🧠:<think>Okay, so I have this problem about creating a spin-off TV show, and I need to figure out two things: the number of possible unique episode combinations and the total budget required. Let me take it step by step.Starting with the first part: determining the number of possible unique episode combinations. The original series has n main characters and m distinct types of scenes. The fan wants to create k episodes, each featuring a unique combination of p characters and q scenes. Hmm, so for each episode, we need to choose p characters out of n and q scenes out of m.I remember that the number of ways to choose p items from n is given by the combination formula, which is "n choose p" or mathematically, that's (binom{n}{p}). Similarly, the number of ways to choose q scenes from m is (binom{m}{q}). Since each episode is a combination of both characters and scenes, the total number of unique episodes for one episode would be the product of these two combinations. So that's (binom{n}{p} times binom{m}{q}).But wait, the problem says the spin-off has exactly k episodes, each with a unique combination. So does that mean we need to choose k unique combinations from all possible combinations? Or is each episode independently chosen with any combination, but ensuring that all k are unique? Hmm, the wording says "exactly k episodes, where each episode features a unique combination." So I think it's the number of ways to choose k unique episodes from all possible unique episode combinations.So first, the total number of possible unique episodes is (binom{n}{p} times binom{m}{q}). Let's denote this as T. Then, the number of ways to choose k unique episodes from T is (binom{T}{k}).Wait, but the problem says "determine the number of possible unique episode combinations for the spin-off show." So maybe it's just asking for the number of possible episodes, not considering the order or selection of k episodes. Hmm, the wording is a bit ambiguous.Looking back: "Determine the number of possible unique episode combinations for the spin-off show." It doesn't specify whether it's the number of ways to choose k episodes or just the total number of possible episodes. But since each episode is a unique combination, and the spin-off has k episodes, each with a unique combination, perhaps it's the number of ways to select k unique episodes from all possible episodes.So, if each episode is a unique combination, then the total number of possible unique episodes is (binom{n}{p} times binom{m}{q}). Let's call this number E. Then, the number of ways to choose k unique episodes from E is (binom{E}{k}).But wait, another interpretation: Maybe each episode is independently chosen, but all k must be unique. So it's the number of injective functions from the set of k episodes to the set of all possible episodes. That would be E × (E - 1) × ... × (E - k + 1), which is equal to (frac{E!}{(E - k)!}). But this is different from (binom{E}{k}), which is the number of combinations without considering order.But the problem doesn't specify whether the order of episodes matters. It just says "k episodes, where each episode features a unique combination." So I think order doesn't matter here. So it's just the number of ways to choose k unique episodes, which is (binom{E}{k}).But wait, hold on. The problem says "the number of possible unique episode combinations for the spin-off show." So maybe it's just the number of possible episodes, which is E, and not considering how many episodes are in the spin-off. Hmm, no, because it specifically mentions k episodes.Wait, maybe I'm overcomplicating. Let me read the question again: "Determine the number of possible unique episode combinations for the spin-off show." So the spin-off has k episodes, each with a