🧠:<think>Okay, so I have this problem about a tennis tournament with 128 players, and I need to calculate the probability that the underdog, who's ranked 128th, wins the tournament. The tournament is single-elimination, which means each loss eliminates a player, and the underdog has to win every match to take the title. First, let me understand the structure. In a single-elimination bracket with 128 players, there are 7 rounds because 2^7 = 128. So, the underdog has to win 7 matches in a row to become the champion. Each match is determined by certain probabilities depending on the opponent's rank.The underdog has a 30% chance of winning against any opponent ranked above them and a 70% chance against opponents ranked below or equal. Since the underdog is ranked 128th, every opponent they face will be ranked above them, right? Because 128 is the lowest rank. So, in every match, the underdog has a 30% chance to win. Wait, hold on. Is that correct? If the underdog is ranked 128th, then all opponents in the bracket are ranked 1 to 127, which are all above them. So, in every match, the probability of the underdog winning is 30%. Therefore, the probability of winning the tournament is (0.3)^7, because they have to win 7 matches in a row.But wait, the problem says that opponents are matched randomly within their bracket section and that the underdog always faces the highest-ranked opponent available. Hmm, that might complicate things. So, does that mean in each round, the underdog is always facing the highest-ranked player remaining in their bracket? Or does it mean that the bracket is structured so that higher-ranked players are more likely to be in certain sections?I think it's the former. So, in each round, the underdog is facing the highest-ranked opponent available in their section. That would mean that in the first round, the underdog is facing the highest-ranked player in their bracket section. But wait, the bracket is single-elimination, so each section is a quarter or an eighth of the bracket? Hmm, I need to clarify.Wait, 128 players, so the bracket is divided into 128 slots. Each round halves the number of players. So, in the first round, 64 matches, second round 32, then 16, 8, 4, 2, and the final. So, the underdog is in a specific section, and in each round, they face the next highest-ranked player in their section.But the problem says opponents are matched randomly within their bracket section. So, does that mean that in each round, the underdog's opponent is randomly selected from the remaining players in their section? Or is it that the bracket is set up in a way that higher-ranked players are more likely to be in certain sections?Wait, the problem says opponents are matched randomly within their bracket section, but the underdog always faces the highest-ranked opponent available. So, in each round, the underdog is matched against the highest-ranked player remaining in their bracket section. So, in the first round, the underdog is matched against the highest-ranked player in their section. If they win, in the next round, they face the next highest-ranked player in their section, and so on.So, in each round, the underdog is facing the highest remaining opponent in their section. Therefore, in each round, the underdog's opponent is the next highest-ranked player, so the probability of winning each subsequent match is 30% each time, because each opponent is ranked above the underdog.Wait, but the underdog is ranked 128th, so all opponents are ranked above them, so each match is a 30% chance. So, regardless of the opponent's rank, as long as they are above the underdog, it's 30%. So, if the underdog has to face 7 opponents, each with a 30% chance, then the total probability is (0.3)^7.But let me double-check. If the underdog is always facing the highest-ranked opponent available, that means in each round, they are facing the next highest-ranked player in their section. So, in the first round, they might face rank 1, then in the second round, rank 2, then rank 3, and so on. But wait, that can't be, because in a single-elimination bracket, higher-ranked players are spread out to avoid facing each other early on. So, actually, the bracket is seeded so that the top players are in different sections.Wait, but the problem says opponents are matched randomly within their bracket section. So, maybe the bracket is not seeded, and the underdog could face any opponent in their section, but the underdog always faces the highest-ranked opponent available. So, in each round, the underdog is matched against the highest-ranked