Free math worksheets

- First determine from the diagram how many rays are displayed. - List every pair of displayed rays systematically. - Do not count the same pair twice in reverse order.

1. The diagram shows four rays labeled \(a\), \(b\), \(c\), and \(d\). 2. Each smaller angle is determined by an unordered pair of rays. 3. The pairs are \((a,b)\), \((a,c)\), \((a,d)\), \((b,c)\), \((b,d)\), and \((c,d)\). 4. There are \(6\) different pairs, so there are \(6\) different smaller angles.

\(6\) angles

- Decide whether changing the order of the same selected projects would create a new group. - Use notation that represents choosing \(k\) objects from \(n\) objects without regard to order.

1. The order of the selected projects does not matter, so use a combination. 2. Choosing 4 projects from 9 is represented by \(\binom{9}{4}\).

\(\binom{9}{4}\)

- Compare the members in the two sets rather than the order in which their names are written. - Ask whether any person's role changes when the names are reordered.

1. No. Both descriptions contain exactly the same three people. 2. A combination does not treat a different ordering of the same selected objects as a new outcome.

No. They are the same committee, so order does not matter in a combination.

- Fix the number of tokens in jar A first, then list what can happen in jars B and C. - Remember that the jars are labeled, so \((2,1,0)\) and \((1,2,0)\) are different. - Check that every listed triple has nonnegative entries that add to \(3\).

1. Group the distributions by the number of tokens in jar A. 2. If \(a=3\): \((3,0,0)\). 3. If \(a=2\): \((2,1,0)\), \((2,0,1)\). 4. If \(a=1\): \((1,2,0)\), \((1,1,1)\), \((1,0,2)\). 5. If \(a=0\): \((0,3,0)\), \((0,2,1)\), \((0,1,2)\), \((0,0,3)\). 6. There are \(10\) distributions. The same count can be represented by arranging three stars and two separators.

\((3,0,0)\), \((2,1,0)\), \((2,0,1)\), \((1,2,0)\), \((1,1,1)\), \((1,0,2)\), \((0,3,0)\), \((0,2,1)\), \((0,1,2)\), \((0,0,3)\); \(10\) distributions

- First determine how many of the five digits must be \(2\). - Once that number is fixed, what choice of positions determines the entire number? - Use an unordered position choice rather than listing cases one at a time.

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 occupied by \(2\). This gives \(\binom{5}{2}=10\) possible numbers.

\(\binom{5}{2}=10\) numbers

- List the pairs systematically so none are repeated or omitted. - Does the order of the two students matter? - Compare your list with what the binomial coefficient counts.

a) The two-student sets are \(\{A,B\}\), \(\{A,C\}\), \(\{A,D\}\), \(\{B,C\}\), \(\{B,D\}\), and \(\{C,D\}\). There are \(6\) selections. b) \(\binom{4}{2}=\frac{4\cdot3}{2\cdot1}=6\), which matches the list because combinations count selections without regard to order. c) 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 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.

- Identify which expressions form symmetry pairs. - Use \(\binom{n}{k}=\binom{n}{n-k}\) before doing unnecessary arithmetic. - Simplify factors before multiplying when a direct calculation remains.

a) \(\binom{18}{2}=\frac{18\cdot17}{2}=153\). b) By symmetry, \(\binom{18}{16}=\binom{18}{2}=153\). c) \(\binom{25}{3}=\frac{25\cdot24\cdot23}{3\cdot2\cdot1}=2300\). d) By symmetry, \(\binom{25}{22}=\binom{25}{3}=2300\).

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

- Evaluate or simplify the binomial coefficients one expression at a time. - Compare the first two results after calculating them. - In part c, look for symmetry before computing either coefficient.

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

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

- Does order matter within each selected group? - Which formula counts \(k\) objects selected from \(n\) without replacement? - For part b, identify the calculator function that evaluates a binomial coefficient.

a) \(\binom{9}{2}=\frac{9\cdot8}{2\cdot1}=36\). b) \(\binom{18}{7}=31{,}824\).

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

- 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 \(0.073\%\).

