- Each interior table cell represents one compound outcome. - For part b, locate the requested spinner row and coin column.

1. The table has \(3\) spinner rows and \(2\) coin columns, so it contains \(3\cdot2=6\) equally likely outcomes. 2. The row for \(2\) and the column for T intersect at \((2, T)\).

a) \(6\) b) \((2, T)\)

- How many flips happen in one trial? - Could the first and second flips give different results?

1. A complete result must state what happens on each of the two flips. 2. HT means heads then tails, while TH means tails then heads. 3. Noor's list does not distinguish those two results.

The list needs both flips in order. For example, HT and TH are different results.

- A full outcome must tell everything that happened in the experiment. - Does each action happen once, or does another random action follow? - Check whether your example records every required result.

1. Part b has a roll and a flip, so a complete result could be rolling \(3\) and getting heads. 2. Part c has two flips, so a complete result could be heads followed by tails. 3. Parts a and d each use one chance action.

b) Roll a number cube and flip a coin: for example, roll \(3\) and get heads. c) Flip a coin twice: for example, heads then tails.

- Start with one card color and pair it with each letter. - Do the same for the other color.

1. With red, the possible pairs are red-X and red-Y. 2. With blue, they are blue-X and blue-Y. 3. These four pairs include every choice from the two boxes.

red-X, red-Y, blue-X, blue-Y.

- Try organizing the choices by flower rather than by the order Amara wrote them. - How many pot choices should each flower have?

1. Organize the list by flower and check the two pots for each flower. 2. The rose-plastic and tulip-clay choices are missing. 3. Three flowers with two pot choices each give \(3\cdot2=6\) combinations, all of which appear after the missing two are added.

rose-plastic and tulip-clay. The complete list contains both pots for each of the three flowers, for \(6\) choices.

- Read the row label and column label for each question mark. - Each cell combines one lunch with one drink.

1. The missing cell in the Taco row and Milk column is taco-milk. 2. The missing cell in the Pasta row and Water column is pasta-water.

taco-milk; pasta-water.

- A path counts only when it reaches its end. - Does every first branch split again?

1. The Stop path is one complete outcome. 2. The Play path splits into H and T, giving two more complete outcomes. 3. There are \(1+2=3\) complete outcomes.

\(3\) complete outcomes.

- Imagine performing each experiment. What could be recorded at the end? - For a multistep experiment, include every ordered combination of the stage outcomes. - Write sample spaces using braces and separate outcomes with commas.

1. a) One spin can result in any of the four colors, so \(S = \{\mathrm{R}, \mathrm{B}, \mathrm{Y}, \mathrm{G}\}\). 2. b) Each outcome records the first and second flips in order. Therefore, \(S = \{(\mathrm{H}, \mathrm{H}), (\mathrm{H}, \mathrm{T}), (\mathrm{T}, \mathrm{H}), (\mathrm{T}, \mathrm{T})\}\). 3. c) The possible outcomes are the labels on the cards, so \(S = \{1, 3, 5, 7\}\).

a) \(S = \{\mathrm{R}, \mathrm{B}, \mathrm{Y}, \mathrm{G}\}\) b) \(S = \{(\mathrm{H}, \mathrm{H}), (\mathrm{H}, \mathrm{T}), (\mathrm{T}, \mathrm{H}), (\mathrm{T}, \mathrm{T})\}\) c) \(S = \{1, 3, 5, 7\}\)

- Use ordered pairs for the two spins. - A product is odd only when both factors are odd. - Relative frequency is successes divided by trials.

1. There are \(10\cdot10=100\) equally likely ordered pairs. The pairs with sum \(10\) are \((1, 9),(2, 8),(3, 7),(4, 6),(5, 5)\) and their reverses, for \(9\) outcomes. Thus, \(P=\frac{9}{100}=0.09\). 2. A product is odd only if both digits are odd. There are \(5\) odd digits, so \(P=\frac{5}{10}\cdot\frac{5}{10}=0.25\). 3. The relative frequency is \(\frac{924}{10000}=0.0924\), which is \(0.0024\) above the theoretical probability \(0.09\).

1. \(0.09\) 2. \(0.25\) 3. \(0.0924\), which is close to \(0.09\)

- Count red sections from the spinner to find the one-spin red probability. - State the complement of “at least one red.” - Find the two-spin complement probability before subtracting from \(1\).

1. The spinner has \(2\) red sections out of \(8\), so \(P(\text{red})=\frac{1}{4}\) and \(P(\text{not red})=\frac{3}{4}\). 2. The probability of no red on either spin is \(\left(\frac{3}{4}\right)^2=\frac{9}{16}\). 3. Therefore, \(P(\text{at least one red})=1-\frac{9}{16}=\frac{7}{16}=43.75\%\).

\(\frac{7}{16}=43.75\%\)

- Count green and red sections from the spinner first. - Multiply the green probability for the two-spin event in part a. - For part b, use the complement of getting no red.

1. From the spinner, \(P(\text{green})=\frac{2}{10}=\frac{1}{5}\) and \(P(\text{red})=\frac{3}{10}\). 2. \(P(\text{green twice})=\frac{1}{5}\cdot\frac{1}{5}=\frac{1}{25}=4\%\). 3. The probability of not red on one spin is \(\frac{7}{10}\), so \(P(\text{at least one red})=1-\left(\frac{7}{10}\right)^2=\frac{51}{100}=51\%\).

a) \(4\%\) b) \(\frac{51}{100}=51\%\)

- Read how many A and N balls are in the bag. - Replacement keeps the one-draw probabilities unchanged. - Multiply the probabilities in the requested order.

1. The bag shows \(3\) A balls, \(2\) N balls, and \(1\) B ball. Thus, \(P(A)=\frac{1}{2}\) and \(P(N)=\frac{1}{3}\). 2. Because the first ball is replaced, the probabilities stay the same on the second draw. 3. \(P(\text{AN})=\frac{1}{2}\cdot\frac{1}{3}=\frac{1}{6}\). 4. \(P(\text{AA})=\frac{1}{2}\cdot\frac{1}{2}=\frac{1}{4}\).

a) \(\frac{1}{6}\) b) \(\frac{1}{4}\)

- Organize the ordered pairs from labels \(1\) through \(5\). - Identify only pairs whose sum is greater than \(8\). - Divide the favorable count by all \(25\) equally likely pairs.

1. There are \(5\cdot5=25\) equally likely ordered pairs. 2. The pairs with a sum greater than \(8\) are \((4,5),(5,4),(5,5)\). 3. Therefore, the probability is \(\frac{3}{25}=12\%\).

\(\frac{3}{25}=12\%\)

- List all ordered outcomes systematically. - Which outcomes contain exactly one H?