unique combination. So the total number of possible such spin-offs is the number of ways to choose k unique episodes from all possible episodes.Therefore, if E is the total number of possible episodes, then the number of spin-offs is (binom{E}{k}).But E is (binom{n}{p} times binom{m}{q}). So substituting, the number of possible spin-offs is (binom{binom{n}{p} times binom{m}{q}}{k}).Alternatively, if the problem is asking for the number of unique episode combinations per episode, not considering the k episodes, then it's just (binom{n}{p} times binom{m}{q}). But the question says "for the spin-off show," which has k episodes. So I think it's the former.But I'm a bit confused because sometimes "number of possible unique episode combinations" could be interpreted as the number of possible episodes, not the number of ways to choose k episodes. Let me check the wording again: "Determine the number of possible unique episode combinations for the spin-off show." Since the spin-off has k episodes, each with a unique combination, it's more likely that they want the number of ways to have k unique episodes, which would be (binom{binom{n}{p} times binom{m}{q}}{k}).But I'm not entirely sure. Maybe I should consider both interpretations.Wait, another angle: If each episode is a unique combination, then the total number of unique episodes is (binom{n}{p} times binom{m}{q}). So if the spin-off has k episodes, each being unique, then the number of possible spin-offs is the number of ways to choose k unique episodes from this total. So yes, that would be (binom{binom{n}{p} times binom{m}{q}}{k}).But let me think if that's the case. Alternatively, maybe each spin-off is a sequence of k episodes, each with a unique combination, but order matters. Then it would be permutations instead of combinations. But the problem doesn't specify order, so I think it's combinations.So, to summarize, the number of possible unique episode combinations for the spin-off show is (binom{binom{n}{p} times binom{m}{q}}{k}).But wait, hold on. Let me think again. The problem says "each episode features a unique combination." So each episode is a unique combination, but the spin-off is a collection of k such episodes. So the spin-off is a set of k unique episodes. So the number of possible spin-offs is the number of ways to choose k unique episodes from all possible episodes.Therefore, the answer is (binom{binom{n}{p} times binom{m}{q}}{k}).But wait, another thought: Maybe the problem is asking for the number of possible ways to assign p characters and q scenes to each of the k episodes, with all k episodes being unique. So it's like arranging k unique episodes, each with p characters and q scenes.In that case, the first episode can be any combination: (binom{n}{p} times binom{m}{q}). The second episode has to be different, so (binom{n}{p} times binom{m}{q} - 1), and so on. So the total number would be (frac{(binom{n}{p} times binom{m}{q})!}{(binom{n}{p} times binom{m}{q} - k)!}). But this is if order matters.But the problem doesn't specify order, so perhaps it's combinations. So the number is (binom{binom{n}{p} times binom{m}{q}}{k}).Alternatively, if the problem is just asking for the number of unique episodes, regardless of how many are in the spin-off, then it's just (binom{n}{p} times binom{m}{q}). But since the spin-off has k episodes, each unique, I think the first interpretation is correct.So, moving on to the second part: finding the total budget required to produce all k episodes, given the cost function (C(p, q) = a cdot p^2 + b cdot q^2 + c cdot p cdot q).Each episode has p characters and q scenes, so each episode's cost is (a p^2 + b q^2 + c p q). Since all k episodes are unique, but each episode has the same p and q, right? Wait, no. Wait, each