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

- How many choices are available at each pulse position? - For a fixed number of positive pulses, think about which positions are positive. - “At least \(6\)” includes \(6\), \(7\), and \(8\).

a) Each of the \(8\) positions has \(2\) choices, so there are \(2^8=256\) sequences. b) Choose the \(3\) positions for positive pulses: \(\binom{8}{3}=56\). c) At least \(6\) positive pulses means \(6\), \(7\), or \(8\). The number of favorable sequences is \(\binom{8}{6}+\binom{8}{7}+\binom{8}{8}=37\). Thus, the probability is \(\frac{37}{256}\approx0.1445\).

a) \(256\) b) \(56\) c) \(\frac{37}{256}\approx0.1445\)

- Can symmetry make any coefficient easier to compute? - Use the factorial formula only as far as needed before canceling. - For part f, evaluate each binomial coefficient before comparing the two sides.

a) \(\binom{9}{3}=84\). b) By symmetry, \(\binom{15}{13}=\binom{15}{2}=105\). c) \(\binom{25}{2}=300\). d) By symmetry, \(\binom{50}{48}=\binom{50}{2}=1225\). e) By symmetry, \(\binom{2000}{1999}=\binom{2000}{1}=2000\). f) The left side is \(\binom{10}{3}+\binom{10}{4}=120+210=330\), and the right side is \(\binom{11}{4}=330\). The two sides are equal.

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

\(A\): This event requires two apples, two pears, and one banana, so it matches expression (2). \(B\): This event requires all five fruits to come from the \(12\) apples and bananas, so it matches expression (1). \(C\): This event 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. - For the symmetry explanation, ask whether any student is favored before the random selection.

a) There are \(\binom{25}{4}\) possible groups. Groups containing Luke are formed by choosing the other \(3\) students from the remaining \(24\), so \(P=\frac{\binom{24}{3}}{\binom{25}{4}}=\frac{4}{25}=0.16\). b) By symmetry, each of the \(25\) students has the same chance of being selected into one of the \(4\) places, so 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.

a) Represent the \(10\) trials by \(10\) labeled positions and place one success marker in each position where a success occurs. b) 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. c) Choose the \(3\) success positions from the \(10\) trial 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

- A disjoint event can use only outcomes outside \(A\). - Count the available outcomes in the complement. - The event is an unordered subset.

1. The complement of \(A\) contains \(10-4=6\) outcomes. 2. A disjoint event \(B\) must choose all \(3\) outcomes from those \(6\). 3. The number is \(\binom{6}{3}=20\).

There are \(\binom{6}{3}=20\) such events.

- Think of the total \(7\) as identical units that must be divided among three variables. - How many separators are needed to create three groups? - Once stars and separators are combined, count which positions can hold the separators.

1. Represent the \(7\) identical units by seven stars. 2. Two separators divide the stars into the amounts assigned to \(x\), \(y\), and \(z\). 3. There are \(7+2=9\) positions containing seven stars and two separators. 4. Choose the two separator positions: \(\binom{9}{2}=36\).

\(36\) solutions

- 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 \(\binom{n}{n-k}\).

a) Since order does not matter, the number of pairs is \(\binom{15}{2}=\frac{15\cdot14}{2}=105\). b) The number of complementary groups of \(13\) students is \(\binom{15}{13}=105\). Every choice of \(2\) selected students uniquely determines the \(13\) not selected, illustrating \(\binom{n}{k}=\binom{n}{n-k}\). Thus, \(\binom{15}{2}=\binom{15}{13}\).

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.

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

a) The number of three-ball selections is \(\binom{7}{3}=35\). b) The number of four-ball selections is \(\binom{7}{4}=35\). The claim is true because 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.