1. The sample space is \(S = \{\mathrm{HH}, \mathrm{HT}, \mathrm{TH}, \mathrm{TT}\}\). 2. The favorable outcomes are \(\mathrm{HT}\) and \(\mathrm{TH}\). 3. Therefore, \(P(\text{exactly one head}) = \frac{2}{4} = \frac{1}{2}\).

\(S = \{\mathrm{HH}, \mathrm{HT}, \mathrm{TH}, \mathrm{TT}\}\); the probability is \(\frac{1}{2}\).

- List all ordered combinations of a color and a coin result. - Do not count the blue-and-tails outcome twice.

1. There are \(3 \cdot 2 = 6\) equally likely ordered outcomes. 2. The favorable outcomes are \((\text{blue}, \text{heads})\), \((\text{blue}, \text{tails})\), \((\text{red}, \text{tails})\), and \((\text{green}, \text{tails})\). 3. Therefore, \(P(\text{blue or tails}) = \frac{4}{6} = \frac{2}{3}\).

\(\frac{2}{3}\)

- Identify every endpoint in the compound experiment that belongs to event \(B\). - Add those endpoint counts before forming a part-to-whole ratio. - Divide by the total number of trials.

1. Event \(B\) occurs in the two endpoint categories with counts \(72\) and \(20\), so its total frequency is \(72+20=92\). 2. The relative frequency of \(B\) is \(\frac{92}{200}=0.46\).

\(0.46=46\%\)

- Identify the choices that satisfy each condition. - Count the endpoints that satisfy both conditions.

1. The six equally likely codes are represented by the six endpoints of the tree. 2. The codes that do not begin with \(A\) and end in \(2\) are \(B2\) and \(C2\), so there are \(2\) favorable outcomes. 3. Therefore, \(P = \frac{2}{6} = \frac{1}{3}\).

\(\frac{1}{3}\)

- Find one description shared by all selected outcomes. - Which single outcome is not selected?

1. The selected outcomes are \(HH\), \(HT\), and \(TH\). 2. A common description is “at least one head.” 3. Its complement is “no heads,” which is the outcome \(TT\).

Event: at least one head Complement: no heads, or \(TT\)

- For each possible first number, list the larger second numbers. - Compare the favorable endpoints with all \(16\) endpoints.

1. The favorable ordered pairs are \((1, 2)\), \((1, 3)\), \((1, 4)\), \((2, 3)\), \((2, 4)\), and \((3, 4)\). 2. There are \(6\) favorable outcomes among \(4 \cdot 4 = 16\) equally likely ordered outcomes. 3. Therefore, \(P = \frac{6}{16} = \frac{3}{8}\).

\(\frac{3}{8} = 37.5\%\)

- Keep the order of the two conditions consistent. - Pair every outcome from the first condition with every outcome from the second. - Check each pair for the exact number of unfavorable conditions.

1. Pair each weather outcome with each train outcome. 2. The sample space is \(\{(S, O),(S, L),(C, O),(C, L)\}\). 3. Exactly one unfavorable condition occurs in \((S, L)\) and \((C, O)\).

Sample space: \(\{(S, O),(S, L),(C, O),(C, L)\}\). Exactly one unfavorable condition: \(\{(S, L),(C, O)\}\).

- Read each complete path from the first inspection to the second. - Keep the inspection order fixed in each ordered pair. - Count the number of passes in every outcome.

1. Following every path gives the sample space \(\{(P, P),(P, F),(F, P),(F, F)\}\). 2. Exactly one pass occurs in \((P, F)\) and \((F, P)\).

Sample space: \(\{(P, P),(P, F),(F, P),(F, F)\}\). Exactly one pass: \(\{(P, F),(F, P)\}\).

- Pair every first-category outcome with every second-category outcome. - Interpret “or” as including outcomes that satisfy either condition. - Check that no qualifying ordered pair is omitted.

1. Pair each altitude with each direction. 2. The sample space is \(\{(L, N),(L, E),(L, S),(H, N),(H, E),(H, S)\}\). 3. “High altitude or east” includes all high-altitude outcomes and the low-east outcome. 4. The event is \(\{(L, E),(H, N),(H, E),(H, S)\}\).

Sample space: \(\{(L, N),(L, E),(L, S),(H, N),(H, E),(H, S)\}\). High altitude or east: \(\{(L, E),(H, N),(H, E),(H, S)\}\).

- Choose each possible first letter and pair it with the remaining letters. - Preserve the order of the letters in each outcome. - Compare the position of the first letter with the second.

1. Starting with \(A\) gives \(AB\) and \(AC\). 2. Starting with \(B\) gives \(BA\) and \(BC\). 3. Starting with \(C\) gives \(CA\) and \(CB\). 4. The sample space is \(\{AB,AC,BA,BC,CA,CB\}\). 5. The alphabetically ordered signals are \(AB,\ AC,\ BC\).

Sample space: \(\{AB,AC,BA,BC,CA,CB\}\). Alphabetical order: \(\{AB,AC,BC\}\).

- Fix the first move and pair it with every possible second move. - Preserve the order of the two intersections. - Compare the two symbols in each completed outcome.

1. Starting with \(L\) gives \(LL,\ LS,\ LR\). 2. Starting with \(S\) gives \(SL,\ SS,\ SR\). 3. Starting with \(R\) gives \(RL,\ RS,\ RR\). 4. The sample space is \(\{LL,LS,LR,SL,SS,SR,RL,RS,RR\}\). 5. The same choice occurs twice in \(LL,\ SS,\ RR\).

Sample space: \(\{LL,LS,LR,SL,SS,SR,RL,RS,RR\}\). Same choice twice: \(\{LL,SS,RR\}\).

- Each row fixes Spinner A's result, and each column fixes Spinner B's result. - Write an ordered pair using the row value first and the column value second. - After completing the table, identify every pair whose coordinates add to \(4\).

1. The missing entry in row \(1\), column \(3\) is \((1, 3)\). 2. The missing entry in row \(2\), column \(2\) is \((2, 2)\). 3. The table contains \(2\cdot3=6\) equally likely outcomes. 4. The outcomes with sum \(4\) are \((1, 3)\) and \((2, 2)\), so the probability is \(\frac{2}{6}=\frac{1}{3}\).

Missing entries: \((1, 3)\) and \((2, 2)\) Probability of a sum of \(4\): \(\frac{1}{3}\)

- What information does a row label give? What does a column label give? - Mark the cells that qualify before making your list. - Check your final list against the table: have you missed a pair or written one twice?

1. Row 2 is 2H, 2T, pairing the spinner's 2 with each coin outcome. 2. The table contains six equally likely outcomes. 3. The pairs with an even number or heads are 1H, 2H, 2T, and 3H. 4. Four of the six pairs work, so the probability is \(\frac{4}{6}=\frac{2}{3}\).