episode is a unique combination of p characters and q scenes, but p and q are fixed per episode. So each episode has exactly p characters and q scenes, but different combinations.So the cost per episode is the same for each episode, because p and q are fixed. So the total cost would be k times the cost per episode.Wait, but hold on. Is p and q fixed for each episode, or can they vary? The problem says "each episode features a unique combination of p characters and q scenes." So p and q are fixed per episode, meaning each episode has exactly p characters and q scenes, but different combinations.Therefore, each episode has the same cost, which is (C(p, q)). So the total budget is (k times C(p, q)), which is (k(a p^2 + b q^2 + c p q)).But wait, is that correct? Because each episode is a unique combination, but the cost is based on p and q, which are fixed. So regardless of the combination, each episode costs the same. So total cost is k times that.But let me think again. Suppose p and q are fixed, so each episode has p characters and q scenes, but different sets. So the cost per episode is the same, so total cost is k times that cost.Yes, that makes sense. So the total budget is (k(a p^2 + b q^2 + c p q)).But wait, another thought: If the cost function depends on p and q, which are fixed per episode, then yes, each episode has the same cost. So total cost is k multiplied by that cost.Therefore, the total budget is (k(a p^2 + b q^2 + c p q)).But let me check the problem statement again: "the cost function is given by (C(p, q) = a cdot p^2 + b cdot q^2 + c cdot p cdot q), where a, b, and c are constants representing various production costs, find the total budget required to produce all k episodes."So yes, each episode has p characters and q scenes, so each episode's cost is C(p, q). Therefore, total cost is k * C(p, q).So, putting it all together:1. The number of possible unique episode combinations is (binom{binom{n}{p} times binom{m}{q}}{k}).2. The total budget is (k(a p^2 + b q^2 + c p q)).But wait, for the first part, is it (binom{binom{n}{p} times binom{m}{q}}{k}) or is it (binom{n}{p} times binom{m}{q}) choose k? Yes, that's what I have.But let me think if there's another way to interpret it. Maybe the problem is asking for the number of ways to assign p characters and q scenes to each of the k episodes, with all combinations being unique. So it's like arranging k unique episodes, each with p characters and q scenes.In that case, the number of ways would be the number of injective functions from the set of k episodes to the set of all possible episodes. So that would be (binom{binom{n}{p} times binom{m}{q}}{k}) multiplied by k! if order matters, but since the episodes are just a set, not ordered, it's just (binom{binom{n}{p} times binom{m}{q}}{k}).Yes, that seems correct.So, to recap:1. The number of possible unique episode combinations is (dbinom{dbinom{n}{p} times dbinom{m}{q}}{k}).2. The total budget is (k(a p^2 + b q^2 + c p q)).I think that's the answer.</think>

🧠:<think>Alright, so I've got this problem about an advertising campaign, and I need to figure out two things. First, when does the campaign reach 75% of its maximum potential reach, and second, when does the engagement rate first reach its maximum value. Let me take this step by step.Starting with the first part: the reach of the campaign is modeled by a logistic growth equation. The equation given is ( R(t) = frac{L}{1 + e^{-k(t - t_0)}} ). The parameters are ( L = 1,000,000 ), ( k = 0.4 ), and ( t_0 = 10 ). I need to find the time ( t ) when the reach is 75% of ( L ). Okay, so 75% of ( L ) is ( 0.75 times 1,000,000 = 750,000 ). So, I need to solve for ( t ) in the equation:( 750,000 = frac{1,000,000}{1 + e^{-0.4(t - 10)}} )Let me write