player remaining in their section.So, in the first round, the underdog's section has 64 players? Wait, no, 128 players, so each section is 64 players? Wait, no, 128 players, so the bracket is divided into 128 slots, but each section is a quarter or an eighth? Wait, maybe it's a 128-player bracket, so each section is 16 players? Because 128 divided by 8 is 16. So, each section has 16 players, and the underdog is in one of those sections.Wait, actually, in a standard single-elimination bracket with 128 players, it's usually divided into 8 sections of 16 players each. So, each section has 16 players, and the winner of each section goes to the quarterfinals. Wait, no, actually, 128 players would have 7 rounds: 64, 32, 16, 8, 4, 2, 1. So, each section is actually 16 players, because 128 divided by 8 is 16. So, each section has 16 players, and the winner of each section advances to the quarterfinals.Wait, but in the first round, each section has 16 players, so 8 matches. Then, the next round has 8 players in the section, then 4, then 2, then 1. So, the underdog is in a section with 15 other players. So, in each round, the underdog is facing someone in their section.But the problem says opponents are matched randomly within their bracket section and that the underdog always faces the highest-ranked opponent available. So, in each round, the underdog is facing the highest-ranked player remaining in their section. So, in the first round, the underdog is matched against the highest-ranked player in their section. If they win, in the next round, they face the next highest-ranked player in their section, and so on.So, in the first round, the underdog is facing rank 1, then in the second round, rank 2, then rank 3, etc., up to rank 7. Wait, but the underdog is rank 128, so the highest-ranked player in their section would be rank 1, then rank 2, etc., but that can't be because in a standard bracket, higher-ranked players are distributed across different sections to prevent them from meeting early.Wait, maybe the bracket is not seeded, so the underdog could be in a section with all the top players. But the problem says opponents are matched randomly within their bracket section, but the underdog always faces the highest-ranked opponent available. So, regardless of the bracket structure, in each round, the underdog is matched against the highest-ranked player remaining in their section.So, if the underdog is in a section with, say, rank 1, rank 2, ..., rank 16, then in the first round, they face rank 1, then if they win, they face rank 2, and so on until they face rank 16 in the final of their section.But wait, in reality, the bracket is structured so that higher-ranked players are in different sections, but the problem says opponents are matched randomly within their bracket section. So, maybe the underdog's section is randomly assigned, but within that section, the underdog always faces the highest-ranked opponent available.So, in the first round, the underdog is in a section with 15 other players, and the highest-ranked player in that section is, say, rank k. Then, the underdog faces rank k in the first round. If they win, they face the next highest-ranked player in the section, and so on.But the problem is that the underdog is ranked 128th, so all opponents are ranked above them. So, in each round, the underdog faces the highest-ranked opponent in their section, which is someone ranked 1 to 127. So, the probability of winning each match is 30%, regardless of the opponent's rank, as long as they are above the underdog.Therefore, the underdog has to win 7 matches in a row, each with a 30% chance. So, the probability is 0.3^7.But wait, let me think again. If the underdog is always facing the highest-ranked opponent available, that means in each round, they are facing the next highest-ranked player in their section. So, in the first round, they face the highest-ranked player in their section, say rank x. Then, in the second round, they face the next highest-ranked player in their section, say rank y, and so on.But the key point is that the underdog is facing the highest-ranked opponent available in their section in each round. So, in each round, the opponent is the highest-ranked remaining in their section, which is someone ranked above the underdog. Therefore, each match has a 30% chance of winning.Therefore, the probability of winning the tournament is 0.3^7.But wait, let me calculate that. 0.3^7 is 0.0002187, which is 0.02187%, which seems extremely low. Is that correct?Alternatively, maybe I'm misunderstanding the structure. Maybe