Section 6.4
Problems 39-46
In order to know which element fits in the set, you must ask whether it meets the
criteria…
Problem 39
…does the element meet the criteria “…the digit is even…”?
“2”, “4” and “6” are even so all elements with these in it MEET the criteria, and
are part of the set.
So the answer is: A: {2a, 2d, 4a, 4g, 6a, 6f}
Problem 40
…does the element meet the criteria “…the letter is a vowel…”?
“a” and “e” are vowels so all elements with these in it MEET the criteria, and are
part of the set.
So the answer is: B: {1a, 2a, 3e, 4a, 5e, 6a}
Problem 41
…does the element meet the criteria “…the digit is odd…”?
“1”, “3” and “5” are odd so all elements with these in it MEET the criteria, and are
part of the set.
So the answer is: C: {1a, 1b, 1c, 3b, 3e, 3f, 5b, 5e}
Problem 42
…does the element meet the criteria “…the letter is not a vowel…”? (which
means it is a consonant) “b”, “c”, “d”, “f” and “g” are consonants (not vowels) so
all elements with these in it MEET the criteria, and are part of the set.
So the answer is: D: {1b, 1c, 2d, 3b, 3f, 4g, 5b, 6f}
Problem 43
…does the element meet the criteria “…divides the digit evenly…”?
“3” and “6” are digits that 3 divides evenly so all elements with these in it MEET
the criteria, and are part of the set.
So the answer is: E: {3b, 3e, 3f, 6a, 6f}
Problem 44
…does the element meet the criteria “…the letter is a consonant…”?
“b”, “c”, “d”, “f” and “g” are consonants so all elements with these in it MEET the
criteria, and are part of the set.
So the answer is: D: {1b, 1c, 2d, 3b, 3f, 4g, 5b, 6f}
Problem 45
…does the element meet the criteria “…3 divides the digit and the letter is a
consonant…”? the “and” means that BOTH things must be true, not just one or
the other. So of those with 3 divides the digit (problem 43) only those that have:
“b”, “c”, “d”, “f” and “g” are consonants so all elements with these in it MEET the
criteria, and are part of the set.
So the answer is: E: {3b, 3f, 6f }
Problem 46
…does the element meet the criteria “…3 divides the digit or the letter is a
consonant…”? the “or” means that one or the other or both is true. So all those
with either a “3” or a “6” plus all those that meet the consonant criteria:
“b”, “c”, “d”, “f” and “g” are consonants so all elements with these in it MEET the
criteria, and are part of the set.
So the answer is: E: {1b, 1c, 2d, 3b, 3e, 3f, 4g, 5b, 6a, 6f }
Problem 47
…event G…terms of events E and F…”where G is defined as “…3 divides the digit
and the letter is a consonant…” The key is the word “and” means that both E and
F are true at the same time, which is the definition of an intersection.
So the way to write that would be: G=EF
Problem 48
…event H…terms of events E and F…”where H is defined as “…3 divides the digit
or the letter is a consonant…” The key is the word “or” means that the set
includes all that is in E and all that is in F or all that is in both E and F, which is the
definition of a union.
So the way to write that would be: H=EF
Problems 49-54
Now look at the answers for problems 39-46 and see which ones contain the
value…
Problem 49
…“2a” is found in sets: A, B
Problem 50
…“6f” is found in sets: A, D, E, F, G, H
Problem 51
…“1c” is found in sets: C, D, F, H
Problem 52
…“5b” is found in sets: C, D, F, H
Problem 53
…“3e” is found in sets: B, C, E, G, H
Problem 54
…“4g” is found in sets: A, D, F, H
Problem 55
The three players could be represented as A, B and C, so they could be lined up as
ABC (an arbitrary line-up)…but based on this I could represent them all using
Titleist by “UUU”…player A uses Titleist, player B uses Titleist, and player C uses
Titleist…and if any of them do not use it we substitute the “u” for an “N”.
