🧠:<think>Alright, so I have this problem about the Mandelbrot set, which I remember is this famous fractal in complex numbers. The artist is using a sequence defined by ( z_0 = c ) and ( z_{n+1} = z_n^2 + c ). The goal is to find the set of ( c = a + bi ) such that the sequence doesn't escape to infinity, meaning it stays bounded with modulus never exceeding 2. Starting with part 1: I need to describe the Mandelbrot set in terms of conditions on ( a ) and ( b ). From what I recall, the Mandelbrot set is the set of all complex numbers ( c ) for which the sequence ( z_n ) doesn't escape to infinity. The key here is that if at any point the modulus of ( z_n ) exceeds 2, the sequence will definitely escape to infinity. So, the condition is that for all ( n ), ( |z_n| leq 2 ).But how do I translate this into conditions on ( a ) and ( b )? I think it's not straightforward because the recursion is quadratic, so each term depends on the square of the previous term plus ( c ). This makes it a bit complex to write a direct condition on ( a ) and ( b ). I remember that the Mandelbrot set is defined by the condition that the sequence remains bounded. So, mathematically, it's the set ( M = { c in mathbb{C} : exists R > 0, forall n in mathbb{N}, |z_n| leq R } ). But since it's known that if ( |z_n| ) ever exceeds 2, it will escape, we can say that ( M ) is the set of ( c ) such that ( |z_n| leq 2 ) for all ( n ).But the problem asks for a mathematical description in terms of ( a ) and ( b ). I think it's not possible to write a simple closed-form condition because the Mandelbrot set is a fractal with a very complex boundary. However, we can describe it as the set of all ( c = a + bi ) where the sequence ( z_n ) does not escape to infinity, which is equivalent to ( |z_n| leq 2 ) for all ( n ). So, in terms of ( a ) and ( b ), the Mandelbrot set ( M ) is the set of all complex numbers ( c = a + bi ) such that the sequence defined by ( z_0 = c ) and ( z_{n+1} = z_n^2 + c ) remains bounded, specifically, ( |z_n| leq 2 ) for all ( n ).Moving on to part 2: Given ( c = frac{1}{4} + frac{sqrt{3}}{4}i ), I need to calculate the first 10 terms of the sequence ( {z_n} ) and determine if this ( c ) belongs to the bounded set.Let me write down the recursion formula:( z_{n+1} = z_n^2 + c ).Given ( c = frac{1}{4} + frac{sqrt{3}}{4}i ), so ( a = frac{1}{4} ) and ( b = frac{sqrt{3}}{4} ).Starting with ( z_0 = c = frac{1}{4} + frac{sqrt{3}}{4}i ).Let me compute each term step by step.First, ( z_0 = frac{1}{4} + frac{sqrt{3}}{4}i ).Compute ( z_1 = z_0^2 + c ).Let me compute ( z_0^2 ):( z_0^2 = left( frac{1}{4} + frac{sqrt{3}}{4}i right)^2 ).Using the formula ( (a + bi)^2 = a^2 - b^2 + 2abi ):( a = frac{1}{4} ), ( b = frac{sqrt{3}}{4} ).So,( a^2 = left( frac{1}{4} right)^2 = frac{1}{16} ).( b^2 = left( frac{sqrt{3}}{4} right)^2 = frac{3}{16} ).So, ( a^2 - b^2 = frac{1}{16} - frac{3}{16} = -frac{2}{16} = -frac{1}{8} ).The imaginary part is ( 2ab = 2 * frac{1}{4} * frac{sqrt{3}}{4} = frac{sqrt{3}}{8} ).So, ( z_0^2 = -frac{1}{8} + frac{sqrt{3}}{8}i ).Now, add ( c ) to this:( z_1 = z_0^2 + c = left( -frac{1}{8} + frac{sqrt{3}}{8}i right) + left( frac{1}{4} + frac{sqrt{3}}{4}i right) ).Combine real and imaginary parts:Real: ( -frac{1}{8} + frac{1}{4} = -frac{1}{8} + frac{2}{8} = frac{1}{8} ).Imaginary: ( frac{sqrt{3}}{8} + frac{sqrt{3}}{4} = frac{sqrt{3}}{8} + frac{2sqrt{3}}{8} = frac{3sqrt{3}}{8} ).So, ( z_1 = frac{1}{8} + frac{3sqrt{3}}{8}i ).Now, compute ( z_2 = z_1^2 + c ).First, ( z_1^2 ):( z_1 = frac{1}{8} + frac{3sqrt{3}}{8}i ).So, ( a = frac{1}{8} ), ( b = frac{3sqrt{3}}{8} ).Compute ( a^2 = left( frac{1}{8} right)^2 = frac{1}{64} ).( b^2 = left( frac{3sqrt{3}}{8} right)^2 = frac{9 * 3}{64} = frac{27}{64} ).So, ( a^2 - b^2 = frac{1}{64} - frac{27}{64} = -frac{26}{64} = -frac{13}{32} ).Imaginary part: ( 2ab = 2 * frac{1}{8} * frac{3sqrt{3}}{8} = frac{6sqrt{3}}{64} = frac{3sqrt{3}}{32} ).So, ( z_1^2 = -frac{13}{32} + frac{3sqrt{3}}{32}i ).Add ( c ):( z_2 = z_1^2 + c = left( -frac{13}{32} + frac{3sqrt{3}}{32}i right) + left( frac{1}{4} + frac{sqrt{3}}{4}i right) ).Convert ( frac{1}{4} ) to 32nds: ( frac{8}{32} ).Convert ( frac{sqrt{3}}{4}i ) to 32nds: ( frac{8sqrt{3}}{32}i ).So,Real: ( -frac{13}{32} + frac{8}{32} = -frac{5}{32} ).Imaginary: ( frac{3sqrt{3}}{32} + frac{8sqrt{3}}{32} = frac{11sqrt{3}}{32} ).Thus, ( z_2 = -frac{5}{32} + frac{11sqrt{3}}{32}i ).Now, compute ( z_3 = z_2^2 + c ).First, ( z_2 = -frac{5}{32} + frac{11sqrt{3}}{32}i ).Compute ( z_2^2 ):( a = -frac{5}{32} ), ( b = frac{11sqrt{3}}{32} ).( a^2 = left( -frac{5}{32} right)^2 = frac{25}{1024} ).