Free Math Worksheets

- List the pairs systematically so none are repeated or omitted. - Does the order of the two students matter? - What does a binomial coefficient count? - How many ways can one member be selected from four?

1. The two-student sets are \(\{A,B\}\), \(\{A,C\}\), \(\{A,D\}\), \(\{B,C\}\), \(\{B,D\}\), and \(\{C,D\}\). There are \(6\) selections. 2. \(\binom{4}{2}=\frac{4\cdot 3}{2\cdot 1}=6\), which matches the list because combinations count selections without regard to order. 3. Selecting one student gives \(\binom{4}{1}=4\) choices.

a) \(\{A,B\}\), \(\{A,C\}\), \(\{A,D\}\), \(\{B,C\}\), \(\{B,D\}\), \(\{C,D\}\) b) \(\binom{4}{2}=6\), matching the six listed sets. c) \(\binom{4}{1}=4\)

- Does the order in which members are selected matter? - Can the same member be selected more than once? - Which formula counts an unordered subset of a larger set?

1. The committee is an unordered selection of \(3\) members from \(22\). 2. The number of committees is \(\binom{22}{3}=\frac{22\cdot 21\cdot 20}{3\cdot 2\cdot 1}=1540\).

There are \(1540\) possible committees.

- Does the order in which the four games are chosen affect the final set? - Are games selected with or without replacement? - Which counting model applies to an unordered selection?

1. The group selects \(4\) different games from \(14\), and order does not matter. 2. The number of selections is \(\binom{14}{4}=\frac{14\cdot 13\cdot 12\cdot 11}{4\cdot 3\cdot 2\cdot 1}=1001\).

There are \(1001\) possible sets of games.

- Does the order in which the athletes are selected matter? - Can an athlete be selected more than once? - Which formula counts subsets of a given size?

1. The team is an unordered selection of \(3\) people from \(10\). 2. The number of teams is \(\binom{10}{3}=\frac{10\cdot 9\cdot 8}{3\cdot 2\cdot 1}=120\).

There are \(120\) possible teams.

- What does symmetry of binomial coefficients state? - Simplify factors before multiplying. - Which expressions form symmetry pairs? - Compare choosing \(k\) objects with choosing the \(n-k\) objects left out.

1. \(\binom{18}{2}=\frac{18\cdot 17}{2}=153\). 2. By symmetry, \(\binom{18}{16}=\binom{18}{2}=153\). 3. \(\binom{25}{3}=\frac{25\cdot 24\cdot 23}{3\cdot 2\cdot 1}=2300\). 4. By symmetry, \(\binom{25}{22}=\binom{25}{3}=2300\).

a) \(153\) b) \(153\) c) \(2300\) d) \(2300\)

- Evaluate the binomial coefficients one at a time. - Compare the answers to parts a) and b). - In part c), use symmetry before calculating large values. - Recall the definition and properties of \(\binom{n}{k}\).

1. \(\binom{9}{2}=36\) and \(\binom{9}{3}=84\), so the sum is \(36+84=120\). 2. \(\binom{10}{3}=120\). 3. By symmetry, \(\binom{11}{9}=\binom{11}{2}\). Therefore, the quotient is \(1\).

a) \(120\) b) \(120\) c) \(1\)

- Does the order in which the students are named affect the group? - Does being selected first or third create a different council? - Which counting model applies to an unordered selection?

1. Order does not matter, so use a combination. 2. \(\binom{24}{3}=\frac{24\cdot 23\cdot 22}{3\cdot 2\cdot 1}=2024\).

There are \(2024\) possible groups.

- Does the order in which toppings are added change the final combination? - How many toppings are available, and how many are selected? - Which formula counts a subset of a larger set?

1. The customer chooses \(4\) toppings from \(10\), and order does not matter. 2. \(\binom{10}{4}=\frac{10\cdot 9\cdot 8\cdot 7}{4\cdot 3\cdot 2\cdot 1}=210\).

There are \(210\) topping combinations.

- Does order matter within the selected team? - Which formula counts \(k\) objects selected from \(n\) without replacement? - Look for the \(n\text{Cr}\) function on your calculator.

1. For part a), \(\binom{9}{2}=\frac{9\cdot 8}{2\cdot 1}=36\). 2. For part b), \(\binom{18}{7}=31{,}824\).

a) \(36\) b) \(31{,}824\)

- Count all groups of four that can be selected from \(12\) people. - The order of selection does not matter. - Count how many groups match the four specified people.

1. The number of possible groups of four is \(\binom{12}{4}=495\). 2. Exactly one of these groups contains the four specified people. 3. Therefore, \(P(\text{specified group})=\frac{1}{495}\).

\(\frac{1}{495}\approx 0.2\%\)

- How many choices are available at each pulse position? - Choose which positions contain positive pulses. - “At least \(6\)” includes \(6\), \(7\), and \(8\). - When all sequences are equally likely, compare favorable sequences with all sequences.

1. Each of the \(8\) positions has \(2\) choices, so there are \(2^8=256\) sequences. 2. Choose the \(3\) positions for positive pulses: \(\binom{8}{3}=56\). 3. At least \(6\) positive pulses means \(6\), \(7\), or \(8\). The number of favorable sequences is \(\binom{8}{6}+\binom{8}{7}+\binom{8}{8}=28+8+1=37\). Thus the probability is \(\frac{37}{256}\approx 0.1445\).

a) \(256\) b) \(56\) c) \(\frac{37}{256}\approx 0.1445\)

- Can symmetry make any of the binomial coefficients easier to compute? - Recall the factorial formula for \(\binom{n}{k}\). - How do \(n\) and \(k\) determine the factors that remain in the numerator and denominator? - In part f), calculate each binomial coefficient before adding.

