🧠:<think>Okay, so I have this problem about a music producer sending tracks to a radio host. There are two parts to the problem. Let me try to tackle them one by one.Starting with part 1: Determine the total probability that at least one track will be selected by the radio host if the producer sends all 50 tracks. The selection of each track is an independent event.Hmm, okay. So, the radio host is considering 50 tracks, each with a unique probability of being selected. The probability for each track is inversely proportional to the order in which they are sent. The first track has a probability of 1/50, and this increases linearly to 1 for the last track.Wait, inversely proportional? So, if the probability is inversely proportional to the order, that would mean that the probability for the nth track is something like k/n, where k is a constant. But the first track has a probability of 1/50, so maybe k is 1/50. So, the probability for the nth track would be (1/50)/n? Wait, but that can't be right because for n=50, it would be (1/50)/50 = 1/2500, which is way less than 1.Wait, the problem says the probability is inversely proportional to the order, starting from the first track with a probability of 1/50 and increasing linearly to the last track with a probability of 1. Hmm, so maybe it's not inversely proportional in the strict mathematical sense, but rather, the probability is inversely related to the order, but in a way that it's linear from 1/50 to 1.Wait, let me parse that again. "The probability of each track being selected is inversely proportional to the order in which they are sent, starting from the first track with a probability of 1/50 and increasing linearly to the last track with a probability of 1."So, inversely proportional usually means that P(n) = k / n, where k is a constant. But here, they specify that the first track has probability 1/50 and the last has probability 1. So, maybe the probabilities form a sequence where the first term is 1/50, the last term is 1, and each term increases linearly. So, it's an arithmetic sequence.Wait, if it's increasing linearly, then the probability for each track is an arithmetic progression starting at 1/50 and ending at 1 over 50 terms. So, the common difference would be (1 - 1/50)/(50 - 1) = (49/50)/49 = 1/50. So, each subsequent track has a probability 1/50 higher than the previous one.Wait, let me check that. If the first term a1 is 1/50, the 50th term a50 is 1, and the common difference d is (a50 - a1)/(n - 1) = (1 - 1/50)/49 = (49/50)/49 = 1/50. So, yes, each track's probability increases by 1/50 each time.So, the probability for the nth track is a_n = a1 + (n - 1)d = 1/50 + (n - 1)(1/50) = (1 + n - 1)/50 = n/50. Wait, that can't be right because then the 50th track would have probability 50/50 = 1, which matches, and the first track is 1/50. So, actually, each track's probability is n/50, where n is the order. So, the first track has 1/50, the second 2/50, ..., the 50th track has 50/50 = 1.Wait, but the problem says the probability is inversely proportional to the order. But if it's n/50, that's directly proportional. Hmm, maybe I misread the problem.Wait, the problem says: "the probability of each track being selected is inversely proportional to the order in which they are sent, starting from the first track with a probability of 1/50 and increasing linearly to the last track with a probability of 1."Wait, so inversely proportional would mean that P(n) = k / n, but starting from 1/50 and increasing linearly to 1. So, maybe it's not a strict inverse proportionality, but just that the probability is inversely related in the sense that higher order tracks have higher probabilities, but it's increasing linearly.So, perhaps the probability for the nth track is (n)/50, as I thought earlier. So, n=1: 1/50, n=50: 50/50=1. So, that seems to fit.So, each track has a probability p_n = n/50 of being selected, and these are independent events.So, for part 1, we need the probability that at least one track is selected. Since each selection is independent, the probability that at least one is selected is 1 minus the probability that none are selected.So, P(at least one) = 1 - P(none selected).Since each track has a probability p_n of being selected, the probability that a track is not selected is 1 - p_n.Therefore, the probability that none are selected is the product of (1 - p_n) for n from 1 to 50.So, P(none) = product_{n=1 to 50} (1 - n/50).Therefore, P(at least one) = 1 - product_{n=1 to 50} (1 - n/50).Hmm, calculating this product might be tricky. Let me see if I can find a pattern or a formula.Wait, let's write out the terms:For n=1: 1 - 1/50 = 49/50n=2: 1 - 2/50 = 48/50n=3: 47/50...n=50: 1 - 50/50 = 0/50 = 0Wait, hold on, when n=50, the probability of not selecting is 0, which means that the 50th track is guaranteed to be selected. Therefore, the probability that none are selected is 0, because the 50th track will always be selected. Therefore, the probability that at least one track is selected is 1.Wait, that seems too straightforward. Let me think again.If the 50th track has a probability of 1, then it's certain to be selected. Therefore, regardless of the other tracks, at least the 50th track will be selected. Therefore, the probability that at least one track is selected is indeed 1.Wait, but that seems counterintuitive because the problem says "the probability of each track being selected is inversely proportional to the order," but if the 50th track is certain to be selected, then the probability of at least one selection is 1.But maybe I misinterpreted the problem. Let me read it again."The probability of each track being selected is inversely proportional to the order in which they are sent, starting from the first track with a probability of 1/50 and increasing linearly to the last track with a probability of 1."Wait, so inversely proportional to the order, but starting at 1/50 and increasing to 1. So, maybe the probability is inversely proportional, but scaled so that the first term is 1/50 and the last is 1.Wait, if it's inversely proportional, then P(n) = k / n, where k is a constant. But we have P(1) = 1/50, so k = 1/50. Therefore, P(n) = (1/50)/n = 1/(50n). But then P(50) would be 1/(50*50) = 1/2500, which is not 1. So that contradicts the problem statement.Therefore, perhaps the problem is not using strict inverse proportionality but rather that the probability is inversely related in the sense that higher order tracks have higher probabilities, but it's increasing linearly from 1/50 to 1.So, in that case, the probability for the nth track is (n)/50, as I thought earlier. So, the 50th track has probability 1, which is certain.Therefore, the probability that at least one track is selected is 1, because the 50th track is always selected.Wait, but that seems too