🧠:<think>First, I'll define variables for each category of books. Let F represent Fiction books, NF represent Non-Fiction books, and R represent Reference books.From the problem, I know that the total number of books is 5,000, so I can write the equation:F + NF + R = 5000It's also given that the number of Fiction books is twice the number of Non-Fiction books, which translates to:F = 2 * NFAdditionally, the number of Reference books is 300 more than the number of Non-Fiction books:R = NF + 300Now, I'll substitute the expressions for F and R from the second and third equations into the first equation:2 * NF + NF + (NF + 300) = 5000Combining like terms gives:4 * NF + 300 = 5000Subtracting 300 from both sides:4 * NF = 4700Dividing both sides by 4:NF = 1175With NF known, I can find F and R:F = 2 * 1175 = 2350R = 1175 + 300 = 1475So, there are 2350 Fiction books, 1175 Non-Fiction books, and 1475 Reference books.For the second part, the librarian increases Fiction books by 10% and Reference books by 5%, while Non-Fiction remains the same. Calculating the new quantities:New Fiction = 2350 * 1.10 = 2585New Reference = 1475 * 1.05 = 1548.75Adding these together with Non-Fiction:Total Books = 2585 + 1175 + 1548.75 = 5308.75Since the number of books should be a whole number, I'll round up to 5309.</think>

🧠:<think>Alright, so I've got these two questions about network graphs modeling a society. Let me try to unpack them one by one. Starting with the first question: It defines a traditional institution as a subgraph H with high clustering and a minimum degree of at least k. If removing H disconnects the graph into at least two components, then H is necessary for stability. The statement to prove or disprove is: If H isn't necessary, then there's a subset S outside H such that removing H and adding S forms a new H' with the same properties, keeping the network connected.Hmm. So, if H isn't necessary, that means removing H doesn't disconnect the graph. So, G - H is still connected. The question is whether we can find another subset S (not in H) such that H' = S has the same properties as H, and when we remove H and add H', the graph remains connected.Wait, actually, the wording is a bit different. It says when H is removed, S can be used to form H' with the same properties, maintaining connectivity. So, maybe it's about whether the remaining graph without H still has enough structure to form another H' that maintains the same properties and connectivity.Since H isn't necessary, G - H is connected. So, in G - H, can we find a subgraph H' with high clustering and minimum degree k? That would mean that even without H, the rest of the graph can support another such subgraph. But is that necessarily true? High clustering and minimum degree k might require certain structural properties. If G - H is connected, but maybe it's just barely connected, like a tree, which has low clustering. So, maybe G - H doesn't have high clustering, so we can't form H' with high clustering.Alternatively, perhaps the original graph G had high connectivity, so even after removing H, the remaining graph still has enough edges to form another H'. But I'm not sure.Wait, the initial society is stable, meaning high connectivity. So, G is highly connected. If H is a subgraph with high clustering and minimum degree k, but it's not necessary, meaning G - H is still connected. So, G - H is connected but maybe not as highly connected as G.But does G - H necessarily contain another subgraph H' with the same properties? Maybe not necessarily. For example, suppose H is a tightly knit group, and G - H is connected but has lower clustering. Maybe it's a long chain or something, which doesn't have high clustering. So, you can't form H' with high clustering from G - H.Therefore, the statement might be false. So, I think the answer is to disprove it. There doesn't necessarily exist such an S.Moving on to