1. \(\binom{9}{3}=\frac{9\cdot 8\cdot 7}{3\cdot 2\cdot 1}=84\). 2. By symmetry, \(\binom{15}{13}=\binom{15}{2}=\frac{15\cdot 14}{2\cdot 1}=105\). 3. \(\binom{25}{2}=\frac{25\cdot 24}{2\cdot 1}=300\). 4. By symmetry, \(\binom{50}{48}=\binom{50}{2}=\frac{50\cdot 49}{2\cdot 1}=1225\). 5. By symmetry, \(\binom{2000}{1999}=\binom{2000}{1}=2000\). 6. On the left, \(\binom{10}{3}=120\) and \(\binom{10}{4}=210\), so the sum is \(120+210=330\). On the right, \(\binom{11}{4}=330\). Therefore, both sides are equal.

a) \(84\) b) \(105\) c) \(300\) d) \(1225\) e) \(2000\) f) Left side: \(120+210=330\); right side: \(330\).

- Identify the total number of fruits. - Translate each event into the number selected from each fruit type. - Check that the selected counts add to five.

1. There are \(18\) fruits total, so every denominator is \(\binom{18}{5}\). 2. Event \(A\) requires two apples, two pears, and one banana, so it matches expression (2). 3. Event \(B\) requires all five fruits to come from the \(12\) apples and bananas, so it matches expression (1). 4. Event \(C\) requires three apples and two bananas, so it matches expression (3).

\(A\) matches (2). \(B\) matches (1). \(C\) matches (3).

- Count all groups and the groups containing the specified student. - Use the fact that all students are treated symmetrically.

1. There are \(\binom{25}{4}\) possible groups. Groups containing Luke are formed by choosing the other \(3\) students from the remaining \(24\), so \(P=\frac{\binom{24}{3}}{\binom{25}{4}}=\frac{4}{25}=0.16\). 2. By symmetry, each of the \(25\) students has the same chance of occupying any of the \(4\) selected positions. Therefore, each student’s selection probability is \(\frac{4}{25}\).

a) \(\frac{4}{25}=0.16\) b) Four of the \(25\) equally likely students are selected, so each student has probability \(\frac{4}{25}\).

- Count unordered selections with a binomial coefficient. - Calculate the number of possible selections for each format. - When exactly one selection wins, how does the total number of selections affect the probability?

1. Format A has \(\binom{42}{6}=5{,}245{,}786\) possible selections. 2. Format B has \(\binom{37}{7}=10{,}295{,}472\) possible selections. 3. Each format has exactly one winning selection. Therefore, the jackpot probabilities are \(\frac{1}{5{,}245{,}786}\) for Format A and \(\frac{1}{10{,}295{,}472}\) for Format B. Since Format A has fewer possible selections, it has the greater jackpot probability.

Format A: \(5{,}245{,}786\) selections Format B: \(10{,}295{,}472\) selections Format A has the greater jackpot probability.

- Picture \(10\) labeled positions and mark \(3\) of them as successes. - Are the success markers distinguishable? - Can one trial contain two successes? - Count the selected positions with a binomial coefficient.

1. Represent the \(10\) trials by \(10\) labeled positions and place one success marker in each position where a success occurs. 2. The success markers are identical, so their order does not matter. A trial can have only one outcome, so no position can contain more than one marker. 3. The count is the number of ways to choose \(3\) of the \(10\) positions: \(\binom{10}{3}=\frac{10\cdot 9\cdot 8}{3\cdot 2\cdot 1}=120\).

a) Use \(10\) labeled trial positions and place \(3\) identical success markers in the positions of the successes. b) The order of the markers does not matter, and each position can hold at most one marker. c) \(\binom{10}{3}=120\) outcome sequences

- How many times must the digit \(2\) appear for the five digits to sum to \(7\)? - After determining the numbers of \(1\)s and \(2\)s, choose the positions of one repeated digit. - List the numbers in increasing order to avoid omissions.

1. Five digits of \(1\) would have sum \(5\). Replacing a \(1\) with a \(2\) increases the sum by \(1\), so exactly two positions must contain \(2\), and the other three positions must contain \(1\). 2. Choose the two positions for the \(2\)s. There are \(\binom{5}{2}=10\) choices. 3. In increasing order, the numbers are \(11{,}122\), \(11{,}212\), \(11{,}221\), \(12{,}112\), \(12{,}121\), \(12{,}211\), \(21{,}112\), \(21{,}121\), \(21{,}211\), and \(22{,}111\).

\(11{,}122\), \(11{,}212\), \(11{,}221\), \(12{,}112\), \(12{,}121\), \(12{,}211\), \(21{,}112\), \(21{,}121\), \(21{,}211\), and \(22{,}111\)

- First choose four students for one group. - Determine whether the remaining group requires another choice. - Account for the fact that the two groups have no labels.

1. Choose four of the eight students for one group: \(\binom{8}{4}=70\). 2. The remaining four students automatically form the other group. 3. Each grouping has been counted twice because choosing either group first produces the same pair of unlabeled groups. Therefore, \(70\div2=35\).

