F 0 )+Pr(E∩F ). So, .55 =Pr(E∩F 0 )+.33, so Pr(E∩F 0 ) = .22.
Problem 21. Will’s House of Pancakes is introducing a new French Toast Platter to its menu. The recipe for the French Toast cost $500 and each platter costs
$4 to produce. If the French Toast Platter is sold for $6.50 each, how many
French Toast Platters must be sold to reach the breakeven point?
Solution. We have a fixed cost of $500, and a unit cost of $4, so the cost
function will be C = 500 + 4x. The revenue function will be R = 6.5x. At the
breakeven point, we have that C = R, so
6.5x = 500 + 4x
2.5x = 500
x = 200
So, WHOP needs to make 200 French Toast Platters to reach the breakeven
point.
Problem 22. Find the value of k that makes the following system consistent.
2x − 5y = 5
−3x + 3y = 9
x + ky = 0
Solution. From equation 1 we can solve for x to get x = 25 y + 52 . We plug this
in to equation 2 to get:
5
5
−3( y + ) + 3y = 9
2
2
9
15
− y−
=9
2
2
9
33
− y=
2
2
33
11
y=−
=−
9
3
So, x = 52 (− 11
3 )+
5
2
= − 20
3 . From the last equation, we have that
−x
y
20/3
=
−11/3
20
=−
11
k=
9
Problem 23. Solve the following system using matrix inversion.
3x1 − 4x2 = 2
x1 + x2 = −3
3 −4
x1
2
Solution. Take A =
,X=
,B=
.
1 1
x2
−3
−1
Notice that
det(A) = (3)(1) − (1)(−4) = 7, so A exists. In fact,
1 4
A−1 = 71
. So, we have that
−1 3
1 4
2
−10
= 17
.
X = A−1 B = 71
−1 3
−3
−11
−11
So, the solution to the system is ( −10
7 , 7 ).
Problem 24. Solve the following system via Gauss-Jordan Elimination.
2x1 + 4x2 + 8x3 = 42
x1 + 2x2 + 3x3 = 16
x1 + 3x2 + 5x3 = 29
Solution.
We aregoing to need to perform
at most 3 pivots
to get into RREF.



2 4 8 42
0 0 2 10
0
0
 (1) 2 3 16  →R10 =R1 −2R2 →  1 2 3 16  →R2 =R2 −2R3 →
R3 =R3 −R2
 0 (1) 2 13 
 1 3 5 29 
0 0 1
0 0 (2) 10
5
0
0
 1 0 −1 −10  →R1 = 21 R1 →  1 0 −1 −10  →R20 =R2 +R1 →
R3 =R3 −2R1
0 1 2
0 1 2
13
13




0 0 1 5
1 0 0 −5
 1 0 0 −5  →R1 ↔R2 ↔R3 →  0 1 0 3 
0 1 0 3
0 0 1 5
Which yields the solution (−5, 3, 5) to our system.
Problem 25. The girl scout troop had a bake sale and 200 people were counted
coming through the line. The entire supply of 150 cakes and 120 pies were sold.
If nobody bought more than one of each.
(a) How many bought both?
(b) Draw a Venn Diagram to illustrate this situation. Shade and indicate the
number of people in the region representing those who bought cakes but not
pies.
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Solution. (a) If we define the set A to be the set of cakes and B to be the set
of pies, noting that we want the number of people who bought both cakes and
pies, that is n(A ∩ B) we can use the Inclusion Exclusion Formula:
n(A ∪ B) = n(A) + n(B) − n(A ∩ B)
200 = 150 + 120 − n(A ∩ B)
70 = n(A ∩ B)
So, we have that 70 people bought both.
(b) We are asked to find n(A∩B 0 ), noting that A∩B 0 and A∩B form a partition
of A, so A = (A ∩ B 0 ) ∪ (A ∩ B), (A ∩ B 0 ) ∩ (A ∩ B) = ∅, then the Inclusion
Exclusion Formula tells us that:
n(A) = n(A ∩ B 0 ) + n(A ∩ B) − n(∅)
150 = n(A ∩ B 0 ) + 70
80 = n(A ∩ B 0 )
Problem 26. Find the consumption matrix C and set the production schedule.
An autobody repairman (AR), and a mechanic, (M) eavh have a company. For
each $1 of business that AR does, he uses $.25 of his own services, and $.25 of
M’s services. For each $1 of business that M does, he uses $.10 of AR’s service
and $.20 of his own. In a certain week, AR receives $690 worth of outside
business, and M receives $460 worth of outside business.
Solution. This is problem #1 from section 3.3.
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