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Fundamental counting principle

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5135947
Jordan is choosing an outfit for school from \(5\) different T-shirts, \(3\) different pairs of pants, and \(2\) different pairs of shoes. a) How many outfits can Jordan make by choosing one item from each category? b) Jordan decides to wear one particular blue T-shirt. How many outfit choices remain?

Hints

- How many successive choices are being made? - Multiply the number of options at each stage. - What changes when one category has only one available choice?

Solution

1. Apply the fundamental counting principle. There are \(5\) choices for the T-shirt, \(3\) choices for the pants, and \(2\) choices for the shoes, so the total is \(5\cdot3\cdot2=30\). 2. When the T-shirt is fixed, there is \(1\) shirt choice, \(3\) pants choices, and \(2\) shoe choices. The number of outfits is \(1\cdot3\cdot2=6\).

Answer

a) \(30\) outfits b) \(6\) outfits
5157727
Four soccer teams, A, B, C, and D, play in a small league. Each team plays every other team twice: once at home and once away. How many games are played altogether?

Hints

- Choose the home team first and then choose one of its opponents. - Make sure each game is counted once, by its home team.

Solution

1. Each of the \(4\) teams hosts one game against each of the other \(3\) teams. 2. Apply the fundamental counting principle: \(4 \cdot 3 = 12\) home games. Every league game has exactly one home team, so this counts each game once.

Answer

\(12\) games
5171157
Lena builds an ice cream sundae by choosing exactly one ice cream flavor and exactly one topping. Flavors: chocolate, vanilla, and strawberry Toppings: whipped cream, sprinkles, and chocolate sauce How many different sundaes can Lena make?

Hints

- Choose a flavor and then a topping. - Each flavor can be paired with every topping. - Multiply the number of choices at the two stages.

Solution

1. There are \(3\) choices for the flavor. 2. For each flavor, there are \(3\) choices for the topping. 3. By the fundamental counting principle, \(3 \cdot 3 = 9\) sundaes are possible.

Answer

\(9\) different sundaes
5172737
A two-digit puzzle code uses only the digits \(5,6,7,\) and \(8\). Digits may repeat. a) Make an organized list of all possible codes. b) How many codes are possible?

Hints

- How many choices are available for the first digit? - For each first digit, how many choices are available for the second digit? - Organize the codes by their first digit. - A table can help ensure that no pair is missed.

Solution

1. Organize the list by the first digit: \(55,56,57,58\); \(65,66,67,68\); \(75,76,77,78\); and \(85,86,87,88\). 2. There are \(4\) choices for the first digit and \(4\) choices for the second digit. By the fundamental counting principle, \(4\cdot4=16\).

Answer

a) \(55,56,57,58,65,66,67,68,75,76,77,78,85,86,87,88\) b) \(16\) codes
5172747
In a coded language, every word has exactly five characters. Each character is either a dot “•” or a dash “—”. How many different words are possible?

Hints

- How many positions must be filled? - How many choices are available at each position? - Does choosing one character change the choices at the next position? - What repeated multiplication represents the total?

Solution

1. Each of the \(5\) positions has \(2\) choices: dot or dash. 2. Apply the fundamental counting principle: \(2\cdot2\cdot2\cdot2\cdot2=2^5=32\).

Answer

\(32\) words
5172757
How many different four-digit numbers can be formed using only the odd digits \(1,3,5,7,\) and \(9\) if digits may repeat? Use the fundamental counting principle.

Hints

- How many odd digits are available? - Think of four empty positions. How many choices can fill each one? - Does allowing repetition keep the same number of choices at every position? - Which operation combines the numbers of choices across the positions?

Solution

1. There are \(5\) available digits. 2. Each of the \(4\) positions can contain any of the \(5\) digits. 3. By the fundamental counting principle, the number of four-digit numbers is \(5\cdot5\cdot5\cdot5=5^4=625\).

Answer

\(625\) four-digit numbers
5198707
Leon is creating a character for a video game. He chooses one of \(3\) character classes, one of \(5\) hairstyles, and one of \(4\) armor colors. How many different characters can he create?

Hints

- How many hairstyle choices are available for each character class? - How many armor colors can be paired with each class-and-hairstyle choice? - Multiply the numbers of choices across the three categories.

Solution

1. There are \(3\) choices for the character class, \(5\) choices for the hairstyle, and \(4\) choices for the armor color. 2. By the fundamental counting principle, the total is \(3\cdot5\cdot4=60\).

Answer

\(60\) characters
5204677
A sports team is designing a logo. Each logo uses one symbol and one background color. The symbol choices are a star, circle, or square. The background choices are red, blue, green, or yellow. a) How many symbol-and-background logos are possible? b) The team adds a border that can be black or white. How many logos are now possible using one symbol, one background color, and one border color?

Hints

- First count the symbol-and-background combinations. - How many backgrounds can be paired with one symbol? - The border adds another independent choice. - Multiply the numbers of choices across the categories.

Solution

1. There are \(3\) symbol choices and \(4\) background choices, so \(3\cdot4=12\) logos are possible. 2. Each of those \(12\) logos can have either of \(2\) border colors. Therefore, \(12\cdot2=24\) logos are possible. 3. Equivalently, apply the fundamental counting principle directly: \(3\cdot4\cdot2=24\).

Answer

a) \(12\) logos b) \(24\) logos
5374397
Four T-shirt colors can each be paired with three pants colors. Each dot in the \(4 \times 3\) array represents one outfit. How many outfits are possible? How many outfits would be possible after adding two more T-shirt colors? How many additional outfits is that?
Figure for problem 537439

Hints

- Each T-shirt color can be paired with each pants color. - The number of pants colors stays \(3\). - Multiply the numbers of choices before and after the change.

Solution

1. With \(4\) T-shirt colors and \(3\) pants colors, there are \(4 \cdot 3 = 12\) outfits. 2. Adding two T-shirt colors gives \(6\) T-shirt colors. Then there are \(6 \cdot 3 = 18\) outfits. 3. The increase is \(18 - 12 = 6\) outfits.

Answer

Initially: \(12\) outfits; after the change: \(18\) outfits; increase: \(6\) outfits
5375027
An ice cream shop offers \(3\) flavors and \(2\) sauces for each flavor. Use the tree diagram to determine how many different orders are possible. Each order contains exactly one flavor and one sauce.
Figure for problem 537502

Hints

- Count the endpoints of the tree diagram. - Check that each endpoint represents exactly one complete order.

Solution

1. Each flavor can be paired with \(2\) sauces. 2. Therefore, the total number of orders is \(3 \cdot 2 = 6\).

Answer

\(6\) different orders
5375127
A stage costume is created by independently choosing from \(2\) tops, \(3\) pairs of pants, and \(2\) pairs of shoes. How many complete costumes are possible?
Figure for problem 537512

Hints

- Count the choices at each stage. - Each endpoint represents one complete combination.

Solution

1. There are \(2 \cdot 3 = 6\) top-and-pants combinations. 2. Including the shoe choice gives \(6 \cdot 2 = 12\) complete costumes.

Answer

\(12\) complete costumes
5135977
A bicycle combination lock has \(3\) number wheels. Each wheel can display any digit from \(0\) through \(9\). a) How many different codes are possible? b) Tim remembers that his code uses only the digits \(1\), \(2\), \(3\), \(4\), and \(5\). How many codes satisfy this condition? c) With no information about the code, what is the probability of guessing it correctly on the first try? Give the result as a fraction and a percent.

Hints

- How many choices are available for each wheel? - How do you combine the numbers of choices for the separate positions? - What changes when only certain digits may be used? - For equally likely codes, probability is favorable outcomes divided by total outcomes.