\(35\) groupings

- Does the order of the two selected students matter? - Once two students are selected, what is determined about the rest of the class? - Compare \(\binom{n}{k}\) with \(\binom{n}{n-k}\).

1. Since order does not matter, the number of pairs is \(\binom{15}{2}=\frac{15\cdot 14}{2}=105\). 2. The number of groups of \(13\) students is \(\binom{15}{13}=105\). 3. Every selection of \(2\) students uniquely determines the complementary group of \(13\) students. This illustrates \(\binom{n}{k}=\binom{n}{n-k}\), so \(\binom{15}{2}=\binom{15}{13}\).

a) \(105\) b) \(105\). The counts are equal because \(\binom{15}{2}=\binom{15}{13}\); choosing the two selected students determines the thirteen not selected.

- Use combinations because order does not matter. - Compute both binomial coefficients. - When \(6\) flavors are selected, how many are left out? - How are choosing and leaving out related?

1. For part a), \(\binom{8}{2}=\frac{8\cdot 7}{2}=28\). 2. For part b), \(\binom{8}{6}=28\). 3. Every choice of \(2\) flavors uniquely determines the \(6\) flavors not chosen, and every choice of \(6\) flavors uniquely determines the \(2\) flavors left out. Therefore, \(\binom{8}{2}=\binom{8}{6}\).

a) \(28\) b) \(28\) c) Choosing \(2\) flavors determines the complementary set of \(6\), so \(\binom{8}{2}=\binom{8}{6}\).

- How many balls remain when \(3\) are selected? - How are the selected and unselected sets related? - Compute both binomial coefficients and compare them.

1. The number of three-ball selections is \(\binom{7}{3}=35\). 2. The number of four-ball selections is \(\binom{7}{4}=35\). 3. The claim is true. Every set of \(3\) selected balls uniquely determines the complementary set of \(4\) balls left behind. Thus \(\binom{7}{3}=\binom{7}{4}\).

a) \(35\) b) The claim is true because \(\binom{7}{3}=\binom{7}{4}=35\). Choosing \(3\) balls determines the \(4\) balls left behind.

- How many groups of \(11\) can be selected from \(15\)? - Does selection order matter? - Choosing \(11\) players is equivalent to choosing which \(4\) are left out. - How many of the possible groups match the predetermined group?

1. The number of possible eleven-player groups is \(\binom{15}{11}=\binom{15}{4}=1365\). 2. Exactly one of these equally likely groups is the predetermined group, so the probability is \(\frac{1}{1365}\approx 0.0007326\), or approximately \(0.073\%\).

\(\frac{1}{1365}\approx 0.073\%\)

- Count all selections of five tickets from \(30\). - For part b, choose all five tickets from the nonwinning tickets. - For part c, include the specified ticket first and choose the remaining four tickets from the others.

1. The total number of selections is \(\binom{30}{5}=142{,}506\). 2. There is one way to select all five winning tickets. Therefore, \(P(\text{all winning})=\frac{1}{142506}\). 3. There are \(25\) nonwinning tickets, so \(P(\text{none winning})=\frac{\binom{25}{5}}{\binom{30}{5}}=\frac{53130}{142506}=\frac{1265}{3393}\approx 0.3728\). 4. If one specified ticket is included, the other four tickets can be chosen from the remaining \(29\): \(P(\text{specified ticket included})=\frac{\binom{29}{4}}{\binom{30}{5}}=\frac{1}{6}\).

a) \(\frac{1}{142506}\) b) \(\frac{1265}{3393}\approx 0.3728\) c) \(\frac{1}{6}\)

- Count all groups of three apples. - For each event, count favorable groups using combinations. - For part d, choose the bruised and unbruised apples separately, then multiply the counts.

1. The total number of selections is \(\binom{20}{3}=1140\). 2. For three bruised apples, \(P(\text{all bruised})=\frac{\binom{4}{3}}{\binom{20}{3}}=\frac{1}{285}\). 3. There are \(16\) apples that are not bruised, so \(P(\text{none bruised})=\frac{\binom{16}{3}}{\binom{20}{3}}=\frac{28}{57}\). 4. If the specified apple is included, choose two more apples from the remaining \(19\): \(P(\text{specified apple included})=\frac{\binom{19}{2}}{\binom{20}{3}}=\frac{3}{20}\). 5. For exactly two bruised apples, choose two of the four bruised apples and one of the sixteen unbruised apples: \(P(\text{exactly two bruised})=\frac{\binom{4}{2}\binom{16}{1}}{\binom{20}{3}}=\frac{8}{95}\).

a) \(\frac{1}{285}\) b) \(\frac{28}{57}\) c) \(\frac{3}{20}\) d) \(\frac{8}{95}\)

- How many people remain after \(9\) are selected? - Is selecting the team equivalent to naming those left out? - Does each team correspond to exactly one complementary group? - Use the binomial coefficient formula to calculate the count.

1. The number of teams is \(\binom{12}{9}\). 2. By symmetry, \(\binom{12}{9}=\binom{12}{3}=\frac{12\cdot 11\cdot 10}{3\cdot 2\cdot 1}=220\). 3. Every nine-person team uniquely determines the three people left out, and every group of three people left out uniquely determines the nine-person team. Therefore, the two counts are equal.

a) \(220\) b) Each selection of \(9\) team members uniquely determines the \(3\) members left out, so \(\binom{12}{9}=\binom{12}{3}\).

