🧠:<think>Okay, so I have this problem about orbital mechanics and delta-v calculations. Let me try to figure it out step by step. I remember that a Hohmann transfer is the most efficient way to move a spacecraft from one circular orbit to another. It involves two engine burns: one to move from the initial orbit into an elliptical transfer orbit, and another to circularize at the target orbit. First, I need to find the delta-v required for each burn and then sum them up to get the total delta-v. The problem gives me the altitude of the low Earth orbit (LEO) as 300 km. The radius of the Earth is 6371 km, so the radius of the LEO is 6371 + 300 = 6671 km. Next, the target is a geostationary orbit (GEO). I think the standard altitude for GEO is about 42,164 km above the Earth's surface, so the radius would be 6371 + 42,164 = 48,535 km. Let me confirm that... Yeah, I think that's correct because GEO is much higher than LEO.Now, the gravitational parameter μ is given as 398,600 km³/s². I remember that the formula for the velocity of a circular orbit is v = sqrt(μ / r). So, I can calculate the velocity at LEO and at GEO.But wait, for the Hohmann transfer, I don't just need the velocities at LEO and GEO; I also need the velocities at the perigee and apogee of the transfer ellipse. The transfer orbit will have a perigee equal to the radius of LEO and an apogee equal to the radius of GEO.So, let me denote the radius of LEO as r1 = 6671 km and the radius of GEO as r2 = 48,535 km. The semi-major axis (a) of the transfer ellipse will be (r1 + r2)/2. Let me calculate that: (6671 + 48535)/2 = (55206)/2 = 27603 km.Now, the velocity at perigee (v1) of the transfer orbit can be found using the vis-viva equation: v = sqrt(μ * (2/r - 1/a)). Plugging in the numbers: sqrt(398600 * (2/6671 - 1/27603)). Let me compute that step by step.First, compute 2/6671: 2 divided by 6671 is approximately 0.0002998. Then, compute 1/27603: that's approximately 0.0000362. Subtracting these gives 0.0002998 - 0.0000362 = 0.0002636. Multiply by μ: 398600 * 0.0002636 ≈ 105.1 km²/s². Taking the square root gives sqrt(105.1) ≈ 10.25 km/s.Wait, that doesn't seem right. Let me double-check my calculations. Maybe I messed up the units or the arithmetic. Let's see:2/6671 is approximately 0.0002998 km⁻¹, and 1/27603 is approximately 0.0000362 km⁻¹. So, 2/6671 - 1/27603 ≈ 0.0002998 - 0.0000362 = 0.0002636 km⁻¹. Then, μ is 398600 km³/s². So, 398600 * 0.0002636 ≈ 105.1 km²/s². The square root of 105.1 is indeed about 10.25 km/s.But wait, the velocity in LEO should be higher than in GEO because lower orbits have higher velocities. Let me check the velocity at LEO using v = sqrt(μ / r1). So, sqrt(398600 / 6671). Let's compute that: 398600 / 6671 ≈ 59.75 km²/s². Square root is about 7.73 km/s. Hmm, that's lower than the transfer orbit's perigee velocity. That can't be right because the transfer orbit's perigee velocity should be higher than the LEO velocity to escape it.Wait, no, actually, the transfer orbit's perigee is at LEO, so the velocity there should be higher than the circular orbit velocity at LEO. Because to enter the transfer orbit, you need to add velocity (delta-v) to the spacecraft. So, the transfer orbit's perigee velocity is higher than LEO's circular velocity.So, my calculation of 10.25 km/s for the transfer orbit's perigee velocity is correct because it's higher than 7.73 km/s. So, the delta-v required for the first burn is the difference between the transfer orbit's perigee velocity and the LEO velocity.So, delta-v1 = v_transfer_perigee - v_LEO = 10.25 - 7.73 ≈ 2.52 km/s.Now, moving on to the second burn. At apogee, the spacecraft needs to circularize into GEO. So, the velocity at apogee of the transfer orbit is v_transfer_apogee = sqrt(μ * (2/r2 - 1/a)). Let's compute that.First, 2/r2 = 2 / 48535 ≈ 0.0000412 km⁻¹. Then, 1/a = 1 / 27603 ≈ 0.0000362 km⁻¹. So, 2/r2 - 1/a ≈ 0.0000412 - 0.0000362 = 0.000005 km⁻¹. Multiply by μ: 398600 * 0.000005 ≈ 1.993 km²/s². Square root is about 1.412 km/s.Wait, that seems too low. The circular velocity at GEO should be lower than the transfer orbit's apogee velocity because the transfer orbit's apogee is at GEO, so the spacecraft needs to add velocity to circularize. Wait, no, actually, the transfer orbit's apogee velocity is lower than the circular velocity at GEO because the spacecraft is moving slower at apogee. So, to circularize, it needs to add velocity.Wait, let me think again. The transfer orbit's apogee is at r2, so the velocity there is lower than the circular velocity at r2. Therefore, the delta-v required is the difference between the circular velocity at GEO and the transfer orbit's apogee velocity.So, first, compute the circular velocity at GEO: v_GEO = sqrt(μ / r2) = sqrt(398600 / 48535). Let's compute that: 398600 / 48535 ≈ 8.21 km²/s². Square root is about 2.866 km/s.Wait, that can't be right because I thought GEO velocity is around 3.07 km/s. Let me check my calculation: 398600 divided by 48535. Let me compute 48535 * 8.21 ≈ 48535 * 8 = 388,280 and 48535 * 0.21 ≈ 10,192.35, so total ≈ 398,472.35, which is close to 398,600. So, sqrt(8.21) ≈ 2.866 km/s. Hmm, that seems lower than I remember. Maybe I'm missing something.Wait, no, I think I made a mistake in the calculation of v_transfer_apogee. Let me recalculate that. The vis-viva equation at apogee is v = sqrt(μ * (2/r2 - 1/a)). So, 2/r2 is 2 / 48535 ≈ 0.0000412 km⁻¹, and 1/a is 1 / 27603 ≈ 0.0000362 km⁻¹. So, 2/r2 - 1/a ≈ 0.0000412 - 0.0000362 = 0.000005 km⁻¹. Multiply by μ: 398600 * 0.000005 = 1.993 km²/s². Square root is about 1.412 km/s.Wait, that still seems too low. Maybe I made a mistake in the formula. Let me check the vis-viva equation again. It's v = sqrt(μ * (2/r - 1/a)). So, at apogee, r = r2, so it's correct. Hmm, but the circular velocity at r2 is sqrt(μ / r2) ≈ 2.866 km/s, as I calculated earlier. So, the transfer orbit's apogee velocity is 1.412 km/s, which is much lower. That seems impossible because the spacecraft would need to add a lot of velocity to reach the circular orbit.Wait, that can't be right. Let me check my numbers again. Maybe I messed up the semi-major axis. The semi-major axis a is (r1 + r2)/2 = (6671 + 48535)/2 = 55206/2 = 27603 km. That's correct.Wait, but 27603 km is the semi-major axis, which is correct because it's the average of r1 and r2. So, the vis-viva equation should be correct. Hmm, maybe I made a mistake in the arithmetic. Let me compute 2/r2 - 1/a again.2/r2 = 2 / 48535 ≈ 0.0000412 km⁻¹.1/a = 1 / 27603 ≈ 0.0000362 km⁻¹.So, 0.0000412 - 0.0000362 = 0.000005 km⁻¹.Multiply by μ: 398600 * 0.000005 = 1.993 km²/s².Square root is sqrt(1.993) ≈ 1.412 km/s.Wait, that's correct. So, the transfer orbit's apogee velocity is 1.412 km/s, and the circular velocity at GEO is 2.866 km/s. Therefore, the delta-v required for the second burn is v_GEO - v_transfer_apogee = 2.866 - 1.412 ≈ 1.454 km/s.So, the total delta-v is delta-v1 + delta-v2 = 2.52 + 1.454 ≈ 3.974 km/s.Wait, that seems a bit high. I thought Hohmann transfer delta-v for LEO to GEO is around 3.27 km/s. Maybe I made a mistake somewhere.Let me check the initial velocity in LEO. I calculated v_LEO = sqrt(μ / r1) = sqrt(398600 / 6671) ≈ sqrt(59.75) ≈ 7.73 km/s. That seems correct.Then, transfer perigee velocity: sqrt(μ * (2/r1 - 1/a)) = sqrt(398600 * (2/6671 - 1/27603)).Let me compute 2/6671 ≈ 0.0002998, 1/27603 ≈ 0.0000362. So, 0.0002998 - 0.0000362 = 0.0002636. Multiply by 398600: 398600 * 0.0002636 ≈ 105.1 km²/s². Square root is about 10.25 km/s. So, delta-v1 = 10.25 - 7.73 ≈ 2.52 km/s.Then, transfer apogee velocity: sqrt(μ * (2/r2 - 1/a)) = sqrt(398600 * (2/48535 - 1/27603)).2/48535 ≈ 0.0000412, 1/27603 ≈ 0.0000362. So, 0.0000412 - 0.0000362 = 0.000005. Multiply by 398600: 1.993 km²/s². Square root is about 1.412 km/s.Circular velocity at GEO: sqrt(μ / r2) = sqrt(398600 / 48535) ≈ sqrt(8.21) ≈ 2.866 km/s.So, delta-v2 = 2.866 - 1.412 ≈ 1.454 km/s.Total delta-v ≈ 2.52 + 1.454 ≈ 3.974 km/s.Hmm, that's higher than I expected. Maybe I made a mistake in the calculation of the transfer orbit's apogee velocity. Let me try another approach.Alternatively, I can use the formula for the delta-v for Hohmann transfer:delta-v_total = sqrt(μ / r1) * (sqrt(2*r1 / (r1 + r2)) - 1) + sqrt(μ / r2) * (1 - sqrt(2*r2 / (r1 + r2)))Wait, is that correct? Let me think. The delta-v for the first burn is the difference between the transfer orbit's perigee velocity and the LEO velocity. The transfer perigee velocity is sqrt(μ * (2/r1 - 1/a)), and a = (r1 + r2)/2. So, let me express it in terms of r1 and r2.Alternatively, I can use the formula:delta-v1 = sqrt(μ / r1) * (sqrt(2*r1 / (r1 + r2)) - 1)Similarly, delta-v2 = sqrt(μ / r2) * (1 - sqrt(2*r2 / (r1 + r2)))Let me try that.First, compute sqrt(2*r1 / (r1 + r2)):r1 = 6671 km, r2 = 48535 km.2*r1 = 13342 km.r1 + r2 = 6671 + 48535 = 55206 km.So, 2*r1 / (r1 + r2) = 13342 / 55206 ≈ 0.2416.sqrt(0.2416) ≈ 0.4916.So, delta-v1 = sqrt(μ / r1) * (0.4916 - 1) = sqrt(398600 / 6671) * (-0.5084). Wait, that can't be right because delta-v1 should be positive. Maybe I have the formula wrong.Wait, no, the formula should be delta-v1 = sqrt(μ / r1) * (sqrt(2*r1 / (r1 + r2)) - 1). But sqrt(2*r1 / (r1 + r2)) is less than 1, so the term in the parentheses is negative, which would give a negative delta-v, which doesn't make sense. I must have the formula wrong.Wait, perhaps it's the other way around. Maybe delta-v1 = sqrt(μ / r1) * (sqrt(2*r1 / (r1 + r2)) - 1). But that would be negative. Alternatively, maybe it's delta-v1 = sqrt(μ / r1) * (sqrt(2*r1 / (r1 + r2)) - 1), but taking the absolute value.Wait, no, let me think again. The transfer orbit's perigee velocity is higher than the LEO velocity, so delta-v1 is positive. So, perhaps the formula is delta-v1 = sqrt(μ / r1) * (sqrt(2*r1 / (r1 + r2)) - 1). But since sqrt(2*r1 / (r1 + r2)) is less than 1, this would give a negative value, which is not correct.Wait, maybe I have the formula inverted. Let me check the correct formula for Hohmann transfer delta-v.I think the correct formula is:delta-v1 = sqrt(μ / r1) * (sqrt(2*r2 / (r1 + r2)) - 1)Wait, let me check that. If I compute sqrt(2*r2 / (r1 + r2)), that would be sqrt(2*48535 / 55206) ≈ sqrt(97070 / 55206) ≈ sqrt(1.758) ≈ 1.326.So, delta-v1 = sqrt(μ / r1) * (1.326 - 1) ≈ sqrt(59.75) * 0.326 ≈ 7.73 * 0.326 ≈ 2.52 km/s. That matches my earlier calculation.Similarly, delta-v2 = sqrt(μ / r2) * (1 - sqrt(2*r1 / (r1 + r2))) ≈ sqrt(8.21) * (1 - 0.4916) ≈ 2.866 * 0.5084 ≈ 1.454 km/s.So, total delta-v ≈ 2.52 + 1.454 ≈ 3.974 km/s.Hmm, that's consistent with my earlier result, but I thought the standard Hohmann transfer from LEO to GEO is around 3.27 km/s. Maybe I'm missing something.Wait, perhaps I made a mistake in the units. Let me check. The gravitational parameter μ is given as 398,600 km³/s². Is that correct? Yes, because μ for Earth is approximately 398600 km³/s².Wait, but sometimes people use meters and seconds, so μ is 3.986e5 km³/s², which is the same as 3.986e14 m³/s². But in this case, we're using km, so it's correct.Wait, another thought: maybe I made a mistake in the calculation of the circular velocity at GEO. Let me recalculate that.v_GEO = sqrt(μ / r2) = sqrt(398600 / 48535). Let me compute 398600 divided by 48535.48535 * 8 = 388,280.48535 * 8.2 = 48535*8 + 48535*0.2 = 388,280 + 9,707 = 397,987.That's very close to 398,600. So, 8.2 * 48535 ≈ 397,987, which is about 613 less than 398,600. So, 398,600 / 48535 ≈ 8.2 + (613 / 48535) ≈ 8.2 + 0.0126 ≈ 8.2126 km²/s².So, sqrt(8.2126) ≈ 2.866 km/s. That's correct.Wait, but I remember that the circular velocity at GEO is about 3.07 km/s. Hmm, maybe I'm confusing it with something else. Let me check online... Oh, no, I can't access external resources, but I think I might have made a mistake in the radius of GEO.Wait, the standard GEO altitude is 42,164 km above Earth's surface, so radius is 6371 + 42164 = 48535 km, which is what I used. So, the calculation seems correct. Maybe the discrepancy is because I'm using a different value for μ or Earth's radius.Wait, Earth's radius is given as 6371 km, which is correct. μ is 398,600 km³/s², which is correct. So, maybe the standard Hohmann transfer delta-v is indeed around 3.97 km/s, and my initial thought of 3.27 km/s was incorrect.Alternatively, perhaps I'm confusing it with the delta-v from LEO to a 400 km transfer orbit or something else.Wait, let me check the delta-v for a Hohmann transfer from LEO (300 km) to GEO (42,164 km). Using the standard formula, the total delta-v is approximately 3.97 km/s, which seems high, but maybe that's correct.Alternatively, maybe I made a mistake in the calculation of the transfer orbit's perigee velocity. Let me recalculate that.v_transfer_perigee = sqrt(μ * (2/r1 - 1/a)).r1 = 6671 km, a = 27603 km.So, 2/r1 = 2 / 6671 ≈ 0.0002998 km⁻¹.1/a = 1 / 27603 ≈ 0.0000362 km⁻¹.So, 2/r1 - 1/a ≈ 0.0002998 - 0.0000362 = 0.0002636 km⁻¹.Multiply by μ: 398600 * 0.0002636 ≈ 105.1 km²/s².Square root is sqrt(105.1) ≈ 10.25 km/s.Yes, that's correct. So, delta-v1 = 10.25 - 7.73 ≈ 2.52 km/s.Similarly, v_transfer_apogee = sqrt(μ * (2/r2 - 1/a)).2/r2 = 2 / 48535 ≈ 0.0000412 km⁻¹.1/a = 0.0000362 km⁻¹.So, 2/r2 - 1/a ≈ 0.0000412 - 0.0000362 = 0.000005 km⁻¹.Multiply by μ: 398600 * 0.000005 = 1.993 km²/s².Square root is sqrt(1.993) ≈ 1.412 km/s.v_GEO = sqrt(μ / r2) ≈ 2.866 km/s.So, delta-v2 = 2.866 - 1.412 ≈ 1.454 km/s.Total delta-v ≈ 2.52 + 1.454 ≈ 3.974 km/s.Hmm, that seems correct. Maybe my initial thought of 3.27 km/s was wrong. So, I think the answer is approximately 3.97 km/s.Now, moving on to part (b), which is about energy efficiency using the rocket equation. The specific impulse I_sp is 450 seconds, and the initial mass is 2000 kg. I need to find the final mass after the Hohmann transfer.The rocket equation is Δv = I_sp * g0 * ln(m_initial / m_final), where g0 is standard gravity, approximately 9.80665 m/s².Wait, but Δv is in km/s, and I_sp is in seconds, so I need to make sure the units are consistent.First, let's convert Δv to m/s: 3.974 km/s = 3974 m/s.I_sp is 450 seconds, so the effective exhaust velocity c = I_sp * g0 = 450 * 9.80665 ≈ 4412.99 m/s.Now, the rocket equation is Δv = c * ln(m_initial / m_final).So, ln(m_initial / m_final) = Δv / c ≈ 3974 / 4412.99 ≈ 0.899.Therefore, m_initial / m_final = e^0.899 ≈ 2.458.So, m_final = m_initial / 2.458 ≈ 2000 / 2.458 ≈ 813.6 kg.So, the final mass is approximately 813.6 kg.Wait, let me double-check the calculations.Δv = 3.974 km/s = 3974 m/s.I_sp = 450 s, so c = 450 * 9.80665 ≈ 4412.99 m/s.Δv / c ≈ 3974 / 4412.99 ≈ 0.899.ln(m_initial / m_final) = 0.899.So, m_initial / m_final = e^0.899 ≈ e^0.9 ≈ 2.4596.So, m_final ≈ 2000 / 2.4596 ≈ 813.3 kg.Yes, that seems correct.So, the final mass is approximately 813 kg.But wait, let me make sure I didn't make a mistake in the rocket equation. The formula is Δv = c * ln(m_initial / m_final), so solving for m_final gives m_final = m_initial * exp(-Δv / c).Wait, no, because Δv = c * ln(m_initial / m_final), so rearranging gives ln(m_initial / m_final) = Δv / c, so m_initial / m_final = exp(Δv / c), so m_final = m_initial / exp(Δv / c).Yes, that's correct. So, m_final = 2000 / exp(0.899) ≈ 2000 / 2.458 ≈ 813.6 kg.So, the final mass is approximately 813.6 kg.I think that's it.</think>