Solution

1. a) Each of the \(3\) positions has \(10\) choices, so the number of codes is \(10 \cdot 10 \cdot 10 = 10^3 = 1000\). 2. b) Each position now has \(5\) choices, so the number of codes is \(5 \cdot 5 \cdot 5 = 5^3 = 125\). 3. c) Exactly one of the \(1000\) equally likely codes is correct. Therefore, \(P = \frac{1}{1000} = 0.001 = 0.1\%\).

Answer

a) \(1000\) codes b) \(125\) codes c) \(\frac{1}{1000} = 0.1\%\)
5135987
A raffle sells tickets with three-digit codes from \(\text{000}\) through \(\text{999}\). Each code appears exactly once. a) How many different ticket codes are there? b) A participant says, “The chance of drawing \(\text{000}\) is much smaller than the chance of drawing \(\text{527}\) because \(\text{000}\) is a special code.” Evaluate this claim mathematically. c) Suppose every ticket was sold. What is the probability that the winning code ends in \(7\)?

Hints

- Count the choices for all three positions, including codes that begin with \(0\). - Does a visible pattern make one equally likely ticket more or less likely than another? - If the last digit must be \(7\), how many choices remain for the first two positions?

Solution

1. a) Each of the three positions has \(10\) choices, so there are \(10 \cdot 10 \cdot 10 = 1000\) ticket codes. 2. b) Every ticket is equally likely to be drawn. Therefore, both \(\text{000}\) and \(\text{527}\) have probability \(\frac{1}{1000}\). A visually unusual pattern does not change its probability. 3. c) The last position is fixed as \(7\), while each of the first two positions has \(10\) choices. Thus, there are \(10 \cdot 10 \cdot 1 = 100\) favorable codes. 4. The probability is \(\frac{100}{1000} = \frac{1}{10} = 10\%\).

Answer

a) \(1000\) ticket codes b) The claim is false. Each code has probability \(\frac{1}{1000}\). c) \(\frac{1}{10} = 10\%\)
5136047
A school cafeteria lets students build a lunch by choosing exactly one appetizer and one main dish. There are \(6\) appetizers and exactly \(42\) possible lunches. How many main dishes are offered? Explain your reasoning.

Hints

- How would you find the total number of lunches if both category counts were known? - Write an equation with the number of main dishes as the unknown. - Which operation undoes multiplication by \(6\)?

Solution

1. Let \(m\) be the number of main dishes. By the fundamental counting principle, \(6m=42\). 2. Divide both sides by \(6\): \(m=42\div6=7\). 3. Check: \(6\cdot7=42\), so the cafeteria offers \(7\) main dishes.

Answer

\(7\) main dishes
5136057
A three-character code for a classroom game is created using this pattern: - First character: one uppercase letter from A through Z (\(26\) choices) - Second character: one digit from \(0\) through \(9\) (\(10\) choices) - Third character: one of the symbols “!”, “?”, “#”, “*”, or “&” (\(5\) choices) a) How many different codes are possible? b) A fourth character is added, and it can be any digit from \(0\) through \(9\). By what factor does the number of possible codes increase? Explain without calculating the new total.

Hints

- Identify the number of choices for each character. - What does one additional stage with \(10\) choices do to the total? - For part b, focus on the multiplier contributed by the new character.

Solution

1. Apply the fundamental counting principle to the three characters: \(26\cdot10\cdot5=1300\) possible codes. 2. The added character has \(10\) possible digits. Each original code can be extended in \(10\) ways. 3. Therefore, the number of possible codes is multiplied by a factor of \(10\).

Answer

a) \(1300\) codes b) The number of codes increases by a factor of \(10\).
5151757
A combination lock has four dials, each labeled with the digits \(0\) through \(9\). A person has forgotten the combination and sets all four dials randomly. Find the probability that the lock opens on the first attempt.

Hints

- Count the possible digits on one dial. - Use the fundamental counting principle to count all four-digit combinations. - Exactly one of those combinations is correct.

Solution

1. Each dial has \(10\) possible digits, and exactly one is correct. 2. There are \(10^4=10{,}000\) equally likely four-digit combinations. 3. Only one combination opens the lock, so the probability is \(\frac{1}{10000}=0.0001\).

Answer

\(\frac{1}{10000}=0.0001=0.01\%\)
5157717
A smoothie bar offers four fruits: apple, banana, cherry, and orange. A two-fruit smoothie always contains exactly two different fruits. How many different two-fruit smoothies are possible? List all of them.

Hints

- First count the choices for the first and second fruits. - Decide whether reversing the order creates a new smoothie. - List the combinations systematically to check your count.

Solution

1. There are \(4\) choices for the first fruit and \(3\) choices for the second fruit, giving \(4 \cdot 3 = 12\) ordered choices. 2. Each smoothie is counted twice because apple-banana and banana-apple are the same combination. Divide by \(2\): \(12 \div 2 = 6\). 3. The combinations are apple-banana, apple-cherry, apple-orange, banana-cherry, banana-orange, and cherry-orange.

Answer

There are \(6\) smoothies: apple-banana, apple-cherry, apple-orange, banana-cherry, banana-orange, and cherry-orange.
5172767
For each condition, determine how many three-digit numbers satisfy it. a) The tens digit is \(4\). b) The tens digit is \(4\), and the ones digit is \(0\). c) The tens digit is \(4\), but the ones digit is not \(0\).

Hints

- How many choices are possible for the hundreds digit of a three-digit number? - Count the choices for the tens and ones digits separately. - Use one factor for each digit position. - For part c, you can also use the results of parts a and b.

Solution

1. For part a, the hundreds digit has \(9\) choices, the tens digit has \(1\) choice, and the ones digit has \(10\) choices. Thus, \(9\cdot1\cdot10=90\). 2. For part b, the hundreds digit has \(9\) choices, while the tens and ones digits are fixed. Thus, \(9\cdot1\cdot1=9\). 3. For part c, the hundreds digit has \(9\) choices, the tens digit is fixed, and the ones digit has \(9\) nonzero choices. Thus, \(9\cdot1\cdot9=81\). Equivalently, \(90-9=81\).

Answer

a) \(90\) numbers b) \(9\) numbers c) \(81\) numbers
5172957
a) List all four-digit numbers that use only the digits \(0\) and \(3\). Remember that a four-digit number cannot begin with \(0\). b) How many numbers are in your list?

Hints

- Which digit must be in the thousands place? - List the numbers in increasing order. - A tree diagram can help organize the choices for the last three positions.

Solution

1. The thousands digit must be \(3\). Each of the other three positions can be either \(0\) or \(3\). 2. The numbers are \(3000\), \(3003\), \(3030\), \(3033\), \(3300\), \(3303\), \(3330\), and \(3333\). 3. There are \(2\) choices for each of the last three positions, so \(2^3=8\) numbers.

Answer

a) \(3000\), \(3003\), \(3030\), \(3033\), \(3300\), \(3303\), \(3330\), \(3333\) b) \(8\) numbers
5198987
A three-digit puzzle code uses only the digits \(2,4,6,\) and \(8\). a) How many codes are possible if digits may repeat? b) How many codes are possible if the first digit must be \(8\) and no digit may repeat?

Hints

- Count the choices for each position. - When repetition is allowed, does the number of choices change from one position to the next? - In part b, the first position is fixed and each chosen digit is removed from the available set. - You can list the six codes to check your count.

Solution

1. With repetition allowed, each of the \(3\) positions has \(4\) choices, so there are \(4^3=64\) codes. 2. In part b, the first digit is fixed as \(8\). There are \(3\) choices for the second digit and \(2\) remaining choices for the third digit. 3. Therefore, the number of codes is \(1\cdot3\cdot2=6\).

Answer

a) \(64\) codes b) \(6\) codes
5199887
A puzzle code uses the digits \(1,2,3,4,\) and \(5\), and digits may repeat. Lucas claims, “A three-digit code has exactly \(5\) times as many possibilities as a two-digit code.” Determine whether Lucas is correct by calculating both numbers of possibilities and explaining the comparison.