the second question: Each individual has a stability score s_i, and the total stability is the sum of all s_i. If traditional institutions are removed and replaced with alternative structures, keeping the same number of vertices and edges, find an expression for the new stability S'(G') in terms of original scores and changes in network properties.So, the new graph G' has the same number of vertices and edges as G, but the structure is different. The stability scores might change because each individual's position in the network affects their stability score.So, the original stability is S(G) = sum_{v_i in V} s_i. After replacing H with some other structure, each individual's s_i might change. Let's denote the new stability score as s'_i. Then, S'(G') = sum_{v_i in V} s'_i.But how does the change in network properties affect s'_i? The problem says the stability score may change based on their new position. So, we need to express S'(G') in terms of the original s_i and the change in network properties.Perhaps the change in network properties affects the stability scores. For example, if someone's degree changes, or their clustering coefficient changes, their s_i might change accordingly.But without knowing the exact relationship between network properties and stability scores, it's hard to write an exact expression. Maybe we can model the change as a function of the change in some network metric.Alternatively, if we assume that the stability score is a function of local network properties like degree or clustering, then S'(G') would be the sum over all nodes of f(d'_i, c'_i), where d'_i and c'_i are the new degree and clustering coefficient.But the problem says to express it in terms of the original stability scores and the change in network properties. So, maybe S'(G') = S(G) + sum_{v_i} [delta s_i], where delta s_i is the change in stability score due to the change in network properties.But to make it more precise, perhaps we can write it as S'(G') = sum_{v_i} [s_i + delta s_i], where delta s_i depends on the change in some network metrics for node i.Alternatively, if we denote the change in some property as delta_p_i, then S'(G') = sum_{v_i} s_i + sum_{v_i} f(delta_p_i), where f is some function relating the change in property to the change in stability.But without more specifics, it's hard to give an exact expression. Maybe the problem expects recognizing that the total stability changes based on how the network properties change, so S'(G') = S(G) + sum_{v_i} (s'_i - s_i), where s'_i depends on the new structure.Alternatively, if we consider that the removal and replacement doesn't change the number of edges or vertices, but redistributes them, perhaps the total stability could be expressed as the original sum plus some function of the new connections.But I think the key is that the total stability is the sum of individual stabilities, which depend on their network position. So, if the network changes, each s_i might change, so S'(G') = sum_{v_i} s'_i, where s'_i is a function of the new graph's properties at node i.But since the problem asks for an expression in terms of the original stability scores and the change in network properties, maybe we can write it as S'(G') = S(G) + sum_{v_i} (s'_i - s_i), where (s'_i - s_i) is the change due to the new network properties.Alternatively, if we denote the change in some network property for node i as delta_p_i, then S'(G') = sum_{v_i} [s_i + f(delta_p_i)], where f is some function.But without knowing the exact relationship between network properties and stability scores, it's difficult to specify f. So, perhaps the answer is that S'(G') is the sum of the new stability scores, which depend on the new network structure, so S'(G') = sum_{v_i} s'_i, where s'_i is determined by the individual's new position in G'.But the question says "in terms of the original stability scores and the change in network