Free Math Worksheets

- 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) \(S = \{1, 3, 5, 7\}\)

- List the possible numbers of nickels from \(0\) through \(4\). - For each case, the rest of the four coins must be pennies. - Calculate the total value for each composition.

1. The number of nickels can be \(0,1,2,3,\) or \(4\). The remaining coins are pennies. 2. With \(0\) nickels and \(4\) pennies, the total is \(\$0.04\). 3. With \(1,2,3,\) or \(4\) nickels, the totals are \(\$0.08\), \(\$0.12\), \(\$0.16\), and \(\$0.20\), respectively. 4. These five cases include every possible composition of four coins.

\(\$0.04\), \(\$0.08\), \(\$0.12\), \(\$0.16\), and \(\$0.20\)

- Consider each possible position for the one red balloon. - Once the red balloon is placed, what colors must fill the remaining positions? - Each path in a tree diagram corresponds to one completed arrangement.

1. The red balloon can occupy the first, second, third, or fourth position. 2. These choices produce the arrangements RBBB, BRBB, BBRB, and BBBR. 3. Therefore, there are \(4\) distinct arrangements.

\(4\) arrangements

- First list the masses measurable with one weight. - Then list every pair of weights. - Include the case that uses all three weights. - Organize the combinations so none are missed.

1. Using one weight gives \(5\,\text{g}\), \(10\,\text{g}\), or \(50\,\text{g}\). 2. Using two weights gives \(5+10=15\,\text{g}\), \(5+50=55\,\text{g}\), or \(10+50=60\,\text{g}\). 3. Using all three weights gives \(5+10+50=65\,\text{g}\). 4. These are all nonempty combinations of the three weights, so the measurable masses are \(5,10,15,50,55,60,\) and \(65\) grams.

\(5\,\text{g}\), \(10\,\text{g}\), \(15\,\text{g}\), \(50\,\text{g}\), \(55\,\text{g}\), \(60\,\text{g}\), and \(65\,\text{g}\)

- Count how many balls show each letter. - Find the probability of each letter on one draw. - Replacement means the probabilities do not change. - Multiply the probabilities in the stated order.

1. There are three balls labeled A, two labeled N, and one labeled B. Thus, \(P(A)=\frac{3}{6}=\frac{1}{2}\) and \(P(N)=\frac{2}{6}=\frac{1}{3}\). 2. Because the first ball is replaced, the probabilities stay the same on the second draw. Therefore, \(P(\text{AN})=\frac{1}{2}\cdot\frac{1}{3}=\frac{1}{6}\). 3. Similarly, \(P(\text{AA})=\frac{1}{2}\cdot\frac{1}{2}=\frac{1}{4}\).

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

- List the ordered outcomes with exactly two red draws. - Because the ball is replaced, the probabilities stay the same on every draw. - Determine how many red and blue results each favorable outcome must contain.

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

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

- List the possible ordered pairs. - Identify the pairs whose sum is greater than \(8\). - Divide the number of favorable pairs by the total number of equally likely pairs.

1. There are \(5\cdot 5=25\) equally likely ordered pairs. 2. The pairs with a sum greater than \(8\) are \((4, 5)\), \((5, 4)\), and \((5, 5)\). 3. Therefore, \(P(\text{sum greater than }8)=\frac{3}{25}=12\%\).

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

- 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 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 \(TT\)

- 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 \(\frac{1}{36}\).

- Count the blue and yellow sections on one spin. - A \(10 \times 10\) outcome grid represents the equally likely ordered section pairs for two spins. - Compare how many section pairs belong to each color combination.

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. a) 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. b) 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.

- Write each combination in the same order, such as greatest to least. - Begin with the greatest possible first value and determine the remaining sum. - Continue systematically so that no combination is missed or counted twice. - Check whether a greatest value of \(2\) could produce a sum of \(7\).

1. Organize each combination from greatest to least to avoid duplicates. 2. If the greatest value is \(5\), the remaining values must sum to \(2\), giving \((5, 1, 1)\). 3. If the greatest value is \(4\), the remaining values must sum to \(3\), giving \((4, 2, 1)\). 4. If the greatest value is \(3\), the remaining values must sum to \(4\), giving \((3, 3, 1)\) and \((3, 2, 2)\). 5. A greatest value of \(2\) or less cannot work because \(2+2+2=6<7\). Therefore, there are exactly \(4\) combinations.

