Free Math Worksheets

- Which person is fixed in the first position? - How many people remain for the other two positions? - List the two possible orders of those people.

1. Lisa is fixed in the first position. 2. Tim and Max can fill the remaining two positions in \(2!=2\) orders. 3. The orders are Lisa–Tim–Max and Lisa–Max–Tim.

\(2\) orders: Lisa–Tim–Max and Lisa–Max–Tim

- Place the cards into the four digit positions one at a time. - How many cards are available before the first choice? - How many remain after each card is used? - Why are the successive numbers of choices multiplied?

1. There are \(4\) choices for the thousands digit. 2. After that choice, there are \(3\) choices for the hundreds digit, then \(2\) choices for the tens digit, and \(1\) choice for the ones digit. 3. Therefore, the total is \(4\cdot3\cdot2\cdot1=4!=24\).

a) \(24\) numbers b) The successive numbers of choices are \(4,3,2,\) and \(1\), so \(4\cdot3\cdot2\cdot1=24\).

- Which cards can be used in the hundreds place if the number must be less than \(600\)? - After one card is used, how many remain for the next position? - Multiply the numbers of choices for the three positions.

1. For the number to be less than \(600\), the hundreds digit must be \(2\) or \(4\), giving \(2\) choices. 2. After choosing the hundreds digit, \(3\) cards remain for the tens digit. 3. Then \(2\) cards remain for the ones digit. 4. By the fundamental counting principle, the total is \(2\cdot3\cdot2=12\).

\(12\) numbers

- Count the choices for each position. - For part b, reduce the available choices after each digit is used. - For part c, count the codes in which one digit is repeated four times.

1. Each of the four positions has \(6\) choices, so the total number of codes is \(6^4=1296\). 2. For four different digits, \(6\cdot5\cdot4\cdot3=360\). 3. There are six codes with all digits equal: \(1111\), \(2222\), through \(6666\). Therefore, \(P(\text{all digits equal})=\frac{6}{1296}=\frac{1}{216}\approx 0.00463\).

a) \(1296\) b) \(360\) c) \(\frac{1}{216}\approx 0.00463\)

- Use a factorial to arrange distinct people in a row. - For part b, count only the people whose positions are not fixed. - For part c, arrange the first three people and the remaining five people separately, then multiply.

1. Eight distinct people can be arranged in \(8!=40{,}320\) ways. 2. If A and B occupy the first two positions in that fixed order, the remaining six people can be arranged in \(6!=720\) ways. 3. A, B, and C can be arranged in the first three positions in \(3!\) ways. The remaining five people can be arranged in \(5!\) ways. Therefore, \(3!\cdot5!=720\).

a) \(40{,}320\) b) \(720\) c) \(720\)

- Count all arrangements of \(8\) distinct objects. - Count the arrangements that are sorted by page count. - Identify the assumption made about the different arrangements.

1. The \(8\) distinct books can be arranged in \(8!=40{,}320\) equally likely orders. 2. Exactly \(2\) orders satisfy the condition: increasing page count and decreasing page count. 3. Therefore, \(P=\frac{2}{8!}=\frac{1}{20160}\approx 0.00004960\). 4. This is a theoretical probability calculation based on equally likely outcomes.

The probability is \(\frac{1}{20160}\approx 0.00004960\). It is a theoretical probability based on equally likely arrangements.

- Decide whether the selected students have distinct roles. - Use an ordered count when roles matter and a combination when they do not. - For part b, count one specified group among all equally likely groups.

1. In case (1), the roles are distinct, so order matters: \(15\cdot14\cdot13=2730\). 2. In case (2), only the group membership matters: \(\binom{15}{3}=455\). 3. Exactly one of the \(455\) groups is the specified group. Therefore, \(P(\text{specified group})=\frac{1}{455}\).

a) Case (1): \(2730\); case (2): \(455\) b) \(\frac{1}{455}\approx 0.22\%\)

- Imagine all tickets arranged in one random order. - Ask whether any position is favored over another.

