- Ask whether the process itself can produce more than one possible outcome under the stated conditions. - Distinguish an outcome that is already fixed but unknown from one generated by a random process. - Check whether the selection procedure in each situation actually uses chance.

1. Tossing the thumbtack is a random experiment because there are multiple possible outcomes and the outcome of a toss is not known in advance. 2. Revealing Max's grade is not a random experiment because the grade was already fixed by the completed test and the grading rule before it was revealed. 3. Ms. Rivera's selection is a random experiment because the random number generator determines which student is selected from several possible outcomes.

a) Yes. The thumbtack toss has an uncertain outcome. b) No. Max's grade is already fixed before it is revealed. c) Yes. The random selection can produce different student outcomes.

- Is the outcome uncertain before the process is carried out? - Are the possible outcomes known? - Distinguish a chance process from a result fixed by a physical condition or mathematical rule.

1. A random experiment has a known set of possible outcomes, but the individual outcome cannot be predicted with certainty before the experiment is performed. 2. a) This is not a random experiment in the intended model. Under the stated physical conditions, the water boils at \(100\,^\circ\text{C}\), so the outcome is treated as predictable. 3. b) This is a random experiment. The possible outcomes are \(1, 2, 3, 4, 5, 6\), but the result of a particular roll is uncertain. 4. c) This is a random experiment. The possible numbered balls are known, but the selected ball is uncertain before the draw. 5. d) This is not a random experiment. The circumference is determined by \(C = 2\pi r\), so the result follows from the given radius.

a) Not a random experiment; the outcome is treated as physically determined. b) Random experiment; the roll is uncertain among known outcomes. c) Random experiment; the selected ball is uncertain among known outcomes. d) Not a random experiment; the result is determined by a formula.

- Could the event happen on one draw and fail on another? - What would have to be true for an event never to happen?

1. Red balls are present, but a blue ball could be drawn instead. Event a) is possible but not certain. 2. No green balls are present, so event b) is impossible. 3. Every ball is either red or blue, so event c) is certain.

a) Possible but not certain b) Impossible c) Certain

- Combine all sections that are not green. - Compare that favorable count with all \(6\) equal sections.

1. There are \(1+2=3\) sections that are not green. 2. Therefore, \(P(\text{not green})=\frac{3}{6}=\frac{1}{2}\).

\(\frac{1}{2}=50\%\)

- Think of \(0\) as no chance within the probability model. - Think of \(1\) as the full chance that the event occurs.

1. Probability \(0\) represents an event that cannot occur in the model. 2. Probability \(1\) represents an event that must occur in the model.

a) An impossible event b) A certain event

- Compare each decimal with the benchmark probabilities \(0\), \(0.5\), and \(1\). - “Roughly equally likely” corresponds to a probability near one-half.

1. A probability close to \(0\) is very unlikely, so \(0.08\) is very unlikely. 2. A probability close to \(0.5\) is roughly equally likely to occur or not occur, so \(0.48\) fits that description. 3. A probability close to \(1\) is very likely, so \(0.93\) is very likely.

\(0.08\): very unlikely \(0.48\): roughly equally likely to occur or not occur \(0.93\): very likely

- Which event has no matching marble? - Which event includes every marble in the bag? - Could the remaining event happen sometimes but not always?

1. There are no green marbles, so a) is impossible. 2. Every marble is red or blue, so b) is certain. 3. Some marbles are red but some are blue, so c) is possible but not certain.

a) Impossible. b) Certain. c) Possible but not certain.

- What probability represents equal chances of happening and not happening? - Find that value on the marked number line.

1. An event equally likely to occur or not occur has probability \(\frac{1}{2}=0.5\). 2. Marker B is at \(0.5\) on the probability scale.

Marker B.

- What values mark impossible and certain events? - Which end of the probability scale is nearer to \(0.85\)?

1. Impossible has probability \(0\), and certain has probability \(1\). 2. The distance from \(0.85\) to \(1\) is \(0.15\), smaller than its distance to \(0\).

It is closer to certain because \(0.85\) is closer to \(1\) than to \(0\).

- Is the recorded result a card's physical identity or its letter? - Does drawing either A card give a different recorded result?

1. The recorded outcome is a letter, not the identity of a physical card. 2. The distinct possible letters are A, B, and C. 3. A sample space lists each possible recorded outcome once, even if several cards have the same letter.

\(\{A,B,C\}\). Each possible letter appears only once in the sample space, even when it occurs on several cards.

- An equally likely model requires every listed outcome to have the same chance. - Look for physical symmetry or equal-size regions. - Ask whether every possible result has exactly the same physical chance. - Decide whether the design could favor one outcome.

1. a) No. A thumbtack is not symmetric, so the two landing positions should not be assumed to have equal probabilities. 2. b) Yes. With a well-shuffled deck, each individual card can reasonably be modeled as having probability \(\frac{1}{52}\). 3. c) Yes. A standard number cube is symmetric, and uniformly rounded edges do not favor one face. Each result can be modeled with probability \(\frac{1}{6}\). 4. d) Yes. The four sections have equal central angles, so each color can be modeled with probability \(\frac{1}{4}\).

a) No; the thumbtack is not symmetric. b) Yes; each card has the same chance of being drawn. c) Yes; the number cube is symmetric. d) Yes; the four spinner sections are equal in size.

- Find the total number of pieces of fruit. - Are there equal numbers of pieces of the three kinds? - For “not an apple,” what other kinds are allowed?

1. There are \(12+8+10=30\) pieces of fruit. 2. a) There are \(8\) pears, so \(P(\text{pear})=\frac{8}{30}=\frac{4}{15}\). 3. b) Individual pieces are equally likely, but the three kinds are not equally common. The chance of each kind depends on how many pieces of that kind are in the bowl. 4. c) There are \(8+10=18\) pieces that are not apples. Therefore, \(P(\text{not an apple})=\frac{18}{30}=\frac{3}{5}\).

a) \(\frac{4}{15}\) b) The three kinds have different numbers of pieces. Equal chances for individual pieces do not mean equal chances for fruit types. c) \(\frac{3}{5}\)

- Count the values from \(1\) through \(150\) that satisfy each condition. - List the relevant multiples systematically rather than checking at random. - Find the largest perfect square no greater than \(150\). - A number ending in \(0\) or \(5\) is divisible by \(5\).

1. a) There are \(10\) multiples of \(15\) from \(1\) through \(150\). Thus, \(P=\frac{10}{150}=\frac{1}{15}\). 2. b) The perfect squares are \(1^2\) through \(12^2\), since \(12^2=144\) and \(13^2=169\). There are \(12\), so \(P=\frac{12}{150}=\frac{2}{25}\). 3. c) Numbers ending in \(0\) or \(5\) are multiples of \(5\). There are \(150\div5=30\), so \(P=\frac{30}{150}=\frac{1}{5}\).

a) \(\frac{1}{15}\) b) \(\frac{2}{25}=8\%\) c) \(\frac{1}{5}=20\%\)

- Compare the number of sections of each color. - For equally likely outcomes, treat each equal-size section as a separate outcome. - Numbering the sections can distinguish them.

1. Each of the \(10\) sections has probability \(\frac{1}{10}\). 2. a) The color probabilities are \(P(\text{blue}) = \frac{5}{10} = 0.5\), \(P(\text{red}) = \frac{3}{10} = 0.3\), and \(P(\text{yellow}) = \frac{2}{10} = 0.2\). Because these probabilities differ, the color outcomes are not equally likely. 3. b) Number the individual sections \(1\) through \(10\). Then \(S' = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}\), and every outcome has probability \(\frac{1}{10}\).

a) \(P(\text{blue}) = 0.5\), \(P(\text{red}) = 0.3\), and \(P(\text{yellow}) = 0.2\), so the outcomes are not equally likely. b) One possible sample space is \(S' = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}\), where each number represents one section.

- Divide each category count by the total number of truffles. - Compare the three probabilities. - Equally likely outcomes must have the same probability.

1. There are \(7 + 4 + 1 = 12\) truffles. 2. a) \(P(\text{milk}) = \frac{7}{12}\), \(P(\text{dark}) = \frac{4}{12} = \frac{1}{3}\), and \(P(\text{white}) = \frac{1}{12}\). 3. b) The three probabilities are different. Therefore, the flavor-category outcomes cannot be modeled as equally likely.

a) \(P(\text{milk}) = \frac{7}{12}\), \(P(\text{dark}) = \frac{1}{3}\), and \(P(\text{white}) = \frac{1}{12}\) b) The outcomes are not equally likely because their probabilities are different.

- List every possible recorded result. - Compare the chances of the outcomes in each sample space. - Equal-size spinner sections represent equally likely outcomes.

1. a) \(S = \{\text{heads}, \text{tails}\}\). Because the coin is fair, both outcomes have probability \(\frac{1}{2}\), so they are equally likely. 2. b) \(S = \{\text{red}, \text{blue}\}\). The probabilities are \(P(\text{red}) = \frac{4}{10} = 0.4\) and \(P(\text{blue}) = \frac{6}{10} = 0.6\), so the outcomes are not equally likely. 3. c) \(S = \{1, 2, 3, 4, 5, 6, 7, 8\}\). The sections are equal in size, so each outcome has probability \(\frac{1}{8}\) and the outcomes are equally likely.

a) \(S = \{\text{heads}, \text{tails}\}\); equally likely b) \(S = \{\text{red}, \text{blue}\}\); not equally likely because the probabilities are \(0.4\) and \(0.6\) c) \(S = \{1, 2, 3, 4, 5, 6, 7, 8\}\); equally likely

- A standard deck has \(52\) cards altogether. - Each suit contains \(13\) cards. - Count all kings and aces without overlap. - Count the ranks from \(2\) through \(10\), then account for four suits.

