Math For Liberal Arts I
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.
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 26×26 = 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 10P4 = 10×9×8×7 = 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 676×5040 = 3,407,040 possible license plates all together.
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 43C7 =43P7 / 7! =(43×42×41×40×39×38×37) / (7×6×5×4×3×2×1) =162,409,534,560 / 5040 = 32,224,114 possible outcomes to the lottery.
SOLUTION: We have three choices to make: bread, condiment, and meat. By the FPoC, the number of possible sandwiches is 3×2×5 =30.
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!×3!) =(14×13×12×11×10×9×8×7×6×5×4×3×2×1) / (6×5×4×3×2×1×3×2×1) =87,178,291,200 / 4320 = 20,180,160.
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 52C4=52P4 / 4! =(52×51×50×49) / (4×3×2×1) =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 13C4 =13P4 / 4! = (13×12×11×10) / (4×3×2×1) = 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
To construct a hand with exactly two clubs, we need to make two choices: Choose two clubs, and then choose two non-clubs.
Similarly there are (13C3)x(39C1 )=286x39=11,154 hands containing exactly three clubs.
Therefore, the number hands with at least two clubs is 57,798+11,154+715=69,667.
d. How many 4-card hands are there containing exactly two kings?
SOLUTION: To construct a hand with exactly two kings, we need to make two choices: Choose two kings, and then choose two non-kings.
There are 4 kings in a standard deck of 52 cards, so there are 4C2 = 4P2 / 2! = (4×3) / (2×1) = 12 / 2 = 6 ways to select two of them.
There are 52 - 4 = 48 non-kings in a standard deck of 52 cards, so there are 48C2 = 48P2 / 2! = (48×47) / (2×1) = 2256 / 2 = 1128 ways to select two of them.
Using the FCP, we multiply together the number of possibilities for two kings times the number of possibilities for two non-kings, to get 6×1128 = 6768 possible hands consisting of exactly two kings.
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 2×2×2×2×2×2×2×2×2×2 =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!×5!) = (10×9×8×7×6×5×4×3×2×1) / (5×4×3×2×1×5×4×3×2×1) =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! =(10×9×8×7×6)/(5×4×3×2×1)=30240/120=252 ways to choose 5 out of the 10 questions to be marked true. Then the other 5 questions are automatically false.
(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.
8. (a) How many ways are there to arrange eight people around a circular table?
(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.
in (4)(3)(3)(2)(2)(1)(1)=4!3! ways.
Each circular arrangement still gives rise to eight arrangements in a row, and
9. (a) How many poker hands (5 cards) are possible from a deck of 52?
(b) How many poker hands are straights--all five cards with consecutive values?
(c) What is the probability that a poker hand will be a straight?
You may leave factorials in your answers.
(a) Order does not matter. So it’s
(b) There are 9 values from which a straight can start: 2, 3, 4, 5, 6, 7, 8, 9, 10, J. One example is 7-8-9-10-J. Once the starting value has been chosen, there are 5 choices for each individual card. Thus the answer is
(c) The probability of a straight is the number of straights divided by the number of hands:
10. ..(a)How many ways are there to choose a committee of 5 people from a group of 35?
(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?
(c)What is the probability that a randomly chosen committee of 5 will have exactly 3 men and 2 women?
You may leave your answers in terms of Ps and/or Cs.
(a) Again, order does not matter. So it’s
(b) Here you choose the men and then the women, and then the Fundamental Principle of Counting tells you to multiply:
(c) As in the preceding problem, the probability is the quotient:
11. (a) What is the probability of a family with three children having exactly two girls?
(b) What is the probability of a family with three children having exactly two girls, if they have at least one boy?
(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. (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?
(b) What is the probability that a roll of two dice will have at least one die a 4 if the total is 8?
(a) Of the eleven rolls with at least one 4, only one is an 8:
(b) Of the five ways of rolling an 8, only one, doubles, has a 4:
13. Find the mean, median and standard deviation of the data below.
3, 4, 7, 9, 12, 15, 14, 20
median = 10.5, the average of the fourth and fifth numbers.
standard deviation =
14. Find the five-number summary for the following set of data:
2 21 16 5 17 1 19 13 2 15 8 18 12 16 11
First rewrite the data in order:
1 2 2 5 8 11 12 13 15 16 16 17 18 19 21
There are 15 numbers. The median is the eighth, or 13. The first quartile is the median of the seven numbers before 13, or 5. Similarly, the third quartile is 17. We know what the minimum and maximum are, so the five-number summary is: 1, 5, 13, 17, 21.
15. The scores of students on a standardized test form a normal distribution with a mean of 250 and a standard deviation of 40. Two thousand students took the test. Find the number of students who scored above 330.
For a normal distribution, 95% of the data is within two standard deviations of the mean. Two standard deviations is 80, so 95% of the students scored between 250 – 80 and 250 + 80; i.e., between 170 and 330. This leaves 5% of the two thousand students, or 100 students outside that range. Half of those scored above 330; the answer is 50.
16. Here is a data set::
x 17 25 31 33 39
y 21 33 37 39 45
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
The standard deviation of the x-values is
The mean of the y-values is
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
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.
The slope of the least-squares regression line is
The y-intercept is 35 - (1.06)29 = 35 – 30.74 = 4.26.