So “UUU” represents all of them using Titleist, but if one of them did not use
Titleist, it could be player A (NUU) or player B (UNU) or player C (UUN)…taking it
another step would be that two players did not use Titleist: A and B (NNU) or A
and C (NUN) or B and C (UNN). Leaving the only case left being that all of them
did not use Titleist (NNN)
So the sample space is the set of all of the above:
S= {UUU, UUN, UNU, NUU, NNU, NUN, UNN, NNN}
Problems 56-61
Using the sample space (answer) for problem 55 and see which ones contain the
value(s)…
Problem 56
…“At least two use Titleist…”; “at least two” means the least value is two, or 2 use
Titleist or more use Titleist; so in his case either 2 or 3 use Titleist…
Giving an answer of: A: {UUU, UUN, UNU, NUU}
Problem 57
…“At least one uses Titleist…”; “at least one” means the least value is one, or 1
use Titleist or more use Titleist; so in his case it is 1, 2 or 3 use Titleist…
Giving an answer of: B: {UUU, UUN, UNU, NUU, NNU, NUN, UNN}
Problem 58
…“Exactly one uses Titleist…”; “exactly one” means one and only one uses
Titleist…
Giving an answer of: C: {NNU, NUN, UNN}
Problem 59
…“At most one uses Titleist…”; “at most one” means one is the greatest number
to use Titleist, or one or less use Titleist, so in this case, 1 or none (zero)…
Giving an answer of: D: {NNU, NUN, UNN, NNN}
Problem 60
…“Second man uses Titleist…”; “second man” is our player B, so whenever he has
a “U”…
Giving an answer of: F: {UUU, UUN, NUU, NUN}
Problem 61
…“Last two don’t use Titleist…”; “last two” are our players B and C, so whenever
they BOTH has an “N”…
Giving an answer of: G: {UNN, NNN}
Problem 62
Two tasks in this sample space “throwing a coin” (C) and “tossing a die” (D), so
any result can be represented by CD. The result of the coin is either heads (H) or
tails (T); while the die can have anything from 1 to 6.
So for any H (head) you could have a die with a 1, 2, 3, 4, 5 or 6; and likewise with
T (tails)…
So the sample space is the set of all of the above:
S= {H1, H2, H3, H4, H5, H6, T1, T2, T3, T4, T5, T6}
Problems 63-66
Using the sample space (answer) for problem 62 and see which ones contain the
value(s)…
Problem 63
…“die shows an odd number…”; “odd number” means “1”, “3” or “5”, so all
elements with those in the element …
Giving an answer of: A: {H1, H3, H5, T1, T3, T5}
Problem 64
…“coin shows a head…”; so all elements with “H” in the element …
Giving an answer of: B: {H1, H2, H3, H4, H5, H6}
Problem 65
…“die shows a number greater than four…”; “greater than four” means any
number larger but not equal to 4, so 5 and 6, in this case, so all elements
containing these …
Giving an answer of: C: {H5, H6, T5, T6}
Problem 66
…“coin shows a head or the die shows a number greater than 4…”; “coin shows a
head” is the answer to problem 64; “dies shows a number greater than 4” is the
answer to problem 65; and the “connector” is an “or” (union), so whatever is in
either answer…
Giving an answer of: D: {H5, H6, T5, T6}
Problem 67
“…judge ranks three platform divers A, B, and C…” ; with the key word being
“ranks” meaning to put in order. So we must shuffle ABC into as many
arrangements as possible (which happens to be 3*2*1 = 6 ways)…
So the sample space is the set of all of the above:
S= {ABC, ACB, BAC, BCA, CAB, CBA}
Problems 68-73
Using the sample space (answer) for problem 67 and see which ones contain the
value(s)…
Problem 68
…“A is the best…”; in other words A is in the first slot, so all elements where A is
first …
Giving an answer of: A: {ABC, ACB}
Problem 69
…“A does not end up last…”; in other words A is NOT in the last slot, so all
elements where A is not in the last slot …
Giving an answer of: B: {ABC, ACB, BAC, CAB}
Problem 70
…“Either A or B takes first…”; in other words either A is in the first slot OR B is in
the first slot, so all elements that meet that criteria (remembering that “or” is a
union so all that meet either criteria…A or B in the first slot)…
Giving an answer of: C: {ABC, ACB, BAC, BCA}
Problem 71
…“ A is first and C is last…”; in other words A has to be in the first slot AND C must
be in the last slot, so all elements that meet BOTH criteria at the same time
(remembering that “and” is an intersection so all that meet both criteria…A first
and C last)…
Giving an answer of: D: {ABC}
Problem 72
…“B does not win…”; in other words B is NOT in the first slot, so all elements in
which B is not in the first slot…
Giving an answer of: F: {ABC, ACB, CAB, CBA}
Problem 73
…“A wins or B is not last…”; in other words either A is in the first slot OR B is in the
not in the last slot OR BOTH being true, so all elements that meet that criteria
(remembering that “or” is a union so all that meet either criteria)…
Giving an answer of: F: {ABC, ACB, BAC, BCA, CBA}
Problem 74
Red Die
Two fair dice…I will send you to page 35 of the book which shows the results (like
what I have below (sum in parentheses):
1
2
3
4
5
6
1
1,1 (2)
2,1 (3)
3,1 (4)
4,1 (5)
5,1 (6)
6,1 (7)
Green Die
2
3
1,2 (3)
1,3 (4)
2,2 (4)
2,3 (5)
3,2 (5)
3,3 (6)
4,2 (6)
4,3 (7)
5,2 (7)
5,3 (8)
6,2 (8)
6,3 (9)
4
1,4 (5)
2,4 (6)
3,4 (7)
4,4 (8)
5,4 (9)
6,4 (10)
5
1,5 (6)
2,5 (7)
3,5 (8)
4,5 (9)
5,5 (10)
6,5 (11)
6
1,6 (7)
2,6 (8)
3,6 (9)
4,6 (10)
5,6 (11)
6,6 (12)
So that table becomes the sample space:
S= {(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (3,1),
(3,2), (3,3), (3,4), (3,5), (3,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (5,1), (5,2), (5,3),
(5,4), (5,5), (5,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)}
Note: the reason that this was displayed as (1,1), (1,2), (1,3) rather than 11, 12, 13
is because of the confusion with actual numbers.
