Solutions to Sample Exam #1
Math For Liberal Arts I
Solutions to Sample Exam #1
The following questions are meant to check your readiness for the first exam, as well as to give you an idea of its length and difficulty.
This sample exam is NOT a template for the actual exam, i.e. there may be questions on the exam which have a style different from any of these questions!
YOU  WILL  NEED  A  CALCULATOR  FOR  THE  FIRST  EXAM.
1.    In
North   Dakota
, a license plate consists of  2  letters (with repetitions allowed) followed by  4  digits (with no repetitions allowed).  How many different license plates are possible in
North Dakota
?
SOLUTION:  To make a license plate, we need to make two choices:  Choose two letters, and then choose four digits.
Because repetitions are allowed in the choice of letters, we can use the FCP to compute that there are 
26
ï¿½26 = 676  possibilities all together for the two letters.
Because repetitions are not allowed in the choice of digits, we note that the digits are a permutation of 4 digits chosen from 10 digits, and therefore there are 
10
ï¿½9ï¿½8ï¿½7 =
5040  possibilities all together for the four digits.
Using the FPoC, we multiply together the number of possibilities for the two letters times the number of possibilities for the four digits, to get 
676
ï¿½5040 = 3,407,040  possible license plates all together.
2.    The Alaska Lottery consists of drawing 7  numbers at random from the numbers  1  to  43.  How many possible outcomes are there to the Alaska Lottery?
SOLUTION:  In a lottery, unless we are told otherwise, the numbers chosen must be different, and the order of selection does not matter.  So, we are counting the number of combinations of 7 numbers chosen from 43, which gives 
43
C7
=43P7 / 7! =(43ï¿½42ï¿½41ï¿½40ï¿½39ï¿½38ï¿½37) / (7ï¿½6ï¿½5ï¿½4ï¿½3ï¿½2ï¿½1) =162,409,534,560 / 5040 = 32,224,114  possible outcomes to the lottery.
3.    The Swampy
Sandwich
Shoppe gives you a choice of rye, whole wheat or pita bread for your sandwich.  You get a choice of butter or mayonnaise, and a choice of roast beef, ham, salami, bologna or pastrami.  How many different sandwiches does Swampy sell?
SOLUTION:  We have three choices to make: bread, condiment, and meat.  By the FPoC, the number of possible sandwiches is 3
ï¿½2ï¿½5 =30
.
4.    How many different arrangements in a line are there of the letters of the word  POSSESSIVENESS?
SOLUTION:  We are counting permutations, but some of the letters are repeated, so we have a special formula.  First we count that there are 14 letters in the word "POSSESSIVENESS," with 6 S's and 3 E's (and one each of P, O, I, V, and N).  So the number of permutations is 
14
! / (6!ï¿½3!) =(14ï¿½13ï¿½12ï¿½11ï¿½10ï¿½9ï¿½8ï¿½7ï¿½6ï¿½5ï¿½4ï¿½3ï¿½2ï¿½1) / (6ï¿½5ï¿½4ï¿½3ï¿½2ï¿½1ï¿½3ï¿½2ï¿½1) =87,178,291,200 / 4320 = 20,180,160.
5.    Consider a standard deck of  52  cards.
a.    How many  4-card hands can be made from the 52  cards?
SOLUTION:  For hands of cards, unless we are told otherwise, the cards dealt must be different, and the order in which they are dealt does not matter.  So, we are counting the number of combinations of 4 cards chosen from 52, which gives 
52
C4
=52P4 / 4! =(52ï¿½51ï¿½50ï¿½49) / (4ï¿½3ï¿½2ï¿½1) =6,497,400 / 24 = 270,725  hands.
b.    How many  4-card hands are there consisting entirely of diamonds?
SOLUTION:  As there are 13 diamonds in a standard deck of 52 cards, we are counting the number of combinations of 4 cards chosen from the 13 diamonds in the deck.  This gives 
13
13ï¿½12ï¿½11ï¿½10) / (4ï¿½3ï¿½2ï¿½1) =
17160 / 24 =
715  hands consisting entirely of diamonds.
c.    How many 4-card hands are there containing at least 2 clubs?
SOLUTION:  "... at least 2 clubs" means "2 or more clubs", i.e. "2, 3, or 4 clubs". So we need to count the number of 4-card hands with exactly 2 clubs, exactly 3 clubs, and all 4 clubs
and add them together.
To construct a hand with exactly two clubs, we need to make two choices:  Choose two clubs, and then choose two non-clubs.
There are 13 clubs in a standard deck,  so there are 
13C2 = 13P2 / 2! =(13x12)/(2x1)=78
to select two of them.
There are 39 non-clubs in a standard deck, so there are
39C2 = 13P2 / 2! =(39x38)/(2x1)= 741
to select two of them.
So there are 78x741=57,798 hands with exactly two clubs.
Similarly there are (
6768  possible hands consisting of exactly two kings.
6.    Consider an exam consisting of  10 true/false questions.
a.    How many different ways could you answer the entire exam?
SOLUTION:  We have 10 choices to make (namely, the answers to each of the questions), and there are 2 possibilities for each answer (T or F), so by the FCP there are 
2ï¿½2ï¿½2ï¿½2ï¿½2ï¿½2ï¿½2ï¿½2ï¿½2ï¿½2 =210=
1024  different ways to answer the entire exam.
b.    How many different ways could you answer the entire exam using true  5  times and false 5  times?
SOLUTION:  This is the same as deciding how many ways we can arrange the letters  TTTTTFFFFF  (5 T's and 5 F's).  Thus, we are counting the number of permutations 10 letters, with 5 T's and 5 F's.  Using the formula, the number of permutations is
10! / (5!ï¿½5!) =
(10ï¿½9ï¿½8ï¿½7ï¿½6ï¿½5ï¿½4ï¿½3ï¿½2ï¿½1) / (5ï¿½4ï¿½3ï¿½2ï¿½1ï¿½5ï¿½4ï¿½3ï¿½2ï¿½1) =3,628,800 / 14400 =
252, which is the number of ways of answering the exam using true 5 times and false 5 times.
A DIFFERENT SOLUTION: Another way to look at this problem is to use combinations. There are
10C5
=
10P5 / 5! =(10ï¿½9ï¿½8ï¿½7ï¿½6)/(5ï¿½4ï¿½3ï¿½2ï¿½1)=30240/120=252 ways to choose 5 out of the 10 questions to be marked true. Then the other 5 questions are automatically false.
7.   Sam has 8 different marbles.
       (a) How many ways can he store 5 marbles in his right pocket and 3 marbles in his left pocket?
       (b) How many ways can he store 5 marbles in one pocket and 3 marbles in the other?
SOLUTION:
(a) There are 
8C5
or 56 ways to choose 5 marbles for the right pocket. The others so automatically into the left pocket.
(b) Here there are two choices to be made:
Task 1: Choose which pocket will have 5 marbles.   This can be done in 2 ways left or right.
Task 2: Choose 5 marbles to go in that pocket.  This can be done in 56 ways as in part (a).
So the combined task of putting 5 marbles in one of the two pockets can be done in 2x56=112 ways.
The other 3 marbles go automatically in the other pocket.
8.   (a)   How many ways are there to arrange eight people around a circular table?
         (b)  How many ways are there to arrange four women and four men around a circular table, if the men and women must sit in alternate seats?
SOLUTION:
(a) There are two slightly different ways to attack this problem.
We can think of seating one of the eight people first, and then count the ways
of arranging the other seven in a row.  This gives an answer of 7!
We can let x be the answer to this problem, and note that each circular
arrangement can be "unglued" into eight different arrangements in a row,
depending on where the cirle is "cut."  Since there are 8! ways to arrange
eight people in a row, we get
8x=8!, so that x= 8!/8, which is 7!
(b) Again, there are two ways.
If we seat a man first, then we can seat the remaining people as
woman-man-woman-man-woman-man-woman
Each circular arrangement still gives rise to eight arrangements in a row, and
there are (8)(4)(3)(3)(2)(2)(1)(1) ways to seat the eight people in a row with
the genders alternating.  When you divide that last number by eight, you again
get 4!3!.
 