1. a) There are \(13\) hearts, so \(P(\text{heart})=\frac{13}{52}=\frac{1}{4}\). 2. b) There are \(4\) kings and \(4\) aces. These categories do not overlap, so \(P(\text{king or ace})=\frac{8}{52}=\frac{2}{13}\). 3. c) There are \(9\) number ranks from \(2\) through \(10\), with \(4\) cards of each rank. Thus, \(P(\text{number card})=\frac{36}{52}=\frac{9}{13}\).

a) \(\frac{1}{4}=25\%\) b) \(\frac{2}{13}\) c) \(\frac{9}{13}\)

- Count the total number of letters, including repeats. - Count how often each relevant letter appears. - Use a complement for “not H.” - For “C or M,” add the counts of the two different letters.

1. MATHEMATICS has \(11\) letters. The relevant counts are E: \(1\), vowels: \(4\), H: \(1\), C: \(1\), and M: \(2\). 2. a) \(P(E) = \frac{1}{11}\). 3. b) \(P(\text{vowel}) = \frac{4}{11}\). 4. c) \(P(\text{not H}) = \frac{10}{11}\). 5. d) \(P(C\text{ or }M) = \frac{1 + 2}{11} = \frac{3}{11}\).

a) \(\frac{1}{11}\) b) \(\frac{4}{11}\) c) \(\frac{10}{11}\) d) \(\frac{3}{11}\)

- List the possible recorded colors. - Divide each color count by \(20\). - An equally likely model requires all recorded outcomes to have equal probabilities. - Compare the probabilities within each case.

1. a) \(S = \{\text{red}, \text{blue}\}\). Both probabilities are \(\frac{10}{20} = 0.5\), so the outcomes are equally likely. 2. b) \(S = \{\text{red}, \text{blue}, \text{green}, \text{yellow}\}\). Every color has probability \(\frac{5}{20} = 0.25\), so the outcomes are equally likely. 3. c) \(S = \{\text{red}, \text{blue}, \text{green}\}\). The probabilities are \(0.1\), \(0.4\), and \(0.5\), respectively, so the outcomes are not equally likely.

a) \(S = \{\text{red}, \text{blue}\}\); equally likely b) \(S = \{\text{red}, \text{blue}, \text{green}, \text{yellow}\}\); equally likely c) \(S = \{\text{red}, \text{blue}, \text{green}\}\); not equally likely

- Count all letters, including repeated letters. - Count N and the vowels separately in each word. - Compare the fractions using decimals or a common denominator.

1. PERCENTAGE has \(10\) letters and contains N once, so \(P_A(N) = \frac{1}{10}\). Its vowels are E, E, A, and E, so \(P_A(\text{vowel}) = \frac{4}{10} = \frac{2}{5}\). 2. INTEREST has \(8\) letters and contains N once, so \(P_B(N) = \frac{1}{8}\). Its vowels are I, E, and E, so \(P_B(\text{vowel}) = \frac{3}{8}\). 3. a) Since \(\frac{1}{8} > \frac{1}{10}\), Word B has the greater probability of N. 4. b) Since \(\frac{2}{5} = 0.4\) and \(\frac{3}{8} = 0.375\), Word A has the greater probability of a vowel.

a) Word A: \(\frac{1}{10}\); Word B: \(\frac{1}{8}\). Word B is greater. b) Word A: \(\frac{2}{5}\); Word B: \(\frac{3}{8}\). Word A is greater.

- Find the total number of marbles. - Not green includes red and blue. - Convert each fraction to a denominator of \(100\).

1. The bag contains \(12 + 8 + 5 = 25\) marbles. 2. a) \(P(\text{red}) = \frac{12}{25} = \frac{48}{100} = 48\%\). 3. b) There are \(12 + 8 = 20\) marbles that are not green. Thus, \(P(\text{not green}) = \frac{20}{25} = \frac{80}{100} = 80\%\).

a) \(48\%\) b) \(80\%\)

- For each bag, divide the sweet count by the total count. - Compare the probabilities using fractions, decimals, or percents. - Do not compare only the sweet counts.

1. Bag A contains \(20\) gummy bears, so \(P_A(\text{sweet}) = \frac{16}{20} = \frac{4}{5} = 80\%\). 2. Bag B contains \(28\) gummy bears, so \(P_B(\text{sweet}) = \frac{21}{28} = \frac{3}{4} = 75\%\). 3. Since \(80\% > 75\%\), Bag A gives the greater probability.

Paul should choose Bag A. Its sweet-gummy-bear probability is \(80\%\), compared with \(75\%\) for Bag B.

- Multiply the total by each known probability to find the corresponding count. - Convert \(45\%\) to \(0.45\). - Subtract the black and white counts from \(60\).

1. The number of black balls is \(\frac{1}{3} \cdot 60 = 20\). 2. The number of white balls is \(0.45 \cdot 60 = 27\). 3. The number of gray balls is \(60 - 20 - 27 = 13\).

There are \(13\) gray balls.

- Count the parts satisfying each condition. - Add counts for nonoverlapping categories. - Use a complement for “not a screw.”

1. For a nut or washer, there are \(40+20=60\) favorable parts, so \(P=\frac{60}{120}=\frac{1}{2}\). 2. Silver parts include \(15\) screws and \(10\) bolts, so \(P=\frac{25}{120}=\frac{5}{24}\). 3. There are \(120-50=70\) parts that are not screws, so \(P=\frac{70}{120}=\frac{7}{12}\).

a) \(\frac{1}{2}\) b) \(\frac{5}{24}\) c) \(\frac{7}{12}\)

- Count sections satisfying each condition. - For gold, count sections by color rather than prize label. - Equal-size sections support an equally likely-outcomes model.

1. There are \(10+2+8=20\) prize sections, so \(P(\text{any prize})=\frac{20}{24}=\frac{5}{6}\). 2. There are \(2+4=6\) gold sections, so \(P(\text{gold})=\frac{6}{24}=\frac{1}{4}\). 3. Excluding \(4\) no-prize and \(2\) grand-prize sections leaves \(18\) sections, so \(P=\frac{18}{24}=\frac{3}{4}\). 4. All \(24\) sections are equal in size, so they are modeled as equally likely outcomes.

a) \(\frac{5}{6}\) b) \(\frac{1}{4}\) c) \(\frac{3}{4}\) d) The equal-size sections are equally likely.

- Use the total number of equal sections as the denominator. - The complement contains every outcome outside the shaded event. - Check that the two complementary probabilities add to \(1\).

1. The spinner has \(10\) equal sections, and \(4\) of them are shaded. 2. Therefore, \(P(E)=\frac{4}{10}=\frac{2}{5}\). 3. The complement \(\overline{E}\) occurs when the spinner lands on an unshaded section. 4. There are \(6\) unshaded sections, so \(P(\overline{E})=\frac{6}{10}=\frac{3}{5}\). Equivalently, \(1-\frac{2}{5}=\frac{3}{5}\).

1. \(P(E)=\frac{2}{5}\) 2. \(\overline{E}\): The spinner lands on an unshaded section. 3. \(P(\overline{E})=\frac{3}{5}\)

- Count the favorable sectors and the total sectors on each spinner. - Write a probability fraction for each choice. - Compare the fractions for the same winning color.

1. For Mia, Spinner A has \(6\) red sectors out of \(12\), so \(P(\text{red}) = \frac{6}{12} = \frac{1}{2}\). Spinner B has \(3\) red sectors out of \(8\), so \(P(\text{red}) = \frac{3}{8}\). Since \(\frac{1}{2} > \frac{3}{8}\), Mia should choose Spinner A. 2. For Ben, Spinner A has \(2\) yellow sectors out of \(12\), so \(P(\text{yellow}) = \frac{2}{12} = \frac{1}{6}\). Spinner B has \(2\) yellow sectors out of \(8\), so \(P(\text{yellow}) = \frac{2}{8} = \frac{1}{4}\). Since \(\frac{1}{4} > \frac{1}{6}\), Ben should choose Spinner B.

a) Spinner A, because \(\frac{1}{2} > \frac{3}{8}\) b) Spinner B, because \(\frac{1}{4} > \frac{1}{6}\)

- For each spinner, compare prize sections with all equal sections. - Convert both probabilities to the same form before comparing them. - Check which percentage is larger.

1. For Spinner A, \(P(\text{win})=\frac{4}{10}=\frac{2}{5}=40\%\). 2. For Spinner B, \(P(\text{win})=\frac{3}{8}=37.5\%\). 3. Since \(40\%>37.5\%\), Spinner A gives the greater chance of winning.

a) Spinner A: \(\frac{2}{5}=40\%\); Spinner B: \(\frac{3}{8}=37.5\%\) b) Spinner A

- Identify every color that lies outside the event “yellow.” - Count all non-yellow balls or use the complement rule. - Check that the event and complement probabilities add to \(1\).

1. The complement occurs when the selected ball is outside event \(E\), so it is red or blue. 2. There are \(4+1=5\) non-yellow balls out of \(12\), so \(P(\overline{E})=\frac{5}{12}\). 3. Equivalently, \(P(\overline{E})=1-\frac{7}{12}=\frac{5}{12}\).

a) The selected ball is red or blue. b) \(P(\overline{E})=\frac{5}{12}\)

- Count the sections of each color on the spinner. - For “not red,” count all sections that are not red. - For “green or yellow,” include both colors, then convert the fraction to a percent.