properties." So, maybe we can write S'(G') = S(G) + sum_{v_i} (s'_i - s_i), where each term (s'_i - s_i) is a function of the change in network properties for node i.Alternatively, if we let delta_i be the change in some property for node i, then S'(G') = sum_{v_i} s_i + sum_{v_i} f(delta_i), where f is the impact function.But since the exact relationship isn't given, perhaps the answer is simply that S'(G') is the sum of the new stability scores, which are functions of the new network structure, so S'(G') = sum_{v_i} s'_i, where s'_i depends on the new position in G'.But the question wants it in terms of the original scores and changes in network properties. So, maybe S'(G') = S(G) + sum_{v_i} (s'_i - s_i), where each (s'_i - s_i) is determined by the change in network properties for node i.Alternatively, if we denote the change in a property like degree or clustering as delta_d_i and delta_c_i, then S'(G') = sum_{v_i} [s_i + a*delta_d_i + b*delta_c_i], where a and b are coefficients.But without knowing the functional form, it's hard. So, perhaps the answer is that S'(G') is the sum over all nodes of their new stability scores, which are functions of their new network positions, so S'(G') = sum_{v_i} s'_i, where s'_i is determined by the new graph G'.But since the question asks to express it in terms of the original scores and changes, maybe it's S'(G') = S(G) + sum_{v_i} (s'_i - s_i), where each (s'_i - s_i) is due to changes in network properties.Alternatively, if we consider that the total stability could change based on how the edges are reconfigured, but since the number of edges is the same, maybe the total stability could be similar, but redistributed.But I think the most precise answer is that S'(G') is the sum of the new stability scores, which depend on the new network structure, so S'(G') = sum_{v_i} s'_i, where each s'_i is a function of the new position in G'.But to tie it back to the original scores and changes, maybe S'(G') = S(G) + sum_{v_i} (s'_i - s_i), where each term accounts for the change in stability due to network changes.Alternatively, if we denote the change in a node's stability as delta_s_i, then S'(G') = S(G) + sum_{v_i} delta_s_i.But without more specifics, I think the answer is that S'(G') is the sum of the new stability scores, which are functions of the new network properties, so S'(G') = sum_{v_i} s'_i, where s'_i depends on the individual's new position in G'.But the question wants it in terms of the original scores and changes, so maybe S'(G') = sum_{v_i} [s_i + f(p'_i - p_i)], where p_i are the original properties and p'_i are the new properties.But I think the answer is that S'(G') is the sum of the new stability scores, which depend on the new network structure, so S'(G') = sum_{v_i} s'_i, where s'_i is determined by the new graph G'.But to express it in terms of the original scores and changes, perhaps S'(G') = S(G) + sum_{v_i} (s'_i - s_i), where each (s'_i - s_i) is due to the change in network properties.Alternatively, if we consider that the total stability is the sum of individual stabilities, and each stability can be expressed as a function of their network properties, then S'(G') = sum_{v_i} f(p'_i), where p'_i are the new properties, and f is the function mapping properties to stability.But since the original S(G) = sum f(p_i), then S'(G') = sum f(p'_i).But the question wants it in terms of the original scores and changes, so maybe S'(G') = S(G) + sum_{v_i} [f(p'_i) - f(p_i)].Yes, that makes sense. So, the new stability is the original stability plus the sum of the changes in stability for each node due to the change in their network properties.So, S'(G') = S(G) + sum_{v_i} [s'_i - s_i], where s'_i = f(p'_i) and s_i = f(p_i).Therefore, the expression is S'(G') = S(G) + sum_{v_i} (s'_i - s_i).But since the problem says "in terms of the original stability scores and the change in network properties," maybe