Problems 75-79
Using the sample space (answer) and the table for problem 74, and see which
elements contain the value(s) or meet the criteria…
Problem 75
…“sum … is 8…”; in other words wherever the number in the parentheses is equal
to 8, …
Giving an answer of: A: {(2,6), (3,5), (4,4), (5,3), (6,2)}
Problem 76
…“sum … is ten or more…”; in other words wherever the number in the
parentheses is equal to 10, 11 or 12, …
Giving an answer of: B: {(4, 6), (5,5), (5,6), (6,4), (6,5), (6,6)}
Problem 77
…“number on the red die is a 5…”; in other words wherever the first number (red
die) is equal to 5, …
Giving an answer of: C: {(5, 1), (5,2), (5,3), (5,4), (5,5), (5,6)}
Problem 78
…“number on the red die is a two more than the number on the green die…”; so
every element in which the first number is two more than the second number
(exactly two, because it does not say two or more or two or less). That means if
the green die is 1, the red is 3 (3,1), or if the green die is 2, the red is 4 (2,4) and so
forth …
Giving an answer of: D: {(3, 1), (4,2), (5,3), (6,4)}
Problem 79
…“green die shows a number which is at least four…”; “at least four” means the
least (or smallest number is 4, so 4 or more: 4, 5, or 6. So all elements in which
the second number is a 4, 5, or 6 …
Giving an answer of: E: {(1, 4), (1,5), (1,6), (2,4), (2,5), (2,6), (3,4), (3,5), (3,6),
(4,4), (4,5), (4,6), (5,4), (5,5), (5,6), (6,4), (6,5), (6,6)}
Problem 80
The easiest way to look at what the results are is to “map” it out, or use a tree
diagram. “One shell is drawn from the box and fired.” And there are only two
possible results it goes off (BANG!) and shows that the shell is good (G) or it is a
Dud, which means the shell was bad (B).
G
B
“If the shell is good, production continues. If the shell is bad, two more shells are
drawn from the box and fired…” In other words, if it is good (G) you go no further
in the diagram above, but if it is bad you fire off two more shells:
G
G
G
B
B
B
G
Note that I did not give a good and a bad on the lowest line of the tree diagram.
That is because the problem says “box containing two bad shells) and if I place a
bad (B) there it would create a place for 3 bad shells (too many).
So going down each path gives the sample space:
S= {G, BGG, BGB, BBG}
Problems 81-84
Using the sample space (answer) for problem 80, and see which elements contain
the value(s) or meet the criteria…
Problem 81
…“one defective…”; so every element with exactly one defective (not more, not
less) …
Giving an answer of: A: {BGG}
Problem 82
…“More than one defective…”; so every element with 1 or two defective shells…
Giving an answer of: B: {BGG, BGB, BBG}
Problem 83
…“… more good shells than bad…”; so the first element works, only good (G) and
BGG works since there are 2 goods and one bad …
Giving an answer of: C: {G, BGG}
Problem 84
…“… second shell tested is defective…”; so not the first element, since a second
shell was not fired, but for all the rest, if the middle letter is B, all these elements
make up the set …
Giving an answer of: D: {BBG}
Problem 85
The easiest way to look at what the results are is to “map” it out, or use a tree
diagram. “…he either kisses her goodnight (K) or shakes her hand (H).”
K
H
“She responds by blushing (B), fainting (F) or sighing (S)…” And these results may
be for either thing that Sammy did:
B
F
K
H
S
B
F
S
So going down each path gives the sample space:
S= {KB, KF, KS, HB, HF, HS}
Problems 86-87
Using the sample space (answer) for problem 85, and see which elements contain
the value(s) or meet the criteria…
Problem 86
…“did not faint…”; so every element that does not contain “F” …
Giving an answer of: A: {KB, KS, HB, HS }
Problem 87
…“rewarded with a blush or a sigh…”; so every element that does not contain “B”
or a “S”…
Giving an answer of: B: {KB, KS, HB, HS }