1. The spinner has \(10\) equal sections: \(4\) red, \(3\) blue, \(2\) green, and \(1\) yellow. 2. For part a, \(10-4=6\) sections are not red, so the probability is \(\frac{6}{10}=60\%\). 3. For part b, \(2+1=3\) sections are green or yellow, so the probability is \(\frac{3}{10}=30\%\). 4. For part c, \(3\) sections are blue, so the probability is \(\frac{3}{10}=30\%\).

a) \(60\%\) b) \(30\%\) c) \(30\%\)

- Think of each ball as one possible result of the draw. - Are the numbers of different colors alike in both bags? - For part b, compare the red count with all the balls in each bag.

1. Bag A has \(3\) blue, \(3\) red, and \(3\) yellow balls, so each color has probability \(\frac{3}{9}=\frac{1}{3}\). Its color outcomes are equally likely. 2. Bag B has \(4\) blue, \(2\) red, and \(3\) yellow balls, so its color probabilities are not equal. 3. The red-ball probabilities are \(P_A(\text{red})=\frac{3}{9}=\frac{1}{3}\) and \(P_B(\text{red})=\frac{2}{9}\).

a) Bag A, because it contains the same number of balls of each color. b) Bag A: \(\frac{1}{3}\); Bag B: \(\frac{2}{9}\).

- What would make an outcome fail to belong to event \(E\)? - Check whether an outcome and its complement cover every section without sharing one. - Use the same check for event \(A\) before counting its outcomes.

1. There are \(3\) red sections among \(8\) equal sections, so \(P(E)=\frac{3}{8}\). 2. The complement is blue or yellow, so \(P(\overline{E})=1-\frac{3}{8}=\frac{5}{8}\). 3. There are \(4\) blue sections; the complement of blue is red or yellow. 4. There are \(3+1=4\) red or yellow sections, so \(P(\overline{A})=\frac{4}{8}=\frac{1}{2}\).

a) \(P(E)=\frac{3}{8}\) and \(P(\overline{E})=\frac{5}{8}\). b) \(\overline{A}\): red or yellow; \(P(\overline{A})=\frac{1}{2}\).

- Write the number of red sectors over the total number of sectors for each spinner. - Compare the two probability fractions, not only the favorable counts.

1. Spinner A has \(3\) red sectors out of \(6\), so its winning probability is \(\frac{3}{6} = \frac{1}{2}\). 2. Spinner B has \(3\) red sectors out of \(8\), so its winning probability is \(\frac{3}{8}\). 3. Jonas compared only the number of red sectors, not the total number of equal sectors. Since \(\frac{1}{2} > \frac{3}{8}\), Spinner A gives the greater probability.

a) The total number of equal sectors also matters: \(3\) out of \(6\) is not the same probability as \(3\) out of \(8\). b) Spinner A, because \(\frac{3}{6} = \frac{1}{2} > \frac{3}{8}\).

- Read the numbered balls in each urn before counting favorable outcomes. - Remember that \(1\) is not prime. - Compare probabilities rather than just favorable counts.

1. Urn A contains labels \(1\) through \(5\). Its prime probability is \(\frac{3}{5}=60\%\), and its even probability is \(\frac{2}{5}=40\%\). 2. Urn B contains labels \(1\) through \(8\). Its prime probability is \(\frac{4}{8}=\frac{1}{2}=50\%\), and its even probability is \(\frac{4}{8}=\frac{1}{2}=50\%\). 3. A prime is more likely from Urn A, while an even number is more likely from Urn B.

a) Urn A: \(\frac{3}{5}=60\%\), compared with \(\frac{1}{2}=50\%\) for Urn B b) Urn B: \(\frac{1}{2}=50\%\), compared with \(\frac{2}{5}=40\%\) for Urn A

- Count the total balls in each urn from the diagram. - For each event, count favorable balls in each urn separately. - Compare the resulting probabilities, not the raw favorable counts.

1. Urn 1 contains \(8\) balls, and Urn 2 contains \(10\) balls. 2. For blue, Urn 1 has probability \(\frac{2}{8}=25\%\), while Urn 2 has probability \(\frac{3}{10}=30\%\). Blue is more likely from Urn 2. 3. The odd labels are \(1\) and \(3\). Urn 1 has \(6\) odd-labeled balls, so the probability is \(\frac{6}{8}=75\%\). Urn 2 has \(7\), so the probability is \(\frac{7}{10}=70\%\). An odd label is more likely from Urn 1.

a) Urn 2: \(30\%\), compared with \(25\%\) for Urn 1 b) Urn 1: \(75\%\), compared with \(70\%\) for Urn 2

- Write each probability as colored sections divided by total equal sections. - Put the two fractions in a comparable form. - Check which probability is larger.

1. Spinner A has probability \(\frac{8}{12}=\frac{2}{3}\). 2. Spinner B has probability \(\frac{7}{10}\). 3. Using denominator \(30\), \(\frac{2}{3}=\frac{20}{30}\) and \(\frac{7}{10}=\frac{21}{30}\). Therefore, Spinner B has the greater probability.

Spinner B: \(\frac{7}{10}>\frac{2}{3}\) for Spinner A.

- Count sections, not just different colors. - For part b, check the condition against each section before counting it. - In part c, could a single section be counted more than once?

1. The spinner has \(8\) equally likely sections. 2. There are \(3\) blue sections, so the probability in part a is \(\frac{3}{8}\). 3. Neither blue nor yellow means red or green. There are \(2\) red and \(2\) green sections, so the probability in part b is \(\frac{4}{8}=\frac{1}{2}\). 4. There are \(2\) red and \(1\) yellow sections, so the probability in part c is \(\frac{3}{8}\).

a) \(\frac{3}{8}\) b) \(\frac{1}{2}\) c) \(\frac{3}{8}\)

- Read the color attached to each numbered section from the spinner. - For part a, identify outcomes satisfying both conditions. - For part b, combine the two favorable sets and count any shared outcome once.

1. The spinner has \(8\) equally likely outcomes. 2. For event \(A\), the outcomes that are both greater than \(4\) and red are \(5\) and \(6\), so \(P(A)=\frac{2}{8}=\frac{1}{4}\). 3. The odd numbers are \(1,3,5,7\), and the blue sections are \(1\) and \(2\). Their union is \(1,2,3,5,7\), so \(P(B)=\frac{5}{8}\).

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

- Count the red balls and total balls in each urn. - Write each winning probability as a fraction. - Simplify or use equivalent fractions to compare them.

1. Urn A has \(2\) red balls out of \(5\), so Lara’s winning probability is \(\frac{2}{5}\). 2. Urn B has \(4\) red balls out of \(10\), so Jonas’s winning probability is \(\frac{4}{10} = \frac{2}{5}\). 3. The probabilities are equal. Urn B has twice as many red balls, but it also has twice as many balls altogether.

No. Both players have a winning probability of \(\frac{2}{5}\), because \(\frac{4}{10} = \frac{2}{5}\).

- Use \(20\) as the total number of equally likely marbles. - Divide each favorable count by the total and simplify. - Convert each simplified fraction to a percent.

1. The probability of selecting a blue marble is \(\frac{8}{20}=\frac{2}{5}=40\%\). 2. The probability of selecting a white marble is \(\frac{12}{20}=\frac{3}{5}=60\%\).

a) \(\frac{2}{5}=40\%\) b) \(\frac{3}{5}=60\%\)

- Equal section sizes make the individual sections equally likely. - Compare the blue-section count with all \(12\) sections. - Blue and white are complementary events here.

1. Because all \(12\) sections have the same size, each individual section has the same probability. 2. There are \(3\) blue sections, so \(P(E_1)=\frac{3}{12}=\frac{1}{4}=25\%\). 3. There are \(9\) white sections, so \(P(E_2)=\frac{9}{12}=\frac{3}{4}=75\%\).

a) All \(12\) sections are the same size, so each section is equally likely. b) \(P(E_1)=\frac{1}{4}=25\%\); \(P(E_2)=\frac{3}{4}=75\%\)

- Count red and total balls in each urn from the diagram. - Form a probability separately for each urn. - Compare the fractions or their equivalent percents.

1. Urn A has \(3\) red balls out of \(8\), so \(P(\text{red from A})=\frac{3}{8}=37.5\%\). 2. Urn B has \(2\) red balls out of \(5\), so \(P(\text{red from B})=\frac{2}{5}=40\%\). 3. Since \(40\%>37.5\%\), selecting a red ball is more likely from Urn B.

Urn B has the greater probability. Urn A: \(\frac{3}{8}=37.5\%\) Urn B: \(\frac{2}{5}=40\%\)

- Use \(12\) as the total number of equally likely outcomes. - Count every section whose label satisfies the condition. - Repeated labels represent separate outcomes. - Simplify each fraction.

1. The spinner has \(12\) equally likely sections. 2. The numbers greater than \(5\) are \(10\) and \(20\). The number \(10\) appears on \(3\) sections, and \(20\) appears on \(2\) sections, for \(5\) favorable outcomes. Thus, the probability in part a is \(\frac{5}{12}\). 3. The even numbers are \(2\), \(10\), and \(20\). They appear on \(4 + 3 + 2 = 9\) sections, so the probability in part b is \(\frac{9}{12} = \frac{3}{4}\).

a) \(\frac{5}{12}\) b) \(\frac{3}{4}\)

- Count the total equal sections and the colors named in part a. - For part b, count the blue sections first, then consider the complement. - Convert each fraction to a percent and follow the rounding instruction.

1. The spinner has \(12\) equally likely outcomes. 2. There are \(5\) red and \(4\) yellow sections, so part a has \(5+4=9\) favorable outcomes. The probability is \(\frac{9}{12}=\frac{3}{4}=75\%\). 3. There is \(1\) blue section. Using the complement gives \(1-\frac{1}{12}=\frac{11}{12}\). As a percent, \(\frac{11}{12}\cdot100\%\approx91.67\%\).

a) \(\frac{3}{4}=75\%\) b) \(\frac{11}{12}\approx91.67\%\)

- Use \(8\) as the total number of equally likely outcomes. - Count sections, not just distinct number labels. - Remember that a prime number has exactly two positive factors. - Divide favorable sections by \(8\). - Simplify each probability and convert it to a percent.