🧠:<think>Alright, so I have these two probability questions to solve. Let me take them one by one.Starting with the first question about the Markov process. Hmm, okay, so the retired football player's prediction success rate is modeled as a Markov chain with three states: low, medium, and high. The transition probabilities are given, and we need to find the probability that after 3 weeks, he's in the high state, given that he started in the medium state.First, I need to recall how Markov chains work. A Markov chain is a system that moves from one state to another based on certain transition probabilities, and the next state depends only on the current state. So, in this case, each week, his prediction accuracy can change based on these probabilities.The transition probabilities are given as:- From low to medium: 0.3- From low to high: 0.1- From medium to low: 0.4- From medium to high: 0.3- From high to low: 0.2- From high to medium: 0.5Wait, hold on. For a Markov chain, the transition probabilities from each state should sum to 1. Let me check that.For the low state: transitions to medium (0.3) and high (0.1). That's 0.4. Wait, that's only 0.4. So, does that mean the probability of staying in low is 0.6? Because 1 - 0.3 - 0.1 = 0.6. Similarly, for medium: transitions to low (0.4) and high (0.3). So, 0.4 + 0.3 = 0.7. Therefore, the probability of staying in medium is 0.3? Wait, no, 1 - 0.4 - 0.3 = 0.3. So, from medium, he can stay in medium with probability 0.3.Similarly, for high: transitions to low (0.2) and medium (0.5). So, 0.2 + 0.5 = 0.7. Therefore, the probability of staying in high is 0.3? Wait, 1 - 0.2 - 0.5 = 0.3. So, yes, from high, he can stay in high with probability 0.3.So, let me write down the transition matrix. Let me denote the states as Low (L), Medium (M), High (H). So, the transition matrix P would be:From L: [0.6, 0.3, 0.1]From M: [0.4, 0.3, 0.3]From H: [0.2, 0.5, 0.3]Wait, let me confirm:From Low:- To Low: 1 - 0.3 - 0.1 = 0.6- To Medium: 0.3- To High: 0.1From Medium:- To Low: 0.4- To Medium: 0.3- To High: 0.3From High:- To Low: 0.2- To Medium: 0.5- To High: 0.3Yes, that seems correct.So, the transition matrix P is:[0.6, 0.3, 0.1][0.4, 0.3, 0.3][0.2, 0.5, 0.3]Now, the initial state is Medium. So, the initial state vector S0 is [0, 1, 0], since he starts in Medium.We need to find the state vector after 3 weeks, which is S3 = S0 * P^3.So, I need to compute P^3 and then multiply it by S0.Alternatively, since S0 is [0,1,0], multiplying by P^3 will just give us the second row of P^3.So, perhaps it's easier to compute P^3 and then take the second row, third column element, which would be the probability of being in High after 3 weeks.But computing P^3 manually might be a bit tedious. Let me see if I can compute it step by step.First, let's compute P^2 = P * P.Let me denote P as:Row 1: [0.6, 0.3, 0.1]Row 2: [0.4, 0.3, 0.3]Row 3: [0.2, 0.5, 0.3]So, to compute P^2, I need to perform matrix multiplication.Let me compute each element of P^2.First, element (1,1): Row 1 of P multiplied by Column 1 of P.0.6*0.6 + 0.3*0.4 + 0.1*0.2 = 0.36 + 0.12 + 0.02 = 0.5Element (1,2): Row 1 * Column 2.0.6*0.3 + 0.3*0.3 + 0.1*0.5 = 0.18 + 0.09 + 0.05 = 0.32Element (1,3): Row 1 * Column 3.0.6*0.1 + 0.3*0.3 + 0.1*0.3 = 0.06 + 0.09 + 0.03 = 0.18So, Row 1 of P^2: [0.5, 0.32, 0.18]Now, Row 2 of P^2:Element (2,1): Row 2 * Column 1.0.4*0.6 + 0.3*0.4 + 0.3*0.2 = 0.24 + 0.12 + 0.06 = 0.42Element (2,2): Row 2 * Column 2.0.4*0.3 + 0.3*0.3 + 0.3*0.5 = 0.12 + 0.09 + 0.15 = 0.36Element (2,3): Row 2 * Column 3.0.4*0.1 + 0.3*0.3 + 0.3*0.3 = 0.04 + 0.09 + 0.09 = 0.22So, Row 2 of P^2: [0.42, 0.36, 0.22]Now, Row 3 of P^2:Element (3,1): Row 3 * Column 1.0.2*0.6 + 0.5*0.4 + 0.3*0.2 = 0.12 + 0.20 + 0.06 = 0.38Element (3,2): Row 3 * Column 2.0.2*0.3 + 0.5*0.3 + 0.3*0.5 = 0.06 + 0.15 + 0.15 = 0.36Element (3,3): Row 3 * Column 3.0.2*0.1 + 0.5*0.3 + 0.3*0.3 = 0.02 + 0.15 + 0.09 = 0.26So, Row 3 of P^2: [0.38, 0.36, 0.26]So, P^2 is:[0.5, 0.32, 0.18][0.42, 0.36, 0.22][0.38, 0.36, 0.26]Now, let's compute P^3 = P^2 * P.Again, let's compute each element.First, Row 1 of P^3:Element (1,1): Row 1 of P^2 * Column 1 of P.0.5*0.6 + 0.32*0.4 + 0.18*0.2 = 0.3 + 0.128 + 0.036 = 0.464Element (1,2): Row 1 * Column 2.0.5*0.3 + 0.32*0.3 + 0.18*0.5 = 0.15 + 0.096 + 0.09 = 0.336Element (1,3): Row 1 * Column 3.0.5*0.1 + 0.32*0.3 + 0.18*0.3 = 0.05 + 0.096 + 0.054 = 0.199 + 0.054? Wait, 0.05 + 0.096 is 0.146, plus 0.054 is 0.200.Wait, 0.5*0.1 = 0.05, 0.32*0.3 = 0.096, 0.18*0.3 = 0.054. So, 0.05 + 0.096 = 0.146 + 0.054 = 0.200.So, Row 1 of P^3: [0.464, 0.336, 0.200]Now, Row 2 of P^3:Element (2,1): Row 2 of P^2 * Column 1 of P.0.42*0.6 + 0.36*0.4 + 0.22*0.2 = 0.252 + 0.144 + 0.044 = 0.44Element (2,2): Row 2 * Column 2.0.42*0.3 + 0.36*0.3 + 0.22*0.5 = 0.126 + 0.108 + 0.11 = 0.344Element (2,3): Row 2 * Column 3.0.42*0.1 + 0.36*0.3 + 0.22*0.3 = 0.042 + 0.108 + 0.066 = 0.216So, Row 2 of P^3: [0.44, 0.344, 0.216]Row 3 of P^3:Element (3,1): Row 3 of P^2 * Column 1 of P.0.38*0.6 + 0.36*0.4 + 0.26*0.2 = 0.228 + 0.144 + 0.052 = 0.424Element (3,2): Row 3 * Column 2.0.38*0.3 + 0.36*0.3 + 0.26*0.5 = 0.114 + 0.108 + 0.13 = 0.352Element (3,3): Row 3 * Column 3.0.38*0.1 + 0.36*0.3 + 0.26*0.3 = 0.038 + 0.108 + 0.078 = 0.224So, Row 3 of P^3: [0.424, 0.352, 0.224]Therefore, P^3 is:[0.464, 0.336, 0.200][0.44, 0.344, 0.216][0.424, 0.352, 0.224]Now, since we started in Medium, which is the second state, we look at the second row of P^3. The third element of that row is the probability of being in High after 3 weeks.So, the probability is 0.216.Wait, let me double-check my calculations because 0.216 seems a bit low. Let me verify the multiplication steps.Starting with P^2:Row 2 of P^2 is [0.42, 0.36, 0.22]Multiplying by P:Element (2,1): 0.42*0.6 + 0.36*0.4 + 0.22*0.20.42*0.6 = 0.2520.36*0.4 = 0.1440.22*0.2 = 0.044Total: 0.252 + 0.144 = 0.396 + 0.044 = 0.44. Correct.Element (2,2): 0.42*0.3 + 0.36*0.3 + 0.22*0.50.42*0.3 = 0.1260.36*0.3 = 0.1080.22*0.5 = 0.11Total: 0.126 + 0.108 = 0.234 + 0.11 = 0.344. Correct.Element (2,3): 0.42*0.1 + 0.36*0.3 + 0.22*0.30.42*0.1 = 0.0420.36*0.3 = 0.1080.22*0.3 = 0.066Total: 0.042 + 0.108 = 0.15 + 0.066 = 0.216. Correct.So, yes, the probability is 0.216.Alternatively, maybe I can compute it using another method, like step-by-step transitions.Starting from Medium.After 1 week: possible states are Low, Medium, High.From Medium, the probabilities are:To Low: 0.4To Medium: 0.3To High: 0.3So, after 1 week, the state vector is [0.4, 0.3, 0.3]After 2 weeks: we need to transition from each of these.From Low: [0.6, 0.3, 0.1]From Medium: [0.4, 0.3, 0.3]From High: [0.2, 0.5, 0.3]So, the state vector after 2 weeks is:Low: 0.4*0.6 + 0.3*0.4 + 0.3*0.2= 0.24 + 0.12 + 0.06 = 0.42Medium: 0.4*0.3 + 0.3*0.3 + 0.3*0.5= 0.12 + 0.09 + 0.15 = 0.36High: 0.4*0.1 + 0.3*0.3 + 0.3*0.3= 0.04 + 0.09 + 0.09 = 0.22So, after 2 weeks: [0.42, 0.36, 0.22]Then, after 3 weeks:From Low: [0.6, 0.3, 0.1]From Medium: [0.4, 0.3, 0.3]From High: [0.2, 0.5, 0.3]So, state vector after 3 weeks:Low: 0.42*0.6 + 0.36*0.4 + 0.22*0.2= 0.252 + 0.144 + 0.044 = 0.44Medium: 0.42*0.3 + 0.36*0.3 + 0.22*0.5= 0.126 + 0.108 + 0.11 = 0.344High: 0.42*0.1 + 0.36*0.3 + 0.22*0.3= 0.042 + 0.108 + 0.066 = 0.216Yes, same result. So, the probability is 0.216.Okay, so that's the first question.Now, moving on to the second question. It's about a Poisson process. The analysis accuracy is modeled by a continuous-time Poisson process with rate λ. The expected number of correct analyses per month is 15. We need to find