- Paths with a fixed number of successes can be counted by choosing the success positions. - A fixed beginning reduces the number of open positions. - How many successes have already occurred in \(F,S,F\)? - Choose the remaining success positions among the remaining stages.

1. Choose the \(4\) success positions among \(10\): \(\binom{10}{4}=210\). 2. In the fixed beginning \(F,S,F\), one success has already occurred. Seven positions remain, and exactly \(3\) more successes are needed. 3. The number of qualifying paths is \(\binom{7}{3}=35\).

a) \(210\) b) \(35\)

- Think about the objects that are not selected. - How does a choice of included objects determine a choice of excluded objects? - Look at the location of the greatest entries in a row of Pascal's triangle. - Use the symmetry of the row.

1. Choosing the \(k\) objects that are included uniquely determines the \(n-k\) objects that are left out. Therefore, the two choices are counted by the same number, so \(\binom{n}{k}=\binom{n}{n-k}\). 2. Choosing \(38\) of \(40\) objects is equivalent to choosing the \(2\) objects that are left out. Thus \(\binom{40}{38}=\binom{40}{2}\). 3. For fixed \(n\), the binomial coefficients increase toward the middle of the row and then decrease symmetrically. Since \(12\) is even, the single middle value occurs at \(k=6\), so \(\binom{12}{6}\) is greatest.

a) Every choice of \(k\) included objects uniquely determines the \(n-k\) objects that are excluded, so the numbers of choices are equal. b) Choosing \(38\) objects to include is equivalent to choosing \(2\) objects to exclude. c) \(k=6\), because the greatest entry in row \(12\) of Pascal's triangle is the middle entry.

- Does the order in which members are chosen change the group? - Can the same member be selected more than once? - In part b), count the choices from each subgroup separately and then combine them. - A draw without replacement models selecting distinct people.

1. Since order does not matter, the number of six-member groups is \(\binom{16}{6}=8008\). 2. Choose \(3\) of the \(9\) women and \(3\) of the \(7\) men. The number of groups is \(\binom{9}{3}\cdot\binom{7}{3}=84\cdot 35=2940\). 3. Use a bag with \(16\) distinct objects, one for each member: \(9\) of one type and \(7\) of another. Draw \(6\) objects without replacement, and ignore the order of the draw.

a) \(8008\) groups b) \(2940\) groups c) Draw \(6\) objects without replacement and without regard to order from a bag containing \(16\) distinct objects, with \(9\) representing women and \(7\) representing men.

- Decide whether order matters in each selection. - For exactly \(2\) defective resistors, how many working resistors must also be selected? - “At least \(3\)” includes which separate cases? - A draw without replacement models selecting distinct components.

1. The total number of four-resistor selections is \(\binom{25}{4}=12{,}650\). 2. For exactly \(2\) defective resistors, choose \(2\) of the \(5\) defective resistors and \(2\) of the \(20\) working resistors: \(\binom{5}{2}\cdot\binom{20}{2}=10\cdot 190=1900\). 3. “At least \(3\)” means exactly \(3\) or exactly \(4\) defective resistors. The number is \(\binom{5}{3}\cdot\binom{20}{1}+\binom{5}{4}\cdot\binom{20}{0}=10\cdot 20+5\cdot 1=205\). 4. Use a bag containing \(25\) distinct objects: \(5\) marked defective and \(20\) marked working. Draw \(4\) objects without replacement, and ignore the order.

a) \(12{,}650\) selections b) \(1900\) selections c) \(205\) selections d) Draw \(4\) objects without replacement and without regard to order from a bag containing \(5\) defective objects and \(20\) working objects.

- Count all groups of four players. - For part a, count groups matching the one specified set. - For part b, choose all four players from those outside the council. - State whether a player can be selected more than once and whether order changes the group.

1. The total number of groups is \(\binom{22}{4}=7315\). 2. Exactly one group contains all four specified council members. Therefore, \(P(\text{all four council members})=\frac{1}{7315}\). 3. For no council members, choose all four players from the other \(18\): \(P(\text{no council members})=\frac{\binom{18}{4}}{\binom{22}{4}}=\frac{612}{1463}\approx 0.4183\). 4. The model is an unordered selection of \(4\) people from \(22\) without replacement.

a) \(\frac{1}{7315}\) b) \(\frac{612}{1463}\approx 0.4183\) Sampling model: select \(4\) of \(22\) without replacement; order does not matter.

- Separate the chocolates into nut-filled and not nut-filled. - Count all groups of four chocolates. - For each part, choose the required number from each category.

1. The total number of selections is \(\binom{15}{4}=1365\). 2. For four nut-filled chocolates, \(P(\text{four nut-filled})=\frac{\binom{5}{4}\binom{10}{0}}{\binom{15}{4}}=\frac{1}{273}\). 3. For no nut-filled chocolates, \(P(\text{none nut-filled})=\frac{\binom{5}{0}\binom{10}{4}}{\binom{15}{4}}=\frac{2}{13}\). 4. The model is an unordered selection of \(4\) items from \(15\) without replacement, with \(5\) items designated as nut-filled.

a) \(\frac{1}{273}\) b) \(\frac{2}{13}\) Sampling model: select \(4\) of \(15\) without replacement; order does not matter.