1. The odd-numbered sections are labeled \(3, 5, 5, 9\), giving \(4\) favorable outcomes. Thus, the probability in part a is \(\frac{4}{8} = \frac{1}{2} = 50\%\). 2. The prime-numbered sections are labeled \(2, 3, 5, 5\), giving \(4\) favorable outcomes. Thus, the probability in part b is \(\frac{4}{8} = \frac{1}{2} = 50\%\). 3. The sections with numbers greater than \(4\) are labeled \(5, 5, 8, 8, 8, 9\), giving \(6\) favorable outcomes. Thus, the probability in part c is \(\frac{6}{8} = \frac{3}{4} = 75\%\). 4. The sections with numbers divisible by \(3\) are labeled \(3\) and \(9\), giving \(2\) favorable outcomes. Thus, the probability in part d is \(\frac{2}{8} = \frac{1}{4} = 25\%\).

a) \(\frac{1}{2} = 50\%\) b) \(\frac{1}{2} = 50\%\) c) \(\frac{3}{4} = 75\%\) d) \(\frac{1}{4} = 25\%\)

- Compare the marked sections with all equal sections. - Simplify the favorable-over-total fraction. - Convert the simplified fraction to a percent.

1. There are \(4\) favorable sections out of \(10\) equally likely sections. 2. The probability is \(\frac{4}{10}=\frac{2}{5}=40\%\).

The probability of winning is \(\frac{2}{5}\), or \(40\%\).

- Use favorable sections divided by all equal sections. - Simplify the fraction completely.

1. Two of the \(6\) equally likely sections are red. 2. The probability is \(\frac{2}{6}=\frac{1}{3}\).

The probability is \(\frac{1}{3}\).

- Form favorable sections divided by total equal sections. - Simplify the resulting fraction.

1. Six of the \(10\) equally likely sections are yellow. 2. The probability is \(\frac{6}{10}=\frac{3}{5}\).

The probability is \(\frac{3}{5}\).

- Write marked sections divided by all equal sections. - Simplify before converting to decimal and percent forms. - Check that the decimal and percent represent the same probability.

1. The probability of a marked section is \(\frac{6}{16}=\frac{3}{8}\). 2. Converting gives \(\frac{3}{8}=0.375=37.5\%\).

The probability is \(\frac{3}{8}=0.375=37.5\%\).

- Use orange-wrapped chocolates over all chocolates for the fraction. - A uniformly random selection gives the same part-to-whole probability. - Convert the fraction to an equivalent percent.

1. The fraction with orange wrappers is \(\frac{9}{25}\), which is already in simplest form. 2. For a random selection, the probability is the same fraction: \(\frac{9}{25}=\frac{36}{100}=36\%\).

a) \(\frac{9}{25}\) b) \(36\%\)

- Count the red sectors and total sectors on each spinner. - Write a probability fraction for each spinner. - Use a common denominator to compare the fractions.

1. On Spinner a), \(1\) of \(4\) sectors is red, so the winning probability is \(\frac{1}{4}\). 2. On Spinner b), \(3\) of \(8\) sectors are red, so the winning probability is \(\frac{3}{8}\). 3. Rewrite \(\frac{1}{4}\) as \(\frac{2}{8}\). Since \(\frac{3}{8} > \frac{2}{8}\), Spinner b) gives the greater probability.

Spinner b), because \(\frac{3}{8} > \frac{1}{4} = \frac{2}{8}\).

- List the different color outcomes without repeating a color. - Count sections for each color to complete the table. - For part c, use the complement of landing on green.

1. The possible color outcomes are green, orange, and purple, so \(S=\{\text{green},\text{orange},\text{purple}\}\). 2. The spinner has \(10\) equal sections: \(2\) green, \(3\) orange, and \(5\) purple. 3. Therefore, \(P(C=\text{green})=\frac{2}{10}=\frac{1}{5}\), \(P(C=\text{orange})=\frac{3}{10}\), and \(P(C=\text{purple})=\frac{5}{10}=\frac{1}{2}\). 4. Using the complement, \(P(\text{not green})=1-\frac{1}{5}=\frac{4}{5}=80\%\).

a) \(S=\{\text{green},\text{orange},\text{purple}\}\) b) Green: \(\frac{1}{5}\); Orange: \(\frac{3}{10}\); Purple: \(\frac{1}{2}\) c) \(80\%\)

- What does drawing one ball give as the equally likely outcomes? - Compare the counts behind the two colors, not just their names. - Can you change the balance between colors without changing how many balls there are?

1. The bag contains \(5\) red balls and \(1\) blue ball, for \(6\) balls total. 2. The colors are not equally likely because their counts differ. The probability of red is \(P(\text{red})=\frac{5}{6}\). 3. With \(6\) balls and two equally likely colors, there must be \(3\) balls of each color. Replace \(2\) red balls with \(2\) blue balls.

a) The claim is incorrect. \(P(\text{red})=\frac{5}{6}\). b) Use \(3\) red and \(3\) blue balls, replacing \(2\) red balls with \(2\) blue balls.

- Count the yellow balls and total balls in each urn. - Write each winning probability as a fraction. - Use a common denominator to compare the fractions.

1. Urn 1 has \(3\) yellow balls out of \(12\), so its winning probability is \(\frac{3}{12} = \frac{1}{4}\). 2. Urn 2 has \(2\) yellow balls out of \(5\), so its winning probability is \(\frac{2}{5}\). 3. Using a common denominator, \(\frac{1}{4} = \frac{5}{20}\) and \(\frac{2}{5} = \frac{8}{20}\). Since \(\frac{8}{20} > \frac{5}{20}\), Urn 2 gives the greater probability.

Urn 2, because \(\frac{2}{5} > \frac{3}{12} = \frac{1}{4}\).

- Count the blue balls and total balls in each jar. - Compare the winning fractions using a common denominator. - An event is certain only when no other outcome is possible.

1. Jar 1 has \(1\) blue ball out of \(4\), so its winning probability is \(\frac{1}{4} = \frac{2}{8}\). 2. Jar 2 has \(3\) blue balls out of \(8\), so its winning probability is \(\frac{3}{8}\). Since \(\frac{3}{8} > \frac{2}{8}\), Jar 2 gives the greater probability. 3. For drawing blue to be certain from Jar 1, no red balls can remain. All \(3\) red balls must be removed.

a) Jar 2, because \(\frac{3}{8} > \frac{1}{4}\) b) Remove all \(3\) red balls.

- Count the balls of each color. - With equally likely balls, which color count is greatest? - To make the red and green probabilities equal, think about their counts.

1. The urn contains \(3\) red, \(4\) blue, and \(5\) green balls. 2. Green has the greatest probability because it has the greatest number of balls, \(5\), and every ball is equally likely to be drawn. 3. Red and green will have equal probabilities when their counts are equal. Add \(5-3=2\) red balls.

a) Green b) Add \(2\) red balls.

- First find the total number of balls. - For each event, identify the favorable color counts. - Divide each favorable count by the total and write the result as a decimal.

1. The urn contains \(4+7+9=20\) balls. 2. \(P(A)=\frac{4}{20}=0.2\). 3. Event \(B\) has \(7+9=16\) favorable balls, so \(P(B)=\frac{16}{20}=0.8\). 4. Event \(C\) has \(4+7=11\) favorable balls, so \(P(C)=\frac{11}{20}=0.55\).

a) \(P(A)=0.2\) b) \(P(B)=0.8\) c) \(P(C)=0.55\)

- Read each ball's number and color from the urn. - Count the balls satisfying each event before forming a probability. - Distinguish equally likely individual balls from events containing different numbers of balls.

1. The numbered sample space is \(S=\{1,2,3,4,5,6,7,8\}\), with \(8\) equally likely physical balls. 2. \(A=\{2,4,6,8\}\), so \(P(A)=\frac{4}{8}=\frac{1}{2}\). 3. The blue balls are \(1,2,3,4\), so \(P(B)=\frac{4}{8}=\frac{1}{2}\). 4. \(C=\{6,7,8\}\), so \(P(C)=\frac{3}{8}\). 5. Each physical ball has the same chance of being drawn, so the eight individual outcomes are equally likely.

\(P(A)=\frac{1}{2}\), \(P(B)=\frac{1}{2}\), and \(P(C)=\frac{3}{8}\). The individual-ball outcomes are equally likely because each ball has probability \(\frac{1}{8}\).

- Count all balls with a matching label, including repeats. - Use the total number of balls for each event's denominator. - Simplify each fraction fully.

1. There are \(12\) equally likely balls. 2. Three balls are labeled \(2\), so \(P(A)=\frac{3}{12}=\frac{1}{4}\). 3. Eight balls have labels greater than \(1\), so \(P(B)=\frac{8}{12}=\frac{2}{3}\). 4. Nine balls have odd labels, so \(P(C)=\frac{9}{12}=\frac{3}{4}\).

a) \(P(A)=\frac{1}{4}\) b) \(P(B)=\frac{2}{3}\) c) \(P(C)=\frac{3}{4}\)

- Compare the sizes of the three sectors rather than just counting their labels. - Think about what fraction of the whole spinner is covered by the favorable sectors. - Check that the probability you obtain matches the combined area of those sectors.

1. The three labeled outcomes are not equally likely because the sections have different sizes. 2. Sections \(A\) and \(B\) do not overlap, so \(P(A\text{ or }B)=\frac{1}{4}+\frac{1}{4}=\frac{1}{2}\).