\((5, 1, 1)\), \((4, 2, 1)\), \((3, 3, 1)\), and \((3, 2, 2)\)

- Consider the possible first digits in order. - For each first digit, determine the required sum of the other two digits. - Remember that every digit must be at least \(1\). - Count ordered pairs, since reversing two digits can create a different code.

1. Organize the codes by their first digit. 2. If the first digit is \(1\), the last two digits must sum to \(5\): \((1, 4)\), \((2, 3)\), \((3, 2)\), and \((4, 1)\). This gives \(4\) codes. 3. If the first digit is \(2\), the last two digits must sum to \(4\), giving \(3\) codes. If the first digit is \(3\), they must sum to \(3\), giving \(2\) codes. If the first digit is \(4\), they must sum to \(2\), giving \(1\) code. 4. A first digit of \(5\) or greater cannot work because each remaining digit is at least \(1\). 5. The total is \(4+3+2+1=10\) codes.

\(10\) codes

- Which die values from \(1\) through \(6\) are factors of \(12\)? - Begin with the greatest possible value and find the required product of the other two. - Check that every product is exactly \(12\). - Write each combination in the same order to avoid duplicates.

1. Write each combination from greatest to least to avoid duplicates. 2. If the greatest value is \(6\), the other two values must have product \(12\div6=2\), giving \((6, 2, 1)\). 3. If the greatest value is \(4\), the other two values must have product \(12\div4=3\), giving \((4, 3, 1)\). 4. If the greatest value is \(3\), the other two values must have product \(12\div3=4\). The only combination with no value greater than \(3\) is \((3, 2, 2)\). 5. A greatest value of \(2\) cannot work because \(2\cdot2\cdot2=8<12\). Therefore, there are \(3\) combinations.

\((6, 2, 1)\), \((4, 3, 1)\), and \((3, 2, 2)\)

- Begin with the greatest possible number of dimes. - For each number of dimes, determine how the remaining amount can be made with nickels and pennies. - Organize the possibilities in a table so none are missed. - Include the possibility that uses only pennies.

1. Let each ordered triple \((d, n, p)\) represent the numbers of dimes, nickels, and pennies. 2. With \(2\) dimes, the only possibility is \((2, 0, 0)\). 3. With \(1\) dime, the remaining \(\$0.10\) can be made as \((1, 2, 0)\), \((1, 1, 5)\), or \((1, 0, 10)\). 4. With no dimes, the possibilities are \((0, 4, 0)\), \((0, 3, 5)\), \((0, 2, 10)\), \((0, 1, 15)\), and \((0, 0, 20)\). 5. Altogether, there are \(1+3+5=9\) ways.

There are \(9\) ways. Using \((d, n, p)\) for dimes, nickels, and pennies, they are \((2, 0, 0)\), \((1, 2, 0)\), \((1, 1, 5)\), \((1, 0, 10)\), \((0, 4, 0)\), \((0, 3, 5)\), \((0, 2, 10)\), \((0, 1, 15)\), and \((0, 0, 20)\).

- Begin with the greatest possible number of quarters. - For each number of quarters, determine how the remaining amount can be made with dimes and nickels. - How many dimes can be used when there are no quarters?

1. Let each ordered triple \((q, d, n)\) represent the numbers of quarters, dimes, and nickels. 2. With \(1\) quarter, the remaining \(\$0.10\) can be made as \((1, 1, 0)\) or \((1, 0, 2)\). 3. With no quarters, the possibilities are \((0, 3, 1)\), \((0, 2, 3)\), \((0, 1, 5)\), and \((0, 0, 7)\). 4. Therefore, there are \(2+4=6\) combinations.

There are \(6\) combinations. Using \((q, d, n)\) for quarters, dimes, and nickels, they are \((1, 1, 0)\), \((1, 0, 2)\), \((0, 3, 1)\), \((0, 2, 3)\), \((0, 1, 5)\), and \((0, 0, 7)\).

- 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) \(P(E^c)=\frac{1}{6}\)

- Use the fundamental counting principle to find the total number of section pairs. - Count the colors that appear on both spinners. - List the ordered number pairs with sum at least \(6\). - Divide favorable equally likely outcomes by all outcomes.

1. Spinner A has \(5\) sections and spinner B has \(4\) sections, so there are \(5\cdot 4=20\) equally likely section pairs. 2. For matching colors, there are \(2\cdot 1=2\) red-red pairs and \(2\cdot 2=4\) 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 \((2, 4)\), \((3, 3)\), \((3, 4)\), \((4, 2)\), \((4, 3)\), \((4, 4)\), \((5, 1)\), \((5, 2)\), \((5, 3)\), and \((5, 4)\). Therefore, \(P(\text{sum}\ge 6)=\frac{10}{20}=\frac{1}{2}=50\%\).