1. The grand-prize ticket is equally likely to be any of the \(100\) tickets drawn first, so the probability is \(\frac{1}{100}\). 2. In a random ordering of all tickets, the grand-prize ticket is equally likely to occupy any position. Therefore, the probability that it is tenth is also \(\frac{1}{100}\). 3. Every permutation of the tickets is equally likely, and the grand-prize ticket appears equally often in each position across all permutations.

a) \(\frac{1}{100}=0.01\) b) \(\frac{1}{100}=0.01\) c) Random ordering makes every position equally likely for the grand-prize ticket.

- How many permutations are there of \(n\) distinct objects? - Multiply the number of arrangements by the time for each arrangement. - Use \(3600\,\text{s}=1\,\text{h}\) to convert the final time.

1. The number of permutations of \(7\) distinct books is \(7!=7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1=5040\). 2. The total time is \(5040\cdot10\,\text{s}=50{,}400\,\text{s}\). 3. Convert seconds to hours: \(50{,}400\div3600=14\).

a) \(5040\) arrangements b) \(14\) hours

- How many total arrangements are possible with \(6\) distinct letters? - If two letters are fixed in the first two positions, how many positions remain to arrange? - For adjacent vowels, treat the two vowels as one object, but remember that the vowels can switch order inside the block.

1. There are \(6\) distinct letters, so the total number of arrangements is \(6! = 720\). 2. a) Fix F in the first position and A in the second position. The remaining \(4\) letters can be arranged in \(4! = 24\) ways. Therefore, \(P = \frac{24}{720} = \frac{1}{30}\). 3. b) Treat A and O as one block. The block and the four consonants form \(5\) objects, which can be arranged in \(5! = 120\) ways. The vowels can be ordered as AO or OA, so there are \(2 \cdot 120 = 240\) favorable arrangements. 4. Therefore, \(P = \frac{240}{720} = \frac{1}{3}\).

a) \(\frac{1}{30}\) b) \(\frac{1}{3}\)

- Treat the two friends who must stay together as one unit. - How many units are then being arranged? - How many orders are possible within the two-person unit?

1. Treat Ava and Ben as one block. Together with Clara, David, and Eli, there are \(4\) units to arrange. 2. The \(4\) units can be arranged in \(4!=24\) ways. 3. Within the block, Ava and Ben can sit in either order, giving \(2!=2\) arrangements. 4. Multiply: \(4!\cdot2!=24\cdot2=48\).

\(48\) seating arrangements

- In part a, how many positions remain after two books are fixed? - In part b, the math book can occupy either end. - For each possible end position, count the permutations of the other three books.

1. In part a, the positions of the math and history books are fixed. The two remaining books can be arranged in \(2!=2\) ways. 2. In part b, the math book can be at the left end or the right end, giving \(2\) disjoint cases. 3. In each case, the other three books can be arranged in \(3!=6\) ways. 4. Therefore, the total is \(2\cdot3!=2\cdot6=12\).

a) \(2\) arrangements b) \(12\) arrangements

- Count the total number of available characters. - For part a, determine the number of choices for each position when repetition is allowed. - For part b, reduce the number of available choices after each position. - For part c, compare the number of all-digit codes with the total number of codes.

1. There are \(10+8=18\) available characters. 2. With repetition allowed, each of the five positions has \(18\) choices: \(18^5=1{,}889{,}568\). 3. Without repetition, the number of codes is \(18\cdot17\cdot16\cdot15\cdot14=1{,}028{,}160\). 4. For an all-digit code, each position has \(10\) choices, so there are \(10^5\) favorable codes. Therefore, \(P(\text{all digits})=\frac{10^5}{18^5}=\left(\frac{5}{9}\right)^5=\frac{3125}{59049}\approx0.0529\).

a) \(1{,}889{,}568\) b) \(1{,}028{,}160\) c) \(\frac{3125}{59049}\approx 5.29\%\)

- Decide whether repetition is allowed in each part. - Decide whether order matters. - Use a product for ordered selections and a combination for an unordered group.