- Count all five-person committees. - Choose the teenagers and adults separately. - Divide the favorable count by the total count.

1. The total number of committees is \(\binom{25}{5}\). 2. A favorable committee contains three of the ten teenagers and two of the fifteen adults, so the number of favorable committees is \(\binom{10}{3}\binom{15}{2}\). 3. Therefore, \(P(\text{exactly three teenagers})=\frac{\binom{10}{3}\binom{15}{2}}{\binom{25}{5}}=\frac{60}{253}\approx 0.2372\).

\(\frac{60}{253}\approx 23.72\%\)

- Does the order of the chosen side dishes matter? - In part b), focus on the dishes that are not selected. - For part c), each dish creates a yes-or-no choice. - Recall the sum of the entries in a row of Pascal's triangle.

1. The number of selections of exactly \(4\) side dishes is \(\binom{12}{4}=495\). 2. Choosing the \(8\) side dishes that are included uniquely determines the \(4\) side dishes that are left out. Therefore, \(\binom{12}{8}=\binom{12}{4}\). 3. Each of the \(12\) side dishes has two possibilities: chosen or not chosen. Thus the total number of selections is \(2^{12}=4096\). Equivalently, \(\sum_{k=0}^{12}\binom{12}{k}=4096\).

a) \(495\) selections b) Choosing \(8\) dishes to include is equivalent to choosing \(4\) dishes to leave out, so \(\binom{12}{8}=\binom{12}{4}\). c) \(4096\) selections

- How many positions must be chosen for the correct answers? - “At least \(9\)” includes which two cases? - For the total, group the answer patterns by their number of correct answers. - Recall the sum of the binomial coefficients in one row of Pascal's triangle.

1. Choose the \(7\) question positions that are answered correctly: \(\binom{10}{7}=\binom{10}{3}=120\). A corresponding selection model uses \(10\) labeled objects, one for each question, and selects \(7\) without replacement; the selected objects represent the correctly answered questions. 2. “At least \(9\)” means exactly \(9\) or exactly \(10\) correct answers. Thus the number of patterns is \(\binom{10}{9}+\binom{10}{10}=10+1=11\). 3. Summing over every possible number of correct answers gives \(\sum_{k=0}^{10}\binom{10}{k}=2^{10}=1024\), using the identity \(\sum_{k=0}^{n}\binom{n}{k}=2^n\).

a) \(120\) answer patterns; select \(7\) of \(10\) labeled question objects without replacement and without regard to order. b) \(11\) answer patterns c) \(\sum_{k=0}^{10}\binom{10}{k}=2^{10}=1024\)

- Count all samples of five drives. - For a fixed number of defective drives, choose defective and nondefective drives separately. - “At most one” combines the cases \(0\) and \(1\). - For “at least one,” use the complement of none.

1. The total number of samples is \(\binom{40}{5}=658{,}008\). 2. For exactly two defective drives, \(P(X=2)=\frac{\binom{6}{2}\binom{34}{3}}{\binom{40}{5}}=\frac{3740}{27417}\approx 0.1364\). 3. At most one defective drive means zero or one defective drive: \(P(X\le 1)=\frac{\binom{6}{0}\binom{34}{5}+\binom{6}{1}\binom{34}{4}}{\binom{40}{5}}=\frac{23188}{27417}\approx 0.8458\). 4. For at least one defective drive, use the complement of no defective drives: \(P(X\ge 1)=1-\frac{\binom{34}{5}}{\binom{40}{5}}=\frac{15823}{27417}\approx 0.5771\).

a) \(\frac{3740}{27417}\approx 13.64\%\) b) \(\frac{23188}{27417}\approx 84.58\%\) c) \(\frac{15823}{27417}\approx 57.71\%\)

- Determine how many students do not speak Spanish. - Count all groups of six students. - For each case, choose Spanish speakers and non-Spanish speakers separately. - A majority in a group of six means \(4\), \(5\), or \(6\).

1. The total number of groups is \(\binom{24}{6}=134{,}596\). 2. For exactly three Spanish speakers, \(P(X=3)=\frac{\binom{10}{3}\binom{14}{3}}{\binom{24}{6}}=\frac{1560}{4807}\approx 0.3245\). 3. For no Spanish speakers, \(P(X=0)=\frac{\binom{10}{0}\binom{14}{6}}{\binom{24}{6}}=\frac{39}{1748}\approx 0.0223\). 4. A majority means \(4\), \(5\), or \(6\) Spanish speakers. Therefore, \(P(X>3)=\frac{\binom{10}{4}\binom{14}{2}+\binom{10}{5}\binom{14}{1}+\binom{10}{6}\binom{14}{0}}{\binom{24}{6}}=\frac{816}{4807}\approx 0.1698\).

a) \(\frac{1560}{4807}\approx 32.45\%\) b) \(\frac{39}{1748}\approx 2.23\%\) c) \(\frac{816}{4807}\approx 16.98\%\)

- Choose members from the two categories separately. - Divide the favorable team count by the total number of six-person teams. - In part c, interpret each combination before interpreting their sum.