\(P(A\text{ or }B)=\frac{1}{2}\), not \(\frac{2}{3}\).

- Because there are \(100\) equally likely tickets, each favorable count is also a percent. - Compare \(2\%\), \(49\%\), and \(98\%\) with \(0\%\), \(50\%\), and \(100\%\).

1. Event A has probability \(\frac{2}{100}=0.02\), which is near \(0\). 2. Event B has probability \(\frac{49}{100}=0.49\), which is near \(\frac{1}{2}\). 3. Event C has probability \(\frac{98}{100}=0.98\), which is near \(1\).

Near \(0\): Event A Near \(\frac{1}{2}\): Event B Near \(1\): Event C

- Which part of the probability scale represents an even chance? - Could the probability be at either endpoint of the interval described? - Reject any option that includes probabilities the event cannot have.

1. “More likely to happen than not happen” means the probability is greater than \(0.5\). 2. “Not certain” means the probability is less than \(1\). 3. Choice b) gives exactly \(0.5<P(E)<1\). Choice d) allows values that do not fit the event, so it is too broad.

b) \(0.5<P(E)<1\)

- Are the regions the same size? - What determines how often a fair spinner lands in a region?

1. The two regions have unequal areas, so they are not equally likely. 2. Region R takes one quarter of the full spinner, so \(P(R)=\frac{1}{4}\). 3. Counting unequal regions as if equally likely is not valid.

No. \(P(R)=\frac{1}{4}\), not \(\frac{1}{2}\), because the regions have unequal areas.

- What is the probability of the event you want to avoid? - Together, the event and its opposite cover all outcomes.

1. The probability of a winning slip is \(\frac{5}{12}\). 2. The event of not winning is its complement, so the probability is \(1-\frac{5}{12}=\frac{7}{12}\).

\(1-\frac{5}{12}=\frac{7}{12}\).

- Count all the balls, not just the distinct letters. - How many balls have the requested label?

1. The picture shows \(8\) balls in all. 2. Exactly \(3\) balls are labeled H. 3. The probability is \(\frac{3}{8}\).

\(\frac{3}{8}\).

- Equal color probabilities require equal numbers of balls of each color. - Use the largest existing color count as the target. - Divide one color count by the new total.

1. a) There are \(12 + 8 + 4 = 24\) balls. The color probabilities are \(P(\text{red}) = \frac{12}{24} = \frac{1}{2}\), \(P(\text{blue}) = \frac{8}{24} = \frac{1}{3}\), and \(P(\text{green}) = \frac{4}{24} = \frac{1}{6}\). They are not equal. 2. b) To add as few balls as possible, increase each smaller color count to \(12\). Add \(12 - 8 = 4\) blue balls and \(12 - 4 = 8\) green balls. Add no red balls. 3. c) The new bag has \(12\) balls of each color and \(36\) balls total. Therefore, the probability of each color is \(\frac{12}{36} = \frac{1}{3}\).

a) The probabilities are \(\frac{1}{2}\), \(\frac{1}{3}\), and \(\frac{1}{6}\), so they are not equal. b) Add \(0\) red balls, \(4\) blue balls, and \(8\) green balls. c) Each color has probability \(\frac{1}{3}\).

- Compare the shapes and areas of the faces. - Consider what symmetry a fair number cube has that this solid does not. - Decide whether the number of faces alone is enough to establish equal likelihood.

1. Each square face has area \(3 \cdot 3 = 9\,\text{cm}^2\). 2. Each rectangular face has area \(3 \cdot 6 = 18\,\text{cm}^2\). 3. The solid does not have the symmetry of a cube. Its faces differ in shape and area, and its center of mass has different positions relative to the possible landing faces. 4. Therefore, having six faces does not imply \(P(1) = \cdots = P(6) = \frac{1}{6}\). Face area alone is also not enough to calculate the exact landing probabilities; experimental data or a physical model would be needed.

The claim is not justified. The rectangular prism lacks the symmetry of a fair number cube, so six faces do not automatically give six equally likely outcomes. Its landing probabilities would need to be estimated experimentally or determined from an appropriate physical model.

- List the numbers from \(1\) through \(25\) that satisfy each condition. - Remember that \(1\) is not prime. - Divide each favorable count by \(25\).

1. There are \(25\) equally likely outcomes. 2. The primes from \(1\) through \(25\) are \(2, 3, 5, 7, 11, 13, 17, 19, 23\). Thus, \(P(A) = \frac{9}{25} = 0.36\). 3. The multiples of \(6\) are \(6, 12, 18, 24\). Thus, \(P(B) = \frac{4}{25} = 0.16\). 4. The perfect squares are \(1, 4, 9, 16, 25\). Thus, \(P(C) = \frac{5}{25} = 0.2\). 5. Since \(0.16 < 0.2 < 0.36\), the order is \(B\), \(C\), \(A\).

\(P(A) = \frac{9}{25} = 36\%\), \(P(B) = \frac{4}{25} = 16\%\), and \(P(C) = \frac{1}{5} = 20\%\) Least to most likely: \(B, C, A\)

- Recall the definition of a prime number and that \(1\) is not prime. - List the positive divisors of \(20\). - Solve the inequality before counting the spinner outcomes. - Count the outcomes satisfying each condition.

1. There are \(20\) equally likely outcomes. 2. a) The primes are \(2, 3, 5, 7, 11, 13, 17, 19\). Thus, \(P = \frac{8}{20} = \frac{2}{5}\). 3. b) The positive factors of \(20\) are \(1, 2, 4, 5, 10, 20\). Thus, \(P = \frac{6}{20} = \frac{3}{10}\). 4. c) Solving gives \(3x < 16\), or \(x < \frac{16}{3}\). The possible spinner results are \(1, 2, 3, 4, 5\), so \(P = \frac{5}{20} = \frac{1}{4}\).

a) \(\frac{2}{5} = 40\%\) b) \(\frac{3}{10} = 30\%\) c) \(\frac{1}{4} = 25\%\)

- A spinner probability is the section angle divided by \(360^\circ\). - Compare the four probabilities. - Divide a full circle equally among four outcomes.

1. a) A full circle measures \(360^\circ\). Therefore, \(P(A) = \frac{120}{360} = \frac{1}{3}\), \(P(B) = \frac{1}{3}\), \(P(C) = \frac{60}{360} = \frac{1}{6}\), and \(P(D) = \frac{1}{6}\). 2. b) The probabilities are not all equal, so the four section outcomes are not equally likely. 3. c) Four equally likely sections must each have probability \(\frac{1}{4}\). Each central angle must be \(\frac{360^\circ}{4} = 90^\circ\).

a) \(P(A) = \frac{1}{3}\), \(P(B) = \frac{1}{3}\), \(P(C) = \frac{1}{6}\), and \(P(D) = \frac{1}{6}\) b) The outcomes are not equally likely because the sections have different sizes. c) Each section should have a central angle of \(90^\circ\).

- For a uniformly random time, use the fraction of the hour occupied by the program type. - Use a complement for no commercial. - Percentage-point change is the difference between the two percentages.

1. a) Music plays for \(27+15=42\) minutes, so \(P(\text{music})=\frac{42}{60}=70\%\). 2. b) Noncommercial programming lasts \(60-7=53\) minutes, so \(P(\text{no commercial})=\frac{53}{60}\). 3. c) Initially, \(P(\text{rock})=\frac{15}{60}=25\%\). After the change, rock lasts \(18\) minutes, so \(P(\text{rock})=\frac{18}{60}=30\%\). The increase is \(30\%-25\%=5\) percentage points.

a) \(70\%\) b) \(\frac{53}{60}\) c) \(5\) percentage points

- Use \(50\) as the denominator. - For not a dog, count all other animals. - Add the rabbit and bird counts before comparing.

1. a) \(P(\text{bird}) = \frac{4}{50} = \frac{2}{25} = 8\%\). 2. b) There are \(50 - 22 = 28\) animals that are not dogs. Thus, \(P(\text{not a dog}) = \frac{28}{50} = \frac{14}{25} = 56\%\). 3. c) \(P(\text{cat}) = \frac{18}{50} = 36\%\). The rabbit-or-bird count is \(6 + 4 = 10\), so \(P(\text{rabbit or bird}) = \frac{10}{50} = 20\%\). Selecting a cat is more likely.

a) \(\frac{2}{25} = 8\%\) b) \(56\%\) c) A cat is more likely: \(36\% > 20\%\).

- What is one outcome of this draw: a digit or a whole numbered ball? - How can you organize the numbers that fit each rule before writing probabilities? - For the last condition, test possible tens digits and see which ones digits work.

1. a) The multiples of \(4\) are \(4, 8, 12, 16, 20, 24, 28\). Thus, \(P=\frac{7}{30}\). 2. b) The numbers containing the digit \(5\) are \(5, 15, 25\). Thus, \(P=\frac{3}{30}=\frac{1}{10}\). 3. c) The two-digit numbers no greater than \(30\) whose digits sum to \(5\) are \(14\) and \(23\). Thus, \(P=\frac{2}{30}=\frac{1}{15}\).

a) \(\frac{7}{30}\) b) \(\frac{1}{10}\) c) \(\frac{1}{15}\)

- How many balls are in the bag altogether? - For each event, identify exactly which balls satisfy the condition. - For an “or” event, check whether any outcomes satisfy both conditions so you do not count them twice. - For a “neither ... nor ...” event, it may help to count the opposite event first.

