7-2 Probability and Counting
1. A fair die is rolled once. Let A be the event of rolling an even number, and let B be the event of rolling a number
greater than 4. Find A ∩ B.
SOLUTION:
The possible outcomes for event A are all of the numbers on a die that are even, or {2, 4, 6}.
The possible outcomes for event B are all of the numbers on a die that are greater than 4, or {5, 6}.
A ∩ B contains all of the outcomes that are in both sample space A and B.
A ∩ B = {6}
ANSWER:
A ∩ B = {6}
2. A fair die is rolled once. Let A be the event of rolling an even number, and let B be the event of rolling an odd
number. Find A ∩ B.
SOLUTION:
The possible outcomes for event A are all of the numbers on a die that are even, or {2, 4, 6}.
The possible outcomes for event B are all of the numbers on a die that are odd, or {1, 3, 5}.
A ∩ B contains all of the outcomes that are in both sample space A and B.
A∩B=Ø
ANSWER:
A∩B=Ø
eSolutions Manual - Powered by Cognero
Page 1
7-2 Probability and Counting
Use the spinner.
3. Let A be the event of the spinner landing on 4 or 10, and let B be the event of the spinner landing on a section with
a number divisible by 4. What are the possible outcomes of each event?
a. A = {___}
b. B = {___}
c. A ∩ B = {___}
SOLUTION:
a. The possible outcomes for event A are all of the numbers on the spinner that are 4 or 10, or {4, 10}.
b. The possible outcomes for event B are all of the numbers on the spinner that are divisible by 4, or {4, 8, 12}.
c. A ∩ B contains all of the outcomes that are in both sample space A and B.
A ∩ B = {4}
ANSWER:
a. 4, 10
b. 4, 8, 12
c. 4
eSolutions Manual - Powered by Cognero
Page 2
7-2 Probability and Counting
4. Let P be the event of the spinner landing on a section with a prime number, and let Q be the event of the spinner
landing on a section with a number that is a multiple of 3. What are the possible outcomes of each event?
a. P = {___}
b. Q = {___}
c. P ∩ Q = {___}
SOLUTION:
a. The possible outcomes for event P are all of the numbers on the spinner that are prime, or {2, 3, 5, 7, 11}.
b. The possible outcomes for event Q are all of the numbers on the spinner that are a multiple of 3, or {3, 6, 9,
12}.
c. P ∩ Q contains all of the outcomes that are in both sample space P and Q.
P ∩ Q = {3}
ANSWER:
a. 2, 3, 5, 7, 11
b. 3, 6, 9, 12
c. 3
eSolutions Manual - Powered by Cognero
Page 3
7-2 Probability and Counting
5. A card is selected from a standard deck of cards. What is the probability that the card is a diamond and is a
seven?
SOLUTION:
Let A be the event of choosing a diamond, and let B be the event of choosing a 7. The total number of outcomes is
the total number of cards in a deck, or 52.
Write the corresponding number of each card in its correct place in the Venn diagram. In the diagram, D stands for
diamonds, H stands for hearts, C stands for clubs, and S stands for spades.
There are 13 cards that are diamonds in a deck of cards and there is only 1 diamond that is also a 7.
The probability that the card is both a diamond and a 7 is
, or about 1.92%.
ANSWER:
or 1.92%
6. A card is selected from a standard deck of cards. What is the probability that the card has a number on it that is
divisible by 2 and is black?
eSolutions Manual - Powered by Cognero
Page 4
7-2 Probability and Counting
SOLUTION:
Let A be the event of choosing a card that has a number on it that is divisible by 2, and let B be the event of
choosing a card that is black. The total number of outcomes is the total number of cards in a deck, or 52.
Write the corresponding number of each card in its correct place in the Venn diagram. In the diagram, D stands for
diamonds, H stands for hearts, C stands for clubs, and S stands for spades.
There are 20 cards that have a number on it that is divisible by 2 and there are only 10 that is also black cards.