1. With repetition allowed and order recorded, the number of sequences is \(20^3=8000\). 2. With no repeated number, the choices for the three positions are \(20\), \(19\), and \(18\): \(20\cdot19\cdot18=6840\). 3. When three distinct numbers are selected and order does not matter, use a combination: \(\binom{20}{3}=1140\).

a) \(8000\) b) \(6840\) c) \(1140\)

- Determine whether order matters in each part. - Determine whether a digit remains available after it is used. - Match each case to an ordered count or a combination.

1. With repetition allowed and order important, \(9^5=59{,}049\). 2. Without repetition and with order important, \(9\cdot8\cdot7\cdot6\cdot5=15{,}120\). 3. Without repetition and with order irrelevant, \(\binom{9}{5}=126\).

a) \(59{,}049\) b) \(15{,}120\) c) \(126\)

- Determine whether the selected people have distinct roles. - Determine whether order matters. - Determine whether a name may be repeated. - Use the appropriate product or combination.

1. The three roles are distinct, and no student can be repeated. Therefore, \(25\cdot24\cdot23=13{,}800\). 2. The four representatives form an unordered group, so \(\binom{25}{4}=12{,}650\). 3. For the four-name sequence, each position has \(25\) choices and repetition is allowed: \(25^4=390{,}625\).

a) \(13{,}800\) b) \(12{,}650\) c) \(390{,}625\)

- Ask whether an item can be selected more than once. - Ask whether changing the order creates a different outcome. - Identify the number of available choices and the number of selections.

1. For the locker code, \(n=10\), \(k=4\), repetition is allowed, and order matters. The number of codes is \(10^4=10{,}000\). 2. For the prizes, \(n=10\), \(k=3\), repetition is not allowed, and order matters. The number of outcomes is \(10\cdot9\cdot8=720\). 3. For the cleanup group, \(n=12\), \(k=2\), repetition is not allowed, and order does not matter. The number of groups is \(\binom{12}{2}=66\).

a) \(n=10\), \(k=4\), repetition allowed, order matters; \(10{,}000\) outcomes b) \(n=10\), \(k=3\), no repetition, order matters; \(720\) outcomes c) \(n=12\), \(k=2\), no repetition, order does not matter; \(66\) outcomes

- Determine whether a selection can be repeated. - Determine whether the positions or order create different outcomes. - Use a power, an ordered product, or a combination as appropriate.

1. For the bit string, \(n=2\), \(k=6\), repetition is allowed, and order matters. The number of strings is \(2^6=64\). 2. For the three-digit numbers, \(n=5\), \(k=3\), repetition is not allowed, and order matters. The number of numbers is \(5\cdot4\cdot3=60\). 3. For the lottery, \(n=25\), \(k=5\), repetition is not allowed, and order does not matter. The number of possible sets is \(\binom{25}{5}=53{,}130\).

a) \(n=2\), \(k=6\), repetition allowed, order matters; \(64\) outcomes b) \(n=5\), \(k=3\), no repetition, order matters; \(60\) outcomes c) \(n=25\), \(k=5\), no repetition, order does not matter; \(53{,}130\) outcomes

- For part a, reduce the number of available digits after each position. - For part b, only the immediately preceding digit is unavailable. - For part c, choose a pair of digits and count the two possible alternating orders.

1. Without repetition, the four positions have \(6\), \(5\), \(4\), and \(3\) choices: \(6\cdot5\cdot4\cdot3=360\). 2. For a valid code, the first position has \(6\) choices. Each later position has \(5\) choices because it cannot match the immediately preceding digit. Thus, \(6\cdot5^3=750\). 3. Choose the two digits in \(\binom{6}{2}=15\) ways. Each pair produces two alternating codes, \(abab\) and \(baba\), so there are \(15\cdot2=30\) favorable codes. 4. Therefore, \(P(\text{alternating two-digit code})=\frac{30}{750}=\frac{1}{25}=0.04\).

a) \(360\) b) \(750\) c) \(\frac{1}{25}=4\%\)

- For part a, count the choices for each square. - For part b, reduce the number of available colors after each square. - For part c, use the complement that all colors are different.