1. There are \(40\) equally likely outcomes. 2. a) Each color has one ball labeled \(5\), so there are \(4\) favorable outcomes. Therefore, \(P = \frac{4}{40} = \frac{1}{10} = 0.1\). 3. b) There are \(10\) yellow balls. Therefore, \(P = \frac{10}{40} = \frac{1}{4} = 0.25\). 4. c) Exactly one ball is the red \(7\). Therefore, \(P = \frac{1}{40} = 0.025\). 5. d) Of the \(10\) blue balls, one is labeled \(10\), so \(9\) are favorable. Therefore, \(P = \frac{9}{40} = 0.225\). 6. e) There are \(10\) green balls and \(3\) additional non-green balls labeled \(2\). Therefore, \(P = \frac{10 + 3}{40} = \frac{13}{40} = 0.325\). 7. f) There are \(10\) red balls and \(4\) balls labeled \(1\), with the red \(1\) counted in both groups. Thus, \(10 + 4 - 1 = 13\) balls are red or labeled \(1\). The remaining \(40 - 13 = 27\) balls satisfy neither condition, so \(P = \frac{27}{40} = 0.675\).

a) \(P = \frac{1}{10} = 0.1\) b) \(P = \frac{1}{4} = 0.25\) c) \(P = \frac{1}{40} = 0.025\) d) \(P = \frac{9}{40} = 0.225\) e) \(P = \frac{13}{40} = 0.325\) f) \(P = \frac{27}{40} = 0.675\)

- Organize the counts by supplier and condition in a small table. - What does “not from Supplier A” tell you about the supplier? - In part c, avoid counting defective components from Supplier B twice.

1. There are \(50\) components total. 2. a) There are \(5 + 2 = 7\) defective components, so \(P(\text{defective}) = \frac{7}{50} = 0.14\). 3. b) Supplier A provided \(30 - 5 = 25\) components that are not defective. Therefore, \(P(\text{A and not defective}) = \frac{25}{50} = \frac{1}{2} = 0.5\). 4. c) All \(20\) components from Supplier B satisfy the event. The \(5\) defective components from Supplier A also satisfy it. Therefore, there are \(20 + 5 = 25\) favorable outcomes, so \(P(\text{B or defective}) = \frac{25}{50} = \frac{1}{2} = 0.5\). 5. d) A component that did not come from Supplier A came from Supplier B. Supplier B provided \(20 - 2 = 18\) components that are not defective. Therefore, \(P = \frac{18}{50} = \frac{9}{25} = 0.36\).

a) \(P = \frac{7}{50} = 0.14\) b) \(P = \frac{1}{2} = 0.5\) c) \(P = \frac{1}{2} = 0.5\) d) \(P = \frac{9}{25} = 0.36\)

- Multiply the total number of letters by each probability. - Check the vowel and M conditions separately. - Both conditions must be true.

1. a) The required number of vowels is \(12 \cdot \frac{1}{3} = 4\). The required number of Ms is \(12 \cdot \frac{1}{6} = 2\). 2. b) MATHEMATICAL has \(12\) letters. Its vowels are A, E, A, I, and A, for \(5\) vowels. It contains \(2\) Ms. 3. The M condition is satisfied because \(\frac{2}{12} = \frac{1}{6}\), but the vowel condition is not because \(\frac{5}{12} \neq \frac{1}{3}\). Therefore, the word does not satisfy both conditions.

a) \(4\) vowels and \(2\) Ms b) No. MATHEMATICAL has \(5\) vowels and \(2\) Ms, so only the M condition is satisfied.

- Write probability as red count divided by total count. - Subtract the red count from the total. - Represent the white count with a variable and the yellow count as twice that variable.

1. a) Let \(T\) be the total. Since \(\frac{12}{T} = 0.25\), \(T = \frac{12}{0.25} = 48\). 2. b) There are \(48 - 12 = 36\) yellow and white gummy bears altogether. 3. c) Let \(x\) be the number of white gummy bears. Then there are \(2x\) yellow gummy bears. The equation \(x + 2x = 36\) gives \(x = 12\), so there are \(24\) yellow gummy bears.

a) \(48\) gummy bears b) \(36\) yellow and white gummy bears altogether c) \(24\) yellow gummy bears

- First find the red probability in Bag A. - Use an equation to find the total in Bag B. - When the bags are combined, update the total but not the red count.

1. The probability of red from Bag A is \(\frac{15}{20}=\frac{3}{4}\). 2. a) Let \(T\) be the total number of counters in Bag B. Then \(\frac{12}{T}=\frac{3}{4}\), so \(T=16\). Therefore, Bag B has \(16-12=4\) yellow counters. 3. b) The combined container has \(20+16=36\) counters, including \(15\) red counters. Thus, \(P(\text{red})=\frac{15}{36}=\frac{5}{12}\).

a) \(4\) yellow counters b) \(\frac{5}{12}\)

- Find the losing-ticket probability, then set losing tickets divided by total tickets equal to that probability. - Multiply the total number of tickets by the grand-prize probability.

1. The losing-ticket probability is \(1 - \frac{1}{50} - \frac{1}{5} = 1 - \frac{11}{50} = \frac{39}{50}\). 2. a) Let \(T\) be the total number of tickets. Since \(\frac{390}{T} = \frac{39}{50}\), \(T = 500\). 3. b) The number of grand-prize tickets is \(500 \cdot \frac{1}{50} = 10\).

a) \(500\) tickets b) \(10\) grand-prize tickets

- How can you account for all three colors in the starting container? - Which results do not count as blue? - For part c, compare the container before and after some red counters are removed. - What must the new fraction of blue counters equal?

1. There are \(40-15-10=15\) green counters. 2. a) \(P(\text{green})=\frac{15}{40}=\frac{3}{8}\). 3. b) There are \(30\) counters that are not blue, so \(P(\text{not blue})=\frac{30}{40}=\frac{3}{4}\). 4. c) Let \(x\) be the number of red counters removed. Then \(\frac{10}{40-x}=\frac{1}{3}\). Solving gives \(30=40-x\), so \(x=10\).

a) \(\frac{3}{8}\) b) \(\frac{3}{4}\) c) \(10\) red counters

- Which numbers from \(1\) through \(10\) are perfect squares? - Which numbers from \(1\) through \(10\) are prime? - After identifying the winning outcomes, which outcomes remain? - For a fair spinner, how is probability calculated from the number of favorable sections?

1. The sample space is \(S = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}\). 2. a) The perfect squares are \(1, 4, 9\), and the prime numbers are \(2, 3, 5, 7\). Therefore, \(G = \{1, 2, 3, 4, 5, 7, 9\}\). 3. b) The outcomes outside \(G\) are \(6, 8, 10\). Thus, \(\overline{G}\) is the event that the spinner lands on \(6\), \(8\), or \(10\). 4. c) There are \(7\) winning outcomes and \(3\) outcomes in the complement. Therefore, \(P(G) = \frac{7}{10} = 0.7\) and \(P(\overline{G}) = \frac{3}{10} = 0.3\). 5. An event and its complement include every outcome without overlap, so \(P(G) + P(\overline{G}) = 1\).

a) \(S = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}\); \(G = \{1, 2, 3, 4, 5, 7, 9\}\) b) \(\overline{G}\): The spinner lands on \(6\), \(8\), or \(10\). c) \(P(G) = 0.7\) and \(P(\overline{G}) = 0.3\). Their sum is \(1\).

- Divide the \(12\) sections equally among the three labels for part a. - Probability equals the number of matching sections divided by \(12\). - For part c, combine the sections labeled \(1\) and \(3\).

1. a) Each of three equally likely labels must appear on \(\frac{1}{3} \cdot 12 = 4\) sections. 2. b) \(P(1) = \frac{2}{12} = \frac{1}{6}\), \(P(2) = \frac{4}{12} = \frac{1}{3}\), and \(P(3) = \frac{6}{12} = \frac{1}{2}\). The probabilities are different, so the outcomes are not equally likely. 3. c) Even occurs on the \(4\) sections labeled \(2\), while odd occurs on the \(2 + 6 = 8\) sections labeled \(1\) or \(3\). Thus, \(P(\text{even}) = \frac{4}{12} = \frac{1}{3}\) and \(P(\text{odd}) = \frac{8}{12} = \frac{2}{3}\). The claim is false.

a) \(4\) sections for each label b) \(P(1) = \frac{1}{6}\), \(P(2) = \frac{1}{3}\), and \(P(3) = \frac{1}{2}\); not equally likely c) The claim is false because \(P(\text{even}) = \frac{1}{3}\) and \(P(\text{odd}) = \frac{2}{3}\).

- List the outcomes from \(1\) through \(12\) that satisfy each condition. - Do not count \(12\) twice in part b. - Find how many favorable outcomes correspond to probability \(\frac{1}{3}\).

1. a) The prime results are \(2, 3, 5, 7, 11\), so \(P = \frac{5}{12}\). 2. b) The results divisible by \(3\) are \(3, 6, 9, 12\), and those divisible by \(4\) are \(4, 8, 12\). Their union is \(\{3, 4, 6, 8, 9, 12\}\), so \(P = \frac{6}{12} = \frac{1}{2}\). 3. c) A probability of \(\frac{1}{3}\) requires \(12 \cdot \frac{1}{3} = 4\) winning outcomes. One possible rule is “win if the result is greater than \(8\),” which gives \(9, 10, 11, 12\).

a) \(\frac{5}{12}\) b) \(\frac{1}{2}\) c) One possible rule is “the result is greater than \(8\).”

- Assign every number from \(1\) through \(20\) to exactly one class. - Remember that \(1\) is neither prime nor composite. - A composite number has more than two positive factors. - Check that the three probabilities add to \(1\).