Hints

- Count the available digits for each position. - Use the fundamental counting principle for the two-digit code. - Repeat for the three-digit code. - Compare the two totals using multiplication.

Solution

1. A two-digit code has \(5\) choices for each position, so there are \(5^2=25\) codes. 2. A three-digit code has \(5\) choices for each position, so there are \(5^3=125\) codes. 3. Since \(25\cdot5=125\), the three-digit code has exactly \(5\) times as many possibilities. Lucas is correct.

Answer

Lucas is correct. There are \(25\) two-digit codes and \(125\) three-digit codes, and \(125=5\cdot25\).
5199897
A game code uses the letters A, B, C, and D, and letters may repeat. a) How many different two-letter codes are possible? b) How many different four-letter codes are possible? c) Without calculating a new total, explain how the number of possibilities changes when the code length increases from four letters to five letters.

Hints

- How many letters are available for each position? - Apply the fundamental counting principle for each code length. - What multiplier is contributed by one additional position?

Solution

1. A two-letter code has \(4\) choices for each position, so there are \(4^2=16\) codes. 2. A four-letter code has \(4\) choices for each position, so there are \(4^4=256\) codes. 3. Adding a fifth position gives \(4\) possible extensions for every four-letter code. Therefore, the number of codes is multiplied by \(4\).

Answer

a) \(16\) codes b) \(256\) codes c) The number of codes is multiplied by \(4\).
5375187
A three-stage selection process offers \(2\) paths, then \(3\) tasks for each path, and finally \(2\) response formats for each task. A student says there are \(7\) final outcomes because \(2 + 3 + 2 = 7\). Correct the student’s reasoning.
Figure for problem 537518

Hints

- Count complete paths rather than individual branches. - How many endpoints result from each first-stage choice?

Solution

1. A complete outcome includes one choice from each stage. 2. The number of complete paths is \(2 \cdot 3 \cdot 2 = 12\). 3. The student added choices from different stages instead of multiplying the choices that combine to form complete outcomes.

Answer

There are \(12\) final outcomes.
5375527
A workshop is assigned one of four rooms and then one of three time blocks. Room D is unavailable during time block \(3\). How many valid room-and-time combinations are there?
Figure for problem 537552

Hints

- Count the endpoints under each room branch. - Exclude the one unavailable combination.

Solution

1. Rooms A, B, and C each have \(3\) available time blocks, giving \(3\cdot 3=9\) combinations. 2. Room D has only \(2\) available time blocks. 3. The total is \(9+2=11\) valid combinations.

Answer

\(11\) valid combinations
5135957
During an elective week, each student chooses exactly one course from each category: - Athletics: soccer, basketball, or volleyball (\(3\) choices) - Arts: painting, ceramics, photography, or theater (\(4\) choices) - Languages: Spanish or French (\(2\) choices) a) How many different three-course schedules are possible? b) One student cannot choose ceramics. How many schedules are available to that student? c) The school adds a fifth arts course. Explain why the total number of schedules increases by more than \(1\).

Hints

- How does the product change when one option is removed from a category? - For part c, count the combinations of athletics and language choices that can be paired with the new arts course. - Adding one option to a category adds one complete group of combinations, not just one finished schedule.

Solution

1. Apply the fundamental counting principle: \(3\cdot4\cdot2=24\) possible schedules. 2. Without ceramics, there are \(3\) arts choices, so the number of schedules is \(3\cdot3\cdot2=18\). 3. With a fifth arts course, there are \(3\cdot5\cdot2=30\) schedules. The increase is \(30-24=6\). The new arts course can be paired with each of the \(3\cdot2=6\) combinations of athletics and language choices.

Answer

a) \(24\) schedules b) \(18\) schedules c) The total increases by \(6\), because the new arts course can be paired with each of the \(6\) athletics-and-language combinations.
5135967
A locker combination lock has \(4\) number wheels. Each wheel can display any digit from \(0\) through \(9\), including \(0\) in the first position. a) How many different four-digit codes are possible? b) How many codes use only the digits \(1\), \(2\), and \(3\)? c) A person tries to guess the code. In which case is a correct first guess more likely? Justify your answer by comparing the numbers of possible codes. - Case 1: The first digit is known to be \(5\). - Case 2: Every position is known to contain an even digit from \(0, 2, 4, 6, 8\).

Hints

- How many choices are available for each position? - How does the number of choices per position change in each part? - Fewer possible codes means a greater chance of guessing the correct one. - How many even digits are there from \(0\) through \(9\)? Remember to include \(0\).

Solution

1. a) Each of the \(4\) positions has \(10\) choices. By the fundamental counting principle, there are \(10 \cdot 10 \cdot 10 \cdot 10 = 10^4 = 10{,}000\) codes. 2. b) Each position has \(3\) choices, so there are \(3 \cdot 3 \cdot 3 \cdot 3 = 3^4 = 81\) codes. 3. c) In Case 1, the first position is fixed and each of the other three positions has \(10\) choices. This gives \(1 \cdot 10 \cdot 10 \cdot 10 = 1000\) possible codes. In Case 2, each of the four positions has \(5\) choices, giving \(5 \cdot 5 \cdot 5 \cdot 5 = 5^4 = 625\) possible codes. Because \(625 < 1000\), a correct first guess is more likely in Case 2.

Answer

a) \(10{,}000\) codes b) \(81\) codes c) Case 2. It has \(625\) possible codes, compared with \(1000\) in Case 1, so each individual code has a greater chance of being guessed.
5136097
Three-digit numbers are formed using the digits \(3\), \(5\), \(7\), and \(9\). a) How many different numbers can be formed if digits may repeat? b) How many different numbers can be formed if no digit may be used more than once? c) For a number selected at random from part a and from part b, find the probability that it is divisible by \(5\).

Hints

- Count the choices available for each position. - What last digit makes a whole number divisible by \(5\)? - For equally likely outcomes, probability is favorable outcomes divided by total outcomes.

Solution

1. a) Each of the \(3\) positions has \(4\) choices, so there are \(4 \cdot 4 \cdot 4 = 4^3 = 64\) numbers. 2. b) There are \(4\) choices for the first digit, \(3\) for the second, and \(2\) for the third. Therefore, there are \(4 \cdot 3 \cdot 2 = 24\) numbers. 3. c) A number is divisible by \(5\) when its last digit is \(5\). In part a, the first two positions each have \(4\) choices, giving \(4 \cdot 4 = 16\) favorable numbers. Thus, \(P = \frac{16}{64} = \frac{1}{4}\). 4. In part b, after fixing \(5\) as the last digit, the first two positions have \(3\) and \(2\) choices. This gives \(3 \cdot 2 = 6\) favorable numbers, so \(P = \frac{6}{24} = \frac{1}{4}\).

Answer

a) \(64\) numbers b) \(24\) numbers c) \(\frac{1}{4} = 25\%\) in both cases
5136107
A fair six-sided die is rolled twice to form a two-digit number. The first roll is the tens digit, and the second roll is the ones digit. a) How many different numbers can be formed? b) How many different numbers can be formed if the two rolls must show different values? c) For the sample spaces in parts a and b, find the probability that the sum of the digits is exactly \(7\).

Hints

- Treat the tens digit and ones digit as separate choices. - List the ordered pairs of values from \(1\) through \(6\) that add to \(7\). - Use the correct total number of outcomes for each sample space.