1. With repetition allowed, each of the five positions has \(12\) choices: \(12^5=248{,}832\). 2. For five different colors, \(12\cdot11\cdot10\cdot9\cdot8=95{,}040\). 3. Use the complement that all five colors are different: \(P(\text{at least one repeat})=1-\frac{95040}{248832}=\frac{89}{144}\approx 0.6181\).

a) \(248{,}832\) b) \(95{,}040\) c) \(\frac{89}{144}\approx 61.81\%\)

- Write the binomial coefficient using factorials. - How do you divide by a fraction? - Expand only enough factorial factors to cancel. - What role does ordering play in the two expressions?

1. \(A=\frac{9!}{5!}=9\cdot 8\cdot 7\cdot 6=3024\). 2. \(B=\binom{9}{4}=\frac{9\cdot 8\cdot 7\cdot 6}{4\cdot 3\cdot 2\cdot 1}=126\). 3. Therefore, \(\frac{A}{B}=\frac{3024}{126}=24=4!\). 4. In general, \(\binom{n}{k}=\frac{n!}{k!(n-k)!}\). Thus \(\frac{A}{B}=\frac{n!}{(n-k)!}\div\frac{n!}{k!(n-k)!}=\frac{n!}{(n-k)!}\cdot\frac{k!(n-k)!}{n!}=k!\).

a) \(A=3024\), \(B=126\) b) \(\frac{A}{B}=24\) c) Substitution of \(\binom{n}{k}=\frac{n!}{k!(n-k)!}\) causes \(n!\) and \((n-k)!\) to cancel, leaving \(k!\).

- Count the number of factors in the product. - Expand \(n!\) and identify the factors contained in \((n-k)!\). - Substitute \(k=n\) and use \(0!=1\).

1. When \(n=10\) and \(k=3\), the product has three factors: \(10\cdot9\cdot8=720\). 2. Expand \(n!\) as \(n!=n(n-1)\cdots(n-k+1)(n-k)!\). Dividing by \((n-k)!\) leaves \(\frac{n!}{(n-k)!}=n(n-1)\cdots(n-k+1)\). 3. If \(k=n\), then \(\frac{n!}{(n-n)!}=\frac{n!}{0!}=n!\) because \(0!=1\).

a) \(720\) b) Expanding \(n!\) and canceling \((n-k)!\) gives \(\frac{n!}{(n-k)!}=n(n-1)\cdots(n-k+1)\). c) \(n!\)

- Count the number of choices available to each teenager in order. - For all different flavors, the number of available choices decreases after each person. - For all the same flavor, first choose the common flavor.

1. There are \(10^6\) equally likely ordered sets of choices. 2. For all different flavors, the teenagers have \(10\), \(9\), \(8\), \(7\), \(6\), and \(5\) choices in order. Thus, \(P=\frac{10\cdot9\cdot8\cdot7\cdot6\cdot5}{10^6}=0.1512\). 3. For all the same flavor, there are \(10\) favorable ordered outcomes, one for each flavor. Thus, \(P=\frac{10}{10^6}=0.00001\).

a) \(0.1512\) b) \(0.00001\)

- Treat the four consecutive songs as one block. - Arrange the block with the other songs, then arrange the songs inside the block. - For part b), choose the band songs and the other songs among the first five positions.

1. Treat the \(4\) band songs as one block. The block and the other \(8\) songs form \(9\) objects, which can be arranged in \(9!\) ways. The \(4\) songs within the block can be arranged in \(4!\) ways. Therefore, the number of play orders is \(9!\cdot4!=8{,}709{,}120\). 2. The first \(5\) songs form a sample without replacement. Thus, \(P=\frac{\binom{4}{2}\binom{8}{3}}{\binom{12}{5}}=\frac{336}{792}=\frac{14}{33}\approx 0.4242\).

a) \(8{,}709{,}120\) play orders b) \(\frac{14}{33}\approx 0.4242\), or about \(42.42\%\)

- List the pairs of positions that have exactly one position between them. - Account for the two possible orders of Anna and Ben. - For part b), count the left-end and right-end cases separately.