 ï¿½ï¿½ï¿½ 9.ï¿½ï¿½ï¿½ï¿½ (a) How many poker hands (5 cards) are possible from a deck of 52?
ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ ï¿½(b) How many poker hands are straights--all five cards with consecutive values?
ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ (c) What is the probability that a poker hand will be a straight?
ï¿½ï¿½ï¿½ You may leave factorials in your answers.
 
SOLUTION:
ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ (a) Order does not matter. So it’s
.
ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ (b) There are 9 values from which a straight can start: 2, 3, 4, 5, 6, 7, 8, 9, 10, J.ï¿½ One example is 7-8-9-10-J.ï¿½ Once the starting value has been chosen, there are 5 choices for each individual card.ï¿½ Thus the answer is
.
ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ (c) The probability of a straight is the number of straights divided by the number of hands:
ï¿½
 
ï¿½ï¿½ï¿½ 10.ï¿½ ..(a)How many ways are there to choose a committee of 5 people from a group of 35?
ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ (b)If 20 of the people are men and 15 are women, how many ways are there to choose a committee with exactly 3 men and 2 women?
ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ (c)What is the probability that a randomly chosen committee of 5 will have exactly 3 men and 2 women?
ï¿½ï¿½ï¿½ You may leave your answers in terms of Ps and/or Cs.
ï¿½
SOLUTION:
 
ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ (a) Again, order does not matter.ï¿½ So it’s
.
 
ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ (b) Here you choose the men and then the women, and then the Fundamental Principle of Counting tells you to multiply:
ï¿½
 
ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ (c) As in the preceding problem, the probability is the quotient:
ï¿½/
.
 
ï¿½ï¿½ï¿½ï¿½ 11.ï¿½ï¿½ (a) What is the probability of a family with three children having exactly two girls?
ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ (b) What is the probability of a family with three children having exactly two girls, if they have at least one boy?
 
ï¿½ï¿½ï¿½ï¿½
SOLUTION:
 
(a)   There are eight ways of distributing the genders of three children; three of them have exactly one boy and two girls:
(b)   There are seven distributions with at least one boy:
 
 
12.ï¿½ï¿½ï¿½ (a) What is the probability that a roll of two dice will result in a total of 8 if at least one die is a 4?
ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ (b) What is the probability that a roll of two dice will have at least one die a 4 if the total is 8?
ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½
 
SOLUTION
 
ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ (a) Of the eleven rolls with at least one 4, only one is an 8:
ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ (b) Of the five ways of rolling an 8, only one, doubles, has a 4:
 
ï¿½ï¿½ 13.ï¿½ï¿½ï¿½ï¿½ Find the mean, median and standard deviation of the data below.
ï¿½ï¿½ï¿½ 3, 4, 7, 9, 12, 15, 14, 20
 
SOLUTION:
ï¿½mean =
ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ median = 10.5, the average of the fourth and fifth numbers.
 
ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ variance =
.
ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ standard deviation =
ï¿½ï¿½ï¿½
ï¿½ï¿½ï¿½ 14.ï¿½ï¿½ï¿½ï¿½ï¿½ Find the five-number summary for the following set of data:
ï¿½ï¿½ï¿½ 2ï¿½ï¿½ 21ï¿½ï¿½ 16ï¿½ï¿½ 5ï¿½ 17ï¿½ï¿½ 1ï¿½ï¿½ 19ï¿½ï¿½ 13ï¿½ï¿½ 2ï¿½ 15ï¿½ï¿½ 8ï¿½ 18ï¿½ 12ï¿½ 16ï¿½ 11
 
SOLUTION:
ï¿½First rewrite the data in order:
1        2ï¿½ 2ï¿½ 5ï¿½ 8ï¿½ 11ï¿½ 12ï¿½ 13ï¿½ 15ï¿½ 16ï¿½ 16ï¿½ 17ï¿½ 18ï¿½ 19ï¿½ 21
There are 15 numbers.ï¿½ The median is the eighth, or 13.ï¿½ The first quartile is the median of the seven numbers before 13, or 5.ï¿½ Similarly, the third quartile is 17.ï¿½ We know what the minimum and maximum are, so the five-number summary is: 1, 5, 13, 17, 21.
ï¿½ï¿½ï¿½ 15.ï¿½ï¿½ï¿½ï¿½ï¿½ The scores of students on a standardized test form a normal distribution with a mean of 250 and a standard deviation of 40.ï¿½ Two thousand students took the test.ï¿½ Find the number of students who scored above 330.
 
SOLUTION:
 
For a normal distribution, 95% of the data is within two standard deviations of the mean.ï¿½ Two standard deviations is 80, so 95% of the students scored between 250 – 80 and 250 + 80; i.e., between 170 and 330.ï¿½ This leaves 5% of the two thousand students, or 100 students outside that range.ï¿½ Half of those scored above 330; the answer is 50.
 
ï¿½ï¿½ï¿½ 16.ï¿½ï¿½ï¿½ï¿½ Here is a data set::
 
xï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ 17ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ 25ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ 31ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ 33ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ 39
yï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ 21ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ 33ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ 37ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ 39ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ 45
ï¿½ï¿½ï¿½ Using the formulas for correlation and least squares regression, find the correlation between x and y and use the regression line to predict the value of y when x is 40.
 
The mean of the x-values is
.
The variance of the x-values is
ï¿½
The standard deviation of the x-values is
 
The mean of the y-values is
ï¿½ï¿½ï¿½
The variance of the y-values is
.
The standard deviation of the y-values is
.
 
Plugging into the formula for correlation, and using the differences we used in the calculation of the variance, we see that
 
ï¿½
 
We are going to simplify our calculation, by using the common denominator—he had the denominator in each term to stress to you the fact that we were calculating using the standard deviation as our unit, so as to make the correlation a number between 0 and 1.