1. Choose four of the fifteen college students and two of the ten working professionals: \(\binom{15}{4}\binom{10}{2}=1365\cdot45=61{,}425\). 2. The total number of six-person teams is \(\binom{25}{6}=177{,}100\). Therefore, \(P(\text{exactly four college students})=\frac{61425}{177100}=\frac{351}{1012}\approx 0.3468\). 3. The term \(\binom{15}{6}\) counts teams made entirely of college students, and \(\binom{10}{6}\) counts teams made entirely of working professionals. Their sum counts teams consisting entirely of one group or the other.

a) \(61{,}425\) b) \(\frac{351}{1012}\approx 34.68\%\) c) It counts six-person teams made entirely of college students or entirely of working professionals.

- Choose books from each genre separately. - “At least seven” includes exactly seven and exactly eight. - Add the favorable counts and divide by the total number of eight-book selections.

1. The factor \(\binom{18}{5}\) chooses five mystery novels, and \(\binom{12}{3}\) chooses three historical novels. Multiplying combines the two separate choices into one eight-book selection. 2. At least seven mystery novels means either seven mystery novels and one historical novel or eight mystery novels and no historical novels. The favorable count is \(\binom{18}{7}\binom{12}{1}+\binom{18}{8}\binom{12}{0}=425{,}646\). 3. The total number of eight-book selections is \(\binom{30}{8}=5{,}852{,}925\). Therefore, \(P(\text{at least seven mystery novels})=\frac{425646}{5852925}=\frac{3638}{50025}\approx 0.0727\).

a) The first factor chooses the mystery novels, the second chooses the historical novels, and the product combines the choices. b) \(\frac{3638}{50025}\approx 7.27\%\)

- Count the choices from the information technology group first. - Count the choices from the marketing group separately. - How should the two independent selections be combined? - Does the order in which team members are named matter?

1. Choose \(3\) of the \(12\) information technology specialists: \(\binom{12}{3}=220\). 2. Choose \(2\) of the \(8\) marketing specialists: \(\binom{8}{2}=28\). 3. Each first choice can be paired with each second choice, so the multiplication principle gives \(220\cdot 28=6160\) teams.

\(6160\) teams

- List the possible apple-and-pear counts that contain more apples than pears. - Count each possible case separately. - The cases do not overlap, so how should their counts be combined? - Selecting the fruits at the same time means order does not matter.

1. With \(4\) fruits total, more apples than pears means either \(3\) apples and \(1\) pear or \(4\) apples and \(0\) pears. 2. The number of selections with \(3\) apples and \(1\) pear is \(\binom{10}{3}\binom{6}{1}=120\cdot 6=720\). 3. The number of selections with \(4\) apples and no pears is \(\binom{10}{4}\binom{6}{0}=210\cdot 1=210\). 4. The cases are disjoint, so the total is \(720+210=930\).

\(\binom{10}{3}\binom{6}{1}+\binom{10}{4}\binom{6}{0}=720+210=930\) selections

- Count all six-student groups. - For part b, choose the girls and boys separately. - For part c, include the two specified students first and choose the remaining four. - State whether students can be selected more than once and whether order matters.

1. The total number of groups is \(\binom{18}{6}=18{,}564\). 2. For only girls, \(P(\text{only girls})=\frac{\binom{10}{6}}{\binom{18}{6}}=\frac{5}{442}\approx 0.0113\). 3. For exactly three girls and three boys, \(P(3\text{ girls and }3\text{ boys})=\frac{\binom{10}{3}\binom{8}{3}}{\binom{18}{6}}=\frac{80}{221}\approx 0.3620\). 4. If the two representatives are included, choose the remaining four students from the other \(16\): \(P(\text{both representatives})=\frac{\binom{16}{4}}{\binom{18}{6}}=\frac{5}{51}\approx 0.0980\). 5. The model is an unordered selection of \(6\) students from \(18\) without replacement.

a) \(\frac{5}{442}\approx 1.13\%\) b) \(\frac{80}{221}\approx 36.20\%\) c) \(\frac{5}{51}\approx 9.80\%\) d) Select \(6\) of \(18\) without replacement; order does not matter.

- If two specified people are included, count the remaining open positions. - For a fixed composition, choose members from each category separately. - Interpret the subtraction in part b as a complement.

1. For event \(A\), Sophie and Leon are fixed members, so choose the other two team members from the remaining \(16\): \(P(A)=\frac{\binom{16}{2}}{\binom{18}{4}}\). 2. For event \(B\), choose two of the ten girls and two of the eight boys: \(P(B)=\frac{\binom{10}{2}\binom{8}{2}}{\binom{18}{4}}\). 3. In part b, \(\binom{8}{4}\) counts teams made only of boys. Subtracting these from all four-person teams counts teams with at least one girl.

a) \(P(A)=\frac{\binom{16}{2}}{\binom{18}{4}}\); \(P(B)=\frac{\binom{10}{2}\binom{8}{2}}{\binom{18}{4}}\) b) The team contains at least one girl.

- When specified people are included, reduce both the remaining team size and the available group. - Multiply selections made from separate groups. - Interpret the subtraction as excluding all-hardware teams.

1. For event \(C\), the two specified people are already included, so the other \(3\) team members are selected from the remaining \(13\). Thus, \(P(C)=\frac{\binom{13}{3}}{\binom{15}{5}}\). 2. For event \(D\), select \(3\) of the \(9\) hardware specialists and \(2\) of the \(6\) software specialists. Thus, \(P(D)=\frac{\binom{9}{3}\binom{6}{2}}{\binom{15}{5}}\). 3. The expression in part b) subtracts teams containing only hardware specialists from all teams. It therefore represents the event that the team contains at least one software specialist.

a) \(P(C)=\frac{\binom{13}{3}}{\binom{15}{5}}\) \(P(D)=\frac{\binom{9}{3}\binom{6}{2}}{\binom{15}{5}}\) b) The team contains at least one software specialist.