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

- Count how many sections carry each number and each color. - Treat the two spins as ordered. - For part a, add the probabilities for matching \(1\)s, matching \(2\)s, and matching \(3\)s. - For part b, identify the possible number pairs before counting section pairs.

1. There are \(6\cdot 6=36\) equally likely section pairs. 2. The number \(1\) appears on \(2\) sections, \(2\) appears on \(3\) sections, and \(3\) appears on \(1\) section. 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 \((2, 3)\), \((3, 2)\), or \((3, 3)\). These represent \(3\cdot 1+1\cdot 3+1\cdot 1=7\) section pairs, so the probability is \(\frac{7}{36}\). 4. There are \(2\) red sections and \(2\) blue sections. Therefore, \(P(\text{red then blue})=\frac{2}{6}\cdot\frac{2}{6}=\frac{1}{9}\).

a) \(\frac{7}{18}\approx 38.9\%\) b) \(\frac{7}{36}\approx 19.4\%\) c) \(\frac{1}{9}\approx 11.1\%\)

- Count all ordered section pairs. - For part a, multiply the red probabilities from the two spinners. - For part b, list all ordered number pairs with sum \(6\). - Simplify each fraction.

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

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

- 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 \(\frac{24}{36}=\frac{2}{3}\).

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

- Find the one-spin probability of each color. - Add the three matching-color path probabilities. - For at least one green, use the complement of no green. - For a specified order, multiply the two probabilities.

1. The one-spin probabilities are \(P(\text{red})=\frac{3}{8}\), \(P(\text{blue})=\frac{4}{8}\), and \(P(\text{green})=\frac{1}{8}\). 2. \(P(\text{same color})=\left(\frac{3}{8}\right)^2+\left(\frac{4}{8}\right)^2+\left(\frac{1}{8}\right)^2=\frac{26}{64}=\frac{13}{32}=40.625\%\). 3. Use the complement of no green: \(P(\text{at least one green})=1-\left(\frac{7}{8}\right)^2=\frac{15}{64}=23.4375\%\). 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) \(\frac{3}{16}=18.75\%\)

- Count the sections of each color. - Multiply the probabilities for the specified order in part a. - For part b, use the complement of landing on no red sections.

1. The one-spin probabilities are \(P(\text{green})=\frac{2}{5}\), \(P(\text{red})=\frac{2}{5}\), 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. The probability of not landing on red is \(\frac{3}{5}\). Therefore, \(P(\text{at least one red})=1-\left(\frac{3}{5}\right)^2=\frac{16}{25}=64\%\).

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

- Count the red sections. - State the complement of at least one red. - Multiply the not-red probability for two spins, then subtract from \(1\).

1. Two of the eight sections are red, so \(P(\text{red})=\frac{2}{8}=\frac{1}{4}\) 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, \(P(\text{at least one red})=1-\frac{9}{16}=\frac{7}{16}=43.75\%\).

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

- Find the probability of each color on one spin. - Add the probabilities of the matching-color paths. - For part b, use the complement of no yellow.

1. The one-spin probabilities are \(P(\text{red})=\frac{4}{8}=\frac{1}{2}\), \(P(\text{green})=\frac{2}{8}=\frac{1}{4}\), and \(P(\text{yellow})=\frac{2}{8}=\frac{1}{4}\). 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. The probability of no yellow on one spin is \(\frac{3}{4}\). Therefore, \(P(\text{at least one yellow})=1-\left(\frac{3}{4}\right)^2=\frac{7}{16}=43.75\%\).

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

- Determine the total number of ordered outcomes for two spins. - Organize the number pairs systematically. - For part a, identify which pairs have sums that meet the condition. - For part b, consider counting the complement: two spins of the same color.

1. There are \(5\cdot 5=25\) equally likely ordered outcomes. 2. The outcomes with a sum of at least \(7\) are \((2, 5)\), \((3, 4)\), \((3, 5)\), \((4, 3)\), \((4, 4)\), \((4, 5)\), \((5, 2)\), \((5, 3)\), \((5, 4)\), and \((5, 5)\). Therefore, \(P(\text{sum}\ge 7)=\frac{10}{25}=\frac{2}{5}=0.40\). 3. For the complement in part b, there are \(2\cdot 2=4\) red-red outcomes, \(2\cdot 2=4\) blue-blue outcomes, and \(1\) green-green outcome. Thus, \(P(\text{same color})=\frac{9}{25}\). 4. Therefore, \(P(\text{different colors})=1-\frac{9}{25}=\frac{16}{25}=0.64\).