easy. Maybe I'm missing something. Let me check.If the 50th track is always selected, then yes, at least one track is always selected. So, the probability is 1.But perhaps the problem meant that the probability is inversely proportional to the order, but not necessarily that the last track has probability 1. Maybe it's inversely proportional, so P(n) = k / n, and we have to find k such that the probabilities sum to something, but the problem says starting from 1/50 and increasing linearly to 1.Wait, perhaps the problem is that the probability is inversely proportional to the order, but scaled such that the first track is 1/50 and the last is 1. So, it's a linear function of n, but inversely related.Wait, maybe it's P(n) = (50 - n + 1)/50? No, that would make the first track 50/50=1, which contradicts.Wait, perhaps it's P(n) = (n)/50, which is directly proportional, but the problem says inversely proportional. So, maybe it's P(n) = (51 - n)/50? Let's see, for n=1, P(1)=50/50=1, which is not 1/50. So that doesn't fit.Wait, maybe the problem is using "inversely proportional" in a non-technical sense, meaning that higher order tracks have higher probabilities, but it's increasing linearly from 1/50 to 1.So, in that case, P(n) = (n)/50, as I thought earlier. So, the 50th track is certain to be selected, making the probability of at least one selection equal to 1.Alternatively, maybe the problem is that the probability is inversely proportional to the order, but the first track has probability 1/50, so P(n) = (1/50)/n, but then the last track would have P(50)=1/(50*50)=1/2500, which is not 1. So, that contradicts.Alternatively, maybe the probability is inversely proportional to the order, but the sum of probabilities is 1. Wait, but the problem doesn't say that. It just says each track has a unique probability, starting at 1/50 and increasing to 1.Wait, perhaps the problem is that the probability is inversely proportional to the order, but adjusted so that the first track is 1/50 and the last is 1. So, maybe P(n) = (1/50) * (50/n). Wait, that would make P(n) = 1/n, but for n=1, P(1)=1, which contradicts.Wait, maybe it's P(n) = (1/50) * (n). So, P(n) = n/50, which is directly proportional, but the problem says inversely proportional. Hmm.Wait, perhaps the problem is using "inversely proportional" incorrectly, and it's actually directly proportional. Because if it's inversely proportional, the first track would have the highest probability, but the problem says the first track has the lowest probability, 1/50, and the last has the highest, 1. So, that suggests that the probability increases with n, which is directly proportional.Therefore, maybe the problem meant directly proportional, but said inversely. Alternatively, maybe it's a translation issue or a misstatement.Given that, perhaps we can proceed under the assumption that P(n) = n/50, as that fits the description of starting at 1/50 and increasing linearly to 1.Therefore, for part 1, since the 50th track has P=1, the probability that at least one track is selected is 1.But let me think again. If the 50th track is always selected, then yes, P(at least one) = 1. So, that's the answer for part 1.Now, moving on to part 2: If the producer wants to maximize the expected number of tracks selected by the radio host, how many tracks should they send, assuming they can send any number of tracks from 1 to 50? Calculate the expected number of tracks selected for the optimal number of tracks sent.Okay, so the producer can send k tracks, where k is from 1 to 50. The expected number of tracks selected is the sum of the probabilities of each track being selected.But wait, the probabilities are dependent on the order in which they are sent. So, if the producer sends k tracks, the first track sent has probability 1/50, the second 2/50, ..., the kth track has probability k/50.Wait, but if the producer can choose which tracks to send, perhaps they can send the tracks with the highest probabilities first? Or is the order fixed?Wait, the problem says the producer sends tracks in a certain order, with the first track having probability 1/50, the second 2/50, etc., up to the kth track with probability k/50.But the problem doesn't specify whether the producer can choose the order or if the order is fixed. Wait, the problem says "the probability of each track being selected is inversely proportional to the order in which they are sent," but the producer is sending them in a certain order.Wait, perhaps the producer can choose the order, so they can send the tracks in the order that maximizes the expected number. But the problem doesn't specify that. It just says the producer sends tracks, each with a probability inversely proportional to their order.Wait, maybe the order is fixed, and the producer can choose how many to send, but not the order. So, if they send k tracks, the first track has probability 1/50, the second 2/50, ..., the kth track has probability k/50.Wait, but that would mean that the expected number of selected tracks is the sum from n=1 to k of n/50.But wait, if the producer can choose the order, they might want to send the tracks with the highest probabilities first. So, instead of sending track 1, track 2, ..., track k, they could send track 50, track 49, ..., track (50 - k + 1). Then, the probabilities would be 50/50=1, 49/50, ..., (50 - k +1)/50.In that case, the expected number would be the sum from n=(50 - k +1) to 50 of n/50.But the problem doesn't specify whether the producer can choose the order or not. It just says the probability is inversely proportional to the order in which they are sent. So, if the producer sends k tracks, the first track has probability 1/50, the second 2/50, ..., the kth track has probability k/50.Wait, but that would mean that the expected number is sum_{n=1}^k (n/50). But that sum is (k(k +1))/2 / 50.Alternatively, if the producer can choose the order, they might send the tracks with the highest probabilities first, so the expected number would be sum_{n=(50 - k +1)}^{50} (n/50).Which would be (sum_{n=1}^{50} n/50) - (sum_{n=1}^{50 - k} n/50) = (50*51/2)/50 - ( (50 - k)(51 - k)/2 )/50.Simplifying, that would be (51/2) - ( (50 - k)(51 - k)/2 )/50.Wait, but I'm not sure if the producer can choose the order or not. The problem says "the probability of each track being selected is inversely proportional to the order in which they are sent." So, if the producer sends them in a certain order, the probabilities are determined by that order.Therefore, if the producer wants to maximize the expected number, they should send the tracks in the order that gives the highest probabilities to the tracks they send. So, they should send the tracks with the highest individual probabilities first.Wait, but each track has a unique probability based on its order. So, if the producer sends