- Count all selections of five tickets from \(30\). - For each event, decide what must be included or excluded from the selected set. - For part c, compare selections containing the specified ticket with all five-ticket selections.

a) There are \(\binom{30}{5}=142{,}506\) equally likely five-ticket selections. Exactly one selection contains all five winning tickets, so \(P=\frac{1}{142506}\). b) There are \(25\) nonwinning tickets. Thus, \(P=\frac{\binom{25}{5}}{\binom{30}{5}}=\frac{1265}{3393}\approx0.3728\). c) If the specified ticket is included, choose the other four tickets from the remaining \(29\). Thus, \(P=\frac{\binom{29}{4}}{\binom{30}{5}}=\frac{1}{6}\).

a) \(\frac{1}{142506}\) b) \(\frac{1265}{3393}\approx0.3728\) c) \(\frac{1}{6}\)

- Treat the three apples as an unordered sample without replacement. - Count favorable samples and compare them with all three-apple samples. - For an exact bruised/unbruised composition, choose from the two categories separately.

a) There are \(\binom{20}{3}\) equally likely three-apple samples. Thus, \(P(\text{all bruised})=\frac{\binom{4}{3}}{\binom{20}{3}}=\frac{1}{285}\). b) There are \(16\) unbruised apples, so \(P(\text{none bruised})=\frac{\binom{16}{3}}{\binom{20}{3}}=\frac{28}{57}\). c) If the specified apple is included, choose two more apples from the remaining \(19\). Thus, \(P=\frac{\binom{19}{2}}{\binom{20}{3}}=\frac{3}{20}\). d) Choose two of the four bruised apples and one of the sixteen unbruised apples. Thus, \(P=\frac{\binom{4}{2}\binom{16}{1}}{\binom{20}{3}}=\frac{8}{95}\).

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

- 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 \(F,S,F\)?

a) Choose the \(4\) success positions among \(10\): \(\binom{10}{4}=210\). b) In the fixed beginning \(F,S,F\), one success has already occurred. Seven positions remain, and exactly \(3\) more successes are needed. Thus, 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? - Use the symmetry and middle of a row of Pascal’s triangle for part c.

a) Choosing the \(k\) included objects uniquely determines the \(n-k\) excluded objects, so the two counts are equal. b) Choosing \(38\) of \(40\) objects is equivalent to choosing the \(2\) objects left out. Thus, \(\binom{40}{38}=\binom{40}{2}\). c) For fixed \(n\), the binomial coefficients increase toward the middle 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 \(n-k\) excluded objects. b) Choosing \(38\) objects to include is equivalent to choosing \(2\) objects to exclude. c) \(k=6\).

- Does the order in which members are chosen change the group? - In part b, count the choices from each subgroup separately. - For part c, identify what each object represents and whether it returns to the bag after selection.

a) Since order does not matter, the number of six-member groups is \(\binom{16}{6}=8008\). b) Choose \(3\) of the \(9\) women and \(3\) of the \(7\) men. The number of groups is \(\binom{9}{3}\binom{7}{3}=2940\). c) 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 \(6\) distinct member-objects without replacement from a bag containing \(9\) objects of one type and \(7\) of another; order does not matter.

- Decide whether order matters in each selection. - For exactly \(2\) defective resistors, determine how many working resistors must also be selected. - “At least \(3\)” includes which separate cases? - For the bag model, represent each physical resistor by a distinct object.

1. Order does not matter, so 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: \(\binom{5}{2}\binom{20}{2}=1900\). 3. “At least \(3\)” means exactly \(3\) or exactly \(4\) defective resistors, giving \(\binom{5}{3}\binom{20}{1}+\binom{5}{4}\binom{20}{0}=205\). 4. A matching bag model uses \(25\) distinct objects: \(5\) marked defective and \(20\) marked working. Draw \(4\) objects without replacement and ignore 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.

- Separate the chocolates into nut-filled and not nut-filled categories. - Count all unordered four-chocolate samples. - For each event, decide how many chocolates must come from each category.

a) There are \(\binom{15}{4}=1365\) equally likely four-chocolate samples. For four nut-filled chocolates, \(P=\frac{\binom{5}{4}\binom{10}{0}}{\binom{15}{4}}=\frac{1}{273}\). b) For no nut-filled chocolates, \(P=\frac{\binom{5}{0}\binom{10}{4}}{\binom{15}{4}}=\frac{2}{13}\). Sampling model: select \(4\) distinct chocolates from \(15\) without replacement; order does not matter.