1. Class \(A\) contains the primes \(2, 3, 5, 7, 11, 13, 17, 19\), for \(8\) outcomes. Thus, \(P(A) = \frac{8}{20} = 0.4\). 2. Class \(B\) contains \(4, 8, 12, 16, 20\), for \(5\) outcomes. Thus, \(P(B) = \frac{5}{20} = 0.25\). 3. Class \(C\) contains the remaining numbers \(1, 6, 9, 10, 14, 15, 18\), for \(7\) outcomes. Thus, \(P(C) = \frac{7}{20} = 0.35\). 4. The probabilities add to \(0.4 + 0.25 + 0.35 = 1\).

\(P(A) = 0.4\) \(P(B) = 0.25\) \(P(C) = 0.35\)

- List each distinct letter only once. - Classify vowels first. - For the blue tiles, compare the remaining consonants with SCIENCE. - Assign every distinct letter to exactly one color.

1. The distinct letters in MATHEMATICS are \(\{M, A, T, H, E, I, C, S\}\), giving \(8\) tiles. 2. The red vowels are \(A, E, I\), giving \(3\) red tiles. Thus, \(P(\text{red}) = \frac{3}{8}\). 3. The consonants shared with SCIENCE are \(C\) and \(S\), giving \(2\) blue tiles. Thus, \(P(\text{blue}) = \frac{2}{8} = \frac{1}{4}\). 4. The remaining letters are \(M, T, H\), giving \(3\) green tiles. Thus, \(P(\text{green}) = \frac{3}{8}\).

\(P(\text{red}) = \frac{3}{8}\) \(P(\text{blue}) = \frac{1}{4}\) \(P(\text{green}) = \frac{3}{8}\)

- Is the outcome produced by a chance mechanism or by deliberate choices? - An outcome can be uncertain without being generated randomly. - What feature of the name drawing makes probabilities meaningful?

1. A random chance experiment uses a chance mechanism to select among known possible outcomes. 2. In an election, students make intentional choices. The result may be uncertain beforehand, but the voting process is not itself a random mechanism. 3. Drawing one name from a well-mixed container is a random experiment because the selection is made by a chance process and each name can be assigned a probability. 4. Therefore, the election is not a random chance experiment in this sense, while the name drawing is.

The election is not a random chance experiment because its outcome is produced by students’ choices rather than a chance mechanism. Drawing a name from a well-mixed container is a random experiment because the selection is made randomly from known possible outcomes.

- List the numbers satisfying each condition. - Include numbers such as \(20\) and \(22\) in event \(B\). - Compare the favorable counts because both events have the same total of \(25\) outcomes.

1. Event \(A\) contains \(4, 8, 12, 16, 20, 24\), for \(6\) outcomes. Thus, \(P(A) = \frac{6}{25} = 24\%\). 2. Event \(B\) contains \(2, 12, 20, 21, 22, 23, 24, 25\), for \(8\) outcomes. Thus, \(P(B) = \frac{8}{25} = 32\%\). 3. Since \(32\% > 24\%\), event \(B\) is more likely.

Event \(B\) is more likely. \(P(A) = 24\%\), and \(P(B) = 32\%\).

- What does “fair” mean in terms of the two winning probabilities? - Which numbers from \(1\) through \(8\) are prime? - For part b, identify the outcomes that satisfy both winning conditions.

1. Tim wins on the prime numbers \(2, 3, 5, 7\), so \(P(\text{Tim wins}) = \frac{4}{8} = \frac{1}{2} = 50\%\). 2. Sarah wins on \(5, 6, 7, 8\), so \(P(\text{Sarah wins}) = \frac{4}{8} = \frac{1}{2} = 50\%\). 3. The two players have equal winning probabilities, so the game is fair by the stated definition. 4. Both players win when the outcome is prime and greater than \(4\). The common outcomes are \(5\) and \(7\), so \(P(\text{both win}) = \frac{2}{8} = \frac{1}{4} = 25\%\).

a) Yes. The game is fair because each player has a winning probability of \(50\%\). b) \(P(\text{both win}) = \frac{1}{4} = 25\%\)

- Compare winning tickets divided by total tickets for each class. - In part b, the number of winning tickets stays \(40\). - Write an equation for the new total that gives the same probability as Class A.

1. Class A has winning probability \(\frac{30}{120}=\frac{1}{4}\). 2. Class B has winning probability \(\frac{40}{150}=\frac{4}{15}\). Since \(\frac{4}{15}>\frac{1}{4}\), Class B has the greater probability. 3. To match Class A, Class B must have \(\frac{40}{x}=\frac{1}{4}\), where \(x\) is its new total number of tickets. Thus \(x=160\). 4. Class B needs \(160-150=10\) additional losing tickets.

a) Class B: \(\frac{4}{15}\), compared with \(\frac{1}{4}\) for Class A. b) \(10\) losing tickets.

- Add the favorable and unfavorable parts to find the total number of parts. - Convert the decimal probability to a fraction. - Read “odds against” carefully. - Generalize the favorable-part fraction.

1. Odds of \(3\) to \(5\) mean \(3\) favorable parts and \(5\) unfavorable parts. Thus, \(P(\text{win})=\frac{3}{3+5}=\frac{3}{8}=0.375\). 2. Since \(0.375=\frac{3}{8}\), the favorable-to-unfavorable ratio is \(3:(8-3)=3:5\). 3. Odds against winning of \(9\) to \(1\) mean \(9\) losses for every \(1\) win. Thus, \(P(\text{win})=\frac{1}{9+1}=0.1\). 4. In general, \(P(\text{win})=\frac{x}{x+y}\).

a) \(\frac{3}{8}=0.375\) b) \(3\) to \(5\) c) \(\frac{1}{10}=0.1\) d) \(P(\text{win})=\frac{x}{x+y}\)

- How many faces meet the game's winning rule? - For odds, what do you compare? For probability, what do you compare? - In the changed game, how many faces must lose?

1. The prime numbers are \(2,3,5,7\), so \(4\) faces win and \(4\) lose. The odds in favor are \(4\) to \(4\), or \(1\) to \(1\). 2. Probability compares winning faces with all \(8\) faces, giving \(\frac{4}{8}=\frac{1}{2}\). Odds compare wins with losses instead. 3. If \(3\) faces win, \(5\) lose. The new odds in favor are \(3\) to \(5\). For example, \(1,2,3\) could be winning numbers.

a) \(1\) to \(1\) b) \(\frac{1}{2}\); probability compares wins to all outcomes, while odds compare wins to losses. c) \(3\) to \(5\); for example, numbers \(1,2,3\) win.

- Compare the number of blue sections with the total number of equal sections. - Simplify the resulting fraction for the one-spin probability. - For the expected count, apply the one-spin probability to all \(150\) spins.

1. There are \(4\) blue sections out of \(12\) equal sections. 2. Therefore, the probability of a grand prize is \(\frac{4}{12}=\frac{1}{3}\). 3. In \(150\) spins, the expected number of grand prizes is \(150\cdot\frac{1}{3}=50\).

a) \(\frac{1}{3}\) b) About \(50\) grand prizes

- Compare the number of blue sections with all equal sections. - Use the complement relationship for part b. - Apply the one-spin probability to \(120\) spins for part c.

1. There are \(3\) blue sections out of \(8\) equal sections, so \(P(\text{blue})=\frac{3}{8}=37.5\%\). 2. The probability of not landing on blue is the complement: \(1-\frac{3}{8}=\frac{5}{8}=62.5\%\). 3. In \(120\) spins, the expected number of blue results is \(120\cdot\frac{3}{8}=45\).

a) \(\frac{3}{8}\), or \(37.5\%\) b) \(\frac{5}{8}\), or \(62.5\%\) c) About \(45\) times

- Read the eight number-and-color outcomes directly from the spinner. - For “or,” include outcomes satisfying either condition and count any overlap once. - For “and,” keep only outcomes satisfying both conditions.

1. The spinner has \(8\) equally likely outcomes. 2. For \(E_1\), the numbers divisible by \(5\) are \(5,15,20,25\), giving \(P(E_1)=\frac{4}{8}=\frac{1}{2}\). 3. For \(E_2\), the prime numbers are \(2,3,5\), and the yellow sections contain \(12\) and \(15\). These are \(5\) distinct outcomes, so \(P(E_2)=\frac{5}{8}\). 4. For \(E_3\), the odd numbers are \(3,5,15,25\). Excluding the blue section labeled \(5\) leaves \(3,15,25\), so \(P(E_3)=\frac{3}{8}\).

a) \(P(E_1)=\frac{1}{2}=50\%\) b) \(P(E_2)=\frac{5}{8}=62.5\%\) c) \(P(E_3)=\frac{3}{8}=37.5\%\)

- Check both the number and the color of each section. - Could a section satisfy more than one condition in part a? - How is “and” different from “or” when choosing sections?

1. The spinner has \(12\) equally likely outcomes. 2. For event \(A\), the red sections contain \(1,5,9,13\), and the prime numbers are \(2,3,5,7,11,13\). Their union is \(1,2,3,5,7,9,11,13\), giving \(8\) favorable outcomes. Thus, \(P(A)=\frac{8}{12}=\frac{2}{3}\). 3. For event \(B\), the yellow sections contain \(3,8,12,15\). The even yellow numbers are \(8\) and \(12\), so \(P(B)=\frac{2}{12}=\frac{1}{6}\).

a) \(P(A)=\frac{2}{3}\) b) \(P(B)=\frac{1}{6}\)

- Read the number and color on each equal section from the spinner. - For part a, keep only outcomes satisfying both conditions. - For part b, combine the two sets without counting an overlap twice.