- Distinct roles make order matter. - A group with no roles is an unordered selection. - For part c, choose members from the two categories separately.

1. A valid question for part a is: “How many ways can the four distinct roles be assigned to four of the \(25\) students?” The roles make order matter. 2. A valid question for part b is: “How many six-person groups can be selected from the \(25\) students?” Group membership, not order, matters. 3. For part c, choose three of the fifteen female students and two of the ten other students: \(P(X=3)=\frac{\binom{15}{3}\binom{10}{2}}{\binom{25}{5}}=\frac{195}{506}\approx 0.3854\).

a) How many ways can the chair, recorder, treasurer, and host roles be assigned to four of the \(25\) students? b) How many six-person groups can be selected from the \(25\) students? c) \(\frac{195}{506}\approx 38.54\%\)

- Count all four-ball selections. - Choose the required number from each color separately. - “At least three blue” includes exactly three and exactly four blue.

1. The total number of selections is \(\binom{30}{4}=27{,}405\). 2. For \(E_1\), \(P(E_1)=\frac{\binom{15}{2}\binom{10}{1}\binom{5}{1}}{\binom{30}{4}}=\frac{50}{261}\approx 0.1916\). 3. For \(E_2\), choose all four balls from the \(15\) nonred balls: \(P(E_2)=\frac{\binom{15}{4}}{\binom{30}{4}}=\frac{13}{261}\approx 0.0498\). 4. For \(E_3\), add the cases of exactly three and exactly four blue balls: \(P(E_3)=\frac{\binom{5}{3}\binom{25}{1}+\binom{5}{4}\binom{25}{0}}{\binom{30}{4}}=\frac{17}{1827}\approx 0.00930\).

\(E_1\): \(\frac{50}{261}\approx 19.16\%\) \(E_2\): \(\frac{13}{261}\approx 4.98\%\) \(E_3\): \(\frac{17}{1827}\approx 0.93\%\)

- Count all six-person teams. - Select members from each subgroup separately, then multiply.

1. There are \(\binom{30}{6}\) possible teams. 2. For all men, \(P=\frac{\binom{18}{6}}{\binom{30}{6}}\approx 0.03126\). 3. For exactly \(3\) women and \(3\) men, \(P=\frac{\binom{12}{3}\binom{18}{3}}{\binom{30}{6}}\approx 0.30234\).

a) Approximately \(0.03126\), or \(3.13\%\) b) Approximately \(0.30234\), or \(30.23\%\)

- Once the participant group is fixed, what is automatically determined? - Use the symmetry identity for binomial coefficients. - Which equivalent binomial coefficient is easier to calculate?

1. Every choice of \(22\) participants uniquely determines the \(3\) students who are excluded, and every choice of \(3\) excluded students uniquely determines the \(22\) participants. Therefore, \(\binom{25}{22}=\binom{25}{3}\). 2. Compute the smaller equivalent binomial coefficient: \(\binom{25}{3}=\frac{25\cdot 24\cdot 23}{3\cdot 2\cdot 1}=2300\).

a) Each group of \(22\) participants corresponds to exactly one group of \(3\) nonparticipants, so \(\binom{25}{22}=\binom{25}{3}\). b) \(2300\) groups

- Count all committees first. - Break “at least \(5\)” into the possible exact cases. - Interpret the numerator by identifying the group from which all members are chosen.

1. There are \(\binom{25}{6}\) possible committees. 2. For exactly \(3\) women and \(3\) men, \(P=\frac{\binom{15}{3}\binom{10}{3}}{\binom{25}{6}}\approx 0.30830\). 3. At least \(5\) men means either \(5\) men and \(1\) woman or \(6\) men. Thus, \(P=\frac{\binom{10}{5}\binom{15}{1}+\binom{10}{6}}{\binom{25}{6}}\approx 0.02253\). 4. The expression in part c) is the probability that all \(6\) selected members are women.

a) Approximately \(0.30830\), or \(30.83\%\) b) Approximately \(0.02253\), or \(2.25\%\) c) The probability that the committee consists entirely of women

- Multiply the number-set choices by the color choices. - Separate matching and nonmatching numbers. - For an incorrect color, count the two colors not chosen by the player.

1. There are \(\binom{30}{4}\) number sets and \(3\) colors, so the total number of results is \(3\binom{30}{4}=82{,}215\). 2. To match exactly \(3\) numbers, choose \(3\) of the player’s \(4\) numbers and \(1\) of the other \(26\) numbers. The color must match. Thus, \(P=\frac{\binom{4}{3}\binom{26}{1}}{3\binom{30}{4}}=\frac{104}{82215}\approx 0.001265\). 3. For no number matches, choose all \(4\) numbers from the other \(26\). There are \(2\) incorrect colors. Thus, \(P=\frac{2\binom{26}{4}}{3\binom{30}{4}}=\frac{29900}{82215}\approx 0.36368\).

a) \(82{,}215\) b) Approximately \(0.001265\), or \(0.1265\%\) c) Approximately \(0.36368\), or \(36.37\%\)

- Count all selections of three chocolates. - For the same filling, count each filling type separately and add. - For three different fillings, choose one chocolate from each category.