Missing cells: 2H, 2T. Favorable pairs: 1H, 2H, 2T, 3H. Probability: \(\frac{2}{3}\).

- What makes a complete result in this two-step game? - Which first-stage branch could lead to the requested result? - How are the chances written on consecutive branches related to the chance of the full path?

1. The probability of blue is \(\frac{2}{3}\). 2. After either ball color, the chance of tails is \(\frac{1}{2}\). 3. Multiply along the blue-then-tails path: \(\frac{2}{3}\cdot\frac{1}{2}=\frac{1}{3}\).

\(\frac{1}{3}\).

- Can a color appear twice if the bag holds only one bead of that color? - The order of the two draws matters.

1. Two red beads can be drawn, so RR is possible. 2. Either a red bead then the blue bead (RB), or the blue bead then a red bead (BR), is possible. 3. BB is impossible because the bag has only one blue bead and the first bead is not replaced.

RR, RB, and BR are possible. BB is impossible because there is only one blue bead.

- What choices make one complete visit to the park? - Does every second-stage route belong to both first-stage activities?

1. The two water routes apply only to canoe riders. 2. The three trails apply only to hikers. 3. The number of complete trips is \(2+3=5\), not \(2\cdot5\).

There are \(5\) trips: \(2\) canoe routes and \(3\) hiking trails. The route choices are not available for both activities.

- Count the ordered pairs that produce each sum. - A \(6 \times 6\) outcome grid contains all ordered pairs. - Distinguish between a sum and the ordered pair that produces it.

1. a) A sum of \(2\) occurs only with \((1, 1)\), so \(P(2) = \frac{1}{36}\). A sum of \(7\) occurs with \((1, 6)\), \((2, 5)\), \((3, 4)\), \((4, 3)\), \((5, 2)\), and \((6, 1)\), so \(P(7) = \frac{6}{36} = \frac{1}{6}\). The sums are not equally likely. 2. b) Each ordered pair combines one of \(6\) results on the first cube with one of \(6\) results on the second. There are \(6 \cdot 6 = 36\) ordered pairs. 3. A \(6 \times 6\) outcome grid has one cell for each ordered pair, and the fair number cubes make these \(36\) cells equally likely. Therefore, each ordered pair has probability \(\frac{1}{36}\).

a) No. \(P(2) = \frac{1}{36}\), while \(P(7) = \frac{6}{36} = \frac{1}{6}\). b) Yes. There are \(36\) ordered pairs, and each has probability \(\frac{1}{36}\).

- Use the spinner to compare the single-spin probabilities of blue and yellow. - For two spins, think of equally likely ordered section pairs rather than only the four color labels. - Ask whether each color combination contains the same number of those equally likely pairs.

1. Each spin can land on one of \(10\) equal sections, so a \(10 \times 10\) outcome grid has \(100\) equally likely ordered section pairs. 2. There are \(7\) blue choices on the first spin and \(7\) blue choices on the second, giving \(7 \cdot 7 = 49\) blue-blue section pairs. Therefore, \(P(\text{blue}, \text{blue}) = \frac{49}{100} = 0.49 = 49\%\). 3. The four color combinations group different numbers of equally likely section pairs. Blue-blue has \(49\) pairs, blue-yellow has \(7 \cdot 3 = 21\), yellow-blue has \(3 \cdot 7 = 21\), and yellow-yellow has \(3 \cdot 3 = 9\). Therefore, the four color combinations are not equally likely.

a) \(0.49 = 49\%\) b) The four combinations are not equally likely because blue and yellow are not equally likely on a single spin.

- Use the experimental proportion of black balls to estimate the proportion in the bag. - Scale the estimated proportion to a total of \(10\) balls. - For part b, include both possible color orders.

1. a) The relative frequency of black is \(\frac{124}{200} = 0.62\). 2. Multiplying this estimated proportion by the \(10\) balls gives \(10 \cdot 0.62 = 6.2\). Because the number of balls must be a whole number, a reasonable estimate is \(6\) black balls. 3. b) With replacement, there are \(10 \cdot 10 = 100\) equally likely ordered pairs of ball draws. 4. Black then white can occur in \(6 \cdot 4 = 24\) ordered outcomes, and white then black can occur in \(4 \cdot 6 = 24\) ordered outcomes. Thus there are \(48\) favorable outcomes, so \(P = \frac{48}{100} = 0.48\).

a) About \(6\) black balls b) \(0.48 = 48\%\)

- How many choices are available for each digit? Use an organized list or tree diagram. - What is the logical opposite of “at least twice”? - Recall the relationship between the probability of an event and the probability of its complement.

1. Listing the choices systematically gives \(S = \{111, 112, 121, 122, 211, 212, 221, 222\}\), so \(|S| = 8\). 2. The outcomes containing two or three digits equal to \(1\) are \(G = \{111, 112, 121, 211\}\). 3. The complement means the number contains the digit \(1\) at most once. Therefore, \(\overline{G} = \{122, 212, 221, 222\}\). 4. Each outcome is equally likely, so \(P(G) = \frac{4}{8} = \frac{1}{2} = 0.5\) and \(P(\overline{G}) = \frac{4}{8} = \frac{1}{2} = 0.5\). Their sum is \(0.5 + 0.5 = 1\).

1. \(S = \{111, 112, 121, 122, 211, 212, 221, 222\}\) 2. \(G = \{111, 112, 121, 211\}\) 3. \(\overline{G}\): The number contains the digit \(1\) at most once. \(\overline{G} = \{122, 212, 221, 222\}\) 4. \(P(G) = 0.5\), \(P(\overline{G}) = 0.5\), and \(0.5 + 0.5 = 1\).

- Use the complement by counting numbers with no \(5\). - Track how many cards remain on the second draw. - Compare the two probabilities after calculating them.

1. a) With replacement, there are \(5 \cdot 5 = 25\) two-digit outcomes. There are \(4 \cdot 4 = 16\) outcomes with no \(5\). Therefore, \(P(\text{at least one 5}) = \frac{25 - 16}{25} = \frac{9}{25} = 0.36\). 2. b) Without replacement, there are \(5 \cdot 4 = 20\) outcomes. There are \(4 \cdot 3 = 12\) outcomes with no \(5\). Therefore, \(P(\text{at least one 5}) = \frac{20 - 12}{20} = \frac{2}{5} = 0.4\). 3. c) The probability is greater without replacement. If the first digit is not \(5\), the chance of drawing \(5\) second increases from \(\frac{1}{5}\) to \(\frac{1}{4}\).

a) \(\frac{9}{25} = 36\%\) b) \(\frac{2}{5} = 40\%\) c) The probability is greater without replacement.

- Translate each event into conditions on the first, second, and third entries. - “Only on the first spin” excludes \(1\) from the other positions. - “First occurs on the third spin” excludes \(1\) from the first two positions. - Use the fundamental counting principle to count outcomes. - All \(27\) ordered triples are equally likely.