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

- 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) \(\frac{1}{6}\)

- Determine how many ordered pairs are possible. - List the ordered pairs that satisfy each condition. - Divide the number of favorable pairs by the total number of equally likely pairs. - Simplify each fraction.

1. There are \(3\cdot 4=12\) equally likely ordered pairs \((a, b)\), where \(a\) is the result from spinner A and \(b\) is the result from spinner B. Each pair has probability \(\frac{1}{12}\). 2. The pairs with sum \(5\) are \((1, 4)\), \((2, 3)\), and \((3, 2)\). Therefore, \(P(\text{sum }5)=\frac{3}{12}=\frac{1}{4}\). 3. The pairs with \(b>a\) are \((1, 2)\), \((1, 3)\), \((1, 4)\), \((2, 3)\), \((2, 4)\), and \((3, 4)\). Therefore, \(P(b>a)=\frac{6}{12}=\frac{1}{2}\). 4. The matching pairs are \((1, 1)\), \((2, 2)\), and \((3, 3)\). Therefore, \(P(\text{same number})=\frac{3}{12}=\frac{1}{4}\).

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

- Decide which one of the three positions can contain yellow. - Find the one-spin probabilities of blue and yellow. - Multiply along each favorable outcome. - Add the probabilities of the favorable outcomes.

1. Exactly one yellow can occur in the first, second, or third position. The outcomes are \((Y, B, B)\), \((B, Y, B)\), and \((B, B, Y)\). 2. The one-spin probabilities are \(P(Y)=\frac{1}{4}\) and \(P(B)=\frac{3}{4}\). 3. Each favorable outcome has probability \(\frac{1}{4}\cdot\frac{3}{4}\cdot\frac{3}{4}=\frac{9}{64}\). 4. Add the three equal probabilities: \(P(E)=3\cdot\frac{9}{64}=\frac{27}{64}\).

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

- Each numbered section has the same probability. - For part a, multiply the probability of blue by the probability of red. - For part b, count the odd numbers on the spinner. - For part c, list ordered pairs systematically whose sum is greater than \(12\).

1. Four of the eight sections are blue and two are red. Therefore, \(P(C)=\frac{4}{8}\cdot\frac{2}{8}=\frac{1}{8}\). 2. Four of the eight numbers are odd. Therefore, \(P(A)=\frac{4}{8}\cdot\frac{4}{8}=\frac{1}{4}\). 3. There are \(8\cdot 8=64\) equally likely ordered pairs. The pairs whose sum is greater than \(12\) are \((5, 8)\), \((6, 7)\), \((6, 8)\), \((7, 6)\), \((7, 7)\), \((7, 8)\), \((8, 5)\), \((8, 6)\), \((8, 7)\), and \((8, 8)\). Thus, \(P(B)=\frac{10}{64}=\frac{5}{32}\).

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

- Determine the number of equally likely ordered pairs of sections. - Count how many section pairs satisfy each required number or color condition. - Remember that each number and each color appears more than once. - For parts c and d, it may be easier to count the complementary outcomes first.

1. The spinner has \(8\) equal sections, so two spins produce \(8 \cdot 8 = 64\) equally likely ordered pairs of sections. 2. Each of the four numbers appears on two sections. For each number, there are \(2 \cdot 2 = 4\) ordered pairs showing that number twice. Thus, part a has \(4 \cdot 4 = 16\) favorable outcomes, and the probability is \(\frac{16}{64} = \frac{1}{4}\). 3. A sum of \(4\) can come from the number pairs \((1, 3)\), \((2, 2)\), and \((3, 1)\). Each number pair corresponds to \(2 \cdot 2 = 4\) section pairs, so there are \(3 \cdot 4 = 12\) favorable outcomes. The probability is \(\frac{12}{64} = \frac{3}{16}\). 4. For each of the four colors, there are \(2 \cdot 2 = 4\) ordered pairs with that same color, so there are \(16\) same-color outcomes. Therefore, \(64 - 16 = 48\) outcomes have different colors, giving \(\frac{48}{64} = \frac{3}{4}\). 5. Six sections are not labeled \(1\). Therefore, \(6 \cdot 6 = 36\) ordered pairs contain no \(1\), so \(64 - 36 = 28\) contain at least one \(1\). The probability is \(\frac{28}{64} = \frac{7}{16}\).