1. The spinner has \(8\) equally likely outcomes, so each section has probability \(\frac{1}{8}\). 2. For part a, the outcomes that are both odd and red are \(1\) and \(5\). Therefore, the probability is \(\frac{2}{8}=\frac{1}{4}\). 3. For part b, the numbers greater than \(4\) are \(5,6,7,8\), and the blue sections contain \(2\) and \(6\). Their union is \(2,5,6,7,8\), so the probability is \(\frac{5}{8}\).

a) \(\frac{1}{4}\), or \(25\%\) b) \(\frac{5}{8}\), or \(62.5\%\)

- Begin with the total number of balls. - For part a, combine the red balls and even-numbered balls, counting any overlap only once. - For part b, keep only balls that satisfy both conditions. - Convert each simplified fraction to a percent.

1. There are \(8\) equally likely balls. 2. For event \(A\), the red balls are red-\(1\), red-\(2\), and red-\(3\). The even-numbered balls are red-\(2\), blue-\(2\), and green-\(2\). Combining them without counting red-\(2\) twice gives \(5\) favorable balls, so \(P(A)=\frac{5}{8}=62.5\%\). 3. For event \(B\), the non-green balls with numbers greater than \(1\) are red-\(2\), red-\(3\), blue-\(2\), and blue-\(3\). Thus, \(P(B)=\frac{4}{8}=\frac{1}{2}=50\%\).

a) \(P(A)=\frac{5}{8}=62.5\%\) b) \(P(B)=\frac{1}{2}=50\%\)

- What two counts are needed for the chance of drawing blue? - Think about what is added to the urn in part b. - What fraction would represent a one-in-four chance after the change?

1. The urn contains \(6\) blue balls and \(9\) gray balls, for a total of \(15\) balls. 2. Therefore, \(P(\text{blue})=\frac{6}{15}=\frac{2}{5}=40\%\). 3. The blue count stays at \(6\). If \(x\) gray balls are added, the new total is \(15+x\), so \(\frac{6}{15+x}=\frac{1}{4}\). 4. Cross-multiplying gives \(24=15+x\), so \(x=9\).

a) \(\frac{2}{5}=40\%\) b) \(9\) gray balls

- Read the number and color of each section from the spinner. - For “and,” keep only outcomes satisfying both conditions. - For “or,” combine the favorable sets and count overlaps once.

1. The green sections are numbered \(3,7,11,15\), and all four are odd, so the probability in part a is \(\frac{4}{16}=\frac{1}{4}=25\%\). 2. The numbers greater than \(10\) are \(11,12,13,14,15,16\), so the probability in part b is \(\frac{6}{16}=\frac{3}{8}=37.5\%\). 3. The red sections are \(1,5,9,13\), and the even-numbered sections are \(2,4,6,8,10,12,14,16\). These do not overlap, so there are \(12\) favorable outcomes. Thus, the probability in part c is \(\frac{12}{16}=\frac{3}{4}=75\%\).

a) \(\frac{1}{4}=25\%\) b) \(\frac{3}{8}=37.5\%\) c) \(\frac{3}{4}=75\%\)

- Read each ball's number and color directly from the urn. - For “and,” keep only outcomes satisfying both conditions. - For “or,” combine the favorable outcomes without double-counting.

1. There are \(12\) equally likely balls. 2. For \(E_1\), the even numbers are \(2,4,6,8,10,12\), so \(P(E_1)=\frac{6}{12}=\frac{1}{2}\). 3. The blue balls are \(2,6,10\). Those greater than \(5\) are \(6\) and \(10\), so \(P(E_2)=\frac{2}{12}=\frac{1}{6}\). 4. The numbers divisible by \(3\) are \(3,6,9,12\), and the red balls are \(1,5,9\). Their union has \(6\) outcomes, so \(P(E_3)=\frac{6}{12}=\frac{1}{2}\).

a) \(P(E_1)=\frac{1}{2}\) b) \(P(E_2)=\frac{1}{6}\) c) \(P(E_3)=\frac{1}{2}\)

- Read all \(10\) color-number outcomes from the urn. - For “or,” combine favorable sets without double-counting. - For “and,” keep only balls satisfying both conditions.

1. There are \(10\) equally likely balls. 2. For part a, the three red balls are favorable, along with the blue-\(2\) and green-\(2\) balls. That gives \(5\) favorable balls, so the probability is \(\frac{5}{10}=\frac{1}{2}=50\%\). 3. For part b, blue-\(1\) and blue-\(3\) are favorable, so the probability is \(\frac{2}{10}=\frac{1}{5}=20\%\). 4. For part c, the favorable balls are blue-\(2\), blue-\(3\), blue-\(4\), green-\(2\), and green-\(3\). Thus, the probability is \(\frac{5}{10}=\frac{1}{2}=50\%\).

a) \(\frac{1}{2}=50\%\) b) \(\frac{1}{5}=20\%\) c) \(\frac{1}{2}=50\%\)

- Read the color of each numbered section from the spinner. - For “or,” combine favorable outcomes without double-counting. - In part d, first translate “not odd” into a number property.

1. The spinner has \(10\) equally likely outcomes. 2. The even numbers are \(2,4,6,8,10\), so part a is \(\frac{5}{10}=\frac{1}{2}=50\%\). 3. The green or yellow sections are \(1,2,5,7,8,10\), so part b is \(\frac{6}{10}=\frac{3}{5}=60\%\). 4. The numbers less than \(6\) are \(1,2,3,4,5\), and blue adds \(9\) beyond those. Thus, part c is \(\frac{6}{10}=\frac{3}{5}=60\%\). 5. “Not odd” means even. Among the even-numbered sections, \(4,6,10\) are not green, so part d is \(\frac{3}{10}=30\%\).

a) \(\frac{1}{2}=50\%\) b) \(\frac{3}{5}=60\%\) c) \(\frac{3}{5}=60\%\) d) \(\frac{3}{10}=30\%\)

- Identify the prime-numbered sections and the blue sections on the spinner. - For “or,” take every outcome satisfying at least one condition. - The complement contains all spinner outcomes not in event \(A\).

1. The prime numbers from \(1\) through \(12\) are \(2,3,5,7,11\). 2. The blue sections are numbered \(9,10,11,12\). Their union gives \(A=\{2,3,5,7,9,10,11,12\}\). 3. The complement consists of outcomes that are neither prime nor blue: \(\overline{A}=\{1,4,6,8\}\). 4. Therefore, \(P(\overline{A})=\frac{4}{12}=\frac{1}{3}\).

a) \(A=\{2,3,5,7,9,10,11,12\}\) b) \(\overline{A}\): The number is not prime and the section is not blue. \(P(\overline{A})=\frac{1}{3}\).

- Read each number's color from the spinner before applying the prize conditions. - List the outcomes satisfying each prize rule. - Combine the winning lists without counting an outcome twice.

1. The spinner has \(10\) equally likely outcomes. 2. The yellow prime-numbered sections are \(2\) and \(7\). 3. The purple even-numbered sections are \(6,8,10\). 4. The consolation-prize sections are \(1\) and \(10\). 5. Combining the winning outcomes without counting \(10\) twice gives \(1,2,6,7,8,10\), so the probability is \(\frac{6}{10}=\frac{3}{5}=60\%\).

The probability of winning at least one prize is \(\frac{3}{5}\), or \(60\%\).

- Count primes in the first numbered sample space and multiples of \(3\) in the second. - Divide each winning count by its spinner's total number of sections. - For part c, translate \(45\%\) of \(20\) into a winning-section count.

1. The primes from \(1\) through \(15\) are \(2,3,5,7,11,13\), so Spinner A has \(6\) winning sections and \(P(A)=\frac{6}{15}=40\%\). 2. The multiples of \(3\) from \(1\) through \(20\) are \(3,6,9,12,15,18\), so Spinner B has \(6\) winning sections and \(P(B)=\frac{6}{20}=30\%\). 3. Since \(40\%>30\%\), Spinner A gives the greater chance of winning. 4. For Spinner B to have winning probability \(45\%\), it needs \(0.45\cdot20=9\) winning sections.

a) Spinner A: \(40\%\); Spinner B: \(30\%\) b) Spinner A, because \(40\%>30\%\) c) \(9\) winning sections in total

- Work within the equally likely outcomes \(1\) through \(20\). - List every number satisfying each event before counting. - Remember that \(1\) is not prime.

1. There are \(20\) equally likely outcomes. 2. \(E_1=\{2,3,5,7,11,13,17,19\}\), so \(P(E_1)=\frac{8}{20}=\frac{2}{5}=0.4\). 3. \(E_2=\{4,8,12,16,20\}\), so \(P(E_2)=\frac{5}{20}=\frac{1}{4}=0.25\). 4. \(E_3=\{1,4,9,16\}\), so \(P(E_3)=\frac{4}{20}=\frac{1}{5}=0.2\).

a) \(E_1=\{2,3,5,7,11,13,17,19\}\); \(P(E_1)=0.4\) b) \(E_2=\{4,8,12,16,20\}\); \(P(E_2)=0.25\) c) \(E_3=\{1,4,9,16\}\); \(P(E_3)=0.2\)

- Find the initial blue and white counts. - Adding blue marbles changes both the favorable count and the total count. - The number of white marbles stays unchanged while blue marbles are added. - Let a variable represent the number added and write a probability equation.

1. Initially, there are \(0.25 \cdot 40 = 10\) blue marbles and \(30\) white marbles. 2. Let \(x\) be the number of blue marbles added. Then the new blue count is \(10 + x\), and the new total is \(40 + x\). 3. Set up the equation \(\frac{10 + x}{40 + x} = 0.4\). 4. Solve: \(10 + x = 0.4(40 + x)\), so \(10 + x = 16 + 0.4x\). Then \(0.6x = 6\), giving \(x = 10\). 5. Check: after adding \(10\) blue marbles, \(\frac{20}{50} = 0.4\).

\(10\) blue marbles must be added.