Solution

1. a) Each roll has \(6\) possible results, so there are \(6 \cdot 6 = 36\) possible two-digit numbers. 2. b) The first roll has \(6\) choices. The second roll then has \(5\) choices because it must differ from the first. Therefore, there are \(6 \cdot 5 = 30\) numbers. 3. c) The ordered pairs with digit sum \(7\) are \((1, 6)\), \((2, 5)\), \((3, 4)\), \((4, 3)\), \((5, 2)\), and \((6, 1)\). There are \(6\) favorable outcomes, and all have different digits. 4. For part a, \(P = \frac{6}{36} = \frac{1}{6}\). For part b, \(P = \frac{6}{30} = \frac{1}{5}\).

Answer

a) \(36\) numbers b) \(30\) numbers c) Part a: \(\frac{1}{6}\); part b: \(\frac{1}{5}\)
5136117
A four-character code uses the letters A, B, C, and D. a) Find the number of possible codes when letters may repeat. b) Find the number of possible codes when each of the four letters is used exactly once. c) Find the probability that a randomly selected code from part a contains no A. d) Find the probability that a randomly selected code from part b begins with the letters AD.

Hints

- If one letter is excluded, how many choices remain for each position? - When every letter must be used once, the number of choices decreases after each position is filled. - In part d, the first two positions are fixed. Which letters remain?

Solution

1. a) Each of the \(4\) positions has \(4\) choices, so there are \(4 \cdot 4 \cdot 4 \cdot 4 = 4^4 = 256\) codes. 2. b) The choices decrease as letters are used: \(4 \cdot 3 \cdot 2 \cdot 1 = 4! = 24\) codes. 3. c) If A is excluded, each position has \(3\) choices. There are \(3^4 = 81\) favorable codes, so \(P = \frac{81}{256}\). 4. d) Fix A in the first position and D in the second. The remaining letters B and C can be arranged in \(2 \cdot 1 = 2\) ways. Therefore, \(P = \frac{2}{24} = \frac{1}{12}\).

Answer

a) \(256\) codes b) \(24\) codes c) \(\frac{81}{256}\) d) \(\frac{1}{12}\)
5136127
A raffle awards a grand prize and a second prize to \(12\) participants. No person can win more than one prize. a) How many possible outcomes are there for awarding the two different prizes? b) Julia and Tim are participants. Find the probability that Julia wins the grand prize and Tim wins the second prize. c) Find the probability that Julia and Tim both win a prize, regardless of who wins which prize.

Hints

- Count the choices for the first prize and then the choices remaining for the second prize. - Part b asks for one specific ordered result. - In part c, how many ways can Julia and Tim receive the two different prizes?

Solution

1. a) There are \(12\) choices for the grand-prize winner and then \(11\) choices for the second-prize winner. Thus, there are \(12 \cdot 11 = 132\) possible outcomes. 2. b) Exactly one outcome has Julia winning the grand prize and Tim winning the second prize. Therefore, \(P = \frac{1}{132}\). 3. c) There are two favorable outcomes: Julia wins the grand prize and Tim wins the second prize, or Tim wins the grand prize and Julia wins the second prize. Therefore, \(P = \frac{2}{132} = \frac{1}{66}\).

Answer

a) \(132\) outcomes b) \(\frac{1}{132}\) c) \(\frac{1}{66}\)
5136137
A combination lock has \(3\) wheels. Each wheel can display a digit from \(1\) through \(6\). a) How many different codes are possible? b) Find the probability that a randomly selected code contains the digits \(1\), \(2\), and \(3\), each exactly once. c) Find the probability that a randomly selected code has three identical digits.

Hints

- Each wheel has the same number of choices. - In part b, how many orders are possible for the fixed digits \(1\), \(2\), and \(3\)? - How many codes consist of one digit repeated three times?

Solution

1. a) Each wheel has \(6\) choices, so there are \(6 \cdot 6 \cdot 6 = 216\) codes. 2. b) The digits \(1\), \(2\), and \(3\) can be arranged in \(3 \cdot 2 \cdot 1 = 6\) orders. Therefore, \(P = \frac{6}{216} = \frac{1}{36}\). 3. c) The favorable codes are \(111, 222, 333, 444, 555, 666\), for a total of \(6\). Therefore, \(P = \frac{6}{216} = \frac{1}{36}\).

Answer

a) \(216\) codes b) \(\frac{1}{36}\) c) \(\frac{1}{36}\)
5136147
Eight runners compete in a \(100\)-meter race in lanes \(1\) through \(8\). Assume there are no ties and that all possible finish orders are equally likely. a) Find the probability that the runners in lanes \(1\), \(2\), and \(3\) win gold, silver, and bronze, respectively, in that exact order. b) Show that the probability that the runners in lanes \(1\), \(2\), and \(3\) take the first three places in any order is exactly six times the probability in part a. c) Find the probability that none of the runners in lanes \(1\), \(2\), or \(3\) finishes in the top three.

Hints

- Count the ordered ways to fill first, second, and third place from \(8\) runners. - In part b, how many orders are possible for the three specified runners? - In part c, how many runners remain eligible for the top three positions?

Solution

1. There are \(8\) choices for first place, \(7\) for second place, and \(6\) for third place, giving \(8 \cdot 7 \cdot 6 = 336\) equally likely ordered medal outcomes. 2. a) Exactly one outcome has lanes \(1\), \(2\), and \(3\) in that exact order, so \(P = \frac{1}{336}\). 3. b) The three specified runners can be arranged in the top three places in \(3 \cdot 2 \cdot 1 = 6\) orders. Thus, \(P = \frac{6}{336} = \frac{1}{56}\), which is six times \(\frac{1}{336}\). 4. c) The top three places must be filled by the other \(5\) runners. There are \(5 \cdot 4 \cdot 3 = 60\) favorable outcomes, so \(P = \frac{60}{336} = \frac{5}{28}\).

Answer

a) \(\frac{1}{336}\) b) \(\frac{6}{336} = \frac{1}{56} = 6 \cdot \frac{1}{336}\) c) \(\frac{5}{28}\)
5136177
A student club has \(12\) members: \(4\) girls and \(8\) boys. The club must fill three different offices: president, treasurer, and secretary. No member may hold more than one office. a) How many different ways can the three offices be filled? b) If the offices are assigned at random, what is the probability that all three offices are filled by girls?

Hints

- Does it matter which person holds which office? - How many choices are available for the first office, and how many remain for each later office? - Repeat the counting process using only the girls as eligible candidates.

Solution

1. a) There are \(12\) choices for president, \(11\) remaining choices for treasurer, and \(10\) remaining choices for secretary. Therefore, there are \(12 \cdot 11 \cdot 10 = 1320\) possible assignments. 2. b) For all three officers to be girls, there are \(4\) choices for the first office, \(3\) for the second, and \(2\) for the third. This gives \(4 \cdot 3 \cdot 2 = 24\) favorable assignments. 3. Therefore, \(P = \frac{24}{1320} = \frac{1}{55} \approx 1.8\%\).

Answer

a) \(1320\) assignments b) \(\frac{1}{55} \approx 1.8\%\)
5136187
A suitcase has a four-digit combination lock. Each wheel displays a digit from \(0\) through \(9\). a) How many different codes are possible? b) A person guesses once at random. What is the probability that the guess is correct? c) Suppose the first digit is known to be \(7\). How does this information change the probability of a correct first guess? Briefly explain. d) What is the probability that a randomly selected code has at least one digit in the wrong position?

Hints

- Count the choices for each position. - How many of all possible codes open the lock? - Knowing one digit reduces the number of unknown positions. - What is the complement of “at least one digit is wrong”?

Solution

1. a) Each of the \(4\) positions has \(10\) choices, so there are \(10^4 = 10{,}000\) possible codes. 2. b) Exactly one code is correct, so \(P(\text{correct}) = \frac{1}{10000} = 0.0001 = 0.01\%\). 3. c) If the first digit is fixed, only the other \(3\) positions are unknown. There are \(10^3 = 1000\) possible codes, so the probability becomes \(\frac{1}{1000} = 0.001 = 0.1\%\). This is ten times the original probability. 4. d) “At least one digit is in the wrong position” is the complement of “all four digits are correct.” Therefore, \(P = 1 - \frac{1}{10000} = \frac{9999}{10000} = 0.9999 = 99.99\%\).