1. There are \(8\) position pairs with exactly one position between them: \((1, 3)\) through \((8, 10)\). Anna and Ben can switch places in \(2\) ways, and the other \(8\) people can be arranged in \(8!\) ways. Thus, the count is \(8\cdot2\cdot8!=645{,}120\). 2. For both people in the first three positions, choose their two positions in \(\binom{3}{2}\) ways, arrange them in \(2!\) ways, and arrange the other people in \(8!\) ways. The same count applies to the last three positions, and the two events are disjoint. 3. Therefore, \(P=\frac{2\binom{3}{2}2!8!}{10!}=\frac{2}{15}\approx 0.1333\).

a) \(645{,}120\) lineups b) \(\frac{2}{15}\approx 0.1333\), or about \(13.33\%\)

- Determine whether the prizes make the order of winners important. - Use a combination for identical prizes and an ordered count for ranked prizes. - The probability of one exact equally likely outcome is the reciprocal of the number of outcomes.

1. In model (1), order does not matter because the prizes are identical. The number of winner groups is \(\binom{20}{4}=4845\). 2. In model (2), the four places are distinct, so order matters. The number of outcomes is \(20\cdot19\cdot18\cdot17=116{,}280\). 3. Each outcome is equally likely, so the probabilities of one exact prediction are \(P_1=\frac{1}{4845}\approx 0.000206\) and \(P_2=\frac{1}{116280}\approx 0.00000860\).

a) Model (1): \(4845\); model (2): \(116{,}280\) b) Model (1): \(\frac{1}{4845}\approx 0.0206\%\); model (2): \(\frac{1}{116280}\approx 0.000860\%\)

- Count all length-\(n\) codes. - Count the codes that use every value exactly once. - Form a probability in terms of \(n\) and test small positive integers.

1. There are \(n^n\) possible codes. 2. A code with all different entries is a permutation of the \(n\) available values, so there are \(n!\) favorable codes. 3. Thus, \(P(\text{all different})=\frac{n!}{n^n}\). 4. Check positive integer values in order: \(n=1\): \(\frac{1!}{1^1}=1\) \(n=2\): \(\frac{2!}{2^2}=0.5\) \(n=3\): \(\frac{3!}{3^3}=\frac{2}{9}\approx 0.222\) \(n=4\): \(\frac{4!}{4^4}=\frac{3}{32}=0.09375\). The first value below \(0.10\) is \(n=4\).

\(n=4\)

- Use the complement that all birth months are different. - Reduce the number of available unused months for each additional student. - Test consecutive values of \(k\) until the probability crosses \(60\%\).

1. Use the complement that all \(k\) students have different birth months. 2. For \(k\le12\), \(P(\text{all different})=\frac{12}{12}\cdot\frac{11}{12}\cdots\frac{12-k+1}{12}\). 3. Check consecutive values near the threshold: For \(k=4\), \(P(\text{at least one shared month})=1-\frac{12\cdot11\cdot10\cdot9}{12^4}=\frac{41}{96}\approx 0.427\). For \(k=5\), \(P(\text{at least one shared month})=1-\frac{12\cdot11\cdot10\cdot9\cdot8}{12^5}=\frac{89}{144}\approx 0.618\). 4. The probability first exceeds \(0.60\) when \(k=5\).

\(k=5\)

- Consider the caps one at a time and count how many unused designs remain. - Use a complement for “at least two match.” - Relative frequencies vary across finite simulations.

1. The probability that all six designs are different is \(\frac{25}{25}\cdot\frac{24}{25}\cdot\frac{23}{25}\cdot\frac{22}{25}\cdot\frac{21}{25}\cdot\frac{20}{25}\approx 0.52229\). 2. The event “at least two match” is the complement, so its probability is \(1-0.52229\approx 0.47771\). 3. A simulation produces a relative frequency that varies randomly from one set of trials to another. With \(1000\) trials, the estimate should often be near the theoretical value but usually will not equal it exactly.

a) Approximately \(0.52229\), or \(52.23\%\) b) Approximately \(0.47771\), or \(47.77\%\) c) Simulation estimates are subject to random variation.