( b^2 = left( frac{11sqrt{3}}{32} right)^2 = frac{121 * 3}{1024} = frac{363}{1024} ).So, ( a^2 - b^2 = frac{25}{1024} - frac{363}{1024} = -frac{338}{1024} = -frac{169}{512} ).Imaginary part: ( 2ab = 2 * (-frac{5}{32}) * frac{11sqrt{3}}{32} = -frac{110sqrt{3}}{1024} = -frac{55sqrt{3}}{512} ).So, ( z_2^2 = -frac{169}{512} - frac{55sqrt{3}}{512}i ).Add ( c ):( z_3 = z_2^2 + c = left( -frac{169}{512} - frac{55sqrt{3}}{512}i right) + left( frac{1}{4} + frac{sqrt{3}}{4}i right) ).Convert ( frac{1}{4} ) to 512nds: ( frac{128}{512} ).Convert ( frac{sqrt{3}}{4}i ) to 512nds: ( frac{128sqrt{3}}{512}i ).So,Real: ( -frac{169}{512} + frac{128}{512} = -frac{41}{512} ).Imaginary: ( -frac{55sqrt{3}}{512} + frac{128sqrt{3}}{512} = frac{73sqrt{3}}{512} ).Thus, ( z_3 = -frac{41}{512} + frac{73sqrt{3}}{512}i ).Compute ( z_4 = z_3^2 + c ).First, ( z_3 = -frac{41}{512} + frac{73sqrt{3}}{512}i ).Compute ( z_3^2 ):( a = -frac{41}{512} ), ( b = frac{73sqrt{3}}{512} ).( a^2 = left( -frac{41}{512} right)^2 = frac{1681}{262144} ).( b^2 = left( frac{73sqrt{3}}{512} right)^2 = frac{5329 * 3}{262144} = frac{15987}{262144} ).So, ( a^2 - b^2 = frac{1681 - 15987}{262144} = frac{-14306}{262144} = -frac{7153}{131072} ).Imaginary part: ( 2ab = 2 * (-frac{41}{512}) * frac{73sqrt{3}}{512} = -frac{2 * 41 * 73 sqrt{3}}{262144} ).Calculate numerator: 2 * 41 * 73 = 2 * 3000 + 2 * 13 = 6000 + 26 = 6026? Wait, no.Wait, 41 * 73: 40*73=2920, 1*73=73, so total 2920+73=2993. Then 2*2993=5986.So, imaginary part: ( -frac{5986sqrt{3}}{262144} ).Simplify: Divide numerator and denominator by 2: ( -frac{2993sqrt{3}}{131072} ).So, ( z_3^2 = -frac{7153}{131072} - frac{2993sqrt{3}}{131072}i ).Add ( c ):( z_4 = z_3^2 + c = left( -frac{7153}{131072} - frac{2993sqrt{3}}{131072}i right) + left( frac{1}{4} + frac{sqrt{3}}{4}i right) ).Convert ( frac{1}{4} ) to 131072 denominator: ( frac{32768}{131072} ).Convert ( frac{sqrt{3}}{4}i ) to 131072 denominator: ( frac{32768sqrt{3}}{131072}i ).So,Real: ( -frac{7153}{131072} + frac{32768}{131072} = frac{32768 - 7153}{131072} = frac{25615}{131072} ).Imaginary: ( -frac{2993sqrt{3}}{131072} + frac{32768sqrt{3}}{131072} = frac{32768 - 2993}{131072}sqrt{3} = frac{29775sqrt{3}}{131072} ).Thus, ( z_4 = frac{25615}{131072} + frac{29775sqrt{3}}{131072}i ).This is getting quite tedious, but let me try to compute a few more terms.Compute ( z_5 = z_4^2 + c ).First, ( z_4 = frac{25615}{131072} + frac{29775sqrt{3}}{131072}i ).Compute ( z_4^2 ):Let me denote ( a = frac{25615}{131072} ), ( b = frac{29775sqrt{3}}{131072} ).Compute ( a^2 ):( a^2 = left( frac{25615}{131072} right)^2 ). This is a very small number, but let me compute it:25615^2 = ?Wait, this is getting too big. Maybe I should use approximate decimal values to make it manageable.Let me approximate ( a ) and ( b ):( a = frac{25615}{131072} approx 0.1954 ).( b = frac{29775sqrt{3}}{131072} approx frac{29775 * 1.732}{131072} approx frac{51527.1}{131072} approx 0.3927 ).So, ( z_4 approx 0.1954 + 0.3927i ).Compute ( z_4^2 ):( (0.1954 + 0.3927i)^2 ).Using ( (x + yi)^2 = x^2 - y^2 + 2xyi ).Compute ( x^2 = 0.1954^2 approx 0.0382 ).Compute ( y^2 = 0.3927^2 approx 0.1542 ).So, ( x^2 - y^2 approx 0.0382 - 0.1542 = -0.116 ).Compute ( 2xy = 2 * 0.1954 * 0.3927 approx 2 * 0.0769 approx 0.1538 ).So, ( z_4^2 approx -0.116 + 0.1538i ).Add ( c approx 0.25 + 0.4330i ).So, ( z_5 approx (-0.116 + 0.25) + (0.1538 + 0.4330)i approx 0.134 + 0.5868i ).Compute modulus of ( z_5 ): ( sqrt{0.134^2 + 0.5868^2} approx sqrt{0.0179 + 0.3443} approx sqrt{0.3622} approx 0.602 ), which is less than 2.Proceeding to ( z_6 = z_5^2 + c ).( z_5 approx 0.134 + 0.5868i ).Compute ( z_5^2 ):( (0.134 + 0.5868i)^2 ).( x = 0.134 ), ( y = 0.5868 ).( x^2 = 0.0179 ).( y^2 = 0.3443 ).( x^2 - y^2 = 0.0179 - 0.3443 = -0.3264 ).( 2xy = 2 * 0.134 * 0.5868 approx 2 * 0.0786 approx 0.1572 ).So, ( z_5^2 approx -0.3264 + 0.1572i ).Add ( c approx 0.25 + 0.4330i ).( z_6 approx (-0.3264 + 0.25) + (0.1572 + 0.4330)i approx -0.0764 + 0.5902i ).Modulus of ( z_6 ): ( sqrt{(-0.0764)^2 + (0.5902)^2} approx sqrt{0.0058 + 0.3483} approx sqrt{0.3541} approx 0.595 ), still less than 2.Compute ( z_7 = z_6^2 + c ).( z_6 approx -0.0764 + 0.5902i ).Compute ( z_6^2 ):( (-0.0764 + 0.5902i)^2 ).( x = -0.0764 ), ( y = 0.5902 ).( x^2 = 0.0058 ).( y^2 = 0.3483 ).( x^2 - y^2 = 0.0058 - 0.3483 = -0.3425 ).( 2xy = 2 * (-0.0764) * 0.5902 approx -2 * 0.0451 approx -0.0902 ).So, ( z_6^2 approx -0.3425 - 0.0902i ).Add ( c approx 0.25 + 0.4330i ).( z_7 approx (-0.3425 + 0.25) + (-0.0902 + 0.4330)i approx -0.0925 + 0.3428i ).Modulus of ( z_7 ): ( sqrt{(-0.0925)^2 + (0.3428)^2} approx sqrt{0.0085 + 0.1175} approx sqrt{0.126} approx 0.355 ), still less than 2.Compute ( z_8 = z_7^2 + c ).( z_7 approx -0.0925 + 0.3428i ).Compute ( z_7^2 ):( (-0.0925 + 0.3428i)^2 ).( x = -0.0925 ), ( y = 0.3428 ).( x^2 = 0.0085 ).( y^2 = 0.1175 ).( x^2 - y^2 = 0.0085 - 0.1175 = -0.109 ).( 2xy = 2 * (-0.0925) * 0.3428 approx -2 * 0.0317 approx -0.0634 ).So, ( z_7^2 approx -0.109 - 0.0634i ).Add ( c approx 0.25 + 0.4330i ).( z_8 approx (-0.109 + 0.25) + (-0.0634 + 0.4330)i approx 0.141 + 0.3696i ).Modulus of ( z_8 ): ( sqrt{0.141^2 + 0.3696^2} approx sqrt{0.0199 + 0.1366} approx sqrt{0.1565} approx 0.395 ), still less than 2.Compute ( z_9 = z_8^2 + c ).( z_8 approx 0.141 + 0.3696i ).Compute ( z_8^2 ):( (0.141 + 0.3696i)^2 ).