Answer

a) \(10{,}000\) codes b) \(\frac{1}{10000} = 0.01\%\) c) \(\frac{1}{1000} = 0.1\%\); the probability is ten times as large. d) \(\frac{9999}{10000} = 99.99\%\)
5136237
A suitcase lock has three wheels: - The first wheel shows the digits \(0\) through \(9\). - The second wheel shows the letters A, B, C, D, and E. - The third wheel shows the four card suits \(\heartsuit\), \(\diamondsuit\), \(\clubsuit\), and \(\spadesuit\). A person has forgotten the correct combination and sets each wheel at random. a) How many different combinations are possible? b) What is the probability that all three positions are wrong?

Hints

- Count the choices available on each wheel. - If one setting on each wheel is correct, how many settings on that wheel are wrong? - Use the fundamental counting principle to count combinations where every wheel is wrong, then compare that count with the total number of combinations.

Solution

1. a) Multiply the number of choices for the three wheels: \(10 \cdot 5 \cdot 4 = 200\) combinations. 2. b) On the first wheel, \(9\) of the \(10\) digits are wrong. On the second, \(4\) of the \(5\) letters are wrong. On the third, \(3\) of the \(4\) suits are wrong. 3. By the fundamental counting principle, there are \(9 \cdot 4 \cdot 3 = 108\) combinations with all three positions wrong. Since all \(200\) combinations are equally likely, \(P(\text{all wrong}) = \frac{108}{200} = \frac{27}{50} = 0.54\).

Answer

a) \(200\) combinations b) \(\frac{27}{50} = 0.54 = 54\%\)
5136247
A building manager has three different keys for three different doors. The manager randomly assigns one key to each lock, using every key exactly once. Find the probability of each event: a) All three keys are assigned to the correct locks. b) No key is assigned to its correct lock. c) Exactly one key is assigned to its correct lock.

Hints

- List all assignments systematically. - How many ways can three objects be assigned to three positions? - For each event, count the listed assignments that satisfy the condition. - If two keys are correctly assigned, what must happen to the third key?

Solution

1. There are \(3 \cdot 2 \cdot 1 = 6\) equally likely assignments of the three keys to the three locks. 2. a) Exactly one assignment matches every key to its correct lock, so \(P = \frac{1}{6}\). 3. b) Label the correct assignment \((1, 2, 3)\). The two assignments with no correct matches are \((2, 3, 1)\) and \((3, 1, 2)\). Therefore, \(P = \frac{2}{6} = \frac{1}{3}\). 4. c) The assignments with exactly one correct match are \((1, 3, 2)\), \((3, 2, 1)\), and \((2, 1, 3)\). Therefore, \(P = \frac{3}{6} = \frac{1}{2}\).

Answer

a) \(\frac{1}{6}\) b) \(\frac{1}{3}\) c) \(\frac{1}{2}\)
5136597
Two explorers must choose paths through two different cave systems. At each fork, they choose randomly among the available paths, so all choices at that fork are equally likely. In each cave, exactly one complete path reaches the goal. - Cave A has \(2\) consecutive forks with \(5\) choices at each fork. - Cave B has \(3\) consecutive forks with \(3\) choices at each fork. Use calculations to determine which cave gives the greater probability of reaching the goal by guessing.

Hints

- Count the total number of complete paths in each cave. - Use the product of the numbers of choices at the successive forks. - When exactly one path succeeds, how does the total number of paths affect the probability?

Solution

1. Cave A has \(5 \cdot 5 = 25\) complete paths. Because exactly one succeeds, \(P(A) = \frac{1}{25} = 0.04\). 2. Cave B has \(3 \cdot 3 \cdot 3 = 27\) complete paths. Because exactly one succeeds, \(P(B) = \frac{1}{27} \approx 0.037\). 3. Since \(\frac{1}{25} > \frac{1}{27}\), Cave A gives the greater probability of reaching the goal.

Answer

Cave A has the greater probability, \(\frac{1}{25}\), compared with \(\frac{1}{27}\) for Cave B.
5141647
Anya wants to join one sports club and one music club. There are \(5\) sports clubs—soccer, basketball, swimming, tennis, and rock climbing—and \(4\) music clubs—choir, orchestra, band, and guitar. a) How many sports-and-music club combinations are possible? b) The swimming club is canceled. How many combinations remain? c) Rock climbing and band meet at the same time, so that one pair cannot be chosen together. How many of the combinations from part b are still possible?

Hints

- A table can show every sports choice paired with every music choice. - How does removing one entire sports option change the product? - How many individual pairs are removed by one specific scheduling conflict?

Solution

1. Apply the fundamental counting principle: \(5\cdot4=20\) combinations. 2. After swimming is canceled, \(4\) sports choices and \(4\) music choices remain, giving \(4\cdot4=16\) combinations. 3. Exactly one of those combinations—rock climbing with band—is not allowed. Therefore, \(16-1=15\) combinations remain.

Answer

a) \(20\) combinations b) \(16\) combinations c) \(15\) combinations
5141657
A four-digit code for a puzzle box uses only the digits \(1,2,3,4,\) and \(5\). a) How many codes are possible if digits may repeat? b) How many codes are possible if no digit may be used more than once? c) How many codes begin with an even digit and end with an odd digit? For this part, digits may repeat.

Hints

- Count the choices available for each position separately. - Without repetition, how does the number of choices change after each digit is used? - Which available digits are even, and which are odd?

Solution

1. With repetition allowed, each of the four positions has \(5\) choices, so there are \(5^4=625\) codes. 2. Without repetition, the successive numbers of choices are \(5,4,3,\) and \(2\). Thus, there are \(5\cdot4\cdot3\cdot2=120\) codes. 3. For part c, the first digit has \(2\) choices, \(2\) or \(4\). Each middle position has \(5\) choices, and the last digit has \(3\) choices, \(1,3,\) or \(5\). Therefore, there are \(2\cdot5\cdot5\cdot3=150\) codes.

Answer

a) \(625\) codes b) \(120\) codes c) \(150\) codes
5141667
A bag contains six balls labeled A, B, C, D, E, and F. Two balls are drawn in order without replacement. Order matters, so \((\mathrm{A}, \mathrm{B})\) and \((\mathrm{B}, \mathrm{A})\) are different outcomes. a) How many possible outcomes are there? b) How many outcomes contain the letter A? c) How many outcomes consist only of vowels?

Hints

- How does drawing without replacement change the number of choices on the second draw? - Count separately the outcomes with A first and the outcomes with A second. - Which letters from A through F are vowels?

Solution

1. a) There are \(6\) choices for the first draw and \(5\) remaining choices for the second draw. Therefore, there are \(6 \cdot 5 = 30\) outcomes. 2. b) A can appear first, followed by any of the other \(5\) letters, or second, preceded by any of the other \(5\) letters. Thus, there are \(5 + 5 = 10\) outcomes containing A. 3. c) The only vowels are A and E. Without replacement, the vowel-only outcomes are \((\mathrm{A}, \mathrm{E})\) and \((\mathrm{E}, \mathrm{A})\), for a total of \(2\).

Answer

a) \(30\) outcomes b) \(10\) outcomes c) \(2\) outcomes
5156087
A clothing store sells T-shirts in \(4\) colors (red, blue, green, and yellow), \(3\) sizes (S, M, and L), and \(2\) neck styles (V-neck and crew neck). a) How many different T-shirt options are available? b) A box contains exactly one T-shirt of each option. If one T-shirt is selected at random, what is the probability that it is red and size L? c) What is the probability that the selected T-shirt has a crew neck?