we can write it as S'(G') = sum_{v_i} s_i + sum_{v_i} (s'_i - s_i) = sum_{v_i} s'_i, but that's just restating it.Alternatively, if we denote the change in network properties for node i as delta_p_i, and assume that the change in stability is proportional to delta_p_i, then S'(G') = S(G) + sum_{v_i} k_i * delta_p_i, where k_i is some constant or function.But without knowing the exact relationship, I think the answer is that S'(G') is the sum of the new stability scores, which depend on the new network properties, so S'(G') = sum_{v_i} s'_i, where s'_i is a function of the new position in G'.But to tie it back to the original scores and changes, perhaps S'(G') = S(G) + sum_{v_i} (s'_i - s_i), where each term accounts for the change in stability due to the network changes.So, I think that's the way to go.</think>

🧠:<think>Okay, so I have this problem about a retired film producer who goes to screenings every day. There are two theaters: Theater A and Theater B. Theater A shows films Monday to Friday, so that's five days, and Theater B shows films on weekends, which is Saturday and Sunday, so two days. Each week, the producer attends one screening per day without skipping. So, in total, he attends seven screenings a week.Now, the films at Theater A are from independent filmmakers, and he has a 70% chance of finding each film interesting. At Theater B, which shows blockbusters, he has an 85% chance of finding each film interesting. The outcomes are independent, meaning whether he likes one film doesn't affect his opinion on another.The first question is: Calculate the probability that, in a given week, the producer finds at least four films interesting.Alright, so I need to find the probability that he finds at least four films interesting in a week. Since the outcomes are independent, I can model this using probability distributions. Specifically, since each screening is a Bernoulli trial (either interesting or not), the total number of interesting films follows a binomial distribution.But wait, hold on. The screenings are split between two theaters with different probabilities. So, actually, it's not a single binomial distribution because the probability of success (finding a film interesting) is different for different days.So, Theater A has five screenings with a 70% chance each, and Theater B has two screenings with an 85% chance each. So, the total number of interesting films is the sum of two independent binomial random variables: one with n=5 and p=0.7, and another with n=2 and p=0.85.Therefore, the total number of interesting films, let's call it X, is X = X_A + X_B, where X_A ~ Binomial(5, 0.7) and X_B ~ Binomial(2, 0.85). Since X_A and X_B are independent, the probability mass function of X can be found by convolving the two distributions.So, to find P(X >= 4), we need to compute the probability that the sum of X_A and X_B is at least 4. That is, P(X_A + X_B >= 4).To compute this, I can consider all possible combinations where X_A + X_B is 4, 5, 6, or 7, since the maximum number of interesting films is 5 + 2 = 7.But since calculating this directly might be a bit tedious, maybe it's easier to compute the cumulative probability for X >= 4 by summing the probabilities for k=4,5,6,7.But since X_A can be from 0 to 5 and X_B can be from 0 to 2, we can compute P(X >=4) by summing over all possible x_A and x_B such that x_A + x_B >=4.Alternatively, we can compute 1 - P(X <=3), which might be simpler because sometimes it's easier to compute the complement.But let's see. Let's list all possible values of X_A and X_B and compute the probabilities.First, let's compute the probability mass functions for X_A and X_B.For X_A ~ Binomial(5, 0.7):The PMF is P(X_A = k) = C(5, k) * (0.7)^k * (0.3)^(5 - k) for k=0,1,2,3,4,5.Similarly, for X_B ~ Binomial(2, 0.85):PMF is P(X_B = m) = C(2, m) * (0.85)^m * (0.15)^(2 - m) for m=0,1,2.So, to find P(X >=4), we