1. The total number of three-chocolate selections is \(\binom{18}{3}=816\). 2. For the same filling, add the selections of three marzipan, three nougat, or three caramel chocolates: \(P(\text{same filling})=\frac{\binom{8}{3}+\binom{6}{3}+\binom{4}{3}}{\binom{18}{3}}=\frac{80}{816}=\frac{5}{51}\). 3. For three different fillings, choose one chocolate of each type: \(P(\text{three different fillings})=\frac{\binom{8}{1}\binom{6}{1}\binom{4}{1}}{\binom{18}{3}}=\frac{192}{816}=\frac{4}{17}\).

1. \(\frac{5}{51}\approx 9.8\%\) 2. \(\frac{4}{17}\approx 23.5\%\)

- Decide whether the selection is made with or without replacement. - Use combinations to count all possible three-student groups. - For the first part, count groups chosen only from the \(10\) women. - For the second part, use the multiplication principle to choose one person from each group.

1. Because order does not matter, the total number of possible groups is \(\binom{20}{3}=1140\). 2. The number of all-women groups is \(\binom{10}{3}=120\). Therefore, \(P(\text{all women})=\frac{120}{1140}=\frac{2}{19}\approx 0.1053\). 3. The number of groups containing one person from each category is \(\binom{10}{1}\binom{7}{1}\binom{3}{1}=210\). Therefore, \(P(\text{one from each group})=\frac{210}{1140}=\frac{7}{38}\approx 0.1842\).

1. \(\frac{2}{19}\approx 0.1053\), or about \(10.5\%\) 2. \(\frac{7}{38}\approx 0.1842\), or about \(18.4\%\)

- Separate the \(25\) numbers into the player’s four choices and the other \(21\) numbers. - For an exact number of matches, choose from both groups. - “At least three” combines the cases of three and four matches.

1. The total number of four-number sets is \(\binom{25}{4}=12{,}650\). 2. For exactly four matches, \(P(X=4)=\frac{\binom{4}{4}\binom{21}{0}}{\binom{25}{4}}=\frac{1}{12650}\). 3. For exactly two matches, \(P(X=2)=\frac{\binom{4}{2}\binom{21}{2}}{\binom{25}{4}}=\frac{126}{1265}\approx 0.0996\). 4. At least three matches means exactly three or exactly four: \(P(X\ge 3)=\frac{\binom{4}{3}\binom{21}{1}+\binom{4}{4}\binom{21}{0}}{\binom{25}{4}}=\frac{17}{2530}\approx 0.00672\).

1. \(\frac{1}{12650}\approx 0.0079\%\) 2. \(\frac{126}{1265}\approx 9.96\%\) 3. \(\frac{17}{2530}\approx 0.672\%\)

- Choose the members of the teams one team at a time. - Track how many students remain after each choice. - Account for the fact that the three teams are unlabeled. - Determine how many orders produce the same three teams.

1. Choose three of the nine students for a first temporary team: \(\binom{9}{3}=84\). 2. Choose three of the remaining six students for a second temporary team: \(\binom{6}{3}=20\). 3. The remaining three students form the third team. This gives \(84\cdot20=1680\) ordered team selections. 4. Because the three teams are unlabeled, each grouping is counted \(3!=6\) times. Therefore, \(1680\div6=280\).

\(280\) groupings

- Choose the positions occupied by the defective components. - Use the binomial probability formula with the given \(n\), \(k\), and \(p\). - For part c), arrange the good components first and identify the available gaps. - How many gaps are created by \(5\) good components?

1. Choose the \(2\) defective positions among \(10\): \(\binom{10}{2}=45\). 2. The binomial probability is \(P(X=2)=\binom{10}{2}(0.05)^2(0.95)^8\approx 0.0746\). 3. For part c), first place the \(5\) good components: \(\_G\_G\_G\_G\_G\_\). This creates \(6\) gaps. To avoid adjacent defects, choose \(3\) distinct gaps for the defective components. The number of sequences is \(\binom{6}{3}=20\).

a) \(45\) b) \(P(X=2)\approx 0.0746\) c) \(20\)

- Count all unordered pairs of balls. - For part a, count pairs within each color. - For part b, list pairs systematically by their label sums. - For part c, use the product of the red and green choices.

1. There are \(\binom{7}{2}=21\) equally likely pairs. 2. Same-color pairs number \(\binom{3}{2}+\binom{2}{2}+\binom{2}{2}=5\), so \(P(\text{same color})=\frac{5}{21}\). 3. The prime sums possible are \(2\), \(3\), \(5\), and \(7\). There is \(1\) pair with sum \(2\), \(4\) pairs with sum \(3\), \(4\) pairs with sum \(5\), and \(3\) pairs with sum \(7\). Thus there are \(1+4+4+3=12\) favorable pairs, so \(P(\text{prime sum})=\frac{12}{21}=\frac{4}{7}\). 4. Choose one of the three red balls and one of the two green balls. There are \(3\cdot2=6\) favorable pairs, so \(P(\text{one red and one green})=\frac{6}{21}=\frac{2}{7}\).

a) \(\frac{5}{21}\) b) \(\frac{4}{7}\) c) \(\frac{2}{7}\)