Hints

- Count the choices for color, size, and neck style. - How many complete options satisfy each condition? - For equally likely options, divide favorable outcomes by total outcomes.

Solution

1. a) Multiply the independent choices: \(4 \cdot 3 \cdot 2 = 24\) T-shirt options. 2. b) A red size-L shirt can have either neck style, so there are \(2\) favorable options. Therefore, \(P = \frac{2}{24} = \frac{1}{12} \approx 8.3\%\). 3. c) A crew-neck shirt can have any of the \(4\) colors and \(3\) sizes, giving \(4 \cdot 3 = 12\) favorable options. Therefore, \(P = \frac{12}{24} = \frac{1}{2} = 50\%\).

Answer

a) \(24\) options b) \(\frac{1}{12} \approx 8.3\%\) c) \(\frac{1}{2} = 50\%\)
5156097
A raffle ticket has a color (blue or white) and a three-digit code. The first digit is \(1\), \(2\), or \(3\). The second and third digits can each be any digit from \(0\) through \(9\). Exactly one ticket exists for every possible combination, and all tickets are equally likely to be drawn. A “super ticket” has identical second and third digits, such as Blue-122 or White-300. How many tickets are there altogether? How many are super tickets? Find the probability of drawing a super ticket.

Hints

- Separate the ticket into color, first digit, second digit, and third digit. - Count the choices for each part. - For a super ticket, how many matching pairs are possible for the final two digits?

Solution

1. There are \(2\) color choices, \(3\) choices for the first digit, and \(10\) choices for each of the last two digits. Therefore, the total number of tickets is \(2 \cdot 3 \cdot 10 \cdot 10 = 600\). 2. For a super ticket, the color and first digit are still unrestricted. The matching pair of final digits can be \(00, 11, \ldots, 99\), giving \(10\) choices. Thus, there are \(2 \cdot 3 \cdot 10 = 60\) super tickets. 3. The probability is \(P = \frac{60}{600} = \frac{1}{10} = 10\%\).

Answer

There are \(600\) tickets, including \(60\) super tickets. The probability of drawing one is \(\frac{1}{10} = 10\%\).
5157577
Five students play a round-robin tennis tournament. Each student plays every other student exactly once. a) How many matches are played altogether? b) Just before the tournament begins, two more students join. How many additional matches must be scheduled so that every pair of students plays once?

Hints

- Count ordered choices for the two players in a match. - Each pair is counted twice, once in each order. - Find the totals for \(5\) and \(7\) students, and then subtract.

Solution

1. For \(5\) students, choose a first player in \(5\) ways and a second player in \(4\) ways. This counts every pair twice, so the number of matches is \(\frac{5 \cdot 4}{2} = 10\). 2. For \(7\) students, the total number of matches is \(\frac{7 \cdot 6}{2} = 21\). 3. The number of additional matches is \(21 - 10 = 11\).

Answer

a) \(10\) matches b) \(11\) additional matches
5167547
Use the digit cards \(1\), \(2\), \(3\), and \(4\) to form two two-digit numbers and multiply them. Use each card exactly once. Reversing the order of the two factors does not create a new problem. a) How many different multiplication problems can you make if the digit \(1\) is in the tens place of one factor? List every problem and calculate each product. b) Which problem has the least product?

Hints

- Choose the ones digit for the factor whose tens digit is \(1\). - Arrange the two remaining digits in both possible orders. - Do not list a product again with its factors reversed.

Solution

1. Put the factor with \(1\) in the tens place first. There are \(3\) choices for its ones digit and then \(2\) orders for the remaining digits, giving \(3 \cdot 2 = 6\) problems. 2. The problems are: \(12 \cdot 34 = 408\) \(12 \cdot 43 = 516\) \(13 \cdot 24 = 312\) \(13 \cdot 42 = 546\) \(14 \cdot 23 = 322\) \(14 \cdot 32 = 448\) 3. The least product is \(312\).

Answer

a) There are \(6\) problems: \(12 \cdot 34 = 408\) \(12 \cdot 43 = 516\) \(13 \cdot 24 = 312\) \(13 \cdot 42 = 546\) \(14 \cdot 23 = 322\) \(14 \cdot 32 = 448\) b) \(13 \cdot 24 = 312\)
5172777
Digits may repeat. a) How many three-digit numbers can be formed using only the digits \(1,2,\) and \(3\)? b) How many three-digit numbers contain no digit \(0\)? c) How many three-digit numbers contain at least one digit \(0\)?

Hints

- Count the choices available for each digit position. - When zero is forbidden, how many digits remain available? - How many three-digit numbers are there from \(100\) through \(999\)? - For part c, count the complement: numbers with no zero.

Solution

1. For part a, each of the three positions has \(3\) choices, so \(3^3=27\). 2. For part b, each position can use any digit from \(1\) through \(9\), giving \(9^3=729\). 3. There are \(900\) three-digit numbers in all, from \(100\) through \(999\). The complement of “at least one zero” is “no zeros.” Therefore, \(900-729=171\).

Answer

a) \(27\) numbers b) \(729\) numbers c) \(171\) numbers
5172787
How many three-digit numbers contain the digit \(8\) exactly twice?

Hints

- Which pairs of positions can contain the two \(8\)s? - The remaining digit cannot also be \(8\). - Remember that the hundreds digit cannot be \(0\). - Count the choices for the free digit in each case.

Solution

1. If the two \(8\)s are in the hundreds and tens places, the ones digit can be any digit except \(8\), giving \(9\) choices. 2. If the two \(8\)s are in the hundreds and ones places, the tens digit can be any digit except \(8\), again giving \(9\) choices. 3. If the two \(8\)s are in the tens and ones places, the hundreds digit can be any nonzero digit except \(8\), giving \(8\) choices. 4. The cases do not overlap, so the total is \(9+9+8=26\).

Answer

\(26\) three-digit numbers
5179567
Use the digits \(4\), \(5\), and \(6\) exactly once each. First use the fundamental counting principle to determine how many three-digit numbers are possible. Then form every possible number, find their sum, and add \(730\).

Hints

- Count the available choices for each place and multiply those numbers. - Choose the hundreds digit first, then list the possible orders of the remaining digits. - Check that each digit appears exactly once in every number.

Solution

1. There are \(3\) choices for the hundreds digit, \(2\) choices for the tens digit, and \(1\) choice for the ones digit, so \(3 \cdot 2 \cdot 1 = 6\) numbers are possible. 2. List the six numbers systematically: \(456\), \(465\), \(546\), \(564\), \(645\), and \(654\). 3. Add them: \(456 + 465 + 546 + 564 + 645 + 654 = 3330\). 4. Add \(730\): \(3330 + 730 = 4060\).

Answer

There are \(6\) possible numbers, and the final result is \(4060\).
5198607
At an ice cream shop, you choose exactly one cup size, one flavor, and one sauce. <table> <tr> <th>Cup size</th> <th>Flavor</th> <th>Sauce</th> </tr> <tr> <td>Small</td> <td>Vanilla</td> <td>Chocolate</td> </tr> <tr> <td>Medium</td> <td>Chocolate</td> <td>Strawberry</td> </tr> <tr> <td>Large</td> <td>Strawberry</td> <td></td> </tr> <tr> <td></td> <td>Banana</td> <td></td> </tr> <tr> <td></td> <td>Mint</td> <td></td> </tr> </table> a) How many size-flavor-sauce combinations are possible? b) The mint and strawberry flavors are sold out. How many combinations remain? c) The shop adds a topping category with \(3\) choices. How many combinations are possible with the full original menu plus the toppings?

Hints

- Count the options in each category. - How does removing two flavors change one factor in the product? - Apply the fundamental counting principle by multiplying the choices across all categories.