( x = 0.141 ), ( y = 0.3696 ).( x^2 = 0.0199 ).( y^2 = 0.1366 ).( x^2 - y^2 = 0.0199 - 0.1366 = -0.1167 ).( 2xy = 2 * 0.141 * 0.3696 approx 2 * 0.0522 approx 0.1044 ).So, ( z_8^2 approx -0.1167 + 0.1044i ).Add ( c approx 0.25 + 0.4330i ).( z_9 approx (-0.1167 + 0.25) + (0.1044 + 0.4330)i approx 0.1333 + 0.5374i ).Modulus of ( z_9 ): ( sqrt{0.1333^2 + 0.5374^2} approx sqrt{0.0178 + 0.2888} approx sqrt{0.3066} approx 0.553 ), still less than 2.Compute ( z_{10} = z_9^2 + c ).( z_9 approx 0.1333 + 0.5374i ).Compute ( z_9^2 ):( (0.1333 + 0.5374i)^2 ).( x = 0.1333 ), ( y = 0.5374 ).( x^2 = 0.0178 ).( y^2 = 0.2888 ).( x^2 - y^2 = 0.0178 - 0.2888 = -0.271 ).( 2xy = 2 * 0.1333 * 0.5374 approx 2 * 0.0717 approx 0.1434 ).So, ( z_9^2 approx -0.271 + 0.1434i ).Add ( c approx 0.25 + 0.4330i ).( z_{10} approx (-0.271 + 0.25) + (0.1434 + 0.4330)i approx -0.021 + 0.5764i ).Modulus of ( z_{10} ): ( sqrt{(-0.021)^2 + (0.5764)^2} approx sqrt{0.0004 + 0.3322} approx sqrt{0.3326} approx 0.576 ), still less than 2.So, after computing up to ( z_{10} ), all terms have modulus less than 2. This suggests that ( c = frac{1}{4} + frac{sqrt{3}}{4}i ) might be in the Mandelbrot set. However, to be certain, we would need to check infinitely many terms, which isn't practical. But since the modulus hasn't exceeded 2 in the first 10 iterations, it's a good sign. I recall that points inside the Mandelbrot set can have orbits that stay bounded indefinitely, while points outside eventually escape. Given that the modulus is decreasing and oscillating but staying below 2, it's likely that this ( c ) is in the set. But wait, let me think again. The initial terms are getting smaller, but sometimes orbits can escape after many iterations. However, since the modulus is consistently below 2 and not increasing, it's a strong indication that it's bounded. Alternatively, I can check if ( c ) is inside the main cardioid of the Mandelbrot set. The main cardioid is defined by ( c = frac{1 - e^{itheta}}{2} ) for ( theta ) in [0, 2π). Let's see if ( c = frac{1}{4} + frac{sqrt{3}}{4}i ) fits this.Express ( c ) in polar form. The modulus of ( c ) is ( sqrt{(frac{1}{4})^2 + (frac{sqrt{3}}{4})^2} = sqrt{frac{1}{16} + frac{3}{16}} = sqrt{frac{4}{16}} = frac{1}{2} ).The argument ( theta ) is ( arctanleft( frac{sqrt{3}/4}{1/4} right) = arctan(sqrt{3}) = frac{pi}{3} ).So, ( c = frac{1}{2} e^{ipi/3} ).Compare this to the main cardioid ( c = frac{1 - e^{itheta}}{2} ). Let me compute ( frac{1 - e^{ipi/3}}{2} ).( e^{ipi/3} = cos(pi/3) + isin(pi/3) = frac{1}{2} + ifrac{sqrt{3}}{2} ).So, ( 1 - e^{ipi/3} = 1 - frac{1}{2} - ifrac{sqrt{3}}{2} = frac{1}{2} - ifrac{sqrt{3}}{2} ).Divide by 2: ( frac{1}{4} - ifrac{sqrt{3}}{4} ).But our ( c ) is ( frac{1}{4} + ifrac{sqrt{3}}{4} ), which is the conjugate. So, it's not exactly on the main cardioid, but it's close. However, the main cardioid is the region where ( |c| leq frac{1}{4} ) and satisfies certain conditions. Wait, no, the main cardioid is actually the set where ( c ) is such that the orbit doesn't escape, and it's parameterized as ( c = frac{1 - e^{itheta}}{2} ). But our ( c ) has modulus ( frac{1}{2} ), which is larger than the maximum modulus of the main cardioid, which is ( frac{1}{2} ) (achieved at ( theta = pi )). Wait, actually, the main cardioid has points with modulus up to 1, but the maximum distance from the origin is 1. Wait, no, the main cardioid is given by ( c = frac{1 - e^{itheta}}{2} ), so the modulus is ( frac{1}{2} sqrt{(1 - costheta)^2 + (sintheta)^2} = frac{1}{2} sqrt{2 - 2costheta} ). The maximum modulus occurs when ( costheta = -1 ), giving modulus ( frac{1}{2} sqrt{4} = 1 ). So, the main cardioid extends from ( c = -1 ) to ( c = frac{1}{4} ).Wait, but our ( c ) is ( frac{1}{4} + frac{sqrt{3}}{4}i ), which is on the boundary of the main cardioid? Because when ( theta = pi/3 ), ( c = frac{1 - e^{ipi/3}}{2} = frac{1}{2} - frac{1}{4} - ifrac{sqrt{3}}{4} = frac{1}{4} - ifrac{sqrt{3}}{4} ). So, it's the conjugate of our ( c ). Therefore, our ( c ) is the reflection across the real axis of the point on the main cardioid. But since the Mandelbrot set is symmetric with respect to the real axis, if ( c ) is in the set, so is its conjugate. Therefore, since ( frac{1}{4} - ifrac{sqrt{3}}{4} ) is on the main cardioid, which is part of the Mandelbrot set, then ( frac{1}{4} + ifrac{sqrt{3}}{4} ) is also in the set.Therefore, ( c ) belongs to the Mandelbrot set.But just to be thorough, let me check if ( c ) is inside the main cardioid or in another part of the set. The main cardioid is the largest connected region of the Mandelbrot set. Points inside the main cardioid are known to be in the set. Since our ( c ) is on the boundary of the main cardioid, it's actually on the boundary, which is part of the set.Wait, no. The main cardioid is the region where the orbit remains bounded because the function ( f_c(z) = z^2 + c ) has an attracting fixed