1. For event \(A\), the first entry is \(1\), and the other entries are \(2\) or \(3\). Thus, \(A=\{(1, 2, 2),(1, 2, 3),(1, 3, 2),(1, 3, 3)\}\). 2. For event \(B\), the first two entries are \(2\) or \(3\), and the third entry is \(1\). Thus, \(B=\{(2, 2, 1),(2, 3, 1),(3, 2, 1),(3, 3, 1)\}\). 3. In event \(C\), the first two spins each have \(3\) choices and the third is fixed, so there are \(3\cdot 3=9\) outcomes. In event \(D\), the third spin has \(2\) choices, so there are \(3\cdot 3\cdot 2=18\) outcomes. 4. There are \(3^3=27\) equally likely outcomes. Therefore, \(P(A)=\frac{4}{27}\), \(P(B)=\frac{4}{27}\), \(P(C)=\frac{9}{27}=\frac{1}{3}\), and \(P(D)=\frac{18}{27}=\frac{2}{3}\).

a) \(A=\{(1, 2, 2),(1, 2, 3),(1, 3, 2),(1, 3, 3)\}\); \(B=\{(2, 2, 1),(2, 3, 1),(3, 2, 1),(3, 3, 1)\}\) b) \(C\): \(9\) outcomes; \(D\): \(18\) outcomes c) \(P(A)=\frac{4}{27}\); \(P(B)=\frac{4}{27}\); \(P(C)=\frac{1}{3}\); \(P(D)=\frac{2}{3}\)

- Use the fundamental counting principle to find the total number of ordered pairs. - Make a table or list of all \(16\) outcomes. - Include equal factors such as \(2\cdot 2\) when checking perfect-square products. - List ordered pairs systematically.

1. There are \(4\cdot 4=16\) equally likely ordered outcomes. 2. The outcomes with sum \(5\) are \((1, 4)\), \((2, 3)\), \((3, 2)\), and \((4, 1)\). Thus, the probability is \(\frac{4}{16}=\frac{1}{4}\). 3. The outcomes with a perfect-square product are \((1, 1)\), \((1, 4)\), \((2, 2)\), \((3, 3)\), \((4, 1)\), and \((4, 4)\). Thus, the probability is \(\frac{6}{16}=\frac{3}{8}\). 4. The outcomes with the first number greater are \((2, 1)\), \((3, 1)\), \((3, 2)\), \((4, 1)\), \((4, 2)\), and \((4, 3)\). Thus, the probability is \(\frac{6}{16}=\frac{3}{8}\).

a) \(\frac{1}{4}=0.25\) b) \(\frac{3}{8}=0.375\) c) \(\frac{3}{8}=0.375\)

- List the number of ordered outcomes that produce each possible sum. - Translate “greater than \(3\) and less than \(11\)” into an inequality. - The endpoints are included in “at most” and “at least.” - An event and its complement have probabilities that add to \(1\).

1. The \(6\times 6\) sample space has \(36\) equally likely ordered outcomes. 2. For \(S\le 4\), the numbers of outcomes for sums \(2\), \(3\), and \(4\) are \(1\), \(2\), and \(3\), for a total of \(6\). For \(S\ge 10\), the counts for sums \(10\), \(11\), and \(12\) are \(3\), \(2\), and \(1\), also totaling \(6\). The two events are disjoint, so the probability is \(\frac{12}{36}=\frac{1}{3}\). 3. The complement of \(3<S<11\) is \(S\le 3\) or \(S\ge 11\). These sums have \(1+2+2+1=6\) outcomes, so \(P(E^c)=\frac{6}{36}=\frac{1}{6}\).

a) \(\frac{1}{3}\) b) \(P(E^c)=\frac{1}{6}\)

- Read each spinner's color-number sections from the image. - Treat a result as an ordered pair, one section from A and one from B. - For part b, list or organize the number pairs meeting the sum condition.

1. There are \(5\cdot4=20\) equally likely ordered section pairs. 2. For matching colors, there are \(2\cdot1=2\) red-red pairs and \(2\cdot2=4\) blue-blue pairs. Thus, \(P(\text{same color})=\frac{6}{20}=\frac{3}{10}=30\%\). 3. The number pairs with sum at least \(6\) are \((2,4),(3,3),(3,4),(4,2),(4,3),(4,4),(5,1),(5,2),(5,3),(5,4)\). Therefore, \(P(\text{sum}\ge6)=\frac{10}{20}=\frac{1}{2}=50\%\).

a) \(\frac{3}{10}=30\%\) b) \(\frac{1}{2}=50\%\)

- Read how often each number and color appears on the spinner. - Treat the two spins as ordered outcomes. - Repeated labels create multiple equally likely section pairs for the same numerical pair.

1. There are \(6\cdot6=36\) equally likely ordered section pairs. 2. The number \(1\) appears on \(2\) sections, \(2\) on \(3\), and \(3\) on \(1\). Thus, \(P(\text{same number})=\frac{2^2+3^2+1^2}{36}=\frac{14}{36}=\frac{7}{18}\). 3. A sum greater than \(4\) comes from number pairs \((2,3),(3,2),(3,3)\). Accounting for repeated labels gives \(3\cdot1+1\cdot3+1\cdot1=7\) favorable section pairs, so the probability is \(\frac{7}{36}\). 4. There are \(2\) red sections and \(2\) blue sections, so \(P(\text{red then blue})=\frac{2}{6}\cdot\frac{2}{6}=\frac{1}{9}\).

a) \(\frac{7}{18}\) b) \(\frac{7}{36}\) c) \(\frac{1}{9}\)

- Read the color-number sections from both spinners. - Count all ordered section pairs before counting favorable ones. - For the sum event, organize the ordered number pairs that total \(6\).

1. There are \(4\cdot6=24\) equally likely ordered section pairs. 2. Spinner A has \(1\) red section and Spinner B has \(3\), so \(P(\text{both red})=\frac{1}{4}\cdot\frac{3}{6}=\frac{1}{8}\). 3. The number pairs with sum \(6\) are \((1,5),(2,4),(3,3),(4,2)\), giving \(P(\text{sum}=6)=\frac{4}{24}=\frac{1}{6}\).

a) \(\frac{1}{8}\) b) \(\frac{1}{6}\)

- Treat the three draws as an ordered sample space without replacement. - For part a, focus on which labels can occupy the hundreds place. - For part b, identify the required set of digits before counting its orders.