The probability that the card has a number on it that is both divisible by 2 and is black is
, or
about 19.23%.
ANSWER:
or 19.23%
eSolutions Manual - Powered by Cognero
Page 5
7-2 Probability and Counting
Use the spinner.
7. Let A be the event that the spinner lands on a vowel. Let B be the event that it lands on the letter J. What are the
possible outcomes
of each event?
a. A = {___}
b. B = {___}
c. A ∪ B = {___}
SOLUTION:
a. The possible outcomes for event A are all of the letters on the spinner that are vowels, or {A, E, O, U}.
b. The possible outcomes for event B are all of the letters on the spinner that are J, or {J}.
c. A ∪ B contains all of the outcomes that are in either sample space(s) A or B.
A ∪ B = {A, E, O, U, J}
ANSWER:
a. A, E, O, U
b. J
c. A, E, O, U, J
eSolutions Manual - Powered by Cognero
Page 6
7-2 Probability and Counting
8. Let X be the event that the spinner lands on a consonant. Let Y be the event that it lands on the letter K. What are
the possible outcomes of each event?
a. X = {___}
b. Y = {___}
c. X ∪ Y = {___}
SOLUTION:
a. The possible outcomes for event X are all of the letters on the spinner that are a consonant, or {K, H, S, J}.
b. The possible outcomes for event Y are all of the letters on the spinner that are K, or {K}.
c. X ∪ Y contains all of the outcomes that are in either sample space(s) X or Y.
X ∪ Y = {K, H, S, J}
ANSWER:
a. K, H, S, J
b. K
c. K, H, S, J
eSolutions Manual - Powered by Cognero
Page 7
7-2 Probability and Counting
9. A random number generator is used to generate one integer between 1 and 20. Let C be the event of generating a
multiple of 5, and let D be the event of generating a number less than 12. What are the possible outcomes of each
event?
a. C = {___}
b. D = {___}
c. C ∪ D = {___}
SOLUTION:
a. The possible outcomes for event C are all of the numbers that are a multiple of 5, or {5, 10, 15, 20}.
b. The possible outcomes for event D are all of the numbers that are less than 12, or {1, 2, 3, 4, 5, 6, 7, 8, 9, 10,
11}.
c. C ∪ D contains all of the outcomes that are in either sample space(s) C or D.
C ∪ D = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 15, 20}
ANSWER:
a. 5, 10, 15, 20
b. 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11
c. 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 15, 20
eSolutions Manual - Powered by Cognero
Page 8
7-2 Probability and Counting
10. A random number generator is used to generate one integer between 1 and 100. Let A be the event of generating a
multiple of 10, and let B be the event of generating a factor of 30. What are the possible outcomes of each event?
a. A = {___}
b. B = {___}
c. A ∪ B = {___}
SOLUTION:
a. The possible outcomes for event A are all of the numbers that are a multiple of 10, or {10, 20, 30, 40, 50, 60,
70, 80, 90, 100}.
b. The possible outcomes for event B are all of the numbers that are a factor of 30, or {1, 2, 3, 5, 6, 10, 15, 30}.
c. A ∪ B contains all of the outcomes that are in either sample space(s) A or B.
A ∪ B = {1, 2, 3, 5, 6, 10, 15, 20, 30, 40, 50, 60, 70, 80, 90, 100}
ANSWER:
a. 10, 20, 30, 40, 50, 60, 70, 80, 90, 100
b. 1, 2, 3, 5, 6, 10, 15, 30
c. 1, 2, 3, 5, 6, 10, 15, 20, 30, 40, 50, 60, 70, 80, 90, 100
eSolutions Manual - Powered by Cognero
Page 9
7-2 Probability and Counting
11. What is the probability of drawing a card from a standard deck and not getting a spade?
SOLUTION:
Let A be the event of choosing a card that is a spade. Then find the probability of the complement of A.