point. Points on the boundary of the main cardioid are the points where the fixed point becomes neutral, meaning the multiplier is 1, and these points are part of the Julia set but not necessarily the Mandelbrot set. Wait, no, the Mandelbrot set includes all points where the orbit doesn't escape, so the boundary is included.But in our case, ( c = frac{1}{4} + frac{sqrt{3}}{4}i ) is on the boundary of the main cardioid, so it's part of the Mandelbrot set.Alternatively, another way to check is that for ( c ) inside the main cardioid, the orbit converges to a fixed point. Let me see if that's the case here.The fixed points satisfy ( z = z^2 + c ), so ( z^2 - z + c = 0 ).For ( c = frac{1}{4} + frac{sqrt{3}}{4}i ), the fixed points are solutions to ( z^2 - z + frac{1}{4} + frac{sqrt{3}}{4}i = 0 ).Using quadratic formula:( z = frac{1 pm sqrt{1 - 4(frac{1}{4} + frac{sqrt{3}}{4}i)}}{2} ).Compute discriminant:( 1 - 4c = 1 - 4*(frac{1}{4} + frac{sqrt{3}}{4}i) = 1 - 1 - sqrt{3}i = -sqrt{3}i ).So, ( sqrt{-sqrt{3}i} ). Let me compute this.Express ( -sqrt{3}i ) in polar form: modulus is ( sqrt{3} ), argument is ( -pi/2 ).So, square root is modulus ( (sqrt{3})^{1/2} = 3^{1/4} ), argument ( -pi/4 ).So, ( sqrt{-sqrt{3}i} = 3^{1/4} (cos(-pi/4) + isin(-pi/4)) = 3^{1/4} left( frac{sqrt{2}}{2} - ifrac{sqrt{2}}{2} right) ).Thus, the fixed points are:( z = frac{1 pm 3^{1/4} left( frac{sqrt{2}}{2} - ifrac{sqrt{2}}{2} right)}{2} ).Compute the modulus of these fixed points to see if they are attracting (modulus of derivative less than 1).The derivative of ( f_c(z) = z^2 + c ) is ( f_c'(z) = 2z ).At the fixed point ( z ), the multiplier is ( 2z ). For the fixed point to be attracting, ( |2z| < 1 ).Compute ( |z| ) for the fixed points.Take ( z = frac{1 + 3^{1/4} (frac{sqrt{2}}{2} - ifrac{sqrt{2}}{2})}{2} ).Compute modulus squared:( |z|^2 = left( frac{1 + 3^{1/4} frac{sqrt{2}}{2}}{2} right)^2 + left( frac{-3^{1/4} frac{sqrt{2}}{2}}{2} right)^2 ).This is getting complicated, but let me approximate ( 3^{1/4} approx 1.316 ).So, ( 3^{1/4} frac{sqrt{2}}{2} approx 1.316 * 0.707 approx 0.931 ).Thus, real part: ( frac{1 + 0.931}{2} approx 0.9655 ).Imaginary part: ( frac{-0.931}{2} approx -0.4655 ).So, modulus squared: ( (0.9655)^2 + (-0.4655)^2 approx 0.932 + 0.216 approx 1.148 ).So, modulus ( |z| approx sqrt{1.148} approx 1.071 ).Thus, ( |2z| approx 2 * 1.071 approx 2.142 > 1 ). So, this fixed point is repelling.Similarly, the other fixed point would have modulus:( z = frac{1 - 3^{1/4} (frac{sqrt{2}}{2} - ifrac{sqrt{2}}{2})}{2} ).Compute real part: ( frac{1 - 0.931}{2} approx 0.0345 ).Imaginary part: ( frac{0.931}{2} approx 0.4655 ).Modulus squared: ( (0.0345)^2 + (0.4655)^2 approx 0.0012 + 0.216 approx 0.217 ).Modulus ( |z| approx sqrt{0.217} approx 0.466 ).Thus, ( |2z| approx 2 * 0.466 approx 0.932 < 1 ). So, this fixed point is attracting.Therefore, the orbit converges to this attracting fixed point, which means the sequence ( z_n ) remains bounded. Hence, ( c ) is indeed in the Mandelbrot set.So, summarizing part 2: After computing the first 10 terms, all have modulus less than 2, and by analyzing the fixed points, we confirmed that the orbit converges to an attracting fixed point, so ( c ) is in the Mandelbrot set.</think>

🧠:<think>Okay, so I'm trying to figure out these two sub-problems about Player A and Player B. Let me take it step by step.First, for Player A, his scoring follows a Poisson distribution with an average of 25 points per game. I need to calculate the probability that he scores exactly 30 points in a game. I remember that the Poisson probability formula is:P(X = k) = (λ^k * e^(-λ)) / k!Where λ is the average rate (which is 25 here), k is the number of occurrences (30 points), and e is the base of the natural logarithm, approximately 2.71828.So, plugging in the numbers:P(X = 30) = (25^30 * e^(-25)) / 30!Hmm, that seems like a big calculation. Maybe I can use a calculator or some software to compute this, but since I'm doing this manually, let me think if there's a way to approximate it or if there's a pattern.Wait, 25^30 is a massive number, and 30! is also huge. Maybe I can use logarithms to simplify the calculation? Or perhaps use the normal approximation to the Poisson distribution since λ is quite large (25). But I think the exact value is needed here, so I should stick with the Poisson formula.Alternatively, maybe I can use the fact that Poisson probabilities can be calculated using factorials and exponentials, but without a calculator, it's going to be tough. Maybe I can use the natural logarithm to compute the log of the probability and then exponentiate it?Let me try that. Taking the natural log of P(X=30):ln(P) = 30*ln(25) - 25 - ln(30!)Compute each term:ln(25) is approximately 3.2189, so 30*3.2189 ≈ 96.567.Then subtract 25: 96.567 - 25 = 71.567.Now, ln(30!) is the sum of ln(1) + ln(2) + ... + ln(30). I remember that ln(n!) can be approximated using Stirling's formula: ln(n!) ≈ n*ln(n) - n + 0.5*ln(2πn). Let's use that.So, ln(30!) ≈ 30*ln(30) - 30 + 0.5*ln(2π*30).Compute each part:30*ln(30): ln(30) ≈ 3.4012, so 30*3.4012 ≈ 102.036.Subtract 30: 102.036 - 30 = 72.036.0.5*ln(2π*30): 2π*30 ≈ 188.4956, ln(188.4956) ≈ 5.24, so 0.5*5.24 ≈ 2.62.Add that to 72.036: 72.036 + 2.62 ≈ 74.656.So ln(30!) ≈ 74.656.Therefore, ln(P) ≈ 71.567 - 74.656 ≈ -3.089.Exponentiate that: e^(-3.089) ≈ e^(-3) * e^(-0.089). e^(-3) ≈ 0.0498, and e^(-0.089) ≈ 0.915. Multiply them: 0.0498 * 0.915 ≈ 0.0456.So the probability is approximately 4.56%. Let me check if that makes sense. Since 30 is higher than the average of 25, the probability should be less than the peak probability at 25. I think that's reasonable.Now, moving on to Player B. His scoring follows a negative binomial distribution with parameters r = 10 and p = 0.5. I need to determine the expected number of games it takes for him to achieve a total of 200 points.Wait, the negative binomial distribution models the number of trials needed to achieve a certain number of successes. In this case, is each point a success? Or is each game a trial where he scores some points?I think in the context of basketball, each game is a trial, and the number of points scored per game follows a negative binomial distribution. But usually, the negative binomial is used for the number of trials needed to achieve a certain number of successes. So if r = 10 and p = 0.5, it's the number of games needed to score 10 points, with each game having a probability p of scoring a point? Hmm, that might not make sense because points are more than just binary successes.Wait, maybe I'm misunderstanding. Perhaps the negative binomial here is being used to model the number of points scored per game, where r is the number of successes (points) and p is the probability of scoring a point. But that might not be the standard interpretation.Alternatively, maybe the negative binomial distribution is being used to model the number of games needed to reach a certain number of points, with each game being a trial where he can score some points. But I'm not sure.Wait, the question says: "the expected number of games it takes for Player B to achieve a total of 200 points, assuming each game's point total follows the given negative binomial distribution."So each game's point total is negative binomial with r=10 and p=0.5. So each game, he scores X points, where X ~ Negative Binomial(r=10, p=0.5). Then, we need to find the expected number of games needed for the total points to reach 200.But wait, the negative binomial distribution models the number of trials needed to achieve a certain number of successes. So if each game is a trial, and each trial can result in a certain number of points, which follows a negative binomial distribution, that might not be the standard case.Alternatively, perhaps the negative binomial is being used to model the number of points per game, where r=10 and p=0.5. So each game, he scores X points, where X ~ Negative Binomial(r=10, p=0.5). Then, the expected number of points per game is r*(1-p)/p. Let me compute that.For a negative binomial distribution, the expected value E[X] = r*(1-p)/p. So here, r=10, p=0.5, so E[X] = 10*(1-0.5)/0.5 = 10*0.5/0.5 = 10. So he scores an average of 10 points per game.Wait, but the question is about the expected number of games to reach 200 points. If he averages 10 points per game, then the expected number of games would be 200 / 10 = 20 games. Is that correct?But wait, the negative binomial distribution is typically used for the number of trials needed to achieve a certain number of successes, but in this case, each game's points are negative binomial. So the total points after n games would be the sum of n independent negative binomial random variables. But the sum of negative binomials isn't straightforward.Wait, actually, if each game's points are Negative Binomial(r=10, p=0.5), then the total points after n games would be the sum of n such variables. But the expectation of the sum is n*E[X] = n*10. So to reach 200 points, we set n*10 = 200, so n=20. Therefore, the expected number of games is 20.But let me double-check. The negative binomial distribution can also be parameterized in terms of the number of successes and the probability. The expectation is indeed r*(1-p)/p. So with r=10, p=0.5, E[X]=10*(0.5)/0.5=10. So each game, he scores on average 10 points. Therefore, to get 200 points, it's 200/10=20 games.Wait, but is this correct? Because the negative binomial distribution is often used for counts, but in this case, it's being used to model points per game. So each game, he scores a number of points following Negative Binomial(r=10, p=0.5). So the expectation per game is 10 points, so over n games, the expectation is 10n. To reach 200, n=20.Yes, that seems correct. So the expected number of games is 20.Wait, but let me think again. The negative binomial distribution is often used for the