the probability that he makes exactly 50 correct analyses in a 3-month period.First, Poisson processes have the property that the number of events in disjoint intervals are independent, and the number of events in any interval follows a Poisson distribution with parameter equal to λ multiplied by the length of the interval.Given that the expected number per month is 15, so λ = 15 per month.But wait, actually, in a Poisson process, the rate λ is the expected number per unit time. So, if the expected number per month is 15, then λ = 15 per month.So, over 3 months, the expected number is 3*15 = 45.Wait, but the question is about exactly 50 correct analyses in 3 months. So, we need to compute P(X = 50), where X ~ Poisson(λ*t) = Poisson(15*3) = Poisson(45).So, the probability mass function of Poisson is:P(X = k) = (λ^k * e^{-λ}) / k!So, plugging in λ = 45 and k = 50.So, P(X = 50) = (45^50 * e^{-45}) / 50!But calculating this directly is computationally intensive because 45^50 is a huge number, and 50! is also a huge number. So, we might need to use approximations or logarithms to compute this.Alternatively, we can use the normal approximation to the Poisson distribution since λ is large (45). For large λ, the Poisson distribution can be approximated by a normal distribution with mean μ = λ and variance σ^2 = λ.So, μ = 45, σ = sqrt(45) ≈ 6.7082.We need to find P(X = 50). But since the normal distribution is continuous, we can use the continuity correction. So, P(X = 50) ≈ P(49.5 < X < 50.5) in the normal approximation.So, we can compute the Z-scores for 49.5 and 50.5.Z1 = (49.5 - 45) / 6.7082 ≈ 4.5 / 6.7082 ≈ 0.671Z2 = (50.5 - 45) / 6.7082 ≈ 5.5 / 6.7082 ≈ 0.820Now, we need to find the area under the standard normal curve between Z1 and Z2.Looking up Z1 = 0.671, the cumulative probability is approximately 0.7486.Looking up Z2 = 0.820, the cumulative probability is approximately 0.7939.So, the area between them is 0.7939 - 0.7486 = 0.0453.So, approximately 4.53%.But wait, let me check the exact value using the Poisson formula. However, calculating 45^50 / 50! is not feasible by hand, but maybe we can use Stirling's approximation for factorials.Stirling's formula: n! ≈ sqrt(2πn) (n / e)^nSo, let's approximate 50! and 45^50.First, ln(P(X=50)) = 50 ln(45) - 45 - ln(50!) + 50 ln(1) - 45 (Wait, no, the formula is:ln(P) = 50 ln(45) - 45 - ln(50!) + 45 ln(1) ?Wait, no, the Poisson PMF is:ln(P) = ln(45^50) - ln(50!) - 45= 50 ln(45) - ln(50!) - 45Using Stirling's approximation for ln(50!):ln(50!) ≈ 50 ln(50) - 50 + 0.5 ln(2π*50)= 50 ln(50) - 50 + 0.5 ln(100π)Compute each term:50 ln(45) ≈ 50 * 3.8067 ≈ 190.335ln(50!) ≈ 50 ln(50) - 50 + 0.5 ln(100π)50 ln(50) ≈ 50 * 3.9120 ≈ 195.60.5 ln(100π) ≈ 0.5 * ln(314.16) ≈ 0.5 * 5.75 ≈ 2.875So, ln(50!) ≈ 195.6 - 50 + 2.875 ≈ 148.475So, ln(P) ≈ 190.335 - 148.475 - 45 ≈ 190.335 - 193.475 ≈ -3.14Therefore, P ≈ e^{-3.14} ≈ 0.0432Which is about 4.32%, which is close to the normal approximation of 4.53%. So, around 4.3% to 4.5%.But let me check if I can get a more accurate value.Alternatively, using the Poisson formula with exact computation is difficult, but maybe using the relationship between Poisson and normal.Alternatively, using the exact value, perhaps using the formula:P(X = k) = e^{-λ} * (λ^k) / k!But again, without a calculator, it's hard. Alternatively, using the ratio of consecutive probabilities.But perhaps the normal approximation is sufficient here, giving around 4.5%.Alternatively, using the Poisson distribution's properties, we can note that the mode is around λ = 45, so 50 is 5 units away, which is about 0.75σ away (since σ ≈ 6.7). So, the probability is not too low, but not too high either.Alternatively, using the exact value, perhaps using the formula:ln(P) = 50 ln(45) - 45 - ln(50!) ≈ 50*3.8067 - 45 - (50*3.9120 - 50 + 0.5*ln(100π))Wait, let me recalculate:50 ln(45) ≈ 50 * 3.80667 ≈ 190.3335ln(50!) ≈ 50 ln(50) - 50 + 0.5 ln(2π*50) ≈ 50*3.9120 - 50 + 0.5*ln(314.16) ≈ 195.6 - 50 + 0.5*5.75 ≈ 148.475So, ln(P) ≈ 190.3335 - 148.475 - 45 ≈ 190.3335 - 193.475 ≈ -3.1415So, P ≈ e^{-3.1415} ≈ 0.0432, which is approximately 4.32%.So, rounding to two decimal places, about 4.32%.But let me see if I can get a better approximation.Alternatively, using the Poisson PMF formula with exact computation, but I think it's beyond manual calculation.Alternatively, using the fact that for Poisson, the PMF can be approximated using the normal distribution with continuity correction, which we did earlier, giving about 4.53%.So, perhaps the answer is approximately 4.3% or 4.5%.But let me see if I can compute it more accurately.Alternatively, using the exact formula:P(X=50) = e^{-45} * (45^50) / 50!But to compute this, we can use logarithms.Compute ln(P) = 50 ln(45) - 45 - ln(50!) ≈ 50*3.80667 - 45 - (50*3.9120 - 50 + 0.5*ln(100π))As before, 50*3.80667 ≈ 190.3335ln(50!) ≈ 148.475So, ln(P) ≈ 190.3335 - 45 - 148.475 ≈ 190.3335 - 193.475 ≈ -3.1415So, P ≈ e^{-3.1415} ≈ 0.0432So, approximately 4.32%.Alternatively, using a calculator, e^{-3.1415} ≈ 0.0432.So, the probability is approximately 4.32%.Alternatively, using the normal approximation with continuity correction, we got about 4.53%.So, the exact value is approximately 4.32%, and the normal approximation gives 4.53%.Given that, perhaps the answer is approximately 4.3%.But let me check if I can find a better approximation.Alternatively, using the Poisson distribution's recursive formula:P(k) = P(k-1) * (λ / k)But starting from P(0) = e^{-λ}, which is e^{-45}, which is extremely small, so not helpful.Alternatively, using the ratio method.But I think the best we can do is approximate it as around 4.3%.Alternatively, using the exact value, perhaps using a calculator, but since I don't have one, I'll stick with the approximation.So, the probability is approximately 4.3%.But wait, let me think again. The normal approximation gave 4.53%, and the exact value is 4.32%, so they are quite close.So, perhaps the answer is approximately 4.3%.Alternatively, if we use the Poisson formula with Stirling's approximation, we get 4.32%, which is about 4.3%.So, I think that's the answer.But let me check if I made any mistakes in the calculations.Wait, in the normal approximation, I calculated Z1 = (49.5 - 45)/6.7082 ≈ 0.671Z2 = (50.5 - 45)/6.7082 ≈ 0.820Then, the cumulative probabilities:For Z=0.67, the cumulative is about 0.7486For Z=0.82, it's about 0.7939Difference is 0.0453, which is 4.53%.But the exact value is 4.32%, so the normal approximation is a bit higher.Alternatively, using the Poisson distribution's exact value, it's about 4.32%.So, perhaps the answer is approximately 4.3%.Alternatively, if we use more precise Z-values.Looking up Z=0.671: more precisely, Z=0.67 corresponds to 0.7486, and Z=0.671 is slightly higher, maybe 0.7488.Similarly, Z=0.82: 0.7939.So, the difference is still about 0.0451, which is 4.51%.So, the normal approximation gives around 4.5%, while the exact value is about 4.3%.So, perhaps the answer is approximately 4.3%.Alternatively, if we use the exact value, it's 4.32%, which is approximately 4.3%.So, I think that's the answer.Final Answer1. The probability is boxed{0.216}.2. The probability is approximately boxed{0.043}.</think>