track 50 first, it has probability 1, track 49 second, probability 49/50, etc.But the problem is that the probability is inversely proportional to the order in which they are sent. So, if the producer sends track 50 first, its probability is 1/50, which is low, not high. Wait, that contradicts.Wait, hold on. If the probability is inversely proportional to the order, then the first track sent has probability 1/50, the second track sent has probability 2/50, ..., the kth track sent has probability k/50.Therefore, if the producer sends track 50 first, its probability is 1/50, which is low. If they send track 1 first, its probability is 1/50, same as track 50.Wait, no, the probability is based on the order of sending, not the inherent quality of the track. So, each track has a unique probability based on its position in the sending order.Therefore, if the producer sends k tracks, the first track has probability 1/50, the second 2/50, ..., the kth track has probability k/50.Therefore, the expected number of selected tracks is sum_{n=1}^k (n/50) = (k(k +1))/2 / 50.So, to maximize the expected number, the producer should send as many tracks as possible, because the expected number increases with k. But wait, that can't be right because when k=50, the expected number is (50*51)/2 /50 = 51/2 = 25.5.But if the producer sends fewer tracks, say k=25, the expected number would be (25*26)/2 /50 = 325/50 = 6.5.Wait, but that's much lower than 25.5. So, the expected number increases as k increases.But wait, that seems counterintuitive because sending more tracks with higher probabilities might lead to a higher expected number. But in this case, the probabilities are increasing with the order, so the later tracks have higher probabilities.Wait, but if the producer sends k tracks, the first track has probability 1/50, the second 2/50, ..., the kth track has probability k/50. So, the expected number is sum_{n=1}^k (n/50) = (k(k +1))/2 /50.Therefore, the expected number is a quadratic function of k, which increases as k increases. So, to maximize the expected number, the producer should send all 50 tracks, resulting in an expected number of 25.5.But that seems odd because the last track is certain to be selected, so the expected number would be at least 1, but actually, it's 25.5.Wait, but let me think again. If the producer sends all 50 tracks, the expected number is sum_{n=1}^{50} (n/50) = (50*51)/2 /50 = 51/2 = 25.5.If they send fewer tracks, say k=25, the expected number is (25*26)/2 /50 = 325/50 = 6.5.So, yes, the expected number increases with k. Therefore, the producer should send all 50 tracks to maximize the expected number, which is 25.5.But wait, that seems counterintuitive because sending more tracks with higher probabilities might lead to a higher expected number, but in this case, the probabilities are increasing with the order, so the later tracks have higher probabilities.Wait, but if the producer sends all 50 tracks, the expected number is 25.5. If they send only the last track, the expected number is 1. If they send the last two tracks, the expected number is 1 + 49/50 ≈ 1.98.Wait, but according to the formula, sending all 50 tracks gives the highest expected number.Wait, but let me think about it differently. The expected number is the sum of the probabilities. So, if the producer sends k tracks, the expected number is sum_{n=1}^k (n/50). So, this is an increasing function of k. Therefore, the maximum expected number is achieved when k=50, giving 25.5.Therefore, the answer to part 2 is that the producer should send all 50 tracks, resulting in an expected number of 25.5 tracks selected.But wait, that seems too straightforward. Let me check if I'm interpreting the problem correctly.The problem says: "the probability of each track being selected is inversely proportional to the order in which they are sent, starting from the first track with a probability of 1/50 and increasing linearly to the last track with a probability of 1."So, if the producer sends k tracks, the first track has probability 1/50, the second 2/50, ..., the kth track has probability k/50.Therefore, the expected number is sum_{n=1}^k (n/50) = (k(k +1))/2 /50.This is a quadratic function in k, which increases as k increases. Therefore, the maximum expected number is achieved when k=50, giving 25.5.Therefore, the producer should send all 50 tracks, and the expected number is 25.5.Wait, but in part 1, we concluded that the probability of at least one track being selected is 1 because the 50th track is certain to be selected. So, that makes sense.Therefore, for part 2, the optimal number is 50 tracks, with an expected number of 25.5.But let me think again. If the producer sends fewer tracks, say k=25, the expected number is 6.5, which is much less than 25.5. So, yes, sending more tracks increases the expected number.Therefore, the answer is to send all 50 tracks, with an expected number of 25.5.Wait, but let me think about the probabilities again. If the producer sends all 50 tracks, the expected number is 25.5. But if they send only the last track, the expected number is 1. If they send the last two tracks, it's 1 + 49/50 ≈ 1.98. So, indeed, the expected number increases with k.Therefore, the producer should send all 50 tracks to maximize the expected number of selected tracks.So, summarizing:1. The probability that at least one track is selected is 1.2. The producer should send all 50 tracks, resulting in an expected number of 25.5 tracks selected.But wait, let me check if the probabilities are correctly interpreted. If the probability is inversely proportional to the order, then P(n) = k/n. But in the problem, P(1)=1/50 and P(50)=1. So, if P(n)=k/n, then P(1)=k=1/50, so P(50)= (1/50)/50=1/2500, which contradicts P(50)=1.Therefore, my earlier assumption that P(n)=n/50 must be incorrect because it contradicts the inverse proportionality.Wait, so perhaps the problem is that the probability is inversely proportional to the order, but scaled such that P(1)=1/50 and P(50)=1.So, let's model it as P(n) = k/n, where k is a constant. Then, P(1)=k/1= k=1/50. Therefore, P(n)=1/(50n). But then P(50)=1/(50*50)=1/2500, which is not 1. So, that doesn't fit.Alternatively, maybe the probability is inversely proportional to the order, but the orders are 1 to 50, so P(n) = k/(n), and we have P(1)=1/50, so k=1/50. Therefore, P(n)=1/(50n). But then P(50)=1/2500, which is not 1.Therefore, the problem's statement is conflicting. It says the probability is inversely proportional to the order, starting at 1/50 and increasing to 1. So, perhaps it's not a strict inverse proportionality but rather that the probability is inversely related in the sense that higher order tracks have higher probabilities, but it's increasing linearly from 1/50 to 1.Therefore, perhaps P(n)= (n)/50, as I thought