Solution

1. The table shows \(3\) sizes, \(5\) flavors, and \(2\) sauces. By the fundamental counting principle, \(3\cdot5\cdot2=30\). 2. With two flavors sold out, \(3\) flavors remain. The number of combinations is \(3\cdot3\cdot2=18\). 3. Adding a topping category with \(3\) choices multiplies the original total by \(3\): \(30\cdot3=90\).

Answer

a) \(30\) combinations b) \(18\) combinations c) \(90\) combinations
5198617
A bicycle company offers one model that customers can customize by choosing one frame color, one seat style, and one handlebar style. There are \(36\) possible bicycles. a) There are \(3\) handlebar styles and \(4\) seat styles. How many frame colors are available? b) The company adds one frame color. How many possible bicycles are there now? c) A customer chooses one specific racing handlebar. Using the original number of frame colors, how many choices remain for the rest of the bicycle?

Hints

- Divide the total by the known category sizes to find the missing factor. - How does adding one frame color change the first factor? - When one category is fixed, multiply only the remaining category sizes.

Solution

1. Let \(c\) be the number of frame colors. The fundamental counting principle gives \(c\cdot4\cdot3=36\), so \(c=36\div12=3\). 2. After one color is added, there are \(4\) frame colors. The new total is \(4\cdot4\cdot3=48\). 3. When the handlebar is fixed, there are \(3\) original frame colors and \(4\) seat styles, giving \(3\cdot4=12\) choices.

Answer

a) \(3\) frame colors b) \(48\) bicycles c) \(12\) choices
5198777
An ice cream cone has three stacked scoops: bottom, middle, and top. Each scoop can be chocolate or vanilla, and flavors may repeat. The order of the scoops matters. a) How many different three-scoop cones are possible? b) How many flavors would need to be available for exactly \(27\) ordered three-scoop cones to be possible?

Hints

- Think of the bottom, middle, and top scoops as three successive choices. - How many choices are available at each position? - For part b, what number multiplied by itself three times equals \(27\)?

Solution

1. Each of the \(3\) scoop positions has \(2\) choices, so the number of cones is \(2^3=8\). 2. Let \(n\) be the number of available flavors. The number of ordered cones is \(n^3\). 3. Solve \(n^3=27\). Since \(3^3=27\), the shop would need \(3\) flavors.

Answer

a) \(8\) cones b) \(3\) flavors
5198997
An ice cream shop has four flavors: strawberry, chocolate, vanilla, and mint. A cone has three stacked scoops, and the order from bottom to top matters. a) How many cones are possible if flavors may repeat? b) How many cones are possible if all three flavors must be different and one scoop must be chocolate?

Hints

- Fill the bottom, middle, and top positions one at a time. - For part b, first choose the position of the chocolate scoop. - The other two scoops must be different from chocolate and from each other. - Multiply the numbers of choices at each stage.

Solution

1. With repetition allowed, each of the \(3\) positions has \(4\) choices, so there are \(4^3=64\) cones. 2. For part b, choose the position of the chocolate scoop in \(3\) ways. 3. Choose the flavor for the first remaining position from the other \(3\) flavors, then choose the flavor for the last position from the remaining \(2\) flavors. 4. The total is \(3\cdot3\cdot2=18\).

Answer

a) \(64\) cones b) \(18\) cones
5320257
An urn contains four balls labeled \(3\), \(5\), \(7\), and \(9\), as shown. Three balls are drawn in order without replacement. The labels form the hundreds, tens, and ones digits of a three-digit number. Find the probability of each event. a) The number is greater than \(700\). b) The sum of the digits is exactly \(15\). c) The number is even.
Figure for problem 532025

Hints

- Count the ordered outcomes for three draws without replacement. - For part a, determine which digits can be in the hundreds place. - For part b, find which set of three digits has sum \(15\), then count its possible orders. - For part c, recall how the ones digit determines whether a number is even.

Solution

1. There are \(4 \cdot 3 \cdot 2 = 24\) equally likely ordered outcomes. 2. For part a, the hundreds digit must be \(7\) or \(9\). For either choice, there are \(3 \cdot 2 = 6\) ways to choose the remaining digits. Thus, there are \(2 \cdot 6 = 12\) favorable outcomes, and the probability is \(\frac{12}{24} = \frac{1}{2}\). 3. For part b, only the digits \(3, 5, 7\) have sum \(15\). They can be arranged in \(3 \cdot 2 \cdot 1 = 6\) orders, so the probability is \(\frac{6}{24} = \frac{1}{4}\). 4. For part c, every available digit is odd, so every possible number is odd. The probability of an even number is \(0\).

Answer

a) \(\frac{1}{2} = 50\%\) b) \(\frac{1}{4} = 25\%\) c) \(0 = 0\%\)
5321557
A spinner has four equal sections: two blue, one red, and one yellow. It is spun three times. Find each probability as a simplified fraction and a decimal. a) The colors occur in the exact order blue, red, yellow. b) The three spins show three different colors.
Figure for problem 532155

Hints

- Find the probability of each color on one spin. - For part a, multiply the probabilities in the stated order. - For part b, count the possible orders of three different colors. - Add the probabilities of all favorable orders.

Solution

1. The one-spin probabilities are \(P(B)=\frac{1}{2}\), \(P(R)=\frac{1}{4}\), and \(P(Y)=\frac{1}{4}\). 2. For the exact order blue, red, yellow, \(P(B, R, Y)=\frac{1}{2}\cdot\frac{1}{4}\cdot\frac{1}{4}=\frac{1}{32}=0.03125\). 3. Three different colors means that blue, red, and yellow each occur once. There are \(3\cdot2\cdot1=6\) possible orders. Each order has probability \(\frac{1}{32}\), so \(P(\text{three different colors})=6\cdot\frac{1}{32}=\frac{3}{16}=0.1875\).

Answer

a) \(\frac{1}{32}=0.03125\) b) \(\frac{3}{16}=0.1875\)
5359647
The spinner shown has four equal sections, one each of red, yellow, blue, and green. It is spun three times. a) Find the probability that the three spins show three different colors. b) Find the probability that exactly two spins land on red.
Figure for problem 535964

Hints

- For part a, count the choices available on the first, second, and third spins when all colors must be different. - For part b, list the possible positions of the one spin that is not red. - Multiply probabilities within each outcome and add the favorable outcomes.

Solution

1. Each color has probability \(\frac{1}{4}\). 2. For three different colors, the first spin can be any color, the second must be one of the other three colors, and the third must be one of the two remaining colors. Thus, \(P(\text{three different colors})=1\cdot\frac{3}{4}\cdot\frac{2}{4}=\frac{3}{8}\). 3. Exactly two red spins can occur in the orders red-red-not red, red-not red-red, or not red-red-red. Each order has probability \(\frac{1}{4}\cdot\frac{1}{4}\cdot\frac{3}{4}=\frac{3}{64}\). Therefore, \(P(\text{exactly two red})=3\cdot\frac{3}{64}=\frac{9}{64}\).

Answer

a) \(\frac{3}{8}\) b) \(\frac{9}{64}\)
5361197
An urn contains four balls labeled \(2\), \(3\), \(5\), and \(8\). All four balls are drawn in order without replacement, and their digits form a four-digit number. Find the probability of each event. a) The number is even. b) The number is greater than \(5000\). c) The number is prime.
Figure for problem 536119

Hints

- Count the possible arrangements of the four distinct digits. - For part a, focus on the last digit. - For part b, focus on the first digit. - For part c, use the divisibility rule for \(3\).