- First decide whether order matters for an unassigned committee. - In part b), how many of the equally likely committees are favorable? - Distinct roles make different arrangements of the same three students count separately.

1. Without assigned roles, order does not matter, so the number of committees is \(\binom{20}{3}=1140\). 2. Exactly one of the \(1140\) equally likely committees contains the three class representatives. Thus the probability is \(\frac{1}{1140}\approx 0.000877\). 3. With three distinct roles, order matters. The number of assignments is \(20\cdot 19\cdot 18=6840\). Equivalently, \(\binom{20}{3}3!=1140\cdot 6=6840\).

a) \(1140\) committees b) \(\frac{1}{1140}\approx 0.000877\), or about \(0.0877\%\) c) \(6840\) role assignments

- For all different results, reduce the number of available new outcomes after each spin. - Use a complement for at least one repeat. - Compare the number of spins with the number of possible results.

1. The probability that all \(4\) results are different is \(\frac{12}{12}\cdot\frac{11}{12}\cdot\frac{10}{12}\cdot\frac{9}{12}=\frac{11880}{20736}\approx 0.57292\). The probability of at least one repeat is the complement, approximately \(0.42708\). Therefore, all different is more likely. 2. There are only \(12\) possible results. By the pigeonhole principle, \(13\) spins must include a repeat, so the probability of all different results is \(0\).

1. All different is more likely: approximately \(0.57292\) versus \(0.42708\). 2. \(0\)

- For all different PINs, reduce the number of available unused PINs at each assignment. - Use the complement for at least one match. - Think about how the number of employee pairs changes.

1. Each of the \(10{,}000\) PINs must be equally likely for every employee, and assignments must be independent. 2. The probability that all \(50\) PINs differ is \(\frac{10000\cdot9999\cdot9998\cdots9951}{10000^{50}}\approx 0.88453\). 3. The complement gives \(P(\text{at least one match})=1-0.88453\approx 0.11547\). 4. With more employees, there are more employee pairs that could match, and each additional PIN must avoid a larger set of previously assigned PINs.

a) PINs are assigned independently and uniformly from the \(10{,}000\) possibilities. b) Approximately \(0.88453\) c) Approximately \(0.11547\), or \(11.55\%\) d) More employees create more possible matching pairs.

- Represent each assignment as a permutation. - Use the complement of no correct assignments. - Consider what remains after three correct assignments. - Small samples can vary noticeably from theoretical probabilities.

1. The permutations with exactly two correct assignments are \((1, 2, 4, 3),(1, 4, 3, 2),(1, 3, 2, 4),(4, 2, 3, 1),(3, 2, 1, 4),(2, 1, 3, 4)\). The identity permutation \((1, 2, 3, 4)\) has four correct assignments, so there are \(7\) permutations with at least two correct assignments. 2. Of the \(4!=24\) permutations, \(9\) have no fixed points. Therefore, \(P(\text{at least one correct})=1-\frac{9}{24}=\frac{5}{8}=0.625\). 3. If three guests receive the correct drinks, the only remaining drink must also belong to the remaining guest, so all four assignments are correct. 4. The observed relative frequency is \(\frac{15}{20}=0.75\). With only \(20\) trials, a difference from \(0.625\) of this size is compatible with random variation and does not contradict the model.

a) \((1, 2, 4, 3),(1, 4, 3, 2),(1, 3, 2, 4),(4, 2, 3, 1),(3, 2, 1, 4),(2, 1, 3, 4),(1, 2, 3, 4)\) b) \(\frac{5}{8}=0.625\) c) Three correct assignments force the fourth assignment to be correct. d) No. A relative frequency of \(0.75\) in only \(20\) trials can result from random variation.

- Count all ordered triples. - For part 1, list the number multisets that sum to \(5\) and count their orders. - A strictly decreasing sequence uses three distinct numbers in one specific order. - For part 3, restrict every draw to the values \(1\) through \(4\).