number of trials needed to achieve r successes, with each trial having a success probability p. So in that case, the expectation is r/p. But in our case, if each game's points are Negative Binomial(r=10, p=0.5), then the expectation is 10*(1-p)/p = 10*(0.5)/0.5=10, as I had before. So yes, each game averages 10 points.Therefore, to reach 200 points, the expected number of games is 20.So, summarizing:1. Player A's probability of scoring exactly 30 points is approximately 4.56%.2. Player B's expected number of games to reach 200 points is 20.Now, comparing the two players:Player A has a Poisson distribution, which is memoryless and has a single parameter λ=25. His scoring is consistent around 25 points per game, with the probability of scoring exactly 30 being about 4.56%, which is lower than the peak at 25 but not extremely rare.Player B, on the other hand, has a negative binomial distribution with r=10 and p=0.5, giving him an average of 10 points per game. To reach 200 points, he needs on average 20 games. This suggests that Player B is less efficient in scoring, averaging only 10 points per game compared to Player A's 25. However, the negative binomial distribution also implies higher variance in scoring, meaning Player B might have games with significantly higher or lower points than the average.In terms of consistency, Player A is more consistent because the Poisson distribution has variance equal to the mean, so his scoring doesn't vary as much as Player B's, whose negative binomial distribution has variance r*(1-p)/p^2 = 10*(0.5)/0.25 = 20, which is higher than Player A's variance of 25. Wait, no, Player A's variance is 25, and Player B's variance per game is 20, but Player B's average is 10, so the coefficient of variation is higher for Player B, indicating less consistency.Wait, actually, variance for Player B per game is r*(1-p)/p^2 = 10*(0.5)/(0.5)^2 = 10*0.5/0.25 = 20. So variance is 20, standard deviation is sqrt(20) ≈ 4.47. For Player A, variance is 25, standard deviation 5. So actually, Player A has a higher variance per game, but his average is higher too. The coefficient of variation (standard deviation / mean) for Player A is 5/25=0.2, while for Player B it's 4.47/10≈0.447. So Player B is less consistent relative to his average.Therefore, in terms of efficiency, Player A is more efficient, scoring more points per game on average, and while he has a higher absolute variance, his relative consistency (lower coefficient of variation) is better than Player B's.So, in conclusion, Player A is more efficient and more consistent in his scoring compared to Player B.</think>

🧠:<think>Okay, so I have this problem where Alice is modeling her emotional state with a differential equation. The equation is:[ frac{dE}{dt} = -a(E - b sin(omega t)) - c sin(E) ]And there are two sub-problems to solve. Let me start with the first one.Sub-problem 1: Conditions for a Stable Periodic SolutionAlright, so the question is asking about the conditions on the parameters ( a ), ( b ), ( c ), and ( omega ) such that the emotional state ( E(t) ) has a stable periodic solution. The equation is a non-autonomous differential equation because of the ( sin(omega t) ) term. So, it's time-dependent.First, I remember that for periodic solutions in differential equations, especially non-autonomous ones, concepts like limit cycles and Poincaré-Bendixson theorem come into play. But since this is a second-order system if we consider time, maybe I should think about it as a first-order non-autonomous equation.Alternatively, I can think of this as a forced oscillator. The term ( -a(E - b sin(omega t)) ) can be seen as a linear damping term with a periodic forcing, and the ( -c sin(E) ) term is a nonlinear restoring force.So, perhaps this is similar to a Duffing oscillator, which is a nonlinear oscillator with a periodic forcing term. The Duffing equation is:[ ddot{x} + delta dot{x} + alpha x + beta x^3 = gamma cos(omega t) ]But in our case, the equation is first-order. Hmm. Maybe I should rewrite the given equation to see if it can be transformed into a more familiar form.Let me write the equation again:[ frac{dE}{dt} = -aE + ab sin(omega t) - c sin(E) ]So, it's a first-order equation with a linear term, a periodic forcing term, and a nonlinear sine term.I think for such equations, the existence of a periodic solution can be approached using the Poincaré-Lindstedt method or by using the concept of averaging. Alternatively, since it's a first-order equation, maybe we can analyze its fixed points or use some kind of phase plane analysis.But since it's non-autonomous, fixed points aren't straightforward. Instead, we might look for periodic solutions where the system returns to the same state after a period ( T = 2pi / omega ).One approach is to consider the equation over one period and see if it's possible for the solution to repeat. For that, maybe we can use the concept of a Poincaré map, which maps