the underdog doesn't have to face the highest-ranked opponent in each round, but rather, in each round, the opponents are randomly matched, but the underdog is always matched against the highest-ranked opponent available in their section.Wait, so in each round, the underdog is matched against the highest-ranked opponent remaining in their section. So, in the first round, they face the highest-ranked player in their section. If they win, in the next round, they face the next highest-ranked player in their section, and so on.So, the underdog's path is facing the top players in their section one after another. So, in the first round, they face rank 1, then rank 2, rank 3, etc., up to rank 7 in the final.But wait, in a standard bracket, the top players are spread out, so the underdog wouldn't face rank 1 in the first round. But the problem says opponents are matched randomly within their bracket section, but the underdog always faces the highest-ranked opponent available.So, perhaps in this case, the underdog's section is randomly assigned, but within that section, the underdog is always facing the highest-ranked opponent. So, the underdog could be in a section with rank 1, rank 2, ..., rank 16, and in each round, they face the highest-ranked player remaining.Therefore, the underdog would have to beat rank 1, then rank 2, then rank 3, etc., up to rank 7 in the final. So, each match is against a higher-ranked player, each with a 30% chance.Therefore, the probability is 0.3^7.But wait, let me think about the number of opponents. In a 128-player bracket, each section has 16 players. So, the underdog has to win 4 matches in their section to become the section winner, then 3 more matches in the final bracket.Wait, no, 128 players, so each section has 16 players, which requires 4 rounds to get to the section final (16, 8, 4, 2, 1). Wait, no, 16 players require 4 rounds: 1st round 8 matches, 2nd round 4, 3rd round 2, 4th round 1. So, 4 rounds to get out of the section. Then, the final bracket has 8 players, requiring 3 more rounds: quarterfinals, semifinals, finals.Wait, no, 128 players, each section has 16, so 8 sections. Each section produces a winner, so 8 winners go to the quarterfinals. Then, quarterfinals (8 players), semifinals (4), finals (2), and the champion.So, the underdog has to win 4 matches in their section, then 3 more in the final bracket, totaling 7 matches.But in each round, the underdog is facing the highest-ranked opponent available in their section. So, in the first round, they face the highest-ranked player in their section, say rank 1. If they win, they face the next highest, rank 2, and so on until they face rank 16 in the section final.Wait, but rank 16 is still above the underdog, who is rank 128. So, each match is a 30% chance.Therefore, the probability is 0.3^7.But let me calculate that: 0.3^7 = 0.0002187, which is 0.02187%.That seems extremely low, but given that the underdog has only a 30% chance against each higher-ranked player, and they have to win 7 times, it's plausible.Alternatively, maybe I'm overcomplicating it. If the underdog has to win 7 matches, each with a 30% chance, then the probability is 0.3^7.But wait, let me think about the structure again. If the underdog is in a section with 16 players, and in each round, they face the highest-ranked opponent available, that means in the first round, they face the top seed in their section, say rank 1. If they win, they face rank 2, then rank 3, etc., up to rank 16 in the section final.But rank 16 is still above the underdog, so each match is a 30% chance. Therefore, to win the section, they need to win 4 matches, each with a 30% chance, so 0.3^4.Then, in the final bracket, they have to face the winners of the other sections. So, in the quarterfinals, they face the winner of another section, which is the highest-ranked player in that section, say rank 17. Then, in the semifinals, they face rank 18, and in the finals, they face rank 19.Wait, no, that doesn't make sense. Because the other sections would have their own highest-ranked players. So, the underdog, after winning their section, would face the next highest-ranked player in the entire bracket, which would be rank 17, then rank 18, etc.But wait, the bracket is structured so that the top 8 seeds are in different sections. So, the underdog's section might have rank 1, then the next section has rank 2, and so on. So, if the underdog wins their section, they would face rank 2 in the quarterfinals, then rank 3 in the semifinals, and rank 4 in the finals.Wait, but that depends on how the bracket