that out:( 750,000 = frac{1,000,000}{1 + e^{-0.4(t - 10)}} )Hmm, I can simplify this equation. Let's divide both sides by 1,000,000 to make it easier:( 0.75 = frac{1}{1 + e^{-0.4(t - 10)}} )Now, taking the reciprocal of both sides:( frac{1}{0.75} = 1 + e^{-0.4(t - 10)} )Calculating ( frac{1}{0.75} ) is approximately 1.3333. So,( 1.3333 = 1 + e^{-0.4(t - 10)} )Subtract 1 from both sides:( 0.3333 = e^{-0.4(t - 10)} )Now, to solve for ( t ), I can take the natural logarithm of both sides:( ln(0.3333) = -0.4(t - 10) )Calculating ( ln(0.3333) ). I remember that ( ln(1/3) ) is approximately -1.0986. So,( -1.0986 = -0.4(t - 10) )Divide both sides by -0.4:( frac{-1.0986}{-0.4} = t - 10 )Calculating that, ( 1.0986 / 0.4 ) is approximately 2.7465. So,( 2.7465 = t - 10 )Adding 10 to both sides:( t = 10 + 2.7465 approx 12.7465 )So, the time ( t ) when the campaign reaches 75% of its maximum potential reach is approximately 12.75 units of time. Since the problem doesn't specify the units, I assume it's in days or weeks, but since it's not specified, I'll just go with the numerical value.Now, moving on to the second part: the engagement rate is modeled by a sinusoidal function ( E(t) = A sin(Bt + C) + D ). The parameters are ( A = 0.05 ), ( B = frac{pi}{6} ), ( C = frac{pi}{4} ), and ( D = 0.1 ). I need to find the first time ( t ) after the campaign launch when the engagement rate reaches its maximum value.First, let me recall that the maximum value of a sine function ( sin(theta) ) is 1. So, the maximum engagement rate ( E_{max} ) is ( A + D = 0.05 + 0.1 = 0.15 ). So, I need to solve for ( t ) when ( E(t) = 0.15 ).Setting up the equation:( 0.15 = 0.05 sinleft(frac{pi}{6} t + frac{pi}{4}right) + 0.1 )Subtract 0.1 from both sides:( 0.05 = 0.05 sinleft(frac{pi}{6} t + frac{pi}{4}right) )Divide both sides by 0.05:( 1 = sinleft(frac{pi}{6} t + frac{pi}{4}right) )So, when does ( sin(theta) = 1 )? That happens at ( theta = frac{pi}{2} + 2pi n ), where ( n ) is an integer. Since we're looking for the first time after the launch, we'll take the smallest positive ( t ), so ( n = 0 ).So,( frac{pi}{6} t + frac{pi}{4} = frac{pi}{2} )Let me solve for ( t ):First, subtract ( frac{pi}{4} ) from both sides:( frac{pi}{6} t = frac{pi}{2} - frac{pi}{4} = frac{pi}{4} )Now, divide both sides by ( frac{pi}{6} ):( t = frac{pi/4}{pi/6} = frac{6}{4} = frac{3}{2} = 1.5 )So, the first time ( t ) when the engagement rate reaches its maximum is at ( t = 1.5 ).Wait, let me double-check that. So, ( frac{pi}{6} t + frac{pi}{4} = frac{pi}{2} ). Subtract ( frac{pi}{4} ) gives ( frac{pi}{6} t = frac{pi}{4} ). Then, ( t = frac{pi/4}{pi/6} = frac{6}{4} = 1.5 ). Yeah, that seems right.So, putting it all together, the first part gives ( t approx 12.75 ) and the second part gives ( t = 1.5 ).But wait, the first part was about reach, which is a logistic growth, so it's a different function. The second part is about engagement rate, which is sinusoidal. So, these are two separate questions. So, I think I did both correctly.Just to recap:1. For the reach, set ( R(t) = 750,000 ), solved the logistic equation, took natural logs, and found ( t approx 12.75 ).2. For the engagement, set ( E(t) = 0.15 ), recognized that sine must be 1, solved for ( t ) and got 1.5.I think that's solid.Final Answer1. The campaign reaches 75% of its maximum potential reach at approximately boxed{12.75}.2. The engagement rate first reaches its maximum value at boxed{1.5}.</think>