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 order matter in a set of chosen side dishes? - In part b, focus on the dishes that are not selected. - For part c, consider the yes-or-no decision for each individual dish.

a) The number of selections of exactly \(4\) side dishes is \(\binom{12}{4}=495\). b) Choosing the \(8\) included dishes uniquely determines the \(4\) excluded dishes, so \(\binom{12}{8}=\binom{12}{4}\). c) Each of the \(12\) side dishes has two possibilities: chosen or not chosen. Thus, there are \(2^{12}=4096\) selections. 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 answer patterns by the number of correct answers.

a) Choose the \(7\) question positions that are answered correctly: \(\binom{10}{7}=120\). A corresponding selection model uses \(10\) labeled question objects and selects \(7\) without replacement; the selected objects represent the correctly answered questions. b) “At least \(9\)” means exactly \(9\) or exactly \(10\) correct answers. Thus, the number of patterns is \(\binom{10}{9}+\binom{10}{10}=11\). c) Grouping all answer patterns by their number of correct answers gives \(\sum_{k=0}^{10}\binom{10}{k}=2^{10}=1024\).

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

- Treat the five drives as an unordered sample without replacement. - For a fixed number of defectives, choose defective and nondefective drives separately. - Use case addition for “at most one” and a complement for “at least one.”

a) There are \(\binom{40}{5}\) equally likely samples. Thus, \(P(X=2)=\frac{\binom{6}{2}\binom{34}{3}}{\binom{40}{5}}=\frac{3740}{27417}\approx0.1364\). b) At most one defective drive means zero or one defective drive. Therefore, \(P(X\le1)=\frac{\binom{6}{0}\binom{34}{5}+\binom{6}{1}\binom{34}{4}}{\binom{40}{5}}=\frac{23188}{27417}\approx0.8458\). c) Use the complement of no defective drives: \(P(X\ge1)=1-\frac{\binom{34}{5}}{\binom{40}{5}}=\frac{15823}{27417}\approx0.5771\).

a) \(\frac{3740}{27417}\approx13.64\%\) b) \(\frac{23188}{27417}\approx84.58\%\) c) \(\frac{15823}{27417}\approx57.71\%\)

- Treat the six students as an unordered sample without replacement. - For each case, choose Spanish speakers and non-Spanish speakers separately. - A majority in a group of six means \(4\), \(5\), or \(6\).

a) There are \(\binom{24}{6}\) equally likely six-student samples. Thus, \(P(X=3)=\frac{\binom{10}{3}\binom{14}{3}}{\binom{24}{6}}=\frac{1560}{4807}\approx0.3245\). b) For no Spanish speakers, \(P(X=0)=\frac{\binom{10}{0}\binom{14}{6}}{\binom{24}{6}}=\frac{39}{1748}\approx0.0223\). c) A majority means \(4\), \(5\), or \(6\) Spanish speakers. Therefore, \(P(X>3)=\frac{\binom{10}{4}\binom{14}{2}+\binom{10}{5}\binom{14}{1}+\binom{10}{6}\binom{14}{0}}{\binom{24}{6}}=\frac{816}{4807}\approx0.1698\).

a) \(\frac{1560}{4807}\approx32.45\%\) b) \(\frac{39}{1748}\approx2.23\%\) c) \(\frac{816}{4807}\approx16.98\%\)

- Choose members from the two categories separately. - For part b, compare the favorable team count with all six-person teams. - In part c, interpret each binomial coefficient before interpreting their sum.

a) Choose four of the fifteen college students and two of the ten working professionals: \(\binom{15}{4}\binom{10}{2}=61{,}425\). b) There are \(\binom{25}{6}=177{,}100\) total six-person teams. Therefore, \(P=\frac{61425}{177100}=\frac{351}{1012}\approx0.3468\). c) \(\binom{15}{6}\) 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) \(\frac{351}{1012}\approx34.68\%\) 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 before comparing with all eight-book selections.

a) The factor \(\binom{18}{5}\) chooses five mystery novels, and \(\binom{12}{3}\) chooses three historical novels. Multiplying combines the two independent category choices into one eight-book selection. b) At least seven mystery novels means either seven mystery and one historical or eight mystery and no historical. The favorable count is \(\binom{18}{7}\binom{12}{1}+\binom{18}{8}\binom{12}{0}=425{,}646\). There are \(\binom{30}{8}=5{,}852{,}925\) total selections, so \(P=\frac{3638}{50025}\approx0.0727\).

a) The first factor chooses the mystery novels, the second chooses the historical novels, and the product combines the choices. b) \(\frac{3638}{50025}\approx7.27\%\)

- Think of the five drawings as five positions. - First choose the positions won by Alex, then choose the position won by Jordan. - What is the probability that a winner is neither Alex nor Jordan? - Find the probability of one specific qualifying order, then account for all such orders.