1. There are \(4\cdot3\cdot2=24\) equally likely ordered outcomes. 2. For part a, the hundreds digit must be \(7\) or \(9\). Each choice allows \(3\cdot2=6\) arrangements of the remaining digits, so 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 \(3\cdot2\cdot1=6\) orders, so the probability is \(\frac{6}{24}=\frac{1}{4}\). 4. Every available digit is odd, so every possible number is odd. The probability of an even number is \(0\).

a) \(\frac{1}{2}=50\%\) b) \(\frac{1}{4}=25\%\) c) \(0=0\%\)

- Count how many sections carry each number. - List the ordered number pairs with sum at least \(6\). - Account for repeated labels by counting section pairs. - Simplify the final fraction.

1. The possible numbers and their probabilities are \(P(1)=\frac{1}{6}\), \(P(2)=\frac{1}{6}\), \(P(3)=\frac{2}{6}\), and \(P(5)=\frac{2}{6}\). 2. The number pairs with sum at least \(6\) are \((1, 5)\), \((2, 5)\), \((3, 3)\), \((3, 5)\), \((5, 1)\), \((5, 2)\), \((5, 3)\), and \((5, 5)\). 3. Counting the corresponding section pairs gives \(2+2+4+4+2+2+4+4=24\) favorable outcomes out of \(36\). 4. Therefore, the probability is \(\frac{24}{36}=\frac{2}{3}\).

\(\frac{2}{3}\)

- Count each color directly from the spinner to get the one-spin probabilities. - For matching colors, consider each possible matching-color path. - For “at least one,” consider whether a complement is easier to count.

1. From the spinner, \(P(\text{red})=\frac{3}{8}\), \(P(\text{blue})=\frac{4}{8}\), and \(P(\text{green})=\frac{1}{8}\). 2. \(P(\text{same color})=\left(\frac{3}{8}\right)^2+\left(\frac{4}{8}\right)^2+\left(\frac{1}{8}\right)^2=\frac{13}{32}=40.625\%\). 3. \(P(\text{at least one green})=1-\left(\frac{7}{8}\right)^2=\frac{15}{64}=23.4375\%\). 4. \(P(\text{red then blue})=\frac{3}{8}\cdot\frac{4}{8}=\frac{3}{16}=18.75\%\).

a) \(\frac{13}{32}=40.625\%\) b) \(\frac{15}{64}=23.4375\%\) c) \(\frac{3}{16}=18.75\%\)

- Read the one-spin color probabilities from the spinner. - Part a specifies an order, so treat the two spins as ordered. - For part b, compare “at least one red” with its complement.

1. From the spinner, \(P(\text{green})=\frac{2}{5}\), \(P(\text{red})=\frac{2}{5}\), and \(P(\text{yellow})=\frac{1}{5}\). 2. \(P(\text{green then yellow})=\frac{2}{5}\cdot\frac{1}{5}=\frac{2}{25}=8\%\). 3. \(P(\text{at least one red})=1-\left(\frac{3}{5}\right)^2=\frac{16}{25}=64\%\).

a) \(\frac{2}{25}=8\%\) b) \(\frac{16}{25}=64\%\)

- Count the color frequencies on the spinner first. - For part a, add the probabilities of the three matching-color cases. - For part b, use the complement of getting no yellow on either spin.

1. From the spinner, \(P(\text{red})=\frac{1}{2}\), \(P(\text{green})=\frac{1}{4}\), and \(P(\text{yellow})=\frac{1}{4}\). 2. \(P(\text{same color})=\left(\frac{1}{2}\right)^2+\left(\frac{1}{4}\right)^2+\left(\frac{1}{4}\right)^2=\frac{3}{8}=37.5\%\). 3. \(P(\text{at least one yellow})=1-\left(\frac{3}{4}\right)^2=\frac{7}{16}=43.75\%\).

a) \(\frac{3}{8}=37.5\%\) b) \(\frac{7}{16}=43.75\%\)

- Read the number and color of each equal section from the spinner. - Organize the ordered number pairs systematically for part a. - For different colors, consider counting the same-color complement.

1. There are \(5\cdot5=25\) equally likely ordered outcomes. 2. The number pairs with sum at least \(7\) are \((2,5),(3,4),(3,5),(4,3),(4,4),(4,5),(5,2),(5,3),(5,4),(5,5)\), so the probability is \(\frac{10}{25}=\frac{2}{5}=40\%\). 3. For the complement in part b, there are \(4\) red-red outcomes, \(4\) blue-blue outcomes, and \(1\) green-green outcome, so \(P(\text{same color})=\frac{9}{25}\). 4. Therefore, \(P(\text{different colors})=1-\frac{9}{25}=\frac{16}{25}=64\%\).

a) \(\frac{2}{5}=40\%\) b) \(\frac{16}{25}=64\%\)

- Find the probability of each color on one spin. - For part a, add the probabilities of the three matching-color outcomes. - For part b, use the complement of no red. - For part c, multiply the probabilities in the stated order. - Simplify each fraction.

1. The one-spin probabilities are \(P(\text{green})=\frac{3}{6}=\frac{1}{2}\), \(P(\text{blue})=\frac{2}{6}=\frac{1}{3}\), and \(P(\text{red})=\frac{1}{6}\). 2. The matching-color outcomes are green-green, blue-blue, and red-red: \(P(\text{same color})=\left(\frac{1}{2}\right)^2+\left(\frac{1}{3}\right)^2+\left(\frac{1}{6}\right)^2=\frac{1}{4}+\frac{1}{9}+\frac{1}{36}=\frac{7}{18}\). 3. Use the complement of no red: \(P(\text{at least one red})=1-\left(\frac{5}{6}\right)^2=1-\frac{25}{36}=\frac{11}{36}\). 4. For the specified order, multiply the probabilities: \(P(\text{green then blue})=\frac{1}{2}\cdot\frac{1}{3}=\frac{1}{6}\).

a) \(\frac{7}{18}\) b) \(\frac{11}{36}\) c) \(\frac{1}{6}\)

- Treat each result as an ordered pair from the two stated number sets. - List the pairs satisfying each condition. - Divide each favorable count by the \(12\) equally likely pairs.

1. There are \(3\cdot4=12\) equally likely ordered pairs \((a,b)\). 2. The pairs with sum \(5\) are \((1,4),(2,3),(3,2)\), so the probability is \(\frac{3}{12}=\frac{1}{4}\). 3. The pairs with \(b>a\) are \((1,2),(1,3),(1,4),(2,3),(2,4),(3,4)\), so the probability is \(\frac{6}{12}=\frac{1}{2}\). 4. The matching pairs are \((1,1),(2,2),(3,3)\), so the probability is \(\frac{3}{12}=\frac{1}{4}\).

a) \(\frac{1}{4}\) b) \(\frac{1}{2}\) c) \(\frac{1}{4}\)

- Read the one-spin blue and yellow probabilities from the spinner. - Place the single yellow in each possible position among the three spins. - Multiply along one favorable sequence, then account for all favorable positions.