🧠:<think>Alright, so I'm trying to help optimize a Python function that processes a large dataset. The function currently goes through a list of integers twice: once to compute the sum and once to compute the sum of squares. My task is to figure out how to do both calculations in a single pass, which should save some time, especially since the list is large.First, let me break down the problem. The list is called L, and it has n elements. Right now, the function does two separate loops: one for sum(L) and another for sum(x2 for x in L). Each loop is O(n), so together they're O(2n), which is still O(n) time complexity. But if I can combine these into one loop, it would be O(n) instead of O(2n), which is a slight improvement, but for very large n, it might make a difference.So, for part 1, I need to derive the mathematical expressions for S (sum) and Q (sum of squares). Well, that's straightforward. S is just the sum of all elements in L, so S = Σx_i for i from 1 to n. Similarly, Q is the sum of each element squared, so Q = Σx_i² for i from 1 to n.To compute both in a single pass, I can initialize two variables, say total_sum and total_squares, to zero. Then, I'll loop through each element in the list once. For each element x, I'll add x to total_sum and x squared to total_squares. That way, after one loop, I have both S and Q.As for the time complexity, since I'm only looping through the list once, it's O(n), which is better than the original O(2n). So the optimized function has a time complexity of O(n), same as before but with a smaller constant factor.Moving on to part 2, I need to find the expected value and variance of S and Q when the elements of L are uniformly distributed random integers between 1 and m. Hmm, okay. Let's start with S.The expected value of S, E[S], is the sum of the expected values of each x_i. Since each x_i is uniformly distributed between 1 and m, the expected value of each x_i is (1 + m)/2. So, E[S] = n * (1 + m)/2.The variance of S, Var(S), is the sum of the variances of each x_i, since the elements are independent. The variance of a single uniform integer between 1 and m is [(m² - 1)/12]. So, Var(S) = n * (m² - 1)/12.Now for Q, which is the sum of squares. The expected value of Q, E[Q], is the sum of the expected values of each x_i². The expected value of x_i² for a uniform distribution between 1 and m is [m(m + 1)(2m + 1)]/(6m) - wait, no, that's the formula for the sum of squares up to m. Actually, the expected value of x_i² is [1² + 2² + ... + m²]/m. The sum of squares formula is m(m + 1)(2m + 1)/6, so E[x_i²] = [m(m + 1)(2m + 1)]/(6m) = (m + 1)(2m + 1)/6.Therefore, E[Q] = n * (m + 1)(2m + 1)/6.For the variance of Q, Var(Q), it's a bit trickier. Since Q is the sum of x_i², and assuming independence, Var(Q) = n * Var(x_i²). The variance of x_i² is E[x_i^4] - (E[x_i²])². First, I need E[x_i^4]. The sum of fourth powers from 1 to m is m(m + 1)(2m + 1)(3m² + 3m - 1)/30. So, E[x_i^4] = [m(m + 1)(2m + 1)(3m² + 3m - 1)]/(30m) = (m + 1)(2m + 1)(3m² + 3m - 1)/30.Then, Var(x_i²) = E[x_i^4] - (E[x_i²])². Plugging in the values:Var(x_i²) = [(m + 1)(2m + 1)(3m² + 3m - 1)/30] - [(m + 1)(2m + 1)/6]^2.Simplifying this expression would give the variance for each x_i², and then Var(Q) = n times that.As for how the range m affects computational efficiency and memory usage, a larger m means each integer is larger, but since we're dealing with Python, which handles big integers well, it shouldn't cause overflow issues unless m is extremely large. However, larger m could mean that the sum S and sum of squares Q become very large, potentially affecting memory if n is also large. But in terms of computational efficiency, since we're still doing O(n) operations, the time complexity remains the same regardless of m. The main factors affecting efficiency are the size of n and the operations per iteration, which are consistent here.Wait, but if m is very large, say in the order of 10^6 or more, then each x_i could be a large number, and squaring it might take slightly more time due to the size of the numbers. However, in