is seeded. If the bracket is seeded, the top 8 seeds are placed in different sections. So, the underdog is in a section with rank 1, then the next section has rank 2, and so on.Therefore, if the underdog wins their section, they would face rank 2 in the quarterfinals, then rank 3 in the semifinals, and rank 4 in the finals.But wait, that's not necessarily the case. The bracket is usually set up so that the top seed is in one section, the second seed in another, etc., so that they don't meet until later rounds.So, if the underdog is in the same section as rank 1, then to win the section, they have to beat rank 1, then in the quarterfinals, they would face the winner of another section, which is the highest-ranked player in that section, say rank 2. Then, in the semifinals, they face rank 3, and in the finals, rank 4.Wait, but that would mean the underdog has to face rank 1, 2, 3, 4 in consecutive matches, each with a 30% chance. So, the probability would be 0.3^4 (to win the section) times 0.3^3 (to win the final bracket), totaling 0.3^7.But wait, if the bracket is seeded, the underdog's section would have rank 1, and the other sections would have ranks 2, 3, 4, etc. So, the underdog would have to face rank 1 in their section, then rank 2 in the quarterfinals, rank 3 in the semifinals, and rank 4 in the finals.But that would mean the underdog has to win 4 matches in their section (against rank 1, 2, 3, 4?), no, wait, in their section, they only have to face rank 1, then the next highest in their section, which might be rank 17, not rank 2.Wait, I'm getting confused. Let me clarify.In a standard seeded bracket with 128 players, the top 8 seeds are placed in different sections. So, each section has one top seed. So, the underdog is in a section with rank 1, then the next section has rank 2, and so on up to rank 8.Therefore, the underdog's section has rank 1, and the other sections have ranks 2-8. So, if the underdog wins their section, they would face rank 2 in the quarterfinals, rank 3 in the semifinals, and rank 4 in the finals.Wait, no, the bracket is structured so that the top seeds are spread out. So, the underdog's section has rank 1, and the other sections have ranks 2, 3, 4, etc. So, the underdog would have to face rank 1 in their section, then in the quarterfinals, they would face the winner of another section, which is rank 2, then in the semifinals, rank 3, and in the finals, rank 4.But that would mean the underdog has to face rank 1, 2, 3, 4 in consecutive matches, each with a 30% chance. So, the probability would be 0.3^4 (to win the section) times 0.3^3 (to win the final bracket), totaling 0.3^7.But wait, in reality, the underdog's section would have rank 1, and the other sections would have ranks 2-8. So, the underdog would have to beat rank 1, then in the quarterfinals, they face rank 2, then in the semifinals, rank 3, and in the finals, rank 4.But that's assuming the bracket is perfectly seeded, which might not be the case here. The problem says opponents are matched randomly within their bracket section, but the underdog always faces the highest-ranked opponent available.So, perhaps the underdog's section is randomly assigned, but within that section, they always face the highest-ranked opponent. So, the underdog could be in a section with any of the top 127 players, but in each round, they face the highest-ranked remaining in their section.Therefore, the underdog's path is facing the highest-ranked players in their section in each round, which could be any of the top 127 players, but each with a 30% chance.So, regardless of the specific opponents, each match is a 30% chance, and they have to win 7 matches in a row. Therefore, the probability is 0.3^7.But let me think again. If the underdog is in a section with, say, rank 1, then in the first round, they face rank 1. If they win, they face the next highest in their section, say rank 2, and so on until they face rank 16 in the section final. Then, in the quarterfinals, they face rank 17, then rank 18, etc.Wait, but that would mean the underdog has to face rank 1, 2, 3, ..., 16 in their section, and then 17, 18, 19 in the final bracket. But that would be 16 matches, which is more than 7. So, that can't be.Wait, no, in a 128-player bracket, each section has 16 players, so the underdog has to win 4 matches in their section (16 to 8 to 4 to 2 to 1), then 3 more matches in the final bracket (quarterfinals, semifinals, finals). So, total 7 matches.Therefore, in each of those 7 matches, the underdog is facing the highest-ranked opponent available in their