need to consider all pairs (k, m) where k + m >=4.So, let's list all possible (k, m) pairs where k + m >=4:Possible k: 0,1,2,3,4,5Possible m: 0,1,2So, for each k from 0 to 5, we can find the m values such that m >=4 -k.Let's go step by step:1. For k=0: m >=4, but m can only be 0,1,2. So, no contribution.2. For k=1: m >=3. Again, m can only be 0,1,2. So, no contribution.3. For k=2: m >=2. So, m=2.4. For k=3: m >=1. So, m=1,2.5. For k=4: m >=0. So, m=0,1,2.6. For k=5: m >= -1, which is always true. So, m=0,1,2.So, the contributing pairs are:- (2,2)- (3,1), (3,2)- (4,0), (4,1), (4,2)- (5,0), (5,1), (5,2)So, now, we can compute the probabilities for each of these pairs and sum them up.First, let's compute the PMFs for X_A and X_B.Compute P(X_A = k) for k=0 to 5:- P(X_A=0) = C(5,0)*(0.7)^0*(0.3)^5 = 1*1*0.00243 = 0.00243- P(X_A=1) = C(5,1)*(0.7)^1*(0.3)^4 = 5*0.7*0.0081 = 5*0.00567 = 0.02835- P(X_A=2) = C(5,2)*(0.7)^2*(0.3)^3 = 10*0.49*0.027 = 10*0.01323 = 0.1323- P(X_A=3) = C(5,3)*(0.7)^3*(0.3)^2 = 10*0.343*0.09 = 10*0.03087 = 0.3087- P(X_A=4) = C(5,4)*(0.7)^4*(0.3)^1 = 5*0.2401*0.3 = 5*0.07203 = 0.36015- P(X_A=5) = C(5,5)*(0.7)^5*(0.3)^0 = 1*0.16807*1 = 0.16807Now, compute P(X_B = m) for m=0,1,2:- P(X_B=0) = C(2,0)*(0.85)^0*(0.15)^2 = 1*1*0.0225 = 0.0225- P(X_B=1) = C(2,1)*(0.85)^1*(0.15)^1 = 2*0.85*0.15 = 2*0.1275 = 0.255- P(X_B=2) = C(2,2)*(0.85)^2*(0.15)^0 = 1*0.7225*1 = 0.7225Now, let's compute the probabilities for each contributing pair:1. (2,2): P(X_A=2)*P(X_B=2) = 0.1323 * 0.7225Let me compute that: 0.1323 * 0.7225 ≈ 0.09562. (3,1): P(X_A=3)*P(X_B=1) = 0.3087 * 0.255 ≈ 0.07873. (3,2): P(X_A=3)*P(X_B=2) = 0.3087 * 0.7225 ≈ 0.22334. (4,0): P(X_A=4)*P(X_B=0) = 0.36015 * 0.0225 ≈ 0.00815. (4,1): P(X_A=4)*P(X_B=1) = 0.36015 * 0.255 ≈ 0.09186. (4,2): P(X_A=4)*P(X_B=2) = 0.36015 * 0.7225 ≈ 0.26037. (5,0): P(X_A=5)*P(X_B=0) = 0.16807 * 0.0225 ≈ 0.00388. (5,1): P(X_A=5)*P(X_B=1) = 0.16807 * 0.255 ≈ 0.04299. (5,2): P(X_A=5)*P(X_B=2) = 0.16807 * 0.7225 ≈ 0.1216Now, let's sum all these probabilities:0.0956 + 0.0787 + 0.2233 + 0.0081 + 0.0918 + 0.2603 + 0.0038 + 0.0429 + 0.1216Let me add them step by step:Start with 0.0956+0.0787 = 0.1743+0.2233 = 0.3976+0.0081 = 0.4057+0.0918 = 0.4975+0.2603 = 0.7578+0.0038 = 0.7616+0.0429 = 0.8045+0.1216 = 0.9261So, the total probability is approximately 0.9261.Wait, that seems high. Let me double-check my calculations because 92.61% seems quite high for at least four films interesting.Alternatively, maybe I made a mistake in the addition.Let me recount:0.0956 (1)+0.0787 = 0.1743 (2)+0.2233 = 0.3976 (3)+0.0081 = 0.4057 (4)+0.0918 = 0.4975 (5)+0.2603 = 0.7578 (6)+0.0038 = 0.7616 (7)+0.0429 = 0.8045 (8)+0.1216 = 0.9261 (9)Hmm, seems consistent. But let's think: the producer has a high probability of liking films, especially at Theater B. So, maybe it's reasonable.But let me cross-verify by computing P(X <=3) and subtracting from 1.Compute P(X <=3) = P(X=0) + P(X=1) + P(X=2) + P(X=3)But computing P(X <=3) would require considering all pairs (k, m) where k + m <=3.Which would be:For k=0: m=0,1,2k=1: m=0,1,2 (since 1+2=3)k=2: m=0,1 (since 2+1=3)k=3: m=0k=4: m cannot be negative, so no contributionSimilarly, k=5: same.So, the pairs are:(0,0), (0,1), (0,2),(1,0), (1,1), (1,2),(2,0), (2,1),(3,0)So, let's compute these:1. (0,0): P(X_A=0)*P(X_B=0) = 0.00243 * 0.0225 ≈ 0.0000546752. (0,1): 0.00243 * 0.255 ≈ 0.000618153. (0,2): 0.00243 * 0.7225 ≈ 0.0017550754. (1,0): 0.02835 * 0.0225 ≈ 0.0006363755. (1,1): 0.02835 * 0.255 ≈ 0.007224756. (1,2): 0.02835 * 0.7225 ≈ 0.0204656257. (2,0): 0.1323 * 0.0225 ≈ 0.002976758. (2,1): 0.1323 * 0.255 ≈ 0.03373659. (3,0): 0.3087 * 0.0225 ≈ 0.00694575Now, sum these up:0.000054675 + 0.00061815 + 0.001755075 + 0.000636375 + 0.00722475 + 0.020465625 + 0.00297675 + 0.0337365 + 0.00694575Let me add step by step:Start with 0.000054675+0.00061815 = 0.000672825+0.001755075 = 0.0024279+0.000636375 = 