the state of the system at time ( t ) to its state at ( t + T ). If the map has a fixed point, then the solution is periodic with period ( T ).But I'm not sure if I can compute that analytically. Maybe I need to consider some approximations or look for conditions where the nonlinear term is small or the forcing is weak.Alternatively, I can think about the equation in terms of its amplitude and phase. Let me consider that ( E(t) ) can be expressed as a sum of a steady-state periodic solution and a transient solution. If the transient solution decays over time, then the steady-state solution will dominate, which would be the periodic solution we're looking for.So, maybe I can assume a solution of the form:[ E(t) = E_p(t) + E_h(t) ]Where ( E_p(t) ) is the particular solution (steady-state) and ( E_h(t) ) is the homogeneous solution (transient).Substituting into the differential equation:[ frac{d}{dt}(E_p + E_h) = -a(E_p + E_h - b sin(omega t)) - c sin(E_p + E_h) ]Expanding this:[ frac{dE_p}{dt} + frac{dE_h}{dt} = -aE_p - aE_h + ab sin(omega t) - c sin(E_p + E_h) ]Now, if ( E_p(t) ) is a particular solution, it should satisfy the equation without the transient term. So, perhaps:[ frac{dE_p}{dt} = -a(E_p - b sin(omega t)) - c sin(E_p) ]And the homogeneous equation would be:[ frac{dE_h}{dt} = -aE_h - c [sin(E_p + E_h) - sin(E_p)] ]Assuming that ( E_h ) is small, we can approximate ( sin(E_p + E_h) approx sin(E_p) + E_h cos(E_p) ) using the Taylor expansion. So, the homogeneous equation becomes:[ frac{dE_h}{dt} approx -aE_h - c E_h cos(E_p) ]Which simplifies to:[ frac{dE_h}{dt} = -[a + c cos(E_p)] E_h ]This is a linear differential equation for ( E_h ). The solution will be:[ E_h(t) = E_h(0) expleft(-int_0^t [a + c cos(E_p(s))] ds right) ]For the transient solution ( E_h(t) ) to decay to zero, the exponent must be negative, which it is because ( a ) and ( c ) are positive constants, and ( cos(E_p) ) is bounded between -1 and 1. So, unless ( cos(E_p) ) is very negative, the exponent will be negative, leading to decay.But wait, ( cos(E_p) ) could be negative if ( E_p ) is in certain ranges. However, since ( a ) is positive and multiplied by the integral of 1, which is just ( t ), the dominant term is ( -a t ), so the exponential will decay regardless. Therefore, the homogeneous solution decays, and the particular solution ( E_p(t) ) becomes the stable periodic solution.Therefore, the system will have a stable periodic solution if the particular solution exists. But does it always exist?Hmm, maybe not. The existence of a particular solution depends on the parameters. For example, if the nonlinear term ( -c sin(E) ) is too strong, it might prevent the system from settling into a periodic solution.Alternatively, if the damping ( a ) is strong enough, it can suppress the nonlinear effects and allow the system to follow the periodic forcing.So, perhaps the condition is that ( a ) is sufficiently large compared to ( c ), so that the linear damping dominates the nonlinear term. Or maybe the forcing amplitude ( b ) is not too large.Alternatively, another approach is to consider the equation as a perturbation from the linear case. If ( c ) is small, then the nonlinear term is a perturbation, and the system will have a periodic solution close to the linear solution.But if ( c ) is not small, the nonlinear term can cause more complex behavior, like amplitude modulation or even chaos.Wait, but the question is about stable periodic solutions. So, I think that if the system is not too nonlinear, meaning ( c ) is not too large, and the damping ( a ) is sufficient, then the system will have a stable periodic solution.Alternatively, maybe we can use the concept of the Melnikov method to determine the conditions for the existence of periodic solutions in perturbed systems. But that might be more complicated.Alternatively, let's consider the case when ( c = 0 ). Then the equation becomes linear:[ frac{dE}{dt} = -aE + ab sin(omega t) ]This has a unique periodic solution which is stable, given by:[ E(t) = frac{ab}{sqrt{a^2 + (a omega)^2}} sin(omega t - phi) ]Where ( phi = arctan(omega) ). So, in this case, the solution is periodic and stable.Now, when we introduce the nonlinear term ( -c sin(E) ), it will perturb this solution. For small ( c ), the solution will still be approximately periodic, but with some amplitude modulation.But for larger ( c ), the nonlinear term can cause the system to have different behaviors, such as period-doubling bifurcations or even chaotic behavior.Therefore, to have a stable periodic solution, the parameter ( c ) should be small enough so that the nonlinear term doesn't destabilize the periodic solution.Alternatively, another way to think about it is to consider the equation in the phase plane. If we can show that all trajectories spiral towards a limit cycle, then that limit cycle is the stable periodic solution.But since it's a first-order equation, the phase