🧠:<think>Okay, so I have two problems to solve here. The first one involves linear algebra, specifically matrix multiplication, and the second one is about probability theory using the Poisson distribution. Let me tackle them one by one.Starting with the first problem: I need to calculate the resulting vector ( mathbf{w} ) after applying the matrix ( A ) to the vector ( mathbf{v} ). The matrix ( A ) is given as:[ A = begin{bmatrix} 2 & 1 & 3 4 & 5 & 6 7 & 8 & 9 end{bmatrix} ]And the vector ( mathbf{v} ) is:[ mathbf{v} = begin{bmatrix} 2 1 3 end{bmatrix} ]So, I need to compute ( mathbf{w} = A mathbf{v} ). Matrix multiplication works by taking the dot product of each row of matrix ( A ) with the vector ( mathbf{v} ). Let me recall how that works.For the first element of ( mathbf{w} ), I take the first row of ( A ) and multiply each element by the corresponding element in ( mathbf{v} ), then sum them up. Similarly, for the second element, I take the second row of ( A ) and do the same, and the same for the third element with the third row.Let me write that out step by step.First element of ( mathbf{w} ):( (2 times 2) + (1 times 1) + (3 times 3) )Calculating each term:- ( 2 times 2 = 4 )- ( 1 times 1 = 1 )- ( 3 times 3 = 9 )Adding them together: ( 4 + 1 + 9 = 14 )So the first element is 14.Second element of ( mathbf{w} ):( (4 times 2) + (5 times 1) + (6 times 3) )Calculating each term:- ( 4 times 2 = 8 )- ( 5 times 1 = 5 )- ( 6 times 3 = 18 )Adding them together: ( 8 + 5 + 18 = 31 )So the second element is 31.Third element of ( mathbf{w} ):( (7 times 2) + (8 times 1) + (9 times 3) )Calculating each term:- ( 7 times 2 = 14 )- ( 8 times 1 = 8 )- ( 9 times 3 = 27 )Adding them together: ( 14 + 8 + 27 = 49 )So the third element is 49.Putting it all together, the resulting vector ( mathbf{w} ) is:[ mathbf{w} = begin{bmatrix} 14 31 49 end{bmatrix} ]Wait, let me double-check my calculations to make sure I didn't make any mistakes.First element: 2*2=4, 1*1=1, 3*3=9. 4+1=5, 5+9=14. Correct.Second element: 4*2=8, 5*1=5, 6*3=18. 8+5=13, 13+18=31. Correct.Third element: 7*2=14, 8*1=8, 9*3=27. 14+8=22, 22+27=49. Correct.Okay, that seems solid.Moving on to the second problem: calculating the probability that "Unique Storytelling" will be a blockbuster using the Poisson probability function.The function is given as:[ f(x) = frac{e^{-lambda} lambda^x}{x!} ]Where ( lambda = 5 ) and ( x = 7 ). So, I need to plug these values into the formula.First, let me note down the values:- ( lambda = 5 )- ( x = 7 )- ( e approx 2.71828 )So, substituting into the formula:[ f(7) = frac{e^{-5} times 5^7}{7!} ]I need to compute each part step by step.First, compute ( e^{-5} ). Since ( e approx 2.71828 ), ( e^{-5} ) is approximately ( 1 / e^5 ).Calculating ( e^5 ):( e^1 = 2.71828 )( e^2 = e times e approx 2.71828 times 2.71828 approx 7.38906 )( e^3 = e^2 times e approx 7.38906 times 2.71828 approx 20.0855 )( e^4 = e^3 times e approx 20.0855 times 2.71828 approx 54.5981 )( e^5 = e^4 times e approx 54.5981 times 2.71828 approx 148.4132 )So, ( e^{-5} approx 1 / 148.4132 approx 0.006737947 )Next, compute ( 5^7 ):( 5^1 = 5 )( 5^2 = 25 )( 5^3 = 125 )( 5^4 = 625 )( 5^5 = 3125 )( 5^6 = 15625 )( 5^7 = 78125 )So, ( 5^7 = 78125 )Now, compute ( 7! ) (7 factorial):( 7! = 7 times 6 times 5 times 4 times 3 times 2 times 1 )Calculating step by step:- 7 × 6 = 42- 42 × 5 = 210- 210 × 4 = 840- 840 × 3 = 2520- 2520 × 2 = 5040- 5040 × 1 = 5040So, ( 7! = 5040 )Now, plug all these back into the formula:[ f(7) = frac{0.006737947 times 78125}{5040} ]First, compute the numerator:0.006737947 × 78125Let me calculate that:First, 0.006737947 × 10000 = 67.37947So, 0.006737947 × 78125 = 0.006737947 × (78 × 1000 + 125)Wait, that might complicate. Alternatively, I can compute 78125 × 0.006737947.Let me compute 78125 × 0.006737947.First, note that 78125 × 0.006 = 468.7578125 × 0.000737947 ≈ ?Compute 78125 × 0.0007 = 54.687578125 × 0.000037947 ≈ ?Compute 78125 × 0.00003 = 2.3437578125 × 0.000007947 ≈ approximately 0.6201171875Adding these up:54.6875 + 2.34375 = 57.0312557.03125 + 0.6201171875 ≈ 57.6513671875So, total numerator ≈ 468.75 + 57.6513671875 ≈ 526.4013671875Wait, that seems a bit off. Let me try another approach.Alternatively, 78125 × 0.006737947 can be calculated as:First, 78125 × 0.006 = 468.75Then, 78125 × 0.000737947.Compute 78125 × 0.0007 = 54.687578125 × 0.000037947 ≈ 78125 × 0.00003 = 2.3437578125 × 0.000007947 ≈ 0.6201171875So, adding up:54.6875 + 2.34375 = 57.0312557.03125 + 0.6201171875 ≈ 57.6513671875So, total numerator ≈ 468.75 + 57.6513671875 ≈ 526.4013671875Wait, that seems high because 0.006737947 is roughly 0.0067, so 78125 × 0.0067 ≈ 78125 × 0.006 + 78125 × 0.0007 ≈ 468.75 + 54.6875 ≈ 523.4375So, approximately 523.4375But my detailed calculation gave me 526.401367, which is a bit higher. Maybe my approximation is a bit off, but let's go with 523.4375 for simplicity.So, numerator ≈ 523.4375Denominator is 5040.So, f(7) ≈ 523.4375 / 5040Compute that division:523.4375 ÷ 5040First, note that 5040 × 0.1 = 504523.4375 - 504 = 19.4375So, 0.1 + (19.4375 / 5040)Compute 19.4375 / 5040:5040 ÷ 1000 = 5.04, so 19.4375 / 5040 ≈ 0.003856So, total ≈ 0.1 + 0.003856 ≈ 0.103856Alternatively, compute 523.4375 ÷ 5040:Divide numerator and denominator by 5:523.4375 ÷ 5 = 104.68755040 ÷ 5 = 1008So, 104.6875 / 1008 ≈ 0.103856So, approximately 0.103856Therefore, f(7) ≈ 0.103856To express this as a probability, it's approximately 0.1039 or 10.39%.Wait, let me check if I did that correctly.Alternatively, perhaps I should use a calculator approach for more precision.Compute numerator: 0.006737947 × 78125Let me compute 78125 × 0.006737947.First, 78125 × 0.006 = 468.7578125 × 0.0007 = 54.687578125 × 0.000037947 ≈ 78125 × 0.00003 = 2.3437578125 × 0.000007947 ≈ 0.6201171875So, adding up:468.75 + 54.6875 = 523.4375523.4375 + 2.34375 = 525.78125525.78125 + 0.6201171875 ≈ 526.4013671875So, numerator ≈ 526.4013671875Denominator: 5040So, 526.4013671875 / 5040 ≈ ?Compute 5040 × 0.1 = 504526.4013671875 - 504 = 22.4013671875So, 0.1 + (22.4013671875 / 5040)Compute 22.4013671875 / 5040 ≈ 0.004444So, total ≈ 0.1 + 0.004444 ≈ 0.104444So, approximately 0.1044 or 10.44%Wait, that's a bit different from my previous 10.39%. Hmm.Alternatively, perhaps I should compute it more precisely.Compute 526.4013671875 / 5040:Divide numerator and denominator by 5:526.4013671875 ÷ 5 = 105.28027343755040 ÷ 5 = 1008So, 105.2802734375 / 1008 ≈ ?Compute 1008 × 0.104 = 104.832105.2802734375 - 104.832 = 0.4482734375So, 0.104 + (0.4482734375 / 1008)Compute 0.4482734375 / 1008 ≈ 0.0004445So, total ≈ 0.104 + 0.0004445 ≈ 0.1044445So, approximately 0.1044445, which is about 0.1044 or 10.44%.Wait, so my initial approximation was 10.39%, then 10.44%. Let me see if I can compute it more accurately.Alternatively, perhaps I should use a calculator for the exact value.But since I don't have a calculator, let me try to compute 526.4013671875 divided by 5040.First, note that 5040 × 0.1 = 5045040 × 0.104 = 5040 × 0.1 + 5040 × 0.004 = 504 + 20.16 = 524.16So, 5040 × 0.104 = 524.16Our numerator is 526.4013671875Subtract 524.16 from 526.4013671875: 526.4013671875 - 524.16 = 2.2413671875So, 2.2413671875 / 5040 ≈ ?Compute 2.2413671875 ÷ 5040 ≈ 0.0004447So, total ≈ 0.104 + 0.0004447 ≈ 0.1044447So, approximately 0.1044447, which is roughly 0.1044 or 10.44%.Therefore, the probability is approximately 10.44%.Wait, but let me check if I did the multiplication correctly.Alternatively, perhaps I should compute ( e^{-5} times 5^7 ) first, then divide by 7!.Let me compute ( e^{-5} times 5^7 ):We have ( e^{-5} approx 0.006737947 )( 5^7 = 78125 )So, 0.006737947 × 78125Let me compute this as:0.006737947 × 78125First, 78125 × 0.006 = 468.7578125 × 0.000737947 ≈ ?Compute 78125 × 0.0007 = 54.687578125 × 0.000037947 ≈ ?Compute 78125 × 0.00003 = 2.3437578125 × 0.000007947 ≈ 0.6201171875So, adding up:54.6875 + 2.34375 = 57.0312557.03125 + 0.6201171875 ≈ 57.6513671875So, total ≈ 468.75 + 57.6513671875 ≈ 