0.003064275+0.00722475 = 0.010289025+0.020465625 = 0.03075465+0.00297675 = 0.0337314+0.0337365 = 0.0674679+0.00694575 = 0.07441365So, P(X <=3) ≈ 0.07441365Therefore, P(X >=4) = 1 - 0.07441365 ≈ 0.92558635Which is approximately 0.9256, which is close to the 0.9261 I got earlier. The slight difference is due to rounding errors in the intermediate steps. So, the accurate value is approximately 0.9256.So, the probability that the producer finds at least four films interesting in a week is approximately 92.56%.But let me check if I considered all pairs correctly.Wait, when I calculated P(X >=4) by summing the pairs, I got approximately 0.9261, and when calculating 1 - P(X <=3), I got approximately 0.9256. The difference is minimal, so it's likely due to rounding during intermediate steps. So, the answer is approximately 0.9256 or 92.56%.But to be precise, maybe I should carry out the calculations with more decimal places.Alternatively, perhaps using generating functions or another method would give a more precise result, but for the purposes of this problem, 0.9256 is a reasonable approximation.So, the answer to part 1 is approximately 0.9256, or 92.56%.Now, moving on to part 2:Given that the producer has found four films interesting in a particular week, what is the probability that exactly two of these interesting films were screened at Theater A?This is a conditional probability problem. We need to find P(exactly two interesting films from A | total interesting films =4).Using Bayes' theorem, this is equal to P(exactly two interesting from A and total interesting=4) / P(total interesting=4)But since if exactly two are from A, then the remaining two must be from B. So, P(exactly two from A and total=4) is equal to P(X_A=2 and X_B=2).Which is P(X_A=2) * P(X_B=2) because they are independent.Wait, but actually, in the context of conditional probability, we have to consider all possible ways to get exactly four interesting films, and among those, the cases where exactly two are from A.So, the numerator is P(X_A=2 and X_B=2), and the denominator is P(X=4).But wait, actually, the total number of interesting films is 4, which can be achieved by different combinations of X_A and X_B.Specifically, X_A can be 2 and X_B=2, or X_A=3 and X_B=1, or X_A=4 and X_B=0. Because 2+2=4, 3+1=4, 4+0=4.So, the denominator P(X=4) is the sum of P(X_A=2, X_B=2) + P(X_A=3, X_B=1) + P(X_A=4, X_B=0).Therefore, the conditional probability is [P(X_A=2) * P(X_B=2)] / [P(X_A=2) * P(X_B=2) + P(X_A=3) * P(X_B=1) + P(X_A=4) * P(X_B=0)]So, let's compute each term:First, compute P(X_A=2) * P(X_B=2) = 0.1323 * 0.7225 ≈ 0.0956P(X_A=3) * P(X_B=1) = 0.3087 * 0.255 ≈ 0.0787P(X_A=4) * P(X_B=0) = 0.36015 * 0.0225 ≈ 0.0081So, the denominator is 0.0956 + 0.0787 + 0.0081 ≈ 0.1824Therefore, the conditional probability is 0.0956 / 0.1824 ≈ 0.524So, approximately 52.4%.Wait, let me check the exact values without rounding:Compute P(X_A=2) * P(X_B=2):0.1323 * 0.7225 = 0.1323 * 0.7225Let me compute 0.1323 * 0.7225:0.1 * 0.7225 = 0.072250.03 * 0.7225 = 0.0216750.0023 * 0.7225 ≈ 0.00166175Adding up: 0.07225 + 0.021675 = 0.093925 + 0.00166175 ≈ 0.09558675Similarly, P(X_A=3) * P(X_B=1):0.3087 * 0.255Compute 0.3 * 0.255 = 0.07650.0087 * 0.255 ≈ 0.0022185Total ≈ 0.0765 + 0.0022185 ≈ 0.0787185P(X_A=4) * P(X_B=0):0.36015 * 0.0225Compute 0.36 * 0.0225 = 0.00810.00015 * 0.0225 = 0.000003375Total ≈ 0.0081 + 0.000003375 ≈ 0.008103375So, denominator:0.09558675 + 0.0787185 + 0.008103375 ≈ 0.182408625Numerator: 0.09558675So, conditional probability ≈ 0.09558675 / 0.182408625 ≈ 0.524So, approximately 52.4%.Therefore, the probability is approximately 52.4%.But let me express this as a fraction or a more precise decimal.0.09558675 / 0.182408625 ≈ 0.524Alternatively, to compute it more precisely:Divide 0.09558675 by 0.182408625First, note that 0.182408625 * 0.5 = 0.0912043125Subtract that from 0.09558675: 0.09558675 - 0.0912043125 ≈ 