1. On each drawing, \(P(A)=\frac{1}{20}\), \(P(J)=\frac{1}{20}\), and the probability that someone else wins is \(P(O)=\frac{18}{20}\). 2. Choose the \(2\) drawings won by Alex and then the \(1\) drawing won by Jordan: \(\binom{5}{2}\binom{3}{1}=10\cdot 3=30\) arrangements. 3. Each qualifying arrangement has probability \(\left(\frac{1}{20}\right)^2\left(\frac{1}{20}\right)\left(\frac{18}{20}\right)^2\). 4. Therefore, the required probability is \(30\left(\frac{1}{20}\right)^3\left(\frac{18}{20}\right)^2=\frac{243}{80{,}000}=0.0030375\).

\(\frac{243}{80{,}000}=0.0030375\), or \(0.30375\%\)

- How many yellow draws must occur? - Find the probability of one specific order, such as red, red, blue, blue, yellow, yellow. - How many ways can the positions of the three colors be assigned? - Replacement makes the draws independent and keeps the color probabilities constant.

1. The color probabilities on each draw are \(P(R)=0.2\), \(P(B)=0.3\), and \(P(Y)=0.5\). 2. If exactly \(2\) draws are red and exactly \(2\) are blue, the remaining \(2\) draws must be yellow. 3. Choose the \(2\) red positions and then the \(2\) blue positions: \(\binom{6}{2}\binom{4}{2}=15\cdot 6=90\) color sequences. 4. Each qualifying sequence has probability \((0.2)^2(0.3)^2(0.5)^2=0.0009\). 5. Therefore, the probability is \(90\cdot 0.0009=0.081\).

\(0.081\), or \(8.1\%\)

- 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 \(\binom{10}{3}\binom{6}{1}=120\cdot6=720\). 3. The number of selections with \(4\) apples and no pears is \(\binom{10}{4}\binom{6}{0}=210\cdot1=210\). 4. The cases are disjoint, so the total is \(720+210=930\).

\(930\) selections

- Find the probability of each of the three result categories on one roll. - Choose the positions of the two \(6\)s first. - After those positions are fixed, how many positions remain for the three \(1\)s? - Multiply the number of qualifying arrangements by the probability of one arrangement.

1. On each roll, \(P(6)=\frac{1}{6}\), \(P(1)=\frac{1}{6}\), and \(P(\text{other})=\frac{4}{6}=\frac{2}{3}\). 2. Choose the \(2\) positions for the \(6\)s and then the \(3\) positions for the \(1\)s: \(\binom{10}{2}\binom{8}{3}=45\cdot 56\) arrangements. 3. Each qualifying arrangement has probability \(\left(\frac{1}{6}\right)^2\left(\frac{1}{6}\right)^3\left(\frac{2}{3}\right)^5\). 4. Therefore, \(P=\binom{10}{2}\binom{8}{3}\left(\frac{1}{6}\right)^5\left(\frac{2}{3}\right)^5=\frac{280}{6561}\approx 0.0427\).

\(\frac{280}{6561}\approx 0.0427\), or about \(4.27\%\)

- Match each probability base with its exponent. - A binomial coefficient can choose positions for one category. - Why are only \(4\) positions available after the first choice?

1. The exponents show that event \(E\) contains exactly \(8\) light packages, \(3\) standard packages, and \(1\) heavy package. 2. The factor \(\binom{12}{8}\) chooses the \(8\) positions occupied by light packages among the \(12\) selections. 3. After those positions are chosen, \(4\) positions remain. The factor \(\binom{4}{3}\) chooses the \(3\) positions occupied by standard packages. The final position must contain the heavy package.