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

- Count the sections of each color. - Multiply the green probability for two spins in part a. - For part b, use the complement of no red.

1. The spinner has ten equal sections: five blue, three red, and two green. Therefore, \(P(\text{green})=\frac{2}{10}=\frac{1}{5}\) and \(P(\text{red})=\frac{3}{10}\). 2. \(P(\text{green twice})=\frac{1}{5}\cdot\frac{1}{5}=\frac{1}{25}=4\%\). 3. Use the complement of no red. The probability of not red on one spin is \(\frac{7}{10}\), so \(P(\text{at least one red})=1-\left(\frac{7}{10}\right)^2=1-\frac{49}{100}=\frac{51}{100}=51\%\).

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

- Find the probability of each color on one spin. - For part a, multiply the probability of not yellow for all three spins. - For part b, list the three possible positions of the spin that is not red. - Rewrite the fractions with a common denominator to compare them.

1. The one-spin probabilities are \(P(\text{red})=\frac{1}{2}\), \(P(\text{blue})=\frac{1}{4}\), and \(P(\text{yellow})=\frac{1}{4}\). 2. The probability of not yellow on one spin is \(\frac{3}{4}\). Therefore, \(P(E_1)=\left(\frac{3}{4}\right)^3=\frac{27}{64}\). 3. Exactly two red spins can occur in three orders. Since both red and not red have probability \(\frac{1}{2}\), \(P(E_2)=3\cdot\left(\frac{1}{2}\right)^2\cdot\frac{1}{2}=\frac{3}{8}=\frac{24}{64}\). 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, \(P(\text{sum }4)=\frac{1}{25}+\frac{1}{25}+\frac{4}{25}=\frac{6}{25}=24\%\).

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

- Find the probability of each color on one spin. - For “exactly,” count the possible positions of the specified color. - For “at least once,” consider using a complement. - For an exact order, multiply the probabilities in that order. - “At most once” includes zero times and one time.

1. The one-spin probabilities are \(P(\text{green})=\frac{3}{8}\), \(P(\text{yellow})=\frac{1}{2}\), and \(P(\text{orange})=\frac{1}{8}\). 2. Exactly two yellow spins can occur in three orders. Since the probability of not yellow is also \(\frac{1}{2}\), \(P(\text{exactly two yellow})=3\cdot\left(\frac{1}{2}\right)^2\cdot\frac{1}{2}=\frac{3}{8}\). 3. Use the complement of no orange: \(P(\text{at least one orange})=1-\left(\frac{7}{8}\right)^3=\frac{169}{512}\). 4. For the specified order, \(P(\text{green, yellow, orange})=\frac{3}{8}\cdot\frac{1}{2}\cdot\frac{1}{8}=\frac{3}{128}\). 5. At most one yellow means zero or one yellow. The probabilities are \(P(\text{zero yellow})=\left(\frac{1}{2}\right)^3=\frac{1}{8}\) and \(P(\text{exactly one yellow})=3\cdot\frac{1}{2}\cdot\left(\frac{1}{2}\right)^2=\frac{3}{8}\). Therefore, \(P(\text{at most one yellow})=\frac{1}{8}+\frac{3}{8}=\frac{1}{2}\).

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

- For each part, determine which numbers are allowed on each spin. - “Only” means the other spins must not have the named property. - List the perfect squares from \(1\) through \(8\). - Multiply probabilities along a specified sequence. - For part d, count the possible repeated-number outcomes.

1. There are four even numbers and four odd numbers from \(1\) through \(8\). Therefore, \(P(\text{even, odd, even})=\frac{4}{8}\cdot\frac{4}{8}\cdot\frac{4}{8}=\frac{1}{8}\). 2. The perfect squares are \(1\) and \(4\), so the probability of a perfect square is \(\frac{2}{8}=\frac{1}{4}\), and the probability of not getting a perfect square is \(\frac{6}{8}=\frac{3}{4}\). Thus, \(P(\text{square, not square, not square})=\frac{1}{4}\cdot\frac{3}{4}\cdot\frac{3}{4}=\frac{9}{64}\). 3. The numbers greater than \(5\) are \(6\), \(7\), and \(8\). Therefore, \(P(\text{all greater than }5)=\left(\frac{3}{8}\right)^3=\frac{27}{512}\). 4. The matching outcomes are \((1, 1, 1)\), \((2, 2, 2)\), through \((8, 8, 8)\). There are eight such outcomes, each with probability \(\left(\frac{1}{8}\right)^3\). Therefore, \(P(\text{same number three times})=8\cdot\left(\frac{1}{8}\right)^3=\frac{1}{64}\).