earlier, even though it's directly proportional. So, the first track has 1/50, the second 2/50, ..., the 50th track has 1.Therefore, the expected number is sum_{n=1}^k (n/50) = (k(k +1))/2 /50.Thus, the maximum expected number is achieved when k=50, giving 25.5.Therefore, despite the initial confusion about inverse proportionality, the answer seems to be that the producer should send all 50 tracks, resulting in an expected number of 25.5 tracks selected.But let me think again. If the probability is inversely proportional to the order, then P(n)=k/n, but in the problem, P(1)=1/50 and P(50)=1. So, if P(n)=k/n, then k=1/50, so P(50)=1/(50*50)=1/2500, which is not 1. Therefore, the problem's description is conflicting.Alternatively, maybe the probability is inversely proportional to the order, but the orders are 1 to 50, so P(n)=k/(n), and we have P(1)=1/50, so k=1/50. Therefore, P(n)=1/(50n). But then P(50)=1/2500, which is not 1.Therefore, perhaps the problem is using "inversely proportional" incorrectly, and it's actually directly proportional. So, P(n)=n/50, which fits the description of starting at 1/50 and increasing to 1.Therefore, under that assumption, the answers are as above.So, to conclude:1. The probability that at least one track is selected is 1.2. The producer should send all 50 tracks, resulting in an expected number of 25.5 tracks selected.But wait, let me check the expected number again. If the producer sends all 50 tracks, the expected number is sum_{n=1}^{50} (n/50) = (50*51)/2 /50 = 51/2 = 25.5. Yes, that's correct.Therefore, the answers are:1. 12. 50 tracks, expected number 25.5</think>

🧠:<think>Alright, so I'm trying to solve this problem about Bayesian networks. It's a bit intimidating because I'm still getting the hang of these concepts, but I think if I break it down step by step, I can figure it out.First, the problem describes a Bayesian network with three nodes: A, B, and C. A is the parent of both B and C, and B is also a parent of C. So, the structure is A -> B -> C and A -> C. That means C depends on both A and B.We have conditional probability tables for each node:1. For node A: - P(A = a) = 0.6 - P(A = ¬a) = 0.42. For node B given A: - If A = a, then P(B = b | A = a) = 0.7 and P(B = ¬b | A = a) = 0.3 - If A = ¬a, then P(B = b | A = ¬a) = 0.2 and P(B = ¬b | A = ¬a) = 0.83. For node C given A and B: - If A = a and B = b, then P(C = c | A = a, B = b) = 0.9 and P(C = ¬c | A = a, B = b) = 0.1 - If A = a and B = ¬b, then P(C = c | A = a, B = ¬b) = 0.4 and P(C = ¬c | A = a, B = ¬b) = 0.6 - If A = ¬a and B = b, then P(C = c | A = ¬a, B = b) = 0.3 and P(C = ¬c | A = ¬a, B = b) = 0.7 - If A = ¬a and B = ¬b, then P(C = c | A = ¬a, B = ¬b) = 0.5 and P(C = ¬c | A = ¬a, B = ¬b) = 0.5The first part asks to calculate P(C = c). So, I need to find the total probability that C is c, considering all possible combinations of A and B.I remember that in Bayesian networks, to find the marginal probability of a node, we can use the law of total probability, summing over all possible combinations of the parent nodes. Since C has two parents, A and B, I need to consider all four combinations of A and B: (a, b), (a, ¬b), (¬a, b), (¬a, ¬b).So, P(C = c) can be calculated as:P(C = c) = P(C = c | A = a, B = b) * P(A = a, B = b) + P(C = c | A = a, B = ¬b) * P(A = a, B = ¬b) + P(C = c | A = ¬a, B = b) * P(A = ¬a, B = b) + P(C = c | A = ¬a, B = ¬b) * P(A = ¬a, B = ¬b)Now, I need to compute each term step by step.First, let's compute the joint probabilities P(A = a, B = b), P(A = a, B = ¬b), P(A = ¬a, B = b), and P(A = ¬a, B = ¬b).Since B depends on A, we can compute these joint probabilities as:P(A = a, B = b) = P(A = a) * P(B = b | A = a) = 0.6 * 0.7 = 0.42P(A = a, B = ¬b) = P(A = a) * P(B = ¬b | A = a) = 0.6 * 0.3 = 0.18Similarly,P(A = ¬a, B = b) = P(A = ¬a) * P(B = b | A = ¬a) = 0.4 * 0.2 = 0.08P(A = ¬a, B = ¬b) = P(A = ¬a) * P(B = ¬b | A = ¬a) = 0.4 * 0.8 = 0.32Now, let's list all the terms:1. P(C = c | A = a, B = b) * P(A = a, B = b) = 0.9 * 0.42 = 0.3782. P(C = c | A = a, B = ¬b) * P(A = a, B = ¬b) = 0.4 * 0.18 = 0.0723. P(C = c | A = ¬a, B = b) * P(A = ¬a, B = b) = 0.3 * 0.08 = 0.0244. P(C = c | A = ¬a, B = ¬b) * P(A = ¬a, B = ¬b) = 0.5 * 0.32 = 0.16Now, adding all these up:0.378 + 0.072 = 0.450.024 + 0.16 = 0.184So, total P(C = c) = 0.45 + 0.184 = 0.634Wait, let me verify that addition again:0.378 + 0.072 is indeed 0.450.024 + 0.16 is 0.184Adding 0.45 and 0.184 gives 0.634. So, P(C = c) is 0.634.Hmm, that seems straightforward. Let me just make sure I didn't make any multiplication errors.Calculating each term:1. 0.9 * 0.42: 0.9 * 0.4 is 0.36, 0.9 * 0.02 is 0.018, so total 0.378. Correct.2. 0.4 * 0.18: 0.4 * 0.1 is 0.04, 0.4 * 0.08 is 0.032, so 0.072. Correct.3. 0.3 * 0.08: 0.024. Correct.4. 0.5 * 0.32: 0.16. Correct.Adding all together: 0.378 + 0.072 = 0.45; 0.024 + 0.16 = 0.184; 0.45 + 0.184 = 0.634. Yep, that's 63.4%.So, that's part 1 done.Moving on to part 2. The computer scientist wants to evaluate the robustness of her model against perturbations in the probabilities. Specifically, she's perturbing P(C = c | A = a, B = b) by ε, so it becomes 0.9 + ε. We need to derive an expression for P(C = c) in terms of ε and compute the sensitivity at ε = 0.Sensitivity, I think, refers to the derivative of P(C = c) with respect to ε. So, we need to find dP(C = c)/dε at ε = 0.First, let's write the expression for P(C = c) when P(C = c | A = a, B = b) is 0.9 + ε.From part 1, we saw that P(C = c) is the sum over all combinations of A and B of P(C = c | A, B) * P(A, B). So, in this case, only the term where A = a and B = b will change because we're perturbing that specific conditional probability.So, the new P(C = c) would be:P(C = c) = (0.9 + ε) * P(A = a, B = b) + 0.4 * P(A = a, B = ¬b) + 0.3 * P(A = ¬a, B = b) + 0.5 * P(A = ¬a, B = ¬b)We already calculated P(A = a, B = b) as 0.42, so substituting:P(C = c) = (0.9 + ε) * 0.42 + 0.4 * 0.18 + 0.3 * 0.08 + 0.5 * 0.32Let me compute each term:1. (0.9 + ε) * 0.42 = 0.9 * 0.42 + ε * 0.42 = 0.378 + 0.42ε2. 0.4 * 0.18 = 0.0723. 0.3 * 0.08 = 0.0244. 0.5 * 0.32 = 0.16Adding all these together:0.378 + 0.42ε + 0.072 + 0.024 + 0.16First, sum the constants:0.378 + 0.072 = 0.450.45 + 0.024 = 0.4740.474 + 0.16 = 0.634So, the expression becomes:P(C = c) = 0.634 + 0.42εTherefore, the sensitivity, which is the derivative of P(C = c) with respect to ε, is simply the coefficient of ε, which is 0.42.So, the sensitivity at ε = 0 is 0.42.Wait, let me double-check. Since P(C = c) is linear in ε, the derivative is just the coefficient. So yes, it's 0.42.Alternatively, if we think about it, the sensitivity is how much P(C = c) changes per unit change in ε. Since only the term involving ε contributes, and that term is 0.42ε, the sensitivity is 0.42.So, summarizing:1. P(C = c) = 0.6342. When perturbing P(C = c | A = a, B = b) by ε, P(C = c) becomes 0.634 + 0.42ε, and the sensitivity is 0.42.I think that's it. Let me just make sure I didn't miss any steps.In part 1, I correctly applied the law of total probability, considering all four combinations of A and B, computed each joint probability, multiplied by the respective conditional probability for C, and summed them up. The result was 0.634.In part 2, I recognized that only the term where A = a and B = b would change with the perturbation ε. So, I substituted 0.9 + ε into that term, expanded the expression, and found that the rest of the terms remained the same as in part 1, which summed to 0.634. The only additional term was 0.42ε, so the total expression is 0.634 + 0.42ε. Therefore, the sensitivity is 0.42.Yes, that makes sense. I don't think I made any mistakes here.Final Answer1. The probability (P(C = c)) is boxed{0.634}).2. The sensitivity of (P(C = c)) with respect to (epsilon) at (epsilon = 0) is boxed{0.42}).</think>