section. So, in the first round, they face the highest-ranked in their section, say rank x. If they win, they face the next highest, rank y, and so on.But the key is that each opponent is ranked above the underdog, so each match is a 30% chance. Therefore, the probability is 0.3^7.So, the answer to part 1 is 0.3^7, which is 0.0002187, or 0.02187%.Now, moving on to part 2. The tournament is modified such that every second match for the underdog involves a "wildcard" rule where their chance of winning doubles. So, their chance of winning doubles in every second match.So, the underdog's matches are as follows: match 1, match 2, match 3, ..., match 7. In every second match (matches 2, 4, 6), their chance of winning doubles. So, instead of 30%, it becomes 60%.So, we need to calculate the new probability of winning the tournament, considering that in matches 2, 4, 6, the underdog has a 60% chance, and in the others, 30%.So, the probability is 0.3 * 0.6 * 0.3 * 0.6 * 0.3 * 0.6 * 0.3.Let me calculate that:0.3 * 0.6 = 0.18Then, 0.18 * 0.3 = 0.0540.054 * 0.6 = 0.03240.0324 * 0.3 = 0.009720.00972 * 0.6 = 0.0058320.005832 * 0.3 = 0.0017496So, the probability is approximately 0.0017496, or 0.17496%.Comparing this to the original probability of 0.02187%, the underdog's chances have increased by a factor of roughly 8 (0.17496 / 0.02187 ≈ 8).Wait, no, 0.17496% is actually higher than 0.02187%. Wait, 0.17496% is approximately 8 times higher than 0.02187%.Wait, 0.02187 * 8 = 0.175, so yes, exactly.So, the underdog's probability of winning the tournament increases by a factor of approximately 8 when every second match has their chance doubled.But let me double-check the calculation:0.3 * 0.6 * 0.3 * 0.6 * 0.3 * 0.6 * 0.3Grouping them:(0.3 * 0.6)^3 * 0.3Because there are three pairs of (0.3 * 0.6) and one extra 0.3.So, (0.18)^3 * 0.30.18^3 = 0.0058320.005832 * 0.3 = 0.0017496Yes, that's correct.So, the original probability was 0.0002187, and the new probability is 0.0017496, which is approximately 8 times higher.Therefore, the expected change is an increase in the underdog's probability of winning the tournament by a factor of about 8.But let me think if there's another way to interpret the "every second match" rule. Maybe it's not matches 2, 4, 6, but every second match starting from the first, so matches 1, 3, 5, 7 have the doubled chance. But the problem says "every second match," which usually means every even-numbered match, i.e., 2, 4, 6.But to be thorough, let's consider both interpretations.If the wildcard applies to every second match starting from the first, so matches 1, 3, 5, 7 have a 60% chance, and matches 2, 4, 6 have 30%. Then, the probability would be:0.6 * 0.3 * 0.6 * 0.3 * 0.6 * 0.3 * 0.6Which is (0.6 * 0.3)^3 * 0.60.18^3 * 0.6 = 0.005832 * 0.6 = 0.0034992So, approximately 0.34992%, which is higher than the previous interpretation.But the problem says "every second match," which typically refers to even-numbered matches, so matches 2, 4, 6. Therefore, the first interpretation is correct.Therefore, the underdog's probability increases from 0.02187% to 0.17496%, which is an 8-fold increase.So, the expected change is an increase in the underdog's probability of winning the tournament by a factor of approximately 8.</think>

🧠:<think>Alright, so I've got this problem about a university professor organizing an international sports conference. The professor has categorized researchers into five different disciplines: Mathematics, Physics, Computer Science, Sports Science, and Economics. Each of these disciplines has a different number of researchers attending, denoted by a, b, c, d, and e respectively. The seating arrangement requires that each table must have exactly one researcher from each discipline, and no two researchers from the same discipline can sit at the same table. There are two sub-problems here. Let me tackle them one by one.Sub-problem 1:The numbers given are:[ a = 5, quad b = 6, quad c = 7, quad d = 8, quad e = 9. ]We need to determine the number of unique tables that can be formed under these constraints.Hmm, okay. So each table must have one researcher from each of the five disciplines. That means for each table, we need one person from Mathematics, one from Physics, and so on. The key here is that no two researchers from the same discipline can sit at the same table. So, for each discipline, the number of researchers available is given, and we need to figure out how many such tables can be formed without violating the constraints.Wait, so if we think about it, each table requires one person from each