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) \(\frac{1}{4}\)

- Count the sections of each color. - For part b, add the probabilities of the three matching-color outcomes. - For part c, use the complement of no yellow. - Compare the resulting percent with \(35\%\).

1. The one-spin probabilities are \(P(\text{blue})=\frac{5}{10}=\frac{1}{2}\), \(P(\text{red})=\frac{3}{10}\), and \(P(\text{yellow})=\frac{2}{10}=\frac{1}{5}\). 2. The matching-color outcomes are blue-blue, red-red, and yellow-yellow. Therefore, \(P(\text{same color})=\left(\frac{1}{2}\right)^2+\left(\frac{3}{10}\right)^2+\left(\frac{1}{5}\right)^2=\frac{19}{50}=38\%\). 3. Use the complement of no yellow. The probability of not yellow on one spin is \(\frac{4}{5}\), so \(P(\text{at least one yellow})=1-\left(\frac{4}{5}\right)^2=\frac{9}{25}=36\%\). Since \(36\%>35\%\), the student 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 \(35\%\).

- Make a table of all \(25\) ordered pairs. - Mark the pairs with a sum of \(6\). - Mark the pairs in which the first number is greater. - For equally likely outcomes, divide favorable outcomes by total outcomes.

1. There are \(5\cdot 5=25\) equally likely ordered outcomes. 2. Event \(A\) contains \((1, 5)\), \((2, 4)\), \((3, 3)\), \((4, 2)\), and \((5, 1)\). Therefore, \(P(A)=\frac{5}{25}=\frac{1}{5}=0.20\). 3. Event \(B\) contains \((2, 1)\), \((3, 1)\), \((3, 2)\), \((4, 1)\), \((4, 2)\), \((4, 3)\), \((5, 1)\), \((5, 2)\), \((5, 3)\), and \((5, 4)\). Therefore, \(P(B)=\frac{10}{25}=\frac{2}{5}=0.40\). 4. Since \(0.40>0.20\), event \(B\) is more likely.

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

- 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.

- 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, \(P = \frac{6}{16} = \frac{3}{8}\).

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

- For part a, first write \(5\) as a sum of three allowed positive digits. - Then arrange each resulting group of digits in every distinct order. - For part b, separate the cases in which the equal pair is first or last. - Exclude \(111\), which creates two overlapping equal adjacent pairs. - List the qualifying codes to verify the count.

1. Ignoring order at first, the only three allowed digits with sum \(5\) are \(1, 1, 3\) and \(1, 2, 2\). 2. Their distinct arrangements are \(113,131,311,122,212,\) and \(221\), giving \(6\) codes. 3. For part b, one case has equal first and second digits: \(11y\), where \(y\) can be \(2,3,\) or \(4\). This gives \(112,113,\) and \(114\). 4. The other case has equal second and third digits: \(1xx\), where \(x\) can be \(2,3,\) or \(4\). This gives \(122,133,\) and \(144\). 5. The two cases are disjoint, so there are \(3+3=6\) codes.

a) \(6\) codes: \(113,131,311,122,212,221\) b) \(6\) codes: \(112,113,114,122,133,144\)

- Add weights that are placed on the same pan opposite the object. - Subtract weights placed beside the object from weights placed on the opposite pan. - Organize the cases by using one weight, two weights, and all three weights. - Remove repeated results from your final list.

1. Using one weight measures \(1\,\text{kg}\), \(2\,\text{kg}\), or \(5\,\text{kg}\). 2. Placing two weights opposite the object gives sums \(1+2=3\), \(1+5=6\), and \(2+5=7\). Placing one weight beside the object gives differences \(2-1=1\), \(5-1=4\), and \(5-2=3\). 3. Using all three weights gives \(1+2+5=8\) when they are together. With weights on both pans, the positive differences are \(5+2-1=6\), \(5+1-2=4\), and \(5-2-1=2\). 4. Combining the distinct results gives every whole-number mass from \(1\,\text{kg}\) through \(8\,\text{kg}\).

\(1\,\text{kg}\), \(2\,\text{kg}\), \(3\,\text{kg}\), \(4\,\text{kg}\), \(5\,\text{kg}\), \(6\,\text{kg}\), \(7\,\text{kg}\), and \(8\,\text{kg}\)