Event \(E\) is “exactly \(8\) light packages, \(3\) standard packages, and \(1\) heavy package.” The factor \(\binom{12}{8}\) chooses the light-package positions, and \(\binom{4}{3}\) chooses the standard-package positions from the remaining \(4\) positions.

- 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 removing one complementary case from all teams.

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

- For part a, think of a situation where four selected people occupy four distinguishable positions. - For part b, think of a selection where only the chosen set matters. - For part c, count favorable groups by choosing from the two categories separately.

1. For part a, one valid question is: “How many ways can four distinct officer roles be assigned to four of the \(25\) students?” The roles are distinct, so exchanging two selected students changes the assignment. This gives \(25\cdot24\cdot23\cdot22\). 2. For part b, one valid question is: “How many six-person groups can be selected from the \(25\) students?” Only membership matters, so the same six students are one group regardless of listing order. This gives \(\binom{25}{6}\). 3. For part c, choose three of the fifteen female students and two of the ten other students. Thus \(P(X=3)=\frac{\binom{15}{3}\binom{10}{2}}{\binom{25}{5}}=\frac{195}{506}\approx0.3854\).

a) One valid example: How many ways can four distinct officer roles be assigned to four of the \(25\) students? Order matters because changing which student has a role changes the assignment. b) One valid example: How many six-person groups can be selected from the \(25\) students? Order does not matter because only group membership matters. c) \(\frac{195}{506}\approx38.54\%\)

- Treat the four balls as an unordered sample without replacement. - Translate each event into the number selected from each color category. - For \(E_3\), identify the separate cases included in “at least three blue.”

\(E_1\): There are \(\binom{30}{4}\) equally likely four-ball samples. Thus, \(P(E_1)=\frac{\binom{15}{2}\binom{10}{1}\binom{5}{1}}{\binom{30}{4}}=\frac{50}{261}\approx0.1916\). \(E_2\): Choose all four balls from the \(15\) nonred balls: \(P(E_2)=\frac{\binom{15}{4}}{\binom{30}{4}}=\frac{13}{261}\approx0.0498\). \(E_3\): Add the cases of exactly three and exactly four blue balls: \(P(E_3)=\frac{\binom{5}{3}\binom{25}{1}+\binom{5}{4}\binom{25}{0}}{\binom{30}{4}}=\frac{17}{1827}\approx0.00930\).

\(E_1\): \(\frac{50}{261}\approx19.16\%\) \(E_2\): \(\frac{13}{261}\approx4.98\%\) \(E_3\): \(\frac{17}{1827}\approx0.93\%\)

- First decide whether order matters for an unassigned committee. - In part b, compare one specified committee with all equally likely committees. - Distinct roles make different arrangements of the same three students count separately.

a) Without assigned roles, order does not matter, so the number of committees is \(\binom{20}{3}=1140\). b) Exactly one of the \(1140\) equally likely committees contains the three class representatives. Thus, the probability is \(\frac{1}{1140}\approx0.000877\). c) With three distinct roles, order matters. The number of assignments is \(20\cdot19\cdot18=6840\). Equivalently, \(\binom{20}{3}3!=6840\).

a) \(1140\) committees b) \(\frac{1}{1140}\approx0.000877\), or about \(0.0877\%\) c) \(6840\) role assignments

- Treat the team as an unordered sample without replacement. - Select the required numbers from the two groups separately. - Compare each favorable count with all six-person samples.

a) There are \(\binom{30}{6}\) equally likely six-person samples. For an all-men team, \(P=\frac{\binom{18}{6}}{\binom{30}{6}}\approx0.03126\). b) For exactly \(3\) women and \(3\) men, \(P=\frac{\binom{12}{3}\binom{18}{3}}{\binom{30}{6}}\approx0.30234\).