Python, integer operations are optimized, so the impact might be minimal unless m is astronomically large.Also, considering memory usage, each element in the list L is an integer, and if m is large, each integer might take up more memory. But again, for a list of n elements, the memory usage is O(n), regardless of m, as each integer's memory footprint is relatively fixed in Python.So, in summary, while m affects the magnitude of the sums and their variances, it doesn't change the time complexity of the optimized function, which remains O(n). Memory usage is proportional to n, not m, so m doesn't directly impact memory unless n is extremely large, which is a separate concern.I think I've covered all the parts. Let me just recap:1. Optimized function computes S and Q in a single pass, O(n) time.2. Derived expected values and variances for S and Q based on uniform distribution.3. Discussed the impact of m on computational efficiency and memory, concluding it doesn't change the time complexity but affects the magnitude of sums and variances.</think>

🧠:<think>Okay, so Carlos is at the Sandia Resort & Casino, playing blackjack. He's using his math skills to try and up his chances. The problem has two parts. Let me tackle them one by one.First, part 1: Calculate the probability that Carlos is dealt exactly two blackjacks in his first four hands. They mention using the binomial distribution, so I remember that the binomial formula is used when there are a fixed number of independent trials, each with two possible outcomes: success or failure. In this case, each hand is a trial, getting a blackjack is a success, and not getting one is a failure.The binomial probability formula is:P(k) = C(n, k) * p^k * (1-p)^(n-k)Where:- P(k) is the probability of exactly k successes,- C(n, k) is the combination of n things taken k at a time,- p is the probability of success on a single trial,- n is the number of trials.Given:- n = 4 hands,- k = 2 blackjacks,- p = 4.83% which is 0.0483 in decimal.So, first, I need to compute C(4, 2). That's the number of ways to choose 2 successes out of 4 trials. The formula for combinations is C(n, k) = n! / (k! * (n - k)!).Calculating that:4! = 24,2! = 2,(4-2)! = 2! = 2.So, C(4, 2) = 24 / (2 * 2) = 24 / 4 = 6.Next, p^k is (0.0483)^2. Let me compute that:0.0483 * 0.0483. Let's see, 0.04 * 0.04 is 0.0016, and 0.0083 * 0.04 is about 0.000332, and similarly, the cross terms. Wait, maybe it's easier to just multiply it directly.0.0483 * 0.0483:First, 483 * 483. Let me compute that:483 * 483:Compute 400*400 = 160,000,400*80 = 32,000,400*3 = 1,200,80*400 = 32,000,80*80 = 6,400,80*3 = 240,3*400 = 1,200,3*80 = 240,3*3 = 9.Wait, that's too detailed. Maybe a better way is to compute 483^2:(400 + 80 + 3)^2 = 400^2 + 2*400*80 + 80^2 + 2*400*3 + 2*80*3 + 3^2.Compute each term:400^2 = 160,000,2*400*80 = 2*32,000 = 64,000,80^2 = 6,400,2*400*3 = 2*1,200 = 2,400,2*80*3 = 2*240 = 480,3^2 = 9.Add them all up:160,000 + 64,000 = 224,000,224,000 + 6,400 = 230,400,230,400 + 2,400 = 232,800,232,800 + 480 = 233,280,233,280 + 9 = 233,289.So, 483^2 = 233,289. Therefore, 0.0483^2 is 0.0483 * 0.0483 = 0.00233289.Wait, let me verify that. 0.0483 * 0.0483:Multiply 0.04 * 0.04 = 0.0016,0.04 * 0.0083 = 0.000332,0.0083 * 0.04 = 0.000332,0.0083 * 0.0083 ≈ 0.00006889.Adding these up:0.0016 + 0.000332 + 0.000332 + 0.00006889 ≈ 0.00233289.Yes, that matches. So, (0.0483)^2 ≈ 0.00233289.Next, (1 - p)^(n - k) = (1 - 0.0483)^(4 - 2) = (0.9517)^2.Compute 0.9517 squared:0.95 * 0.95 = 0.9025,0.0017 * 0.95 = 0.001615,0.95 * 0.0017 = 0.001615,0.0017 * 0.0017 ≈ 0.00000289.Wait, perhaps a better way is:(0.9517)^2 = (1 - 0.0483)^2 = 1 - 2*0.0483 + (0.0483)^2 ≈ 1 - 0.0966 + 0.00233289 ≈ 1 - 0.0966 = 0.9034 + 0.00233289 ≈ 