0.0043824375Now, 0.182408625 * 0.024 ≈ 0.004377807So, 0.5 + 0.024 ≈ 0.524So, approximately 0.524, which is 52.4%.Therefore, the probability is approximately 52.4%.But to express it as a fraction, let's compute the exact value:Numerator: 0.09558675Denominator: 0.182408625Divide numerator and denominator by 0.00000025 to make it manageable:Numerator: 0.09558675 / 0.00000025 = 382347Denominator: 0.182408625 / 0.00000025 = 729634.5Wait, that might not be helpful. Alternatively, express both as fractions:0.09558675 = 9558675 / 1000000000.182408625 = 182408625 / 1000000000But this might get too messy. Alternatively, recognize that 0.09558675 / 0.182408625 = (9558675 / 100000000) / (182408625 / 1000000000) = (9558675 / 100000000) * (1000000000 / 182408625) = (9558675 * 10) / 182408625 = 95586750 / 182408625Simplify numerator and denominator by dividing numerator and denominator by 75:95586750 ÷ 75 = 1274490182408625 ÷ 75 = 2432115So, 1274490 / 2432115Divide numerator and denominator by 15:1274490 ÷15=849662432115 ÷15=162141So, 84966 / 162141Divide numerator and denominator by 3:84966 ÷3=28322162141 ÷3=54047So, 28322 / 54047Check if they have a common divisor. Let's see, 28322 ÷2=14161, which is a prime? Maybe. 54047 ÷2 is not integer. 28322 is even, 54047 is odd. So, no further common divisors. So, 28322/54047 ≈0.524So, approximately 52.4%.Therefore, the probability is approximately 52.4%.So, summarizing:1. The probability of finding at least four films interesting is approximately 92.56%.2. Given that four films were interesting, the probability that exactly two were from Theater A is approximately 52.4%.Final Answer1. The probability is boxed{0.9256}.2. The probability is boxed{0.524}.</think>

🧠:<think>Okay, so I have two problems here about palm lines modeled by parametric equations and a piecewise function. Let me tackle them one by one.Starting with the first problem: the Heart Line. The parametric equations given are:[ x(t) = a cos^3(t) ][ y(t) = b sin^3(t) ]And I need to find the area enclosed by this curve. Hmm, I remember that for parametric equations, the area can be found using the formula:[ text{Area} = frac{1}{2} int_{t_1}^{t_2} (x frac{dy}{dt} - y frac{dx}{dt}) dt ]So, I should compute the derivatives of x and y with respect to t, plug them into this formula, and integrate from 0 to 2π.First, let's find dx/dt and dy/dt.For x(t) = a cos³(t), the derivative dx/dt is:[ frac{dx}{dt} = a cdot 3 cos^2(t) cdot (-sin(t)) = -3a cos^2(t) sin(t) ]Similarly, for y(t) = b sin³(t), the derivative dy/dt is:[ frac{dy}{dt} = b cdot 3 sin^2(t) cdot cos(t) = 3b sin^2(t) cos(t) ]Now, plug these into the area formula:[ text{Area} = frac{1}{2} int_{0}^{2pi} [x cdot frac{dy}{dt} - y cdot frac{dx}{dt}] dt ]Substituting x(t) and y(t):[ text{Area} = frac{1}{2} int_{0}^{2pi} [a cos^3(t) cdot 3b sin^2(t) cos(t) - b sin^3(t) cdot (-3a cos^2(t) sin(t))] dt ]Simplify each term:First term: ( a cdot 3b cos^3(t) sin^2(t) cos(t) = 3ab cos^4(t) sin^2(t) )Second term: ( -b cdot (-3a) sin^3(t) cos^2(t) sin(t) = 3ab sin^4(t) cos^2(t) )So, the integrand becomes:[ 3ab cos^4(t) sin^2(t) + 3ab sin^4(t) cos^2(t) ]Factor out 3ab cos²(t) sin²(t):[ 3ab cos^2(t) sin^2(t) [cos^2(t) + sin^2(t)] ]But wait, cos²(t) + sin²(t) = 1, so this simplifies to:[ 3ab cos^2(t) sin^2(t) ]So now, the area integral is:[ text{Area} = frac{1}{2} cdot 3ab int_{0}^{2pi} cos^2(t) sin^2(t) dt ]Simplify constants:[ text{Area} = frac{3ab}{2} int_{0}^{2pi} cos^2(t) sin^2(t) dt ]Hmm, integrating cos²(t) sin²(t) over 0 to 2π. I think there's a trigonometric identity to simplify this. Let me recall:We know that sin(2t) = 2 sin(t) cos(t), so sin²(2t) = 4 sin²(t) cos²(t). Therefore, sin²(t) cos²(t) = (1/4) sin²(2t).So, substituting:[ cos^2(t) sin^2(t) = frac{1}{4} sin^2(2t) ]Thus, the integral becomes:[ frac{3ab}{2} cdot frac{1}{4} int_{0}^{2pi} sin^2(2t) dt = frac{3ab}{8} int_{0}^{2pi} sin^2(2t) dt ]Now, sin²(u) can