discipline. So, the number of tables we can form is limited by the discipline with the fewest number of researchers because each table needs one from each. But hold on, actually, no. Because each researcher can only be at one table, right? So, if one discipline has only 5 people, then we can have at most 5 tables, each with one person from that discipline. But the other disciplines have more people, so they can supply more, but the limiting factor is the smallest number.Wait, but let me think again. If we have 5 Mathematics researchers, 6 Physics, 7 Computer Science, 8 Sports Science, and 9 Economics. Each table needs one from each. So, the maximum number of tables we can form is limited by the smallest number of researchers in any discipline. Because if we have only 5 Mathematics researchers, we can't have more than 5 tables, each with one Mathematics researcher. The other disciplines have more, so they can supply the extra people for those tables.But wait, is that correct? Let me think. Suppose we have 5 tables, each with one Mathematics researcher. Then, for each of those tables, we need one Physics, one Computer Science, etc. So, the Physics discipline has 6, which is more than 5, so we can assign one to each table and still have one left over. Similarly, Computer Science has 7, which is more than 5, so we can assign one to each table and have two left over. Same with Sports Science (8) and Economics (9). So, the limiting factor is indeed the smallest number, which is 5.But wait, hold on. Is that the case? Because each table must have one from each discipline, but the number of tables isn't necessarily just the minimum of the numbers. Because if you have more people in other disciplines, you could potentially form more tables if you can somehow arrange the seating without overlapping. But no, because each table must have one from each discipline, and each researcher can only be at one table. So, if we have 5 Mathematics researchers, we can only form 5 tables, each with one Mathematics researcher, and each table must have one from each of the other disciplines as well. So, the number of tables is indeed the minimum of a, b, c, d, e.Wait, but let me think again. Suppose we have 5 tables, each with one Mathematics, one Physics, etc. Then, we would use up 5 Physics, 5 Computer Science, 5 Sports Science, and 5 Economics. But we have more in the other disciplines. So, is there a way to form more tables? But no, because we don't have more Mathematics researchers. So, we can't form more than 5 tables because we don't have more Mathematics people.Therefore, the number of unique tables that can be formed is 5.Wait, but hold on. Let me think of it as a matching problem. Each table is a combination of one from each discipline. So, the total number of possible tables is the product of the number of choices from each discipline. But that would be 5*6*7*8*9, which is a huge number. But that's the total number of possible unique tables if we consider all possible combinations. But in reality, the seating arrangement is such that each researcher can only be at one table. So, it's not about the number of possible unique tables, but rather the maximum number of tables that can be formed without overlapping researchers.So, in that case, it's similar to finding the maximum matching in a bipartite graph, but in this case, it's a multi-dimensional matching problem. Each table is a 5-tuple, one from each discipline, and each researcher can only be in one tuple.In such cases, the maximum number of tables is indeed the minimum number of researchers in any discipline. Because if you have a bottleneck in one discipline, you can't exceed that number. So, in this case, since Mathematics has only 5, that's the limiting factor.Wait, but let me think about it differently. Suppose we have 5 tables. Each table will have one Mathematics researcher, one Physics, one Computer Science, etc. So, for each table, we need one from each discipline. So, the maximum number of tables is the minimum number across all disciplines. So, 5 is correct.But wait, another way to think about it is that each table is a combination of one from each discipline, so the number of tables is the minimum of a, b, c, d, e. So, 5 is the answer.Wait, but hold on. Suppose we have 5 tables, each with one Mathematics researcher. Then, we can assign 5 Physics, 5 Computer Science, etc. But we have more in the other disciplines. So, is there a way to form more tables? But no, because we don't have more Mathematics researchers. So, 5 is indeed the maximum number of tables.Therefore, the number