🧠:<think>Alright, so I have this problem about a clinical trial for a new medication. There are two parts: calculating the odds ratio and then using a Bayesian approach to find the posterior probability. Let me try to work through each step carefully.Starting with part 1: calculating the odds ratio. I remember that the odds ratio is a measure of association between an exposure (in this case, the medication) and an outcome (improvement in symptoms). It's calculated by dividing the odds of the outcome in the exposed group by the odds in the unexposed group.The trial had 300 patients split into two groups of 150 each. The medication group had 90 improved, and the placebo group had 45 improved. So, I need to set up a 2x2 contingency table.Let me write it out:| | Improved | Not Improved | Total ||---------------|----------|--------------|-------|| Medication | 90 | 60 | 150 || Placebo | 45 | 105 | 150 || Total | 135 | 165 | 300 |Okay, so for the medication group, the odds of improvement are 90/60, and for the placebo, it's 45/105. The odds ratio (OR) is then (90/60) divided by (45/105). Let me compute that.First, 90 divided by 60 is 1.5. Then, 45 divided by 105 is approximately 0.4286. So, the odds ratio is 1.5 divided by 0.4286. Let me calculate that: 1.5 / 0.4286 ≈ 3.5. Hmm, that seems high. Wait, let me verify.Alternatively, another way to compute the odds ratio is (a*d)/(b*c), where a=90, b=60, c=45, d=105. So, (90*105)/(60*45). Let me compute that: 90*105 is 9450, and 60*45 is 2700. So, 9450 / 2700 is indeed 3.5. Okay, so the odds ratio is 3.5. That means the odds of improvement are 3.5 times higher in the medication group compared to the placebo group.But wait, the question is whether the medication is significantly more effective. The odds ratio alone doesn't tell us significance; we might need a confidence interval or a p-value. However, since the question only asks to calculate the odds ratio, maybe that's sufficient for this part. I'll note that an odds ratio greater than 1 suggests the medication is more effective, but significance would require further analysis.Moving on to part 2: Bayesian approach. The prior probability that the medication is effective is 70%, or 0.7. We need to calculate the posterior probability using the data from the trial.Bayes' theorem states that:Posterior = (Likelihood * Prior) / EvidenceIn this case, the likelihood is the probability of the data given the medication is effective, and the complement of the placebo improvement rate is used as the likelihood when the medication is ineffective.Wait, let me parse that again. The likelihood of improvement given the medication's effectiveness is consistent with the observed data. The complement of the placebo improvement rate is used as the likelihood of improvement given the medication's ineffectiveness.So, the improvement rate in the placebo group is 45/150 = 0.3. Therefore, the complement is 1 - 0.3 = 0.7. So, if the medication is ineffective, the likelihood of improvement is 0.7? That seems high because the placebo group had a 30% improvement rate, so the complement would be 70% not improved. Wait, maybe I'm misunderstanding.Wait, the problem says: "use the complement of the placebo improvement rate as the likelihood of improvement given the medication's ineffectiveness." Hmm, the complement of the improvement rate is the non-improvement rate. So, if the improvement rate in placebo is 0.3, the complement is 0.7, which is the non-improvement rate. But the likelihood of improvement given ineffectiveness would be the improvement rate under the placebo, which is 0.3, not the complement.Wait, maybe I need to clarify. If the medication is ineffective, then the improvement rate should be similar to the placebo, which is 0.3. So, the likelihood of improvement given ineffectiveness is 0.3, not the complement. But the problem says to use the complement of the placebo improvement rate as the likelihood of improvement given ineffectiveness. That would be 1 - 0.3 = 0.7. That seems contradictory because if the medication is ineffective, the improvement should be similar to placebo, which is 0.3, not 0.7.Wait, perhaps I'm overcomplicating. Let's read the problem again: "use the complement of the placebo improvement rate as the likelihood of improvement given the medication's ineffectiveness." So, improvement given ineffectiveness is 1 - 0.3 = 0.7. That seems odd because if the medication is ineffective, the improvement rate shouldn't be higher than the placebo. Maybe it's a typo or misunderstanding.Alternatively, perhaps the complement is used for non-improvement. Wait, the complement of improvement is non-improvement. So, if the improvement rate is 0.3, the non-improvement rate is 0.7. But the likelihood of improvement given ineffectiveness should be 0.3, not 0.7. So, maybe the problem is phrased incorrectly, or I'm misinterpreting.Alternatively, perhaps the likelihood of improvement given ineffectiveness is the same as the placebo improvement rate, which is 0.3, and the complement is used elsewhere. Hmm.Wait, let's think about Bayesian terms. We have two hypotheses: H1 (medication is effective) and H2 (medication is ineffective). The prior probabilities are P(H1) = 0.7 and P(H2) = 0.3.The likelihoods are P(Data | H1) and P(Data | H2). The data is that in the medication group, 90 out of 150 improved, and in the placebo group, 45 out of 150 improved.But the problem says: "assume a likelihood of improvement given the medication's effectiveness is consistent with the observed data, and use the complement of the placebo improvement rate as the likelihood of improvement given the medication's ineffectiveness."Wait, so for H1, the likelihood is consistent with the observed data, which is 90/150 = 0.6 improvement rate. For H2, the likelihood is the complement of the placebo improvement rate, which is 1 - 0.3 = 0.7. But