be expressed using another identity: sin²(u) = (1 - cos(2u))/2.Let me apply that:[ sin^2(2t) = frac{1 - cos(4t)}{2} ]So, substituting back:[ frac{3ab}{8} cdot frac{1}{2} int_{0}^{2pi} [1 - cos(4t)] dt = frac{3ab}{16} int_{0}^{2pi} [1 - cos(4t)] dt ]Now, split the integral:[ frac{3ab}{16} left[ int_{0}^{2pi} 1 dt - int_{0}^{2pi} cos(4t) dt right] ]Compute each integral:First integral: ∫₀²π 1 dt = 2πSecond integral: ∫₀²π cos(4t) dt. Let me compute this. The integral of cos(kt) is (1/k) sin(kt). So:[ int_{0}^{2pi} cos(4t) dt = left[ frac{sin(4t)}{4} right]_0^{2pi} = frac{sin(8pi)}{4} - frac{sin(0)}{4} = 0 - 0 = 0 ]So, the second integral is zero.Therefore, the area becomes:[ frac{3ab}{16} cdot 2pi = frac{3ab}{8} pi ]Wait, hold on, let me check my steps again because I might have made a mistake in the constants.Wait, no, let's retrace:We had:Area = (3ab/8) ∫₀²π sin²(2t) dtThen, sin²(2t) = (1 - cos(4t))/2, so:(3ab/8) * (1/2) ∫₀²π (1 - cos(4t)) dt = (3ab/16) [∫₀²π 1 dt - ∫₀²π cos(4t) dt]Which is (3ab/16)(2π - 0) = (3ab/16)(2π) = (3ab/8) πSo, yes, that seems correct.But wait, I remember that for a parametric curve, sometimes the area can be calculated using another formula, but I think the formula I used is correct.Alternatively, I can think of this curve as an astroid. Wait, the parametric equations are x = a cos³(t), y = b sin³(t). If a = b, it's an astroid, which has area (3/8)πa². So, in this case, since a and b are different, the area would be scaled accordingly.Wait, let me check that. If a = b, then the area is (3/8)πa². So, if a ≠ b, perhaps the area is (3/8)πab? Because in the standard astroid, x = a cos³(t), y = a sin³(t), so if a ≠ b, it's stretched differently. So, yes, the area would be (3/8)πab.So, that matches what I got earlier. So, the area enclosed by the Heart Line is (3πab)/8.Alright, that seems solid.Moving on to the second problem: the Fate Line, which is a piecewise function:[ f(x) = begin{cases} kx & text{if } 0 leq x leq 1 k(2 - x) & text{if } 1 < x leq 2 end{cases}]We need to calculate the arc length from x = 0 to x = 2.Arc length for a function f(x) from a to b is given by:[ L = int_{a}^{b} sqrt{1 + [f'(x)]^2} dx ]Since the function is piecewise, we can split the integral into two parts: from 0 to 1 and from 1 to 2.First, let's compute f'(x) for each piece.For 0 ≤ x ≤ 1: f(x) = kx, so f'(x) = k.For 1 < x ≤ 2: f(x) = k(2 - x), so f'(x) = -k.So, the arc length integral becomes:[ L = int_{0}^{1} sqrt{1 + k^2} dx + int_{1}^{2} sqrt{1 + (-k)^2} dx ]But since (-k)² = k², both integrands are the same:[ L = int_{0}^{1} sqrt{1 + k^2} dx + int_{1}^{2} sqrt{1 + k^2} dx ]Which simplifies to:[ L = sqrt{1 + k^2} left( int_{0}^{1} dx + int_{1}^{2} dx right) ]Compute the integrals:∫₀¹ dx = 1 - 0 = 1∫₁² dx = 2 - 1 = 1So, total integral is 1 + 1 = 2.Therefore, the arc length is:[ L = 2 sqrt{1 + k^2} ]Wait, that seems straightforward. Let me just visualize the function. From x=0 to x=1, it's a straight line with slope k, and from x=1 to x=2, it's a straight line with slope -k. So, the graph is a V-shape with the vertex at (1, k). Each segment is a straight line, so the arc length is just the sum of the lengths of these two line segments.Alternatively, we can compute the length of each line segment.From x=0 to x=1: the line goes from (0,0) to (1, k). The distance between these points is sqrt[(1-0)^2 + (k - 0)^2] = sqrt(1 + k²).Similarly, from x=1 to x=2: the line goes from (1, k) to (2, 0). The distance is sqrt[(2 - 1)^2 + (0 - k)^2] = sqrt(1 + k²).So, total arc length is sqrt(1 + k²) + sqrt(1 + k²) = 2 sqrt(1 + k²), which matches the integral result.Therefore, the arc length is 2√(1 + k²).So, summarizing:1. The area enclosed by the Heart Line is (3πab)/8.2. The arc length of the Fate Line is 2√(1 + k²).Final Answer1. The area enclosed by the Heart Line is boxed{dfrac{3pi ab}{8}}.2. The arc length of the Fate Line is boxed{2sqrt{1 + k^2}}.</think>