of unique tables is 5.But wait, hold on. Let me think again. Suppose we have 5 Mathematics, 6 Physics, 7 Computer Science, 8 Sports Science, and 9 Economics. Each table requires one from each. So, the number of tables is the minimum of these numbers, which is 5. So, 5 tables can be formed, each with one from each discipline.But wait, another perspective: the number of tables is the minimum number of researchers in any discipline because each table needs one from each. So, if one discipline only has 5, you can't have more than 5 tables. So, yes, 5 is correct.Wait, but hold on. Let me think of it as a constraint satisfaction problem. Each table must have one from each discipline, and each researcher can only be in one table. So, the maximum number of tables is the minimum of the sizes of the disciplines because each table requires one from each, and you can't have more tables than the smallest group.Yes, that makes sense. So, the answer is 5.Wait, but let me think of an example. Suppose we have 2 Mathematics, 3 Physics, 4 Computer Science, 5 Sports Science, 6 Economics. Then, the maximum number of tables is 2, because we only have 2 Mathematics researchers. Each table needs one from each, so we can't have more than 2 tables.Similarly, in our original problem, since Mathematics has 5, that's the limiting factor. So, 5 tables.Therefore, the answer to Sub-problem 1 is 5.Sub-problem 2:Assuming the total number of tables formed is T, which we found to be 5, the professor wants to assign each table a unique identifier number ranging from 1 to T. We need to determine the probability that a randomly chosen table has an identifier number that is a prime number.So, T is 5. The identifiers are 1, 2, 3, 4, 5. We need to find how many of these are prime numbers.First, let's list the numbers from 1 to 5: 1, 2, 3, 4, 5.Now, prime numbers are numbers greater than 1 that have no positive divisors other than 1 and themselves.So, let's check each number:1: Not prime (by definition).2: Prime (only divisors are 1 and 2).3: Prime (only divisors are 1 and 3).4: Not prime (divisors are 1, 2, 4).5: Prime (only divisors are 1 and 5).So, the prime numbers in this range are 2, 3, and 5. That's 3 numbers.Therefore, there are 3 prime numbers out of 5 total tables.So, the probability is the number of prime-numbered tables divided by the total number of tables, which is 3/5.Therefore, the probability is 3/5.But wait, let me double-check. The numbers are 1, 2, 3, 4, 5. Primes are 2, 3, 5. So, 3 primes. So, yes, 3/5.So, the probability is 3/5.Final AnswerSub-problem 1: boxed{5}Sub-problem 2: boxed{dfrac{3}{5}}</think>

🧠:<think>Alright, so I have this problem about translating an ancient manuscript with encrypted numerical codes. It mentions something about an n-dimensional hypercube and eigenvalues of a transformation matrix. Hmm, okay, let me try to break this down step by step.First, the problem states that the numerical codes form an n-dimensional hypercube, and each vertex represents a significant event. The total number of events is 1024. I remember that in a hypercube, the number of vertices is 2^n because each dimension adds a binary choice (0 or 1). So, if the number of vertices is 1024, that should be equal to 2^n. Let me write that down:Number of vertices = 2^n = 1024.Okay, so I need to solve for n. I know that 2^10 is 1024 because 2^10 = 1024. So, n must be 10. That seems straightforward. Let me just verify:2^1 = 22^2 = 42^3 = 82^4 = 162^5 = 322^6 = 642^7 = 1282^8 = 2562^9 = 5122^10 = 1024Yep, that's correct. So, the dimension n is 10.Alright, moving on to the second part. The transformation matrix A is given as A = 2I - J, where I is the identity matrix and J is an n x n matrix with all entries equal to 1. I need to find the eigenvalues of A.Hmm, eigenvalues. I remember that for a matrix, the eigenvalues can be found by solving the characteristic equation det(A - λI) = 0. But since A is defined in terms of I and J, maybe there's a smarter way to find its eigenvalues without getting into messy calculations.I recall that J is a rank-1 matrix because all its rows are the same. That might be useful. Also, I remember that if a matrix has rank 1, it has only one non-zero eigenvalue, which is equal to the trace of the matrix. The trace of J is n because all the diagonal entries are 1, and there are n of them. So, the eigenvalues of J are n and 0 (with multiplicity n-1).Now, since A = 2I - J, perhaps I can use the properties of eigenvalues under matrix operations. Specifically, if I know the eigenvalues of J, I can find the eigenvalues of A by