1. Exactly one yellow can occur in the first, second, or third position, giving \((Y,B,B),(B,Y,B),(B,B,Y)\). 2. From the spinner, \(P(Y)=\frac{1}{4}\) and \(P(B)=\frac{3}{4}\). 3. Each favorable sequence has probability \(\frac{1}{4}\cdot\frac{3}{4}\cdot\frac{3}{4}=\frac{9}{64}\). 4. Therefore, \(P(E)=3\cdot\frac{9}{64}=\frac{27}{64}\).

a) \((Y,B,B)\), \((B,Y,B)\), and \((B,B,Y)\) b) \(P(E)=\frac{27}{64}\)

- Read the one-spin color probabilities from the spinner. - For part a, multiply in the exact stated order. - For part b, account for every order in which the three different colors can occur.

1. From the spinner, \(P(B)=\frac{1}{2}\), \(P(R)=\frac{1}{4}\), 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 blue, red, and yellow each occur once. There are \(3\cdot2\cdot1=6\) orders, each with probability \(\frac{1}{32}\), so the probability is \(\frac{3}{16}=0.1875\).

a) \(\frac{1}{32}=0.03125\) b) \(\frac{3}{16}=0.1875\)

- Read the color attached to each numbered section from the spinner. - Treat the two spins as ordered outcomes. - For the sum event, list high-number pairs systematically.

1. The spinner shows \(4\) blue sections and \(2\) red sections, so \(P(C)=\frac{4}{8}\cdot\frac{2}{8}=\frac{1}{8}\). 2. Four of the eight labels are odd, so \(P(A)=\frac{4}{8}\cdot\frac{4}{8}=\frac{1}{4}\). 3. There are \(64\) equally likely ordered number pairs. The pairs with sum greater than \(12\) are \((5,8),(6,7),(6,8),(7,6),(7,7),(7,8),(8,5),(8,6),(8,7),(8,8)\), so \(P(B)=\frac{10}{64}=\frac{5}{32}\).

a) \(P(C)=\frac{1}{8}\) b) \(P(A)=\frac{1}{4}\) c) \(P(B)=\frac{5}{32}\)

- Read how often each number and color appears on the spinner. - Treat the two spins as ordered section pairs. - For parts c and d, a complement may reduce the counting.

1. Two spins produce \(8\cdot8=64\) equally likely ordered section pairs. 2. Each of the four numbers appears on two sections. For each number, \(2\cdot2=4\) pairs show that number twice, so part a has \(16\) favorable outcomes and probability \(\frac{1}{4}\). 3. A sum of \(4\) can come from \((1,3),(2,2),(3,1)\). Each numerical pair corresponds to \(4\) section pairs, so part b is \(\frac{12}{64}=\frac{3}{16}\). 4. Each of the four colors has two sections, so there are \(4\cdot4=16\) same-color outcomes. Thus, part c is \(\frac{48}{64}=\frac{3}{4}\). 5. Six sections are not labeled \(1\), so \(36\) ordered pairs contain no \(1\). Thus, part d is \(\frac{64-36}{64}=\frac{7}{16}\).

a) \(\frac{1}{4}=25\%\) b) \(\frac{3}{16}=18.75\%\) c) \(\frac{3}{4}=75\%\) d) \(\frac{7}{16}=43.75\%\)

- Read the four equally likely colors from the spinner. - For part a, track how many color choices remain after each distinct color appears. - For part b, consider the possible position of the one spin that is not red.

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, and the third one of the remaining two. Thus, the probability is \(1\cdot\frac{3}{4}\cdot\frac{2}{4}=\frac{3}{8}\). 3. Exactly two red spins can occur in three positions. Each favorable order has probability \(\frac{1}{4}\cdot\frac{1}{4}\cdot\frac{3}{4}=\frac{3}{64}\), so the total is \(3\cdot\frac{3}{64}=\frac{9}{64}\).

a) \(\frac{3}{8}\) b) \(\frac{9}{64}\)

- Read the one-spin color probabilities from the spinner. - For part a, think about the probability of not yellow on all three spins. - For part b, consider the possible positions of the one spin that is not red.

1. From the spinner, \(P(\text{red})=\frac{1}{2}\), \(P(\text{blue})=\frac{1}{4}\), and \(P(\text{yellow})=\frac{1}{4}\). 2. \(P(E_1)=\left(\frac{3}{4}\right)^3=\frac{27}{64}\). 3. Exactly two red spins can occur in three positions. Since \(P(\text{red})=P(\text{not red})=\frac{1}{2}\), \(P(E_2)=3\cdot\left(\frac{1}{2}\right)^2\cdot\frac{1}{2}=\frac{3}{8}=\frac{24}{64}\). 4. Since \(\frac{27}{64}>\frac{24}{64}\), event \(E_1\) is more likely.

a) \(P(E_1)=\frac{27}{64}\) b) \(P(E_2)=\frac{3}{8}\) c) Event \(E_1\) is more likely because \(\frac{27}{64}>\frac{24}{64}\).

- Find the probability of each number on one spin. - List the ordered pairs whose sum is \(4\). - Multiply within each ordered pair and add the favorable probabilities.

1. The number \(1\) appears once, \(2\) appears twice, \(3\) appears once, and \(4\) appears once. Thus, \(P(1)=\frac{1}{5}\), \(P(2)=\frac{2}{5}\), and \(P(3)=\frac{1}{5}\). 2. The ordered pairs with sum \(4\) are \((1, 3)\), \((3, 1)\), and \((2, 2)\). 3. Their probabilities are \(\frac{1}{5}\cdot\frac{1}{5}=\frac{1}{25}\), \(\frac{1}{5}\cdot\frac{1}{5}=\frac{1}{25}\), and \(\frac{2}{5}\cdot\frac{2}{5}=\frac{4}{25}\). 4. Therefore, \(P(\text{sum }4)=\frac{1}{25}+\frac{1}{25}+\frac{4}{25}=\frac{6}{25}=24\%\).

\(\frac{6}{25}=24\%\)

- For each part, identify which of the labels \(1\) through \(8\) are allowed on each spin. - The word “only” requires the other spins not to have the named property. - For the all-same event, count the possible repeated-number triples.

1. Four of the eight labels are even and four are odd, so part a is \(\frac{4}{8}\cdot\frac{4}{8}\cdot\frac{4}{8}=\frac{1}{8}\). 2. The perfect squares are \(1\) and \(4\), so part b is \(\frac{2}{8}\cdot\frac{6}{8}\cdot\frac{6}{8}=\frac{9}{64}\). 3. The labels greater than \(5\) are \(6,7,8\), so part c is \(\left(\frac{3}{8}\right)^3=\frac{27}{512}\). 4. There are eight matching triples \((1,1,1)\) through \((8,8,8)\), so part d is \(8\cdot\left(\frac{1}{8}\right)^3=\frac{1}{64}\).

a) \(\frac{1}{8}\) b) \(\frac{9}{64}\) c) \(\frac{27}{512}\) d) \(\frac{1}{64}\)

- Find the probability of each number and each color on one spin. - For part a, list the ordered pairs whose sum is \(3\). - For part b, count the sections labeled with a number greater than \(1\). - For part c, add the probabilities of blue-blue, red-red, and the three other matching-color outcomes.