that would mean that if the medication is ineffective, the improvement rate is 0.7, which is higher than the placebo. That doesn't make sense because if the medication is ineffective, the improvement should be similar to placebo.Alternatively, perhaps the complement is used for non-improvement. So, for H2, the likelihood of improvement is 0.3, and the complement (non-improvement) is 0.7. But then, how does that factor into the likelihood?Wait, maybe the problem is saying that for H2, the likelihood of improvement is the complement of the placebo improvement rate, which is 0.7. But that would imply that under H2, the improvement rate is 0.7, which is higher than the placebo. That seems contradictory.Alternatively, perhaps the complement is used in the sense that the likelihood of improvement given H2 is the same as the likelihood of non-improvement given H1? That might not make sense either.Wait, perhaps I need to model this differently. Let's consider that under H1, the improvement rate is p1, and under H2, the improvement rate is p2. The prior is P(H1)=0.7, P(H2)=0.3.The observed data is 90 improved out of 150 in the medication group, and 45 out of 150 in the placebo group. But in Bayesian terms, we might be considering the likelihood of the medication group's improvement given H1 and H2.Wait, but the problem says: "use the complement of the placebo improvement rate as the likelihood of improvement given the medication's ineffectiveness." So, if the placebo improvement rate is 0.3, the complement is 0.7. So, under H2 (ineffective), the improvement rate is 0.7? That seems odd because the placebo group had 0.3 improvement, so if the medication is ineffective, it should be similar to placebo, not higher.Alternatively, maybe the complement is used for the non-improvement rate. So, under H2, the improvement rate is 0.3 (same as placebo), and the non-improvement rate is 0.7. But then, how does that affect the likelihood?Wait, perhaps the problem is using the complement in a different way. Let me think. If the medication is effective, the improvement rate is higher than the placebo. If it's ineffective, the improvement rate is the same as the placebo. So, perhaps the likelihood for H1 is the observed improvement rate in the medication group, and for H2, it's the improvement rate of the placebo group.But the problem says: "use the complement of the placebo improvement rate as the likelihood of improvement given the medication's ineffectiveness." So, complement of 0.3 is 0.7, so improvement given H2 is 0.7? That doesn't make sense because if the medication is ineffective, the improvement rate shouldn't be higher than the placebo.Wait, maybe the complement is used in the sense that the likelihood of improvement given H2 is the same as the likelihood of non-improvement given H1? That might not be the case.Alternatively, perhaps the problem is using the complement in the sense that the likelihood of improvement given H2 is 1 - the improvement rate of the placebo. So, 1 - 0.3 = 0.7. But that would mean that under H2, the improvement rate is 0.7, which is higher than the placebo. That seems contradictory.Wait, maybe I'm overcomplicating. Let's try to proceed with the given instructions, even if it seems counterintuitive.So, under H1, the likelihood of improvement is consistent with the observed data, which is 90/150 = 0.6. Under H2, the likelihood of improvement is the complement of the placebo improvement rate, which is 1 - 0.3 = 0.7.So, P(Data | H1) = 0.6^90 * (1 - 0.6)^60P(Data | H2) = 0.7^90 * (1 - 0.7)^60Wait, but the data is 90 improved out of 150. So, the likelihood is the probability of observing 90 successes in 150 trials given the improvement rate.But in Bayesian terms, we might model this using the binomial likelihood.So, for H1, the likelihood is Binomial(90; 150, 0.6)For H2, the likelihood is Binomial(90; 150, 0.7)But wait, if H2 is the medication being ineffective, the improvement rate should be similar to placebo, which is 0.3, not 0.7. So, perhaps the problem is incorrectly stating to use the complement, but maybe it's a typo and should be the same as the placebo improvement rate.Alternatively, perhaps the complement is used for the non-improvement rate. So, under H2, the improvement rate is 0.3, and the non-improvement rate is 0.7. But then, the likelihood would still be based on the improvement rate.Wait, maybe the problem is saying that the likelihood of improvement given H2 is the complement of the placebo improvement rate, meaning 1 - 0.3 = 0.7. But that would mean that under H2, the improvement rate is 0.7, which is higher than the placebo. That seems incorrect.Alternatively, perhaps the complement is used in the sense that the likelihood of improvement given H2 is the same as the likelihood of non-improvement given H1. But that might not be the case.Wait, perhaps I need to proceed with the given instructions, even if it seems counterintuitive.So, assuming that under H2, the improvement rate is 0.7, which is the complement of the placebo improvement rate.Then, the likelihoods are:P(Data | H1) = C(150,90) * (0.6)^90 * (0.4)^60P(Data | H2) = C(150,90) * (0.7)^90 * (0.3)^60But then, the Bayes factor would be P(Data | H1) / P(Data | H2). However, since we have prior probabilities, we can compute the posterior odds.Posterior odds = Prior odds * Bayes factorPrior odds = P(H1)/P(H2) = 0.7 / 0.3 ≈ 2.333Bayes factor = [C(150,90)*(0.6^90*0.4^60)] / [C(150,90)*(0.7^90*0.3^60)] = (0.6/0.7)^90 * (0.4/0.3)^60Simplify:(0.6/0.7)^90 = (6/7)^90 ≈ (0.8571)^90(0.4/0.3)^60 = (4/3)^60 ≈ (1.3333)^60Calculating these exponents is going to be a bit tricky, but let's see.First, ln(6/7) ≈ ln(0.8571) ≈ -0.1542So, ln((6/7)^90) = 90*(-0.1542) ≈ -13.878Similarly, ln(4/3) ≈ 0.2877So, ln((4/3)^60) = 60*0.2877 ≈ 17.262Total ln(Bayes factor) ≈ -13.878 + 17.262 ≈ 3.384So, Bayes factor ≈ e^3.384 ≈ 29.36Therefore, posterior odds = prior odds * Bayes factor ≈ 2.333 * 29.36 ≈ 68.45Then, posterior probability P(H1 | Data) = posterior odds / (1 + posterior odds) ≈ 68.45 / (1 + 68.45) ≈ 68.45 / 69.45 ≈ 0.986, or 98.6%.But wait, this seems very high, and it's because we assumed that under H2, the improvement rate is 0.7, which is higher than the observed data. But the observed data in the medication group was 0.6, which is lower than 0.7. So, the likelihood under H2 is actually lower than under H1, which makes the Bayes factor greater than 1, favoring H1.But this result seems counterintuitive because if the medication is ineffective, the improvement rate should be similar to placebo (0.3), not 0.7. So, perhaps the problem intended to use the placebo improvement rate as the likelihood under H2, not the complement.Alternatively, maybe the complement is used for the non-improvement rate. So, under H2, the improvement rate is 0.3, and the non-improvement rate is 0.7. But