There are 13 cards that are spades.
There are 52 cards in a deck.
The probability of the complement of A is P(A') = 1 – P(A).
The probability of not getting a spade is
or
or 0.75.
ANSWER:
or
or 0.75
eSolutions Manual - Powered by Cognero
Page 10
7-2 Probability and Counting
12. What is the probability of flipping a coin and not landing on tails?
SOLUTION:
Let A be the event of a coin landing on tails. Then find the probability of the complement of A.
There is 1 side of a coin that is a tail.
There are 2 sides of a coin.
The probability of the complement of A is P(A') = 1 – P(A).
The probability of flipping a coin and not landing on tails is
or 0.5.
ANSWER:
or 0.5
eSolutions Manual - Powered by Cognero
Page 11
7-2 Probability and Counting
13. Carmela purchased 10 raffle tickets. If 250 were sold, what is the probability that one of Carmela’s tickets will not
be drawn?
SOLUTION:
Let A be the event of one of Carmela’s tickets being drawn. Then find the probability of the complement of A.
There are 10 raffle tickets that are Carmela’s.
There were 250 raffle tickets sold.
The probability of the complement of A is P(A') = 1 – P(A).
The probability that one of Carmela’s tickets will not be drawn is
or 0.96.
ANSWER:
or 0.96
eSolutions Manual - Powered by Cognero
Page 12
7-2 Probability and Counting
14. What is the probability of spinning a spinner numbered 1 to 6 and not landing on 5?
SOLUTION:
Let A be the event of spinning a spinner numbered 1 to 6 and landing on 5. Then find the probability of the
complement of A.
There is 1 section on the spinner labeled 5.
There are 6 sections on the spinner.
The probability of the complement of A is P(A') = 1 – P(A).
The probability spinning a spinner numbered 1 to 6 and not landing on 5 is
or about 0.83.
ANSWER:
or about 0.83
eSolutions Manual - Powered by Cognero
Page 13
7-2 Probability and Counting
15. STATISTICS A survey found that about 90% of the junior class is right-handed. If 1 junior is chosen at random
out of 100 juniors, what is the probability that he or she is left-handed?
SOLUTION:
Let A be the event of choosing a right-handed junior. Then find the probability of the complement of A.
There are 90 juniors that are right-handed because 90% of 100 is 0.9 × 100 = 90.
There are 100 juniors.
The probability of the complement of A is P(A') = 1 – P(A).
The probability of choosing a junior that is not right-handed, or that is left-handed, is
or
or 0.10.
ANSWER:
or
or 0.10
eSolutions Manual - Powered by Cognero
Page 14
7-2 Probability and Counting
16. RAFFLE Raul bought 24 raffle tickets out of 1545 tickets sold. What is the probability that Raul will not win the
grand prize of the raffle?
SOLUTION:
Let A be the event of one of Raul will win the grand prize of the raffle. Then find the probability of the complement
of A.
Raul bought 24 raffle tickets.
There were 1545 raffle tickets sold.
The probability of the complement of A is P(A') = 1 – P(A).
The probability that Raul will not win the grand prize of the raffle is
or 0.98.
ANSWER:
or 0.98
eSolutions Manual - Powered by Cognero
Page 15
7-2 Probability and Counting
17. MASCOT At Riverview High School, 120 students were asked whether they prefer a lion or a timber wolf as the
new school mascot. What is the probability that a randomly selected student will have voted for a lion as the new
school mascot?
Lion
Timber Wolf
Total
Votes
78
42
120
SOLUTION:
Let A be the event of a student preferring a lion. The total number of outcomes is the total number of votes, or
120.
There were 78 students that prefer a lion.
The probability that a randomly selected student will have voted for a lion as the new school mascot is
, or
0.65.
ANSWER:
or 0.65
eSolutions Manual - Powered by Cognero
Page 16
7-2 Probability and Counting
18. COLLEGE In Evan’s senior class of 240 students, 85% are planning to attend college after graduation. What is
the probability that a senior chosen at random is not planning to attend college after graduation?