526.4013671875So, that's correct.Now, divide by 7! = 5040:526.4013671875 / 5040 ≈ 0.1044447So, approximately 0.1044 or 10.44%.So, the probability is approximately 10.44%.Wait, but let me check if I can compute this more accurately.Alternatively, perhaps I can use logarithms or another method, but since I don't have a calculator, I'll stick with this approximation.So, to summarize:- ( e^{-5} approx 0.006737947 )- ( 5^7 = 78125 )- ( 7! = 5040 )- Numerator: 0.006737947 × 78125 ≈ 526.4013671875- Denominator: 5040- Probability: 526.4013671875 / 5040 ≈ 0.1044447 ≈ 0.1044 or 10.44%Therefore, the probability that "Unique Storytelling" will be a blockbuster is approximately 10.44%.Wait, but let me check if I made any miscalculations in the multiplication.Alternatively, perhaps I should compute 0.006737947 × 78125 more accurately.Let me compute 0.006737947 × 78125:First, note that 78125 × 0.006 = 468.7578125 × 0.000737947:Compute 78125 × 0.0007 = 54.687578125 × 0.000037947:Compute 78125 × 0.00003 = 2.3437578125 × 0.000007947 ≈ 0.6201171875So, adding up:54.6875 + 2.34375 = 57.0312557.03125 + 0.6201171875 ≈ 57.6513671875So, total ≈ 468.75 + 57.6513671875 ≈ 526.4013671875Yes, that's correct.So, the calculation seems consistent.Therefore, the probability is approximately 0.1044 or 10.44%.Wait, but let me check if I can express this as a fraction or a more precise decimal.Alternatively, perhaps I can use more precise values for ( e^{-5} ).Wait, ( e^{-5} ) is approximately 0.006737947, but let me check if that's accurate.Using a calculator, ( e^{-5} ) is approximately 0.006737947, so that's correct.So, 0.006737947 × 78125 = 526.4013671875Divided by 5040 gives approximately 0.1044447.So, rounding to four decimal places, that's 0.1044, or 10.44%.Alternatively, if we want to express it as a fraction, 0.1044 is approximately 1044/10000, which simplifies to 261/2500, but that's not necessary unless specified.So, in conclusion, the probability is approximately 10.44%.Wait, but let me check if I can compute this more accurately without a calculator.Alternatively, perhaps I can use the fact that 526.4013671875 / 5040 is equal to (526.4013671875 ÷ 5040).Let me compute this division step by step.First, note that 5040 × 0.1 = 504526.4013671875 - 504 = 22.4013671875So, 0.1 + (22.4013671875 / 5040)Now, compute 22.4013671875 / 5040.First, note that 5040 × 0.004 = 20.1622.4013671875 - 20.16 = 2.2413671875So, 0.004 + (2.2413671875 / 5040)Now, compute 2.2413671875 / 5040.5040 × 0.0004 = 2.0162.2413671875 - 2.016 = 0.2253671875So, 0.0004 + (0.2253671875 / 5040)Compute 0.2253671875 / 5040 ≈ 0.0000447So, adding up:0.1 + 0.004 + 0.0004 + 0.0000447 ≈ 0.1044447So, that's consistent with my previous calculation.Therefore, the probability is approximately 0.1044447, which is about 10.44%.So, rounding to four decimal places, 0.1044, or 10.44%.Alternatively, if we want to express it as a fraction, 1044/10000, but that's not necessary unless specified.Therefore, the probability that "Unique Storytelling" will be a blockbuster is approximately 10.44%.Wait, but let me check if I can express this as a percentage with two decimal places, which is standard.So, 0.1044447 is approximately 10.44%.Alternatively, if I round to three decimal places, it's 10.444%, which is approximately 10.44%.So, I think that's a reasonable approximation.Therefore, the probability is approximately 10.44%.Wait, but let me check if I can compute this more accurately.Alternatively, perhaps I can use a calculator for more precision, but since I don't have one, I'll stick with this.So, in conclusion, the probability is approximately 10.44%.Wait, but let me check if I made any mistake in the calculation of ( e^{-5} ).Wait, ( e^{-5} ) is approximately 0.006737947, which is correct.So, 0.006737947 × 78125 = 526.4013671875Divided by 5040 gives approximately 0.1044447.Yes, that seems correct.Therefore, the probability is approximately 10.44%.Wait, but let me check if I can compute this more accurately.Alternatively, perhaps I can compute it as:( f(7) = frac{e^{-5} times 5^7}{7!} )We can compute this using logarithms.First, compute the natural logarithm of the numerator and denominator.But that might complicate things without a calculator.Alternatively, perhaps I can use the fact that the Poisson probability mass function can be computed step by step.But in any case, I think my approximation is sufficient.Therefore, the probability is approximately 10.44%.Wait, but let me check if I can compute it more accurately.Alternatively, perhaps I can use the fact that 526.4013671875 / 5040 is equal to approximately 0.1044447.Yes, that's correct.So, in conclusion, the probability is approximately 10.44%.Therefore, the probability that "Unique Storytelling" will be a blockbuster is approximately 10.44%.Wait, but let me check if I can express this as a fraction.Alternatively, perhaps I can write it as a decimal with more precision.But without a calculator, it's difficult.Alternatively, perhaps I can accept that the approximate value is 0.1044 or 10.44%.Therefore, I think that's the answer.So, to recap:1. The resulting vector ( mathbf{w} ) after applying matrix ( A ) to vector ( mathbf{v} ) is:[ mathbf{w} = begin{bmatrix} 14 31 49 end{bmatrix} ]2. The probability that "Unique Storytelling" will be a blockbuster is approximately 10.44%.I think that's it.Final Answer1. The resulting vector ( mathbf{w} ) is boxed{begin{bmatrix} 14 31 49 end{bmatrix}}.2. The probability that "Unique Storytelling" will be a blockbuster is approximately boxed{0.1044}.</think>

🧠:<think>Alright, so I'm trying to figure out this problem about the Women's Professional Soccer team's statistics. It's been a while since I did some algebra, but let me see if I can work through this step by step.First, the problem says that the team played a total of 22 games. They won 35% more games than they lost, and the rest were draws. I need to set up a system of equations to find out how many games they won, lost, and drew. Then, there's a second part about calculating a probability related to goals scored, but I'll tackle that after I figure out the first part.Okay, let's start with the first part. I need to define variables for the number of wins, losses, and draws. Let me denote:- ( W ) = number of games won- ( L ) = number of games lost- ( D ) = number of games drawnThe total number of games played is 22, so that gives me my first equation:( W + L + D = 22 )That seems straightforward. Now, the next piece of information is that the team won 35% more games than they lost. Hmm, okay. So, they won 35% more than they lost. I need to translate that into an equation.When something is 35% more than another, it means it's 100% + 35% = 135% of the other. So, if they lost ( L ) games, they won 135% of ( L ) games. In mathematical terms, that would be:( W = 1.35L )So, that's my second equation.Now, I have two equations:1. ( W + L + D = 22 )2. ( W = 1.35L )But I have three variables here: ( W ), ( L ), and ( D ). So, I need a third equation to solve for all three variables. Wait, the problem doesn't mention anything else about draws, just that they drew the remaining games. So, perhaps I can express ( D ) in terms of ( W ) and ( L ). From the first equation, I can rearrange it to solve for ( D ):( D = 22 - W - L )So, that's my third equation. Now, I can substitute the second equation into this to express ( D ) in terms of ( L ) only.Substituting ( W = 1.35L ) into ( D = 22 - W - L ):( D = 22 - 1.35L - L )( D = 22 - 2.35L )So, now I have expressions for both ( W ) and ( D ) in terms of ( L ). But I still have three variables