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

- Count all committees first. - Break “at least \(5\)” into the possible exact cases. - For part c, interpret the numerator before comparing it with the denominator.

a) There are \(\binom{25}{6}\) possible committees. For exactly \(3\) women and \(3\) men, \(P=\frac{\binom{15}{3}\binom{10}{3}}{\binom{25}{6}}\approx0.30830\). b) At least \(5\) men means either \(5\) men and \(1\) woman or \(6\) men. Thus, \(P=\frac{\binom{10}{5}\binom{15}{1}+\binom{10}{6}}{\binom{25}{6}}\approx0.02253\). c) The expression is the probability that all \(6\) selected committee 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

- Combine the number-set choices with the separate color choice. - For part b, separate the player’s chosen numbers into matches and nonmatches. - For part c, count both nonmatching number sets and nonmatching colors.

a) There are \(\binom{30}{4}\) number sets and \(3\) colors, so the total number of results is \(3\binom{30}{4}=82{,}215\). b) To match exactly \(3\) numbers, choose \(3\) of the player’s \(4\) numbers and \(1\) of the other \(26\) numbers. The color must also match. Thus, \(P=\frac{\binom{4}{3}\binom{26}{1}}{3\binom{30}{4}}=\frac{104}{82215}\approx0.001265\). c) For no number matches, choose all \(4\) numbers from the other \(26\). There are \(2\) incorrect colors. Thus, \(P=\frac{2\binom{26}{4}}{3\binom{30}{4}}=\frac{29900}{82215}\approx0.36368\).

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

- Treat the three chocolates as an unordered sample without replacement. - For the same filling, count each filling type separately and add. - For three different fillings, choose one chocolate from each category.

There are \(\binom{18}{3}=816\) equally likely three-chocolate samples. 1. For the same filling, add the samples of three marzipan, three nougat, or three caramel chocolates: \(P=\frac{\binom{8}{3}+\binom{6}{3}+\binom{4}{3}}{\binom{18}{3}}=\frac{5}{51}\). 2. For three different fillings, choose one chocolate of each type: \(P=\frac{\binom{8}{1}\binom{6}{1}\binom{4}{1}}{\binom{18}{3}}=\frac{4}{17}\).

1. \(\frac{5}{51}\approx9.8\%\) 2. \(\frac{4}{17}\approx23.5\%\)

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

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

- Count unordered selections with combinations. - For exactly \(3\) matches, choose matched numbers from the player's selections and unmatched numbers from the remaining numbers. - Relative frequency is the number of successful trials divided by the total number of trials.

1. There are \(\binom{49}{6}=13{,}983{,}816\) possible drawings. To obtain exactly \(3\) matches, choose \(3\) of the player's \(6\) numbers and \(3\) of the other \(43\) numbers. Thus, \(P(\text{exactly 3 matches})=\frac{\binom{6}{3}\binom{43}{3}}{\binom{49}{6}}=\frac{246820}{13983816}\approx 0.0176504\). 2. The relative frequency is \(\frac{17612}{1000000}=0.017612\). The absolute difference is \(\left|0.017612-0.0176504\right|\approx 0.0000384\), or about \(0.00384\) percentage points. 3. The law of large numbers states that, as the number of independent trials increases, the relative frequency tends to stabilize near the theoretical probability. The small difference after one million trials is consistent with this behavior.

1. \(P(\text{exactly 3 matches})\approx 0.0176504\), or about \(1.765\%\) 2. Relative frequency \(0.017612\); absolute difference approximately \(0.0000384\), or \(0.00384\) percentage points 3. With many trials, the relative frequency tends to approach the theoretical probability.

- Model each element by an include/exclude decision. - After fixing one required element, choose the remaining member of the subset. - Compare a subset's size with \(5\) minus that size.

a) Each of five elements is either included or excluded, giving \(2^5=32\) subsets. b) If \(1\) must be included and the subset has size \(2\), choose one of the other four elements. There are \(4\) favorable subsets, so the probability is \(\frac{4}{32}=\frac{1}{8}\). c) A five-element set cannot be split into a subset and complement of equal integer size, so the probability is \(0\).

a) \(32\) outcomes. b) \(\frac{1}{8}\). c) \(0\).

- Condition on the set of possible head positions. - Count position pairs rather than full flip strings. - Identify which pairs are consecutive.

1. The head positions are uniformly one of \(\binom{4}{2}=6\) pairs. 2. Adjacent pairs are \(\{1,2\}\), \(\{2,3\}\), and \(\{3,4\}\), giving \(3\) favorable pairs. 3. The conditional probability is \(\frac{3}{6}=\frac{1}{2}\).