1. The number probabilities are \(P(1)=\frac{2}{8}=\frac{1}{4}\), \(P(2)=\frac{3}{8}\), and \(P(3)=P(4)=P(5)=\frac{1}{8}\). 2. A sum of \(3\) occurs with \((1, 2)\) or \((2, 1)\). Therefore, \(P(\text{sum }3)=\frac{2}{8}\cdot\frac{3}{8}+\frac{3}{8}\cdot\frac{2}{8}=\frac{3}{16}\). 3. The probability that the first number is \(1\) is \(\frac{2}{8}\). Six of the eight sections have numbers greater than \(1\). Thus, \(P(\text{first }1\text{, second greater than }1)=\frac{2}{8}\cdot\frac{6}{8}=\frac{3}{16}\). 4. Blue appears on two sections, red on three sections, and green, yellow, and orange each on one section. Therefore, \(P(\text{same color})=\left(\frac{2}{8}\right)^2+\left(\frac{3}{8}\right)^2+3\cdot\left(\frac{1}{8}\right)^2=\frac{1}{4}\).

a) \(\frac{3}{16}\) b) \(\frac{3}{16}\) c) \(\frac{1}{4}\)

- Read the single-spin probabilities from the equal sections of the spinner. - For matching colors, consider the separate ways the two spins can match. - For “at least once,” decide whether direct cases or the opposite event gives the cleaner calculation.

1. From the spinner, \(P(\text{blue})=\frac{1}{2}\), \(P(\text{red})=\frac{3}{10}\), and \(P(\text{yellow})=\frac{1}{5}\). 2. \(P(\text{same color})=\left(\frac{1}{2}\right)^2+\left(\frac{3}{10}\right)^2+\left(\frac{1}{5}\right)^2=\frac{19}{50}=38\%\). 3. The probability of no yellow is \(\left(\frac{4}{5}\right)^2\), so \(P(\text{at least one yellow})=1-\left(\frac{4}{5}\right)^2=\frac{9}{25}=36\%\). Inês is correct.

a) \(P(\text{blue})=\frac{1}{2}\), \(P(\text{red})=\frac{3}{10}\), and \(P(\text{yellow})=\frac{1}{5}\) b) \(\frac{19}{50}=38\%\) c) Yes. The probability is \(\frac{9}{25}=36\%\), which is greater than \(35\%\).

- Organize all ordered pairs from \(\{1,2,3,4,5\}\times\{1,2,3,4,5\}\). - Count the pairs meeting the sum condition. - Separately count the pairs in which the first coordinate is greater.

1. There are \(5\cdot5=25\) equally likely ordered outcomes. 2. Event \(A\) contains \((1,5),(2,4),(3,3),(4,2),(5,1)\), so \(P(A)=\frac{5}{25}=\frac{1}{5}=20\%\). 3. Event \(B\) has \(10\) ordered pairs with the first number greater than the second, so \(P(B)=\frac{10}{25}=\frac{2}{5}=40\%\). 4. Therefore, event \(B\) is more likely.

Event \(B\) is more likely. \(P(A)=\frac{1}{5}=20\%\), and \(P(B)=\frac{2}{5}=40\%\).

- Treat the four draws as equally likely ordered arrangements without replacement. - For part a, focus on the last digit; for part b, focus on the first digit. - For part c, use a divisibility test before trying to inspect individual arrangements.

1. The four distinct digits can be arranged in \(4\cdot3\cdot2\cdot1=24\) equally likely orders. 2. For part a, the last digit must be \(2\) or \(8\). There are \(2\cdot6=12\) favorable outcomes, so the probability is \(\frac{1}{2}\). 3. For part b, the first digit must be \(5\) or \(8\). Again there are \(12\) 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 \(3\). All are greater than \(3\), so none is prime and the probability is \(0\).

a) \(\frac{1}{2}\) b) \(\frac{1}{2}\) c) \(0\)

- List the possible positions of the one blue result. - Replacement keeps the red and blue probabilities unchanged on each draw. - Find the probability of one favorable order, then account for all favorable orders.

1. On each draw, \(P(\text{red})=\frac{2}{5}=0.4\) and \(P(\text{blue})=\frac{3}{5}=0.6\). 2. The favorable ordered outcomes are red-red-blue, red-blue-red, and blue-red-red. 3. Each has probability \(0.4\cdot0.4\cdot0.6=0.096\). 4. Therefore, the total probability is \(3\cdot0.096=0.288\).

\(0.288\), or \(28.8\%\)

- Find all final nodes labeled “2 makes.” - Trace those paths backward to the totals after the second shot. - If the player already has two makes, determine what the last result must be. - If the player has only one make, determine what the last result must be.

1. The paths with exactly two makes are \(MMX\), \(MXM\), and \(XMM\). There are \(3\) paths. 2. After the second shot, the player must have either \(1\) make or \(2\) makes. 3. With \(1\) make after two shots, the third shot must be made. With \(2\) makes after two shots, the third shot must be missed. With \(0\) makes after two shots, the goal is no longer possible.

a) \(3\) paths: \(MMX\), \(MXM\), and \(XMM\) b) \(1\) make or \(2\) makes c) From \(1\) make, the third shot must be made; from \(2\) makes, the third shot must be missed.

- Choose a possible first tile and pair it with every remaining tile. - Remember that the second tile cannot match the first. - Scan the completed outcomes for the required letter.

1. Starting with M gives \(MA,\ MT,\ MH\). 2. Starting with A gives \(AM,\ AT,\ AH\). 3. Starting with T gives \(TM,\ TA,\ TH\). 4. Starting with H gives \(HM,\ HA,\ HT\). 5. The outcomes containing A are \(MA,\ AM,\ AT,\ AH,\ TA,\ HA\).

Sample space: \(\{MA,MT,MH,AM,AT,AH,TM,TA,TH,HM,HA,HT\}\). Outcomes containing A: \(\{MA,AM,AT,AH,TA,HA\}\).

- Treat the available second choices separately for each first choice. - Do not assume every door has the same number of branches. - Filter the completed outcomes by the button condition.

1. Door A gives \(A1\) and \(A2\). 2. Door B gives \(B3\). 3. Door C gives \(C4,\ C5,\ C6\). 4. The sample space is \(\{A1,A2,B3,C4,C5,C6\}\). 5. The outcomes with even-numbered buttons are \(A2,\ C4,\ C6\).