SOLUTION:
Let A be the event of a senior planning to attend college after graduation. Then find the probability of the
complement of A.
There are 204 seniors planning to attend college after graduation because 85% of 240 is 0.85 × 240 = 204.
There are 240 seniors.
The probability of the complement of A is P(A') = 1 – P(A).
The probability that a senior chosen at random is not planning to attend college after graduation is
or
0.15.
ANSWER:
or
or 0.15
19. DRAMA CLUB The Venn diagram shows the cast members who are in Acts I and II of a school play. One of
the students will be chosen at random to attend a statewide performing arts conference. Let A be the event that a
cast member is in Act I of the play and let B be the event that a cast member is in Act II of the play.
a. Find A ∩ B.
b. What is the probability that the student who is chosen to attend the conference is a cast member in only one of
eSolutions Manual - Powered by Cognero
Page 17
7-2 Probability and Counting
the two Acts of the play.
SOLUTION:
a. The possible outcomes for event A are all of the cast members in Act I of the play, or {Kate, Paige, Brad,
Garrett, Amy, Alex}.
The possible outcomes for event B are all of the cast members in Act II of the play, or {Tyrone, Walter, Fran, Amy,
Alex}.
A ∩ B contains all of the outcomes that are in both sample space A and B.
A ∩ B = {Amy, Alex}
b. Let C be the event of a cast member in both Acts of the play. Then find the probability of the complement of A.
There are 2 cast members in both Acts of the play.
There are 9 cast members.
The probability of the complement of C is P(C') = 1 – P(C).
The probability that the student who is chosen to attend the conference is a cast member in only one of the two
Acts of the play is
or about 0.78.
ANSWER:
a. {Amy, Alex}
b.
eSolutions Manual - Powered by Cognero
Page 18
7-2 Probability and Counting
20. GAMES LaRae is playing a game that uses a spinner. What is the probability that the spinner will land on a prime
number on her next spin?
SOLUTION:
Let A be the event of spinning a prime number, and let B be the event of spinning a composite number. The total
number of outcomes is the total number of sections on the spinner, or 7.
There are 7 sections on the spinner that are prime numbers. None of the sections on the spinner are composite
numbers.
The probability that the spinner will land on a prime number on her next spin is 1.
ANSWER:
1
eSolutions Manual - Powered by Cognero
Page 19
7-2 Probability and Counting
21. SHOPPING Raya asks 40 people outside the mall whether or not they visited for shopping or dining. She
records the results in a Venn diagram. One person will be chosen at random to be interviewed on the local evening
news. Find the probability that the person chosen will be someone who visited the mall for shopping and dining.
SOLUTION:
Let A be the event of a person who visited the mall for shopping, and let B be the event of a person who visited the
mall for dining. The total number of outcomes is the total number of visitors to the mall, or 40.
There were 20 people who visited the mall for dining and there were only 11 that is also visited the mall for
shopping.
The probability that the person chosen will be someone who visited the mall for shopping and dining is
, or
0.275.
ANSWER:
0.275
22. PERSEVERE Let A be the possible integer side measures of the rectangle with perimeter P = 52. Let B
represent the possible integer measures of
in △XYZ.
eSolutions Manual - Powered by Cognero
Page 20
7-2 Probability and Counting
a. Find A ∩ B.
b. Find A ∪ B.
SOLUTION:
The perimeter of the rectangle must be equal to 52.
The possible outcomes for event A are all of the possible side measures of the rectangle, or {10}.
Triangle Inequality Theorem states that the sum of the lengths of any two sides of a triangle is greater than the length
of the third side. The length of the first side is 6 and the length of the second side is 9. Let y represent the length of
the third side of the triangle.
The sum of the lengths of the first and second sides must be greater than the length of the third side.