The probability is \(\frac{1}{2}\).

- A match is determined by two teams, and reversing their order does not create a new match. - Express the number of unordered pairs in terms of the unknown team count. - Keep only a positive integer solution that can represent a number of teams.

1. With \(n\) teams, each match is determined by choosing an unordered pair, so \(\binom{n}{2}=45\). 2. Then \(\frac{n(n-1)}{2}=45\), so \(n(n-1)=90\). 3. Rearranging gives \(n^2-n-90=0\), which factors as \((n-10)(n+9)=0\). 4. The feasible team count is \(n=10\).

\(10\) teams

- 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

- For parts a and b, think about which positions are occupied by the defective components. - Connect the count of success positions with the probability of any one success/failure pattern. - For part c, look for a way to place the good components so that you can see where a defective component could be inserted without creating adjacency.

a) Choose the \(2\) defective positions among \(10\): \(\binom{10}{2}=45\). b) The binomial probability is \(P(X=2)=\binom{10}{2}(0.05)^2(0.95)^8\approx0.0746\). c) Place the \(5\) good components first. They determine \(6\) possible separation positions around and between them. Choosing \(3\) of those positions for the defective components gives \(\binom{6}{3}=20\) sequences.

a) \(45\) b) \(P(X=2)\approx0.0746\) c) \(20\)

- Count all unordered pairs represented in the urn. - For part a, organize pairs by color. - For part b, list pairs systematically by their label sums. - For part c, count choices from the two required colors.

a) There are \(\binom{7}{2}=21\) equally likely pairs. Same-color pairs number \(\binom{3}{2}+\binom{2}{2}+\binom{2}{2}=5\), so \(P(\text{same color})=\frac{5}{21}\). b) 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 \(12\) favorable pairs, so \(P(\text{prime sum})=\frac{12}{21}=\frac{4}{7}\). c) Choose one of the three red balls and one of the two green balls. There are \(3\cdot 2=6\) 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) \(\frac{2}{7}\)

- Count all committees that remain possible after imposing the condition. - For the favorable committees, fix the named person first. - Use the same type of committee count in the numerator and denominator.

1. The number of committees containing at least one senior is \(\binom{12}{5}-\binom{9}{5}=792-126=666\). 2. Every committee containing Maya satisfies the condition, and there are \(\binom{11}{4}=330\) such committees. 3. The conditional probability is \(\frac{330}{666}=\frac{55}{111}\).

The probability is \(\frac{55}{111}\approx 0.495\).

- Under the condition, focus first on the possible pairs of face cards. - Separate favorable pairs by color. - Decide whether the selection of the remaining cards affects the ratio.

1. There are \(6\) red face cards and \(6\) black face cards. 2. The two face cards can be any pair from the \(12\), giving \(\binom{12}{2}=66\) possibilities. 3. Same-color pairs number \(2\cdot\binom{6}{2}=30\). 4. The three non-face cards are selected in the same number of ways for every face-card pair, so that factor cancels. 5. The conditional probability is \(\frac{30}{66}=\frac{5}{11}\).

The probability is \(\frac{5}{11}\approx 0.455\).

- Shift the hundreds digit down by \(1\) so all three variables can be treated as nonnegative. - Use stars and bars for the full conditioned sample space and again after setting one of the last two digits to zero. - The two zero-digit cases overlap once; account for that overlap before forming the probability.

1. Let the hundreds digit be \(h\ge1\), and write \(h'=h-1\). Then \(h'+t+u=8\) in nonnegative integers. Because the sum is only \(8\), no digit upper bound is exceeded. 2. By stars and bars, the number of conditioned integers is \(\binom{10}{2}=45\). 3. If \(t=0\), then \(h'+u=8\), giving \(9\) solutions. If \(u=0\), there are likewise \(9\) solutions. 4. The integer \(900\) is counted in both cases, so inclusion-exclusion gives \(9+9-1=17\) favorable integers. 5. Therefore, the conditional probability is \(\frac{17}{45}\).

\(\frac{17}{45}\)