and only two equations. Wait, maybe I can find another relationship? Hmm, the problem doesn't provide more information, so perhaps I need to find integer solutions because the number of games won, lost, and drawn must be whole numbers.So, let's think about this. Since ( W = 1.35L ), both ( W ) and ( L ) must be integers. Therefore, ( 1.35L ) must be an integer. Let me write that as:( W = frac{27}{20}L )Because 1.35 is equal to 27/20. So, ( W = frac{27}{20}L ). For ( W ) to be an integer, ( L ) must be a multiple of 20. Because 27 and 20 are coprime (they have no common divisors other than 1), so ( L ) must be a multiple of 20 to make ( W ) an integer.But wait, the total number of games is 22. If ( L ) is a multiple of 20, the smallest multiple is 20, but 20 is already more than half of 22, which would make ( W = 27 ) games, which is impossible because the total games are only 22. So, that can't be right.Hmm, maybe I made a mistake in interpreting the 35% more. Let me double-check. The problem says the team won 35% more games than they lost. So, if they lost ( L ) games, they won 35% more than that, which is ( L + 0.35L = 1.35L ). So, that part seems correct.But if ( W = 1.35L ), and both ( W ) and ( L ) must be integers, then ( L ) must be such that 1.35L is an integer. Let me express 1.35 as a fraction: 1.35 = 27/20. So, ( W = (27/20)L ). Therefore, ( L ) must be a multiple of 20 to make ( W ) an integer. But as I saw before, 20 is too large because the total games are only 22.Wait, maybe I don't need ( L ) to be a multiple of 20, but rather, ( 27L ) must be divisible by 20. So, ( L ) must be such that 27L is divisible by 20. Since 27 and 20 are coprime, ( L ) must be a multiple of 20. But again, 20 is too big.Alternatively, maybe I can think of ( W ) and ( L ) as integers, so ( 1.35L ) must be integer. Let me write 1.35 as 27/20, so ( W = (27/20)L ). Therefore, ( L ) must be a multiple of 20/ gcd(27,20). Since gcd(27,20)=1, so ( L ) must be a multiple of 20. But 20 is too big.Wait, maybe I'm overcomplicating this. Let's try plugging in possible integer values for ( L ) and see if ( W ) comes out as an integer.Let me denote ( L = x ). Then, ( W = 1.35x ). So, 1.35x must be integer. Let's try x=20: W=27, but total games would be 27+20=47, which is way over 22.x=10: W=13.5, which is not integer.x=4: W=5.4, not integer.x=5: W=6.75, not integer.x=2: W=2.7, not integer.x=1: W=1.35, not integer.Hmm, none of these are working. Maybe I made a mistake in interpreting the 35% more.Wait, perhaps it's 35% more than the number of losses, but in terms of ratio. Maybe it's not 1.35 times, but rather, the ratio of wins to losses is 135:100, which simplifies to 27:20. So, the ratio of wins to losses is 27:20.Therefore, ( W/L = 27/20 ), so ( W = (27/20)L ). So, same as before. So, same problem.But since 27 and 20 are coprime, the smallest possible integer values are W=27, L=20, but that's 47 games, which is way over 22.Wait, maybe the problem is that the total games are 22, so W + L + D =22, and W =1.35L.So, substituting, 1.35L + L + D =22 => 2.35L + D =22.But D must be non-negative, so 2.35L <=22 => L <=22/2.35 ≈9.36. So, L must be less than or equal to 9.But L must be such that 1.35L is integer. So, let's try L=10: 1.35*10=13.5, not integer.L=9: 1.35*9=12.15, not integer.L=8: 1.35*8=10.8, not integer.L=7: 1.35*7=9.45, not integer.L=6: 1.35*6=8.1, not integer.L=5: 1.35*5=6.75, not integer.L=4: 1.35*4=5.4, not integer.L=3: 1.35*3=4.05, not integer.L=2: 1.35*2=2.7, not integer.L=1: 1.35*1=1.35, not integer.Wait, none of these are integers. That's a problem. Maybe I misinterpreted the 35% more.Wait, perhaps it's 35% more in terms of total games, not in terms of losses. Wait, the problem says "won 35% more games than they lost." So, it's 35% more than the number of losses. So, W = L + 0.35L =1.35L.But as we saw, that leads to fractional games, which is impossible.Alternatively, maybe it's 35% more in terms of total games. So, W = 0.35*(W + L). Let me see.Wait, the problem says "won 35% more games than they lost." So, it's 35% more than the number of losses. So, W = L + 0.35L =1.35L.But as we saw, that leads to fractional games. So, maybe the problem is that the numbers are approximate, or perhaps I need to consider that 35% more could be in terms of ratio, not exact percentage.Alternatively, maybe the problem is that the number of wins is 35% higher than the number of losses, but in terms of ratio, not exact percentage. So, perhaps W/L = 135/100 =27/20. So, W=27k, L=20k, for some integer k.But then, W + L + D=22. So, 27k +20k + D=22 =>47k + D=22. Since k must be at least 1, 47k would be 47, which is more than 22. So, that's impossible.Wait, so maybe the problem is that the percentage is approximate? Or perhaps I need to consider that the number of wins is 35% higher than the number of losses, but in terms of the total games.Wait, maybe it's 35% more wins than losses, but in terms of the total games. So, W =0.35*(W + L). Let me try that.So, W =0.35*(W + L). Let's solve for W.W =0.35W +0.35LW -0.35W =0.35L0.65W =0.35LDivide both sides by 0.05:13W =7LSo, W = (7/13)LHmm, that's a different ratio. So, W/L=7/13.So, W=7k, L=13k, for some integer k.Then, total games: W + L + D=7k +13k + D=20k + D=22.So, D=22 -20k.Since D must be non-negative, 22 -20k >=0 =>k<=1.1.So, k=1.Therefore, W=7, L=13, D=22 -20=2.Wait, let's check: W=7, L=13, D=2.Total games:7+13+2=22. That works.Now, check if W is 35% more than L.Wait, W=7, L=13.So, 7 is what percentage more than 13?The difference is 7 -13= -6, which is negative, so that doesn't make sense. Wait, that can't be.Wait, hold on, if W=7 and L=13, then W is actually less than L, which contradicts the statement that they won 35% more games than they lost.So, that can't be right.Wait, so perhaps my interpretation is wrong. Let me go back.The problem says "won 35% more games than they lost." So, W = L +0.35L=1.35L.But as we saw earlier, that leads to fractional games.Alternatively, if I interpret it as W =1.35L, but then W and L must be integers, so perhaps the numbers are approximate, or perhaps the problem allows for fractional games, but that doesn't make sense in reality.Wait, maybe the 35% is not exact, but approximate. So, perhaps we can round the numbers.Let me try L=5.Then, W=1.35*5=6.75, which is approximately 7.So, W=7, L=5.Then, D=22 -7 -5=10.So, total games:7+5+10=22.Check if W is 35% more than L.35% of L=5 is 1.75, so 5 +1.75=6.75. So, W=6.75, which is approximately 7. So, rounding up, W=7, L=5.Alternatively, if L=4, W=1.35*4=5.4≈5.Then, D=22 -5 -4=13.But 5 is not 35% more than 4: 35% of 4 is 1.4, so 4 +1.4=5.4≈5.But again, not exact.Alternatively, maybe the problem expects us to use exact values, even if they are fractional, but that doesn't make sense in the context of games.Wait, perhaps the problem is in the wording. Maybe it's 35% more wins than losses relative to the total games. So, W =0.35*(W + L). Let's try that again.So, W =0.35*(W + L)Multiply both sides by 100: 100W =35W +35L65W =35LDivide both sides by 5:13W=7LSo, W=7k, L=13k.But then, W + L + D=22 =>7k +13k + D=20k + D=22.So, D=22 -20k.Since D must be non-negative, k=1, D=2.So, W=7, L=13, D=2.But as before, W=7, L=13, which means they lost more games than they won, which contradicts the statement that they won 35% more games than they lost.Wait, so that can't be.Hmm, this is confusing. Maybe the problem is that the wording is ambiguous. Let me read it again."The team won 35% more games than they lost and drew the remaining games."So, "won 35% more games than they lost." So, W = L +0.35L=1.35L.But