0.90573289.Alternatively, compute 0.9517 * 0.9517:Let me compute 9517 * 9517:But that's too time-consuming. Alternatively, approximate:0.9517 * 0.9517 ≈ (0.95 + 0.0017)^2 = 0.95^2 + 2*0.95*0.0017 + (0.0017)^2 ≈ 0.9025 + 0.00323 + 0.00000289 ≈ 0.90573289.So, approximately 0.90573289.So, putting it all together:P(2) = C(4,2) * (0.0483)^2 * (0.9517)^2 ≈ 6 * 0.00233289 * 0.90573289.First, compute 6 * 0.00233289:6 * 0.00233289 = 0.01399734.Then, multiply that by 0.90573289:0.01399734 * 0.90573289 ≈ Let's compute that.0.01 * 0.90573289 = 0.0090573289,0.00399734 * 0.90573289 ≈ Approximately 0.00399734 * 0.9 ≈ 0.003597606, and 0.00399734 * 0.00573289 ≈ ~0.0000229.Adding up: 0.0090573289 + 0.003597606 ≈ 0.012654935, plus 0.0000229 ≈ 0.012677835.So, approximately 0.012677835.So, the probability is approximately 1.2677835%.Wait, let me check my calculations again because I might have messed up somewhere.Wait, 6 * 0.00233289 = 0.01399734.Then, 0.01399734 * 0.90573289.Let me compute 0.01399734 * 0.9 = 0.012597606,0.01399734 * 0.00573289 ≈ 0.0000800.So, total ≈ 0.012597606 + 0.00008 ≈ 0.0126776.So, approximately 0.0126776, which is 1.26776%.So, about 1.27%.Wait, but let me use a calculator approach:Compute 6 * 0.00233289 = 0.01399734.Then, 0.01399734 * 0.90573289.Let me compute 0.01399734 * 0.9 = 0.012597606,0.01399734 * 0.00573289 ≈ 0.0000800.So, total ≈ 0.012597606 + 0.00008 ≈ 0.0126776, which is 1.26776%.So, approximately 1.27%.Wait, but let me compute it more accurately:0.01399734 * 0.90573289.Let me write it as:0.01399734 * 0.90573289 = ?Let me compute 0.01399734 * 0.9 = 0.012597606,0.01399734 * 0.005 = 0.0000699867,0.01399734 * 0.00073289 ≈ 0.00001025.Adding these up:0.012597606 + 0.0000699867 ≈ 0.0126675927,Plus 0.00001025 ≈ 0.0126778427.So, approximately 0.01267784, which is 1.267784%.So, rounding to four decimal places, 1.2678%.But perhaps we can write it as 1.27% approximately.Alternatively, if we use more precise calculations, maybe it's 1.27%.But let me check with another method.Alternatively, using the exact formula:P(2) = 6 * (0.0483)^2 * (0.9517)^2.Compute each part:(0.0483)^2 = 0.00233289,(0.9517)^2 ≈ 0.90573289,So, 6 * 0.00233289 = 0.01399734,0.01399734 * 0.90573289 ≈ 0.0126778.So, yes, 0.0126778, which is 1.26778%.So, approximately 1.27%.Wait, but let me check if I made a mistake in calculating (0.9517)^2.Wait, 0.9517 * 0.9517:Compute 0.95 * 0.95 = 0.9025,0.95 * 0.0017 = 0.001615,0.0017 * 0.95 = 0.001615,0.0017 * 0.0017 ≈ 0.00000289.So, total is 0.9025 + 0.001615 + 0.001615 + 0.00000289 ≈ 0.9025 + 0.00323 + 0.00000289 ≈ 0.90573289.Yes, that's correct.So, the probability is approximately 1.27%.Alternatively, if I use more precise decimal places, maybe it's 1.2678%, which we can round to 1.27%.So, that's part 1.Now, part 2: Given that Carlos is dealt exactly two blackjacks in his first four hands, he decides to bet higher on his next hand. He bets 100, and the payout for a blackjack is 3:2. What is the expected value of his next hand? Assume the probability of being dealt a blackjack remains 4.83%, and the probability of winning a hand without being dealt a blackjack is 42.42% with a 1:1 payout.Okay, so expected value is calculated as the sum of (probability * payout) for each outcome.First, let's identify all possible outcomes and their probabilities and payouts.Outcomes:1. Blackjack: probability 4.83% (0.0483), payout 3:2 on a 100 bet. So, if he gets a blackjack, he wins 1.5 * 100 = 150.2. Win without blackjack: probability 42.42% (0.4242), payout 1:1, so he wins 100.3. Lose: probability is the remaining, which is 1 - 0.0483 - 0.4242 = 1 - 0.4725 = 0.5275, so 52.75%. In this case, he loses his 100 bet.Wait, but in blackjack, there's also the possibility of a push (tie), but the problem doesn't mention it. It just mentions