Sample space: \(\{A1,A2,B3,C4,C5,C6\}\). Even-numbered button: \(\{A2,C4,C6\}\).

- Start with all pairings, then apply the restriction. - Keep the weather entry first in every ordered pair. - Select the remaining outcomes with the requested travel mode.

1. In sunny weather, the possible outcomes are \((S, B)\), \((S, K)\), and \((S, W)\). 2. In rainy weather, the possible outcomes are \((R, B)\) and \((R, W)\); \((R, K)\) is excluded. 3. The sample space is \(\{(S, B),(S, K),(S, W),(R, B),(R, W)\}\). 4. The walking outcomes are \((S, W)\) and \((R, W)\).

Sample space: \(\{(S, B),(S, K),(S, W),(R, B),(R, W)\}\). Walking outcomes: \(\{(S, W),(R, W)\}\).

- Does a leaf name the whole result, or only the last decision on that path? - Can paths with different numbers of letters still be complete outcomes?

1. Finishing immediately gives \(F\). 2. Continuing and choosing left gives \(CL\). 3. Continuing, choosing right, and then choosing upper or lower gives \(CRU\) and \(CRD\). 4. The terminal sample space is \(\{F,CL,CRU,CRD\}\). 5. The outcomes that include a right choice are \(CRU\) and \(CRD\).

Terminal sample space: \(\{F,CL,CRU,CRD\}\). Includes a right choice: \(\{CRU,CRD\}\).

- Pair each possible result of the first spinner with a result of the second. - Count outcomes containing one green result, not two. - Compare that count with the total number of equally likely outcomes.

1. The sample space is \(\{(R, R),(R, Y),(R, G),(Y, R),(Y, Y),(Y, G),(G, R),(G, Y),(G, G)\}\). 2. Exactly one green occurs in \((R, G),(Y, G),(G, R),(G, Y)\). 3. There are \(4\) favorable outcomes out of \(9\) equally likely outcomes. 4. The probability is \(\frac{4}{9}\).

Sample space: \(\{(R, R),(R, Y),(R, G),(Y, R),(Y, Y),(Y, G),(G, R),(G, Y),(G, G)\}\). The probability that exactly one spinner lands on green is \(\frac{4}{9}\).

- First list the combinations before applying the restriction. - Remove only the specifically unavailable outcome. - Filter the remaining outcomes by the number condition.

1. Without the restriction, the outcomes are \(C1O,\ C1T,\ C2O,\ C2T,\ Q1O,\ Q1T,\ Q2O,\ Q2T\). 2. Remove the unavailable outcome \(Q2T\). 3. The sample space is \(\{C1O,C1T,C2O,C2T,Q1O,Q1T,Q2O\}\). 4. The outcomes with number \(2\) are \(C2O,\ C2T,\ Q2O\).

Sample space: \(\{C1O,C1T,C2O,C2T,Q1O,Q1T,Q2O\}\). Number \(2\): \(\{C2O,C2T,Q2O\}\).

- Check whether each row-column combination appears exactly once. - Use the row and column headings together to inspect the final cell. - After correcting the grid, count the equally likely outcomes in which Spinner B shows \(2\).

1. The final cell is in row B and column \(3\), so it should be \((B, 3)\), not a second copy of \((B, 2)\). 2. The corrected table has \(6\) distinct equally likely outcomes. 3. Spinner B shows \(2\) in exactly two outcomes: \((A, 2)\) and \((B, 2)\). 4. Therefore, the probability is \(\frac{2}{6}=\frac{1}{3}\).

a) The last entry should be \((B, 3)\), not \((B, 2)\). b) \(\frac{1}{3}\)

- Count the ordered results of three rolls, then count the ones with three different numbers. - The complement of “at least one repeat” is “all three are different.” - Compare the model probability with the observed frequency using a common denominator. - Many independent repetitions make an experimental estimate more stable, not exact every time.

1. There are \(12^3=1728\) ordered three-roll results. 2. To have no repeats, there are \(12\) choices for the first roll, \(11\) for the second, and \(10\) for the third, giving \(12\cdot11\cdot10=1320\) results. 3. The probability of no repeats is \(\frac{1320}{1728}=\frac{55}{72}\). Therefore, the probability of at least one repeat is \(1-\frac{55}{72}=\frac{17}{72}\). 4. The experimental probability is \(\frac{241}{1000}\). The absolute difference is \(\frac{241}{1000}-\frac{17}{72}=\frac{11}{2250}\). 5. With many independent trials, the relative frequency tends to stabilize near the theoretical probability, although finite results still vary.

a) \(1320\) ordered results have no repeated number; \(P(\text{at least one repeat})=\frac{17}{72}\). b) Relative frequency: \(\frac{241}{1000}\); absolute difference: \(\frac{11}{2250}\). c) With more trials, the relative frequency tends to approach the theoretical probability.

- Read the one-spin color probabilities from the spinner. - For “exactly,” account for the possible positions of the specified color. - For “at least once,” consider a complement; for an exact order, multiply in that order.

1. From the spinner, \(P(\text{green})=\frac{3}{8}\), \(P(\text{yellow})=\frac{1}{2}\), and \(P(\text{orange})=\frac{1}{8}\). 2. Exactly two yellow spins can occur in three positions, so part a is \(3\cdot\left(\frac{1}{2}\right)^2\cdot\frac{1}{2}=\frac{3}{8}\). 3. For part b, use the complement: \(1-\left(\frac{7}{8}\right)^3=\frac{169}{512}\). 4. For the exact order in part c, \(\frac{3}{8}\cdot\frac{1}{2}\cdot\frac{1}{8}=\frac{3}{128}\). 5. At most one yellow means zero or one yellow. Those probabilities are \(\frac{1}{8}\) and \(\frac{3}{8}\), totaling \(\frac{1}{2}\).

a) \(\frac{3}{8}\) b) \(\frac{169}{512}\) c) \(\frac{3}{128}\) d) \(\frac{1}{2}\)

- Read the four equally likely color outcomes from the spinner. - For part b, count one repeated-color sequence for each color. - For part c, choose two colors and count sequences that actually use both; for part d, consider a complement.

1. There are \(4^3=64\) equally likely ordered color sequences. 2. The sequence red-blue-yellow is one outcome, so part a is \(\frac{1}{64}\). 3. There are four same-color sequences, so part b is \(\frac{4}{64}=\frac{1}{16}\). 4. There are \(6\) possible pairs of colors. For each pair, \(2^3-2=6\) sequences use both colors, so part c is \(\frac{36}{64}=\frac{9}{16}\). 5. For part d, the complement has no yellow and therefore \(3^3=27\) sequences. Thus, the probability is \(1-\frac{27}{64}=\frac{37}{64}\).

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