as we saw, that leads to fractional games, which is impossible.Alternatively, maybe it's 35% more in terms of total games. So, W =0.35*(W + L). But that leads to W=7, L=13, which contradicts the statement.Alternatively, maybe it's 35% more wins than losses relative to the total games. So, W =0.35*(W + L + D). But that seems odd.Wait, let's try that.If W =0.35*(22), then W=7.7≈8.Then, L=?But the problem says W=1.35L, so L= W /1.35≈8 /1.35≈5.925≈6.Then, D=22 -8 -6=8.But let's check: W=8, L=6.Is 8 equal to 1.35*6=8.1? Close, but not exact.Alternatively, maybe W=8.1, L=6, D=7.9. But again, fractional games.Hmm, this is tricky.Wait, maybe the problem is that 35% more is in terms of the number of wins compared to losses, but not necessarily in exact terms. So, perhaps we can use the ratio 27:20 for W:L.So, W=27k, L=20k.Total games:27k +20k + D=47k + D=22.So, 47k <=22 =>k=0.468, which is not an integer.So, that's not possible.Alternatively, maybe the ratio is 7:5, since 7/5=1.4, which is close to 1.35.So, W=7k, L=5k.Total games:7k +5k + D=12k + D=22.So, D=22 -12k.To have D>=0, 12k <=22 =>k<=1.833.So, k=1.Thus, W=7, L=5, D=10.Check if W=1.35L: 1.35*5=6.75≈7. So, approximately.So, maybe the problem expects us to round to the nearest whole number.So, W=7, L=5, D=10.That seems plausible.Alternatively, maybe the problem expects us to use exact fractions, even if they result in fractional games, but that seems odd.Wait, let me think differently. Maybe the 35% more is in terms of the total games. So, total games won is 35% more than total games lost.Wait, that would mean W =1.35L.But again, same problem.Alternatively, maybe the 35% is in terms of the difference between wins and losses. So, W - L =0.35L => W=1.35L.Same as before.Hmm, I'm going in circles here.Wait, maybe the problem is that the number of wins is 35% higher than the number of losses, but in terms of the total games. So, W = L +0.35*(W + L). Let's try that.So, W = L +0.35*(W + L)Multiply out:W = L +0.35W +0.35LBring terms with W to the left:W -0.35W = L +0.35L0.65W =1.35LDivide both sides by 0.05:13W =27LSo, W=27k, L=13k.Total games:27k +13k + D=40k + D=22.So, 40k <=22 =>k=0.55, which is not an integer.So, that doesn't work.Alternatively, maybe the 35% is in terms of the total games. So, W =0.35*22=7.7≈8.Then, L=?Since W=1.35L, L= W /1.35≈8 /1.35≈5.925≈6.Then, D=22 -8 -6=8.So, W=8, L=6, D=8.Check: W=8, L=6.Is 8 equal to 1.35*6=8.1? Close, but not exact.Alternatively, maybe the problem expects us to use exact fractions, even if they result in fractional games, but that seems odd.Wait, maybe the problem is that the 35% is not exact, but approximate. So, perhaps we can accept that W=7, L=5, D=10, even though W=7 is only approximately 1.4 times L=5, which is 35% more.Wait, 5*1.35=6.75, which is approximately 7. So, maybe that's the intended answer.So, perhaps the answer is W=7, L=5, D=10.Let me check the total:7+5+10=22. That works.And 7 is approximately 35% more than 5: 5*1.35=6.75≈7.So, maybe that's the intended solution.Alternatively, maybe the problem expects us to use exact fractions, even if they result in fractional games, but that seems odd.Wait, let me try solving the equations without worrying about integer constraints first, and then see if the numbers make sense.So, we have:1. W + L + D =222. W =1.35LFrom equation 2, substitute into equation 1:1.35L + L + D =222.35L + D =22So, D=22 -2.35LNow, since D must be non-negative, 22 -2.35L >=0 => L <=22/2.35≈9.36.So, L<=9.Also, since W=1.35L must be non-negative, L>=0.So, L can be from 0 to9.But we need W, L, D to be integers.So, let's try L=5:W=1.35*5=6.75, not integer.L=4:W=5.4, not integer.L=3:W=4.05, not integer.L=2:W=2.7, not integer.L=1:W=1.35, not integer.L=6:W=8.1, not integer.L=7:W=9.45, not integer.L=8:W=10.8, not integer.L=9:W=12.15, not integer.So, none of these give integer values for W.Hmm, so maybe the problem is designed in such a way that we have to accept fractional games, but that seems odd.Alternatively, maybe the 35% is not exact, and we can use approximate values.So, if we take L=5, W=6.75≈7, D=10.So, W=7, L=5, D=10.That seems to be the closest integer solution.Alternatively, maybe the problem expects us to use exact fractions, even if they result in fractional games, but that seems odd.Wait, maybe the problem is that the 35% is in terms of the total games. So, W=35% more than L, but in terms of the total games.Wait, that would mean W = L +0.35*(W + L).So, W = L +0.35W +0.35LSo, W -0.35W = L +0.35L0.65W =1.35LSo, W = (1.35/0.65)L ≈2.0769LSo, W≈2.0769LSo, let's try L=5:W≈10.3846≈10Then, D=22 -10 -5=7.Check if W≈2.0769L: 10≈2.0769*5=10.3845≈10. So, close.Alternatively, L=4:W≈8.307≈8D=22 -8 -4=10Check:8≈2.0769*4=8.307≈8.So, that's another possible solution.But the problem says "won 35% more games than they lost," which is a specific statement, so perhaps the first interpretation is correct, even if it leads to fractional games.Alternatively, maybe the problem is designed to have W=7, L=5, D=10, even though it's an approximation.Given that, I think the intended answer is W=7, L=5, D=10.So, moving on to the second part.The number of goals scored by the team in the games won follows a Poisson distribution with a mean of 3.5 goals per game. Calculate the probability that in a randomly selected game that the team won, they scored exactly 5 goals.Okay, so Poisson distribution formula is:P(X=k) = (λ^k * e^-λ) / k!Where λ is the mean, which is 3.5, and k is the number of goals, which is 5.So, plugging in the numbers:P(X=5) = (3.5^5 * e^-3.5) / 5!Let me compute that step by step.First, calculate 3.5^5.3.5^1=3.53.5^2=12.253.5^3=42.8753.5^4=150.06253.5^5=525.21875Next, e^-3.5. e is approximately 2.71828.So, e^-3.5 ≈1 / e^3.5 ≈1 / 33.115 ≈0.0302.Now, 5! =120.So, putting it all together:P(X=5)= (525.21875 * 0.0302) / 120First, multiply 525.21875 *0.0302:525.21875 *0.03=15.7565625525.21875 *0.0002=0.10504375So, total≈15.7565625 +0.10504375≈15.86160625Now, divide by 120:15.86160625 /120≈0.13218So, approximately 0.1322, or 13.22%.Alternatively, using a calculator for more precision:3.5^5=525.21875e^-3.5≈0.030197383So, 525.21875 *0.030197383≈15.86115.861 /120≈0.132175So, approximately 0.1322, or 13.22%.So, the probability is approximately 13.22%.Alternatively, using more precise calculation:Let me compute 3.5^5:3.5^1=3.53.5^2=12.253.5^3=42.8753.5^4=150.06253.5^5=525.21875e^-3.5≈0.030197383So, 525.21875 *0.030197383:Let me compute 525.21875 *0.03=15.7565625525.21875 *0.000197383≈525.21875 *0.0002≈0.10504375So, total≈15.7565625 +0.10504375≈15.86160625Divide by 120:15.86160625 /120≈0.13218So, approximately 0.1322, or 13.22%.So, the probability is approximately 13.22%.Alternatively, using a calculator:Poisson(3.5,5)= (3.5^5 * e^-3.5)/5!≈0.13218So, approximately 13.22%.Therefore, the probability is approximately 13.22%.But let me check using a calculator function for Poisson probability.Using a calculator, Poisson PMF with λ=3.5, k=5:P(X=5)= (3.5^5 * e^-3.5)/5!≈0.13218So, yes, approximately 13.22%.Therefore, the probability is approximately 13.22%.So, summarizing:1. The team won 7 games, lost 5, and drew 10.2. The probability of scoring exactly 5 goals in a won game is approximately 13.22%.But wait, in the first part, I had to approximate because the exact solution didn't give integer values. So, maybe the problem expects us to use the exact fractional values, even if they result in fractional games, but that seems odd.Alternatively, maybe the problem is designed to have W=7, L=5, D=10, even though it's an approximation.Given that, I think that's the intended answer.</think>