winning without blackjack and losing. So, perhaps we can assume that the probabilities are only for win and loss, excluding pushes. But let me check.Wait, the problem says: "the probability of winning a hand without being dealt a blackjack is 42.42% with a 1:1 payout." So, that implies that the remaining probability is losing, which is 1 - 0.0483 - 0.4242 = 0.5275, as I calculated.So, the expected value (EV) is:EV = (Probability of Blackjack * Payout Blackjack) + (Probability of Win without Blackjack * Payout Win) + (Probability of Lose * Payout Lose)But the payouts are net gains, so:- Blackjack: +150,- Win: +100,- Lose: -100.Alternatively, sometimes expected value is calculated as the expected net gain, so we can compute it as:EV = (0.0483 * 150) + (0.4242 * 100) + (0.5275 * (-100)).Let me compute each term:First term: 0.0483 * 150 = 7.245.Second term: 0.4242 * 100 = 42.42.Third term: 0.5275 * (-100) = -52.75.Now, sum them up:7.245 + 42.42 = 49.665,49.665 - 52.75 = -3.085.So, the expected value is -3.085 per 100 bet.Alternatively, as a percentage, it's -3.085%.But let me check the calculations again.0.0483 * 150:0.04 * 150 = 6,0.0083 * 150 = 1.245,Total: 6 + 1.245 = 7.245.0.4242 * 100 = 42.42.0.5275 * (-100) = -52.75.Adding up: 7.245 + 42.42 = 49.665,49.665 - 52.75 = -3.085.Yes, that's correct.So, the expected value is -3.085, which is approximately -3.09.Alternatively, if we want to express it as a decimal, it's -0.03085, or -3.085%.But since the question asks for the expected value of his next hand, and he's betting 100, the expected value in dollars is -3.085, which we can round to -3.09.Alternatively, if we want to be precise, it's -3.085, which is approximately -3.09.Wait, but let me check if I interpreted the payouts correctly.When you get a blackjack, the payout is 3:2, meaning for every 2 you bet, you get 3 profit. So, for a 100 bet, the payout is (3/2)*100 = 150 profit, so total return is 250, but net gain is 150.Similarly, a win without blackjack pays 1:1, so you get your 100 back plus 100 profit, so net gain is 100.A loss means you lose your 100, so net gain is -100.So, yes, the calculations are correct.Alternatively, sometimes expected value is expressed as the expected net gain, which is what we calculated.So, the expected value is -3.085, approximately -3.09.Alternatively, if we want to express it as a percentage of the bet, it's -3.085%, but since the question asks for the expected value of his next hand, and he's betting 100, it's better to express it in dollars.So, the expected value is approximately -3.09.Alternatively, if we want to keep more decimal places, it's -3.085, which is approximately -3.09.So, that's part 2.Wait, but let me double-check the probabilities:Blackjack: 4.83%,Win without blackjack: 42.42%,Lose: 100% - 4.83% - 42.42% = 52.75%.Yes, that's correct.So, the expected value is:(0.0483 * 150) + (0.4242 * 100) + (0.5275 * (-100)) = 7.245 + 42.42 - 52.75 = -3.085.Yes, that's correct.So, summarizing:1. The probability of exactly two blackjacks in four hands is approximately 1.27%.2. The expected value of his next hand is approximately -3.09.Wait, but let me check if the problem mentions anything about multiple deck games or single deck. It says "standard single-deck blackjack game," but in part 2, it just says "the probability of being dealt a blackjack remains 4.83%." So, perhaps the composition of the deck changes, but since it's given that the probability remains 4.83%, we don't need to adjust for that.Also, in part 2, it's about the next hand, so it's a separate event, independent of the previous four hands, so the probability remains 4.83%.So, I think the calculations are correct.Final Answer1. The probability is boxed{1.27%}.2. The expected value is boxed{-3.09}.</think>