1. There are \(10^3=1000\) equally likely ordered triples. 2. The triples with sum \(5\) are the three permutations of \((1, 1, 3)\) and the three permutations of \((1, 2, 2)\), for a total of six. Thus, \(P(\text{sum }5)=\frac{6}{1000}=\frac{3}{500}\). 3. A strictly decreasing triple is determined by choosing three distinct numbers; each chosen set has exactly one decreasing order. Therefore, \(P(\text{strictly decreasing})=\frac{\binom{10}{3}}{1000}=\frac{3}{25}\). 4. If the greatest number is at most \(4\), every draw must be one of \(1,2,3,4\). Thus, \(P(\max\le 4)=\frac{4^3}{10^3}=\frac{8}{125}\).

1. \(\frac{3}{500}=0.006\) 2. \(\frac{3}{25}=0.12\) 3. \(\frac{8}{125}=0.064\)

- Count all ordered outcomes from four runs. - Find the probability that each new result differs from all earlier results. - Use the complement of “all four are different.” - For part c), count the outcomes in which one value repeats four times.

1. For \(n=6\), there are \(6^4\) ordered outcomes. The number with all different values is \(6\cdot5\cdot4\cdot3\). Thus, \(P=\frac{6\cdot5\cdot4\cdot3}{6^4}=\frac{5}{18}\approx 0.2778\). 2. The probability that all four numbers are different is \(\frac{n}{n}\cdot\frac{n-1}{n}\cdot\frac{n-2}{n}\cdot\frac{n-3}{n}=\frac{(n-1)(n-2)(n-3)}{n^3}\). The event “at least two are equal” is its complement, giving the stated expression. 3. There are \(n\) ordered outcomes in which all four numbers are the same, out of \(n^4\) total outcomes. Thus, \(\frac{1}{n^3}=0.008\), so \(n^3=125\) and \(n=5\).

a) \(\frac{5}{18}\approx 0.2778\) b) \(P(\text{at least two equal})=1-\frac{(n-1)(n-2)(n-3)}{n^3}\) c) \(n=5\)

- For all different cards, count how many choices remain for each pack. - Use the complement of “all three are different.” - Solve the inequality and then check the neighboring integer values.

1. For \(n=10\), \(P(\text{all different})=\frac{10}{10}\cdot\frac{9}{10}\cdot\frac{8}{10}=0.72\). 2. For general \(n\), \(P(\text{all different})=\frac{n(n-1)(n-2)}{n^3}=\frac{(n-1)(n-2)}{n^2}\). Therefore, \(P(\text{at least two identical})=1-\frac{(n-1)(n-2)}{n^2}=\frac{3n-2}{n^2}\). 3. Solve \(\frac{(n-1)(n-2)}{n^2}>0.98\). This is equivalent to \(0.02n^2-3n+2>0\). The relevant boundary is approximately \(149.33\), so the least integer satisfying the inequality is \(n=150\). Indeed, the probability is less than \(0.98\) at \(n=149\) and greater than \(0.98\) at \(n=150\).

a) \(0.72\) b) \(P(\text{at least two identical})=\frac{3n-2}{n^2}\) c) \(n=150\)

- How many ordered sequences are possible in \(5\) rolls? - Count the choices for each roll when all results must be different. - “At least one repeat” is the complement of “all different.” - Does the chance of repeating a previous result stay constant from roll to roll?

1. There are \(8^5\) ordered sequences of results. For all results to be different, there are \(8\cdot 7\cdot 6\cdot 5\cdot 4\) valid sequences. 2. Therefore, \(P(\text{all different})=\frac{8\cdot 7\cdot 6\cdot 5\cdot 4}{8^5}=\frac{105}{512}\approx 0.2051\). 3. The event in part b) is the complement, so its probability is \(1-\frac{105}{512}=\frac{407}{512}\approx 0.7949\). 4. The student's claim is false. A “success” defined as repeating any previous result does not have a constant probability: it depends on how many distinct values have already appeared. A binomial model with \(p=\frac{1}{8}\) would instead count occurrences of one specified number.

a) \(\frac{105}{512}\approx 0.2051\) b) \(\frac{407}{512}\approx 0.7949\) c) The claim is false because the probability of a repeat changes with the previous results; it is not a constant \(\frac{1}{8}\).