The sum of the lengths of the first and third sides must be greater than the length of the second side.
The sum of the lengths of the second and third sides must be greater than the length of the first side.
The possible outcomes for event B are possible measures of
in △XYZ, or 3 < y < 15 or {4, 5, 6, 7, 8, 9, 10,
11, 12, 13, 14}.
a. A ∩ B contains all of the outcomes that are in both sample space A and B.
A ∩ B = {10}
eSolutions Manual - Powered by Cognero
Page 21
7-2 Probability and Counting
b. A ∪ B contains all of the outcomes that are in either sample space(s) A or B.
A ∪ B = {4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14}
ANSWER:
a. A ∩ B = {10}
b. A ∪ B = {4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14}
eSolutions Manual - Powered by Cognero
Page 22
7-2 Probability and Counting
23. CREATE Let A be the months of the year with 31 days and let B be the months of the year that begin with the
letter J. Create a Venn diagram to display this data.
SOLUTION:
The possible outcomes for event A are all of the months of the year with 31 days, or {January, March, May, July,
August, October, December}.
The possible outcomes for event B are all of the months of the year that begin with the letter J, or {January, June,
July}.
A ∩ B contains all of the outcomes that are in both sample space A and B.
A ∩ B = {January, July}
The outcomes for event A only are March, May, August, October, and December.
The outcome for event B only is June.
The outcomes for event A and B are January and July.
ANSWER:
eSolutions Manual - Powered by Cognero
Page 23
7-2 Probability and Counting
24. WRITE Suppose you need to explain the concept of intersections and unions to someone with no knowledge
of the topic. Write a brief description of your explanation.
SOLUTION:
When two events occur, their intersection consists of all the outcomes in their sample spaces that they have in
common.
When two events occur, their union consists of all of the outcomes in their sample spaces combined.
ANSWER:
Sample answer: When two events occur, their intersection consists of all the outcomes in their sample spaces that
they have in common; whereas their union consists of all of the outcomes in their sample spaces combined.
25. ANALYZE Determine if the following statement is sometimes, always, or never true. Justify your argument.
The union of two sets has more elements than the intersection of two sets.
SOLUTION:
While it is usually the case that the union of two sets will consist of more elements than their intersection, there are
exceptions.
Let A be the days of the week that end in day, and let B be the days of the week that begin with the letters F, M, S,
T, or W.
A = {Monday, Tuesday, Wednesday, Thursday, Friday, Saturday, Sunday}
B = {Monday, Tuesday, Wednesday, Thursday, Friday, Saturday, Sunday}
Both sets have the same number of items.
So, the statement is sometimes true.
ANSWER:
Sometimes; sample answer: While it is usually the case that the union of two sets will consist of more elements than
their intersection, there are exceptions. For example, let A be the days of the week that end in day, and let B be the
days of the week that begin with the letters F, M, S, T, or W. Because the two sets have the same list of items, their
intersection and union will also be this same list.
eSolutions Manual - Powered by Cognero
Page 24
7-2 Probability and Counting
26. FIND THE ERROR Let A be the event that the spinner lands on a vowel. Let B be the event that it lands on the
letter J. Truc says A ∪ B is {A, E, O, U, J}, and Alan says A ∪ B is Ø. Who is correct? Explain.
SOLUTION:
The possible outcomes for event A are all of the letters that are a vowel, or {A, E, O, U}.
The possible outcomes for event B are all of the letters that are a J, or {J}.
A ∪ B contains all of the outcomes that are in either sample space(s) A or B.
A ∪ B = {A, E, O, U, J}
Alan is probably stating the empty set because he mistakenly found the intersection of the two sets.
The set that Truc listed is the union of sets A and B. So, Truc is correct.
ANSWER:
Truc; sample answer: The set that Truc listed is the union of sets A and B. Alan is probably stating the empty set
because he mistakenly found the intersection of the two sets.
eSolutions Manual - Powered by Cognero
Page 25