then, the likelihood would still be based on the improvement rate.Wait, perhaps the problem is saying that the likelihood of improvement given H2 is the complement of the placebo improvement rate, meaning 1 - 0.3 = 0.7. But that would mean that under H2, the improvement rate is 0.7, which is higher than the observed data (0.6). So, the likelihood under H2 would be lower than under H1, making the Bayes factor greater than 1, favoring H1.But this seems contradictory because if the medication is ineffective, the improvement rate should be similar to placebo, not higher. So, perhaps the problem has a typo, and it should be the same as the placebo improvement rate, not the complement.Alternatively, maybe the complement is used in the sense that the likelihood of non-improvement given H2 is the complement of the placebo improvement rate. So, non-improvement rate under H2 is 1 - 0.3 = 0.7, which would mean improvement rate is 0.3, which makes sense.So, if under H2, the improvement rate is 0.3, then the likelihood is Binomial(90; 150, 0.3). Let's recalculate with this assumption.So, P(Data | H1) = C(150,90)*(0.6)^90*(0.4)^60P(Data | H2) = C(150,90)*(0.3)^90*(0.7)^60Then, Bayes factor = [C(150,90)*(0.6^90*0.4^60)] / [C(150,90)*(0.3^90*0.7^60)] = (0.6/0.3)^90 * (0.4/0.7)^60 = (2)^90 * (4/7)^60Calculating the log:ln(2^90) = 90*ln(2) ≈ 90*0.6931 ≈ 62.38ln((4/7)^60) = 60*ln(4/7) ≈ 60*(-0.5596) ≈ -33.58Total ln(Bayes factor) ≈ 62.38 - 33.58 ≈ 28.8So, Bayes factor ≈ e^28.8 ≈ 7.7 x 10^12That's an extremely large Bayes factor, which would make the posterior odds extremely high, almost 1. But this seems too extreme, and it's because the observed data (0.6) is much higher than the expected under H2 (0.3). So, the data strongly support H1.But wait, the problem specifically said to use the complement of the placebo improvement rate as the likelihood of improvement given H2. So, if the complement is 0.7, then the improvement rate under H2 is 0.7, which is higher than the observed data. But that leads to a lower likelihood under H2, making the Bayes factor greater than 1, favoring H1.But this seems contradictory because if the medication is ineffective, the improvement rate should be similar to placebo, not higher. So, perhaps the problem intended to use the same improvement rate as the placebo under H2, not the complement.Alternatively, maybe the complement is used in the sense that the likelihood of improvement given H2 is the same as the non-improvement rate in the placebo group. So, non-improvement rate in placebo is 0.7, so improvement rate under H2 is 0.3. That makes sense.So, perhaps the problem meant that the likelihood of improvement given H2 is the same as the improvement rate in the placebo group, which is 0.3, and the complement (non-improvement) is 0.7. So, under H2, improvement rate is 0.3.In that case, the likelihoods are:P(Data | H1) = Binomial(90; 150, 0.6)P(Data | H2) = Binomial(90; 150, 0.3)Then, the Bayes factor is [P(Data | H1)] / [P(Data | H2)] = [C(150,90)*(0.6^90*0.4^60)] / [C(150,90)*(0.3^90*0.7^60)] = (0.6/0.3)^90 * (0.4/0.7)^60 = (2)^90 * (4/7)^60As calculated earlier, this gives a Bayes factor of approximately e^28.8 ≈ 7.7 x 10^12, which is extremely large, indicating overwhelming evidence in favor of H1.But this seems too extreme, and perhaps the problem intended to use the complement differently. Alternatively, maybe the problem is using the complement in the sense that the likelihood of improvement given H2 is the same as the non-improvement rate in the placebo group, which is 0.7, but that would mean improvement rate under H2 is 0.7, which is higher than the observed data.Wait, the observed data in the medication group is 0.6, which is lower than 0.7. So, if under H2, the improvement rate is 0.7, then the likelihood of observing 90/150 is lower than under H1 (0.6). So, the Bayes factor would be greater than 1, favoring H1.But this is a bit confusing because if H2 is the medication being ineffective, the improvement rate should be similar to placebo, which is 0.3, not 0.7. So, perhaps the problem has a typo, and it should be the same as the placebo improvement rate, not the complement.Alternatively, maybe the complement is used in the sense that the likelihood of improvement given H2 is the same as the non-improvement rate in the placebo group, which is 0.7. But that would mean that under H2, the improvement rate is 0.7, which is higher than the observed data, making the likelihood under H2 lower than under H1, which still favors H1.But this seems contradictory because if the medication is ineffective, the improvement rate should be similar to placebo, not higher. So, perhaps the problem intended to use the same improvement rate as the placebo under H2, not the complement.Given the confusion, perhaps I should proceed with the assumption that under H2, the improvement rate is the same as the placebo, which is 0.3, and the complement is used for non-improvement. So, let's proceed with that.So, P(Data | H1) = Binomial(90; 150, 0.6)P(Data | H2) = Binomial(90; 150, 0.3)Then, the likelihood ratio is P(Data | H1)/P(Data | H2) = (0.6^90 * 0.4^60) / (0.3^90 * 0.7^60) = (0.6/0.3)^90 * (0.4/0.7)^60 = (2)^90 * (4/7)^60Calculating the log:ln(2^90) = 90*ln(2) ≈ 62.38ln((4/7)^60) = 60*ln(4/7) ≈ 60*(-0.5596) ≈ -33.58Total ln(Bayes factor) ≈ 62.38 - 33.58 ≈ 28.8So, Bayes factor ≈ e^28.8 ≈ 7.7 x 10^12Then, prior odds = 0.7 / 0.3 ≈ 2.333Posterior odds = prior odds * Bayes factor ≈ 2.333 * 7.7 x 10^12 ≈ 1.8 x 10^13Then, posterior probability P(H1 | Data) = posterior odds / (1 + posterior odds) ≈ 1.8 x 10^13 / (1 + 1.8 x 10^13) ≈ 1, or 100%.But this seems too extreme. Maybe I made a mistake in the calculation.Wait, let's recalculate the Bayes factor more carefully.First, compute the ratio of the likelihoods:(0.6/0.3)^90 = (2)^90 ≈ 1.237 x 10^27(0.4/0.7)^60 ≈ (0.5714)^60 ≈ 1.17 x 10^-8So, Bayes factor ≈ (1.237 x 10^27) * (1.17 x 10^-8) ≈ 1.447 x 10^19Then, prior odds = 0.7 / 0.3 ≈ 2.333Posterior odds = 2.333 * 1.447 x 10^19 ≈ 3.38 x 10^19Posterior probability ≈ 3.38 x 10^19 / (1 + 3.38 x 10^19) ≈ 1So, the posterior probability is effectively 1, or 100%. That seems too high, but given the data, the improvement in the medication group is much higher than what would be expected under H2 (0.3), so the evidence is overwhelming in favor of H1.But wait, the problem said to use the complement of the placebo improvement rate as the likelihood of improvement given H2. So, if the complement is 0.7, then the improvement rate under H2 is 0.7, which is higher than the observed data. So, let's recalculate with that assumption.So, P(Data | H1) = Binomial(90; 150, 0.6)P(Data | H2) = Binomial(90; 150, 0.7)Then, Bayes factor = [0.6^90 * 0.4^60] / [0.7^90 * 0.3^60] = (0.6/0.7)^90 * (0.4/0.3)^60Calculating the log:ln((0.6/0.7)^90) = 90*ln(0.8571) ≈ 90*(-0.1542) ≈ -13.878ln((0.4/0.3)^60) = 60*ln(1.3333) ≈ 60*0.2877 ≈ 17.262Total ln(Bayes factor) ≈ -13.878 + 17.262 ≈ 3.384So, Bayes factor ≈ e^3.384 ≈ 29.36Then, prior odds = 0.7 / 0.3 ≈ 2.333Posterior odds = 2.333 * 29.36 ≈ 68.45Posterior probability ≈ 68.45 / (1 + 68.45) ≈ 0.986, or 98.6%So, approximately 98.6% posterior probability that the medication is effective.This seems more reasonable, although still very high. The reason is that the observed improvement rate in the medication group (0.6) is higher than the improvement rate under H2 (0.7). Wait, no, 0.6 is lower than 0.7. So, the likelihood under H2 is actually lower than under H1, making the Bayes factor greater than 1, favoring H1.Wait, no, if under H2, the improvement rate is 0.7, and the observed data is 0.6, which is lower than 0.7, then the likelihood under H2 is lower than under H1. So, the Bayes factor is greater than 1, favoring H1.But this is counterintuitive because if the medication is ineffective, the improvement rate should be similar to placebo, not higher. So, perhaps the problem intended to use the same improvement rate as the placebo under H2, not the complement.But given the problem's instructions, I have to proceed with using the complement of the placebo improvement rate as the likelihood of improvement given H2, which is 0.7.Therefore, the posterior probability is approximately 98.6%.But let me double-check the calculations.First, the likelihood ratio:P(Data | H1) / P(Data | H2) = (0.6^90 * 0.4^60) / (0.7^90 * 0.3^60)= (0.6/0.7)^90 * (0.4/0.3)^60= (6/7)^90 * (4/3)^60Calculating the natural log:ln(6/7) ≈ -0.1542, so 90*(-0.1542) ≈ -13.878ln(4/3) ≈ 0.2877, so 60*0.2877 ≈ 17.262Total ln(Bayes factor) ≈ -13.878 + 17.262 ≈ 3.384Bayes factor ≈ e^3.384 ≈ 29.36Prior odds = 0.7 / 0.3 ≈ 2.333Posterior odds = 2.333 * 29.36 ≈ 68.45Posterior probability ≈ 68.45 / (1 + 68.45) ≈ 0.986So, approximately 98.6% posterior probability that the medication is effective.Therefore, despite the initial confusion about the complement, following the problem's instructions leads to a posterior probability of about 98.6%.But I'm still a bit uneasy because using the complement of the placebo improvement rate as the likelihood under H2 seems to imply that the improvement rate under H2 is higher than the observed data, which might not make sense. However, given the instructions, this is the way to proceed.So, to summarize:1. Odds ratio is 3.5, indicating the medication is more effective than placebo.2. Posterior probability using Bayesian approach is approximately 98.6%.</think>

🧠:<think>Okay, so I need to solve this optimization problem for Alex's judo training. Let me try to break it down step by step.First, the problem says that Alex's performance is modeled by the function ( P(t, s) = k left( frac{t^2 s}{t + s} right) ), where ( t ) is hours spent on technique training and ( s ) is hours on strength training. The constant ( k ) depends on talent and coaching. In the first part, ( k = 3 ) and Alex has a maximum of 20 hours per week. So, I need to maximize ( P(t, s) ) given that ( t + s = 20 ). Hmm, okay, so since ( t + s = 20 ), I can express one variable in terms of the other. Let me solve for ( s ): ( s = 20 - t ). Then, substitute this into the performance function.So, substituting ( s ) into ( P(t, s) ), we get:[P(t) = 3 left( frac{t^2 (20 - t)}{t + (20 - t)} right)]Simplify the denominator: ( t + 20 - t = 20 ). So, the denominator is just 20.So, ( P(t) = 3 left( frac{t^2 (20 - t)}{20} right) )Simplify that: ( P(t) = frac{3}{20} t^2 (20 - t) )Let me write that as:[P(t) = frac{3}{20} (20 t^2 - t^3)]So, ( P(t) = frac{3}{20} times 20 t^2 - frac{3}{20} t^3 )Simplify further: ( P(t) = 3 t^2 - frac{3}{20} t^3 )Now, to find the maximum, I need to take the derivative of ( P(t) ) with respect to ( t ) and set it equal to zero.Compute ( P'(t) ):[P'(t) = 6t - frac{9}{20} t^2]Set ( P'(t) = 0 ):[6t - frac{9}{20} t^2 = 0]Factor out ( t ):[t left(6 - frac{9}{20} t right) = 0]So, solutions are ( t = 0 ) or ( 6 - frac{9}{20} t = 0 ).Solving ( 6 - frac{9}{20} t = 0 ):[frac{9}{20} t = 6 t = 6 times frac{20}{9} t = frac{120}{9} t = frac{40}{3} approx 13.333]So, ( t = frac{40}{3} ) hours, which is approximately 13.333 hours.Then, ( s = 20 - t = 20 - frac{40}{3} = frac{60}{3} - frac{40}{3} = frac{20}{3} approx 6.666 ) hours.Wait, let me check if this is a maximum. I can use the second derivative test.Compute ( P''(t) ):[P''(t) = 6 - frac{18}{20} t = 6 - frac{9}{10} t]At ( t = frac{40}{3} ):[P''left( frac{40}{3} right) = 6 - frac{9}{10} times frac{40}{3} = 6 - frac{360}{30} = 6 - 12 = -6]Since ( P''(t) < 0 ), this critical point is a local maximum. So, it's indeed the maximum.Therefore, the optimal values are ( t = frac{40}{3} ) hours and ( s = frac{20}{3} ) hours.Wait, let me just verify the calculations once more to be sure.Starting from the derivative:( P'(t) = 6t - frac{9}{20} t^2 )Setting to zero:( 6t = frac{9}{20} t^2 )Divide both sides by t (assuming t ≠ 0):( 6 = frac{9}{20} t )Multiply both sides by 20:( 120 = 9t )Divide by 9:( t = frac{120}{9} = frac{40}{3} ). Yep, that's correct.So, moving on to the second part. If Alex's performance needs to exceed ( P_{min} = 75 ), we need to find the minimum value of ( k ) under the optimal allocation found in part 1.So, from part 1, the optimal allocation is ( t = frac{40}{3} ) and ( s = frac{20}{3} ).So, plug these into the performance function ( P(t, s) = k left( frac{t^2 s}{t + s} right) ).First, compute ( t^2 s ):( t^2 = left( frac{40}{3} right)^2 = frac{1600}{9} )( s = frac{20}{3} )So, ( t^2 s = frac{1600}{9} times frac{20}{3} = frac{32000}{27} )Compute ( t + s = frac{40}{3} + frac{20}{3} = frac{60}{3} = 20 )So, ( frac{t^2 s}{t + s} = frac{32000}{27} div 20 = frac{32000}{27 times 20} = frac{32000}{540} = frac{3200}{54} = frac{1600}{27} approx 59.259 )So, ( P(t, s) = k times frac{1600}{27} )We need this to be at least 75:[k times frac{1600}{27} geq 75]Solve for ( k ):[k geq 75 times frac{27}{1600}]Compute that:First, multiply 75 and 27:75 * 27: 70*27=1890, 5*27=135, so total 1890 + 135 = 2025So, ( k geq frac{2025}{1600} )Simplify the fraction:Divide numerator and denominator by 25:2025 ÷25=81, 1600 ÷25=64So, ( k geq frac{81}{64} )Convert to decimal: 81 ÷ 64 = 1.265625So, the minimum value of ( k ) is ( frac{81}{64} ) or approximately 1.2656.Let me double-check the calculations:Compute ( t^2 s ):( t = 40/3 ), so ( t^2 = 1600/9 )Multiply by ( s = 20/3 ): ( (1600/9)*(20/3) = 32000/27 )Divide by ( t + s = 20 ): ( 32000/27 /20 = 1600/27 )So, ( P = k*(1600/27) geq 75 )Thus, ( k geq 75*(27/1600) = (75*27)/1600 = 2025/1600 = 81/64 ). Yep, that's correct.So, the minimum ( k ) is ( 81/64 ).Final Answer1. The optimal values are ( t = boxed{dfrac{40}{3}} ) hours and ( s = boxed{dfrac{20}{3}} ) hours.2. The minimum value of ( k ) required is ( boxed{dfrac{81}{64}} ).</think>