applying the same operations to the eigenvalues.Let me think. If J has eigenvalues n and 0 (with multiplicity n-1), then 2I would have eigenvalues 2 and 2 (since multiplying by a scalar affects all eigenvalues). Then, subtracting J from 2I would subtract the eigenvalues of J from those of 2I.So, the eigenvalues of A should be 2 - n and 2 - 0, right? Wait, hold on. Let me clarify.If J has eigenvalues λ, then 2I - J will have eigenvalues 2 - λ. So, for each eigenvalue λ of J, the corresponding eigenvalue of A is 2 - λ.Given that J has eigenvalues n (with multiplicity 1) and 0 (with multiplicity n-1), then A will have eigenvalues 2 - n (with multiplicity 1) and 2 - 0 = 2 (with multiplicity n-1).Therefore, the eigenvalues of A are 2 - n and 2, where 2 - n has multiplicity 1 and 2 has multiplicity n - 1.Since we already found that n = 10, let's plug that in. The eigenvalues of A are 2 - 10 = -8 (with multiplicity 1) and 2 (with multiplicity 9). So, A has eigenvalues -8 and 2, with -8 occurring once and 2 occurring nine times.Let me just verify if this makes sense. If A = 2I - J, then J is a matrix of all ones. So, for example, in the case of n=2, J would be:[1 1][1 1]Then, A would be:[2-1, 2-1] = [1, 1][2-1, 2-1] = [1, 1]Wait, no, that's not correct. Wait, A = 2I - J, so for n=2, I is:[1 0][0 1]So, 2I is:[2 0][0 2]Then, subtracting J:[2-1, 2-1] = [1, 1][2-1, 2-1] = [1, 1]Wait, so A would be:[1 1][1 1]But that can't be right because if n=2, the eigenvalues should be 2 - n = 0 and 2, but in this case, A is a rank-1 matrix with eigenvalues 2 and 0. Hmm, but according to my earlier reasoning, for n=2, A should have eigenvalues 2 - 2 = 0 and 2 (with multiplicity 1). But in reality, A is a rank-1 matrix, so it should have eigenvalues 2 (trace) and 0 (with multiplicity 1). Wait, that seems conflicting.Wait, no, hold on. For n=2, J has eigenvalues 2 and 0. So, A = 2I - J would have eigenvalues 2 - 2 = 0 and 2 - 0 = 2. So, eigenvalues are 0 and 2, which matches the actual calculation. So, in n=2, A has eigenvalues 0 and 2, which is consistent with the earlier reasoning.Similarly, for n=3, J has eigenvalues 3 and 0 (with multiplicity 2). Then, A = 2I - J would have eigenvalues 2 - 3 = -1 and 2 - 0 = 2 (with multiplicity 2). So, eigenvalues are -1 and 2 (twice). That seems correct.Therefore, going back to our problem where n=10, the eigenvalues of A should be 2 - 10 = -8 and 2 (with multiplicity 9). So, that seems correct.Just to make sure, let me think about another way. The matrix J is a matrix of all ones, which is a rank-1 matrix. So, it has one non-zero eigenvalue equal to its trace, which is n. The other eigenvalues are zero. Therefore, when we compute A = 2I - J, the eigenvalues will be 2 - n and 2 - 0 = 2, with the same multiplicities as J's eigenvalues. So, 2 - n has multiplicity 1, and 2 has multiplicity n - 1.Yes, that seems consistent. So, for n=10, eigenvalues are -8 and 2, with -8 once and 2 nine times.Therefore, I think I have the answers:1. The dimension n is 10.2. The eigenvalues of A are -8 and 2, with -8 occurring once and 2 occurring nine times.I should probably write this up neatly.Final AnswerThe dimension ( n ) of the hypercube is boxed{10}, and the eigenvalues of matrix ( A ) are boxed{-8} and boxed{2} (with multiplicity 9).</think>

🧠:<think>First, I'll calculate the relative frequency of the adjective in Heather's novel by dividing the number of times the adjective is used by the total word count. So, for Heather, that's 600 divided by 120,000, which equals 0.005.Next, I'll do the same for her friend's novel. The friend uses the adjective 700 times out of 140,000 words, so dividing 700 by 140,000 gives 0.005 as well.Now, to find the ratio of Heather's relative frequency to her friend's, I'll divide Heather's relative frequency by her friend's. That's 0.005 divided by 0.005, which equals 1. This means their relative frequencies are the same.Moving on to the average sentence length, I'll start with Heather's novel. She has 120,000 words spread over 6,000 sentences. Dividing the total words by the number of sentences gives an average sentence length of 20 words per sentence.For her friend, there are 140,000 words across 7,000 sentences. Dividing 140,000 by 7,000 results in an average sentence length of 20 words per sentence as well.Finally, to determine the percentage difference between Heather's and her friend's average sentence lengths, I'll compare the two. Since both averages are 20 words per sentence, the percentage difference is 0%. This indicates that their average sentence lengths are identical.</think>