Solution

1. The four distinct digits can be arranged in \(4 \cdot 3 \cdot 2 \cdot 1 = 24\) equally likely orders. 2. For part a, the last digit must be \(2\) or \(8\). There are \(2\) choices for the last digit and \(3 \cdot 2 \cdot 1 = 6\) arrangements of the remaining digits, giving \(12\) favorable outcomes. Thus, the probability is \(\frac{12}{24} = \frac{1}{2}\). 3. For part b, the first digit must be \(5\) or \(8\). Again, there are \(2 \cdot 6 = 12\) favorable outcomes, so the probability is \(\frac{1}{2}\). 4. Every possible number has digit sum \(2 + 3 + 5 + 8 = 18\), so every number is divisible by \(3\). Since every possible number is greater than \(3\), none is prime. The probability is \(0\).

Answer

a) \(\frac{1}{2}\) b) \(\frac{1}{2}\) c) \(0\)
5135707
A four-digit lock has digits \(0\) through \(9\) at each position. One code is selected at random. Find the probability of each event. a) All four digits are identical. b) The code begins with \(19\). c) The sum of the four digits is exactly \(2\).

Hints

- Use the fundamental counting principle for all possible codes. - In part a, choose the one digit that repeats. - In part b, the fixed first two digits leave choices only for the remaining positions. - In part c, separate the possible digit patterns that sum to \(2\).

Solution

1. There are \(10^4 = 10{,}000\) possible codes. 2. a) There are \(10\) codes with four identical digits: \(0000, 1111, \ldots, 9999\). Thus, \(P = \frac{10}{10000} = \frac{1}{1000}\). 3. b) The first two digits are fixed, and each of the final two positions has \(10\) choices. There are \(10 \cdot 10 = 100\) favorable codes, so \(P = \frac{100}{10000} = \frac{1}{100}\). 4. c) The digit sum is \(2\) either when one position contains \(2\) and the others contain \(0\), giving \(4\) codes, or when two positions contain \(1\) and the others contain \(0\). There are \(6\) ways to choose the two positions for the \(1\)s, so there are \(10\) favorable codes in all. Thus, \(P = \frac{10}{10000} = \frac{1}{1000}\).

Answer

a) \(\frac{1}{1000} = 0.1\%\) b) \(\frac{1}{100} = 1\%\) c) \(\frac{1}{1000} = 0.1\%\)
5135997
A safe manufacturer offers two keypad-lock systems: System A: A four-digit code using only the digits \(1, 2, 3, 4, 5, 6\). System B: A three-digit code using any digit from \(0\) through \(9\). a) Find the number of possible codes for each system. b) Which system is more secure against a random guess? Justify your answer using the numbers of possible codes. c) What is the minimum number of positions System A would need to allow more than \(50{,}000\) different codes?

Hints

- Use the fundamental counting principle to count the codes for each system. - How does a larger number of possible codes affect the chance of a correct random guess? - For part c, compare successive powers of \(6\) with \(50{,}000\).

Solution

1. a) System A has \(6\) choices for each of \(4\) positions, so it has \(6 \cdot 6 \cdot 6 \cdot 6 = 6^4 = 1296\) codes. 2. System B has \(10\) choices for each of \(3\) positions, so it has \(10 \cdot 10 \cdot 10 = 10^3 = 1000\) codes. 3. b) System A is more secure against one random guess because \(1296 > 1000\), so \(\frac{1}{1296} < \frac{1}{1000}\). 4. c) Check successive powers of \(6\): \(6^5 = 7776\), \(6^6 = 46{,}656\), and \(6^7 = 279{,}936\). Because \(6^6 < 50{,}000 < 6^7\), System A needs at least \(7\) positions.

Answer

a) System A: \(1296\) codes; System B: \(1000\) codes b) System A, because it has more possible codes and therefore a smaller probability of a correct random guess. c) \(7\) positions
5136267
Four friends—Anna, Ben, Clara, and David—sit at random in a row on a bench with four seats. Find the probability of each event: a) Anna sits in one of the two end seats. b) Anna and Ben sit next to each other. c) The four friends sit in exact alphabetical order from left to right: Anna, Ben, Clara, David.

Hints

- How many total orders are possible for four people in four seats? - If Anna is fixed in an end seat, how many ways can the other three people be arranged? - For adjacent people, treat them as a block and account for both possible orders inside the block. - How many seating orders are exactly alphabetical?

Solution

1. There are \(4! = 4 \cdot 3 \cdot 2 \cdot 1 = 24\) possible seating orders. 2. a) Anna can sit in either end seat. For each choice, the other three friends can be arranged in \(3! = 6\) ways. Thus, there are \(2 \cdot 6 = 12\) favorable orders, so \(P = \frac{12}{24} = \frac{1}{2}\). 3. b) Treat Anna and Ben as a block. The block can occupy seats \(1\)-\(2\), \(2\)-\(3\), or \(3\)-\(4\). Anna and Ben can switch order, and the other two friends can switch order. Therefore, there are \(3 \cdot 2 \cdot 2 = 12\) favorable orders, so \(P = \frac{12}{24} = \frac{1}{2}\). 4. c) Exactly one seating order is alphabetical, so \(P = \frac{1}{24}\).

Answer

a) \(\frac{1}{2}\) b) \(\frac{1}{2}\) c) \(\frac{1}{24}\)
5198627
A costume team builds an outfit by choosing one top, one pair of pants, and one hat. The team has \(4\) tops, \(3\) pairs of pants, and \(2\) hats. a) How many outfits are possible? b) One particular yellow top cannot be worn with one particular green pair of pants. All other combinations are allowed. How many outfits remain possible? c) Create two different sets of clothing categories and category sizes that would each produce exactly \(24\) outfits.

Hints

- How many complete outfits contain the one forbidden top-and-pants pair? - For part c, find different factorizations of \(24\). - Assign each factor to the number of choices in a clothing category.

Solution

1. The total number of outfits is \(4\cdot3\cdot2=24\). 2. The forbidden yellow-top-and-green-pants pair can be combined with either of the \(2\) hats, so it eliminates \(2\) outfits. Therefore, \(24-2=22\) outfits remain. 3. For part c, choose category sizes whose product is \(24\). For example, \(2\) shoe choices, \(3\) scarf choices, and \(4\) jacket choices give \(2\cdot3\cdot4=24\). Another example is \(2\) belt choices, \(2\) glove choices, and \(6\) cape choices, giving \(2\cdot2\cdot6=24\).

Answer

a) \(24\) outfits b) \(22\) outfits c) Example 1: \(2\) shoe choices, \(3\) scarf choices, and \(4\) jacket choices Example 2: \(2\) belt choices, \(2\) glove choices, and \(6\) cape choices
5360747
The spinner shown has four equal sections: red, blue, green, and yellow. It is spun three times, and the ordered color sequence is recorded. Find the probability of each event: a) The colors occur in the order red, blue, yellow. b) All three spins show the same color. c) Exactly two different colors occur. d) Yellow occurs at least once.
Figure for problem 536074

Hints

- Use the fundamental counting principle to find the total number of sequences. - For part b, count one repeated-color sequence for each color. - For part c, first choose the two colors, then count sequences that use both. - For part d, count sequences with no yellow and use the complement.

Solution

1. There are \(4^3=64\) equally likely ordered color sequences. 2. The sequence red-blue-yellow is one outcome, so the probability is \(\frac{1}{64}\). 3. There are four same-color sequences, one for each color. Thus, the probability is \(\frac{4}{64}=\frac{1}{16}\). 4. There are six possible pairs of colors. For each pair, there are \(2^3-2=6\) sequences that use both colors. Therefore, the probability is \(\frac{6\cdot 6}{64}=\frac{9}{16}\). 5. The complement of at least one yellow is no yellow. There are \(3^3=27\) sequences with no yellow, so the probability is \(1-\frac{27}{64}=\frac{37}{64}\).

Answer

a) \(\frac{1}{64}\) b) \(\frac{1}{16}\) c) \(\frac{9}{16}\) d) \(\frac{37}{64}\)

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