plane is 1-dimensional, so it's just the E-axis. So, maybe not as straightforward.Alternatively, perhaps we can use the concept of Lyapunov functions. If we can find a function that decreases along the trajectories except at the periodic solution, then that would indicate stability.But I'm not sure how to construct such a function here.Alternatively, maybe we can use numerical methods to explore the behavior for different parameter values.But since the question asks for conditions on the parameters, perhaps we can reason that for the system to have a stable periodic solution, the damping ( a ) should be sufficiently large to counteract the nonlinear term ( c sin(E) ), and the forcing amplitude ( b ) should not be too large to cause the system to escape the periodic behavior.Alternatively, maybe the ratio of ( a ) to ( c ) is important. If ( a ) is much larger than ( c ), then the linear damping dominates the nonlinear term, leading to a stable periodic solution.Similarly, the frequency ( omega ) might play a role in resonance. If ( omega ) is near a natural frequency of the system, it might amplify the response, but since it's a first-order system, the natural frequency is determined by ( a ). So, if ( omega ) is near ( a ), maybe it causes resonance, but I'm not sure.Wait, in the linear case, the frequency response is determined by the damping. The amplitude of the steady-state solution is:[ frac{ab}{sqrt{a^2 + (a omega)^2}} ]So, as ( omega ) increases, the amplitude decreases. So, higher frequencies lead to smaller amplitudes.But when we add the nonlinear term, the behavior might change. For example, if the amplitude is small, the nonlinear term ( -c sin(E) ) is approximately ( -c E ), so it's like adding another linear damping term. So, the total damping would be ( a + c ), making the system more stable.But if the amplitude is large, the nonlinear term can cause more complex behavior.Therefore, perhaps if the amplitude of the forcing ( b ) is small, the system remains in a regime where the nonlinear term doesn't cause instability.So, putting it all together, the conditions for a stable periodic solution are likely:1. The damping ( a ) is sufficiently large compared to the nonlinear parameter ( c ), so that the linear term dominates.2. The forcing amplitude ( b ) is not too large, so that the response amplitude doesn't become too big, which would make the nonlinear term significant.3. The frequency ( omega ) is such that it doesn't cause resonance or other instabilities, but since it's a first-order system, resonance is less of an issue compared to second-order systems.Alternatively, maybe the key condition is that ( a ) is large enough to ensure that the homogeneous solution decays, and the nonlinear term doesn't disrupt the periodicity.But I'm not entirely sure. Maybe I should look for some references or similar problems.Wait, I recall that for the equation ( frac{dE}{dt} = -aE + f(t) - c sin(E) ), the existence of a periodic solution can be shown using the Poincaré-Bohl theorem if the function is periodic and some conditions on the forcing are met.Alternatively, maybe we can use the concept of the implicit function theorem, treating the nonlinear term as a perturbation.But perhaps I should consider the equation in the form:[ frac{dE}{dt} + aE = ab sin(omega t) - c sin(E) ]This is a linear nonhomogeneous equation with a nonlinear term. If we can write this as:[ frac{dE}{dt} + aE + c sin(E) = ab sin(omega t) ]Then, if the left-hand side is a dissipative operator, meaning that it tends to bring the system to a stable state, then the forcing on the right-hand side can lead to a periodic solution.But I'm not sure about the exact conditions.Alternatively, maybe we can use the concept of the existence of a unique periodic solution when the nonhomogeneous term is periodic and the homogeneous equation is dissipative.In our case, the homogeneous equation is:[ frac{dE}{dt} + aE + c sin(E) = 0 ]Which can be written as:[ frac{dE}{dt} = -aE - c sin(E) ]This is an autonomous equation. Let's analyze its fixed points.Fixed points occur when ( frac{dE}{dt} = 0 ), so:[ -aE - c sin(E) = 0 ][ aE + c sin(E) = 0 ]This equation has solutions where ( E = 0 ) is a solution since ( a*0 + c sin(0) = 0 ). Are there other solutions?Let me consider the function ( f(E) = aE + c sin(E) ). The derivative is ( f'(E) = a + c cos(E) ). Since ( a > 0 ) and ( c > 0 ), ( f'(E) ) is always positive because ( cos(E) ) is between -1 and 1, so the minimum value of ( f'(E) ) is ( a - c ). If ( a > c ), then ( f'(E) > 0 ) for all ( E ), meaning ( f(E) ) is strictly increasing, so the only solution is ( E = 0 ).If ( a = c ), then ( f'(E) ) could be zero at some points, but ( f(E) ) would still be increasing overall because ( a = c ), so ( f'(E) = a(1 + cos(E)/a) ). Wait, no, ( f'(E) = a + c cos(E) ). If ( a = c ), then ( f'(E) = a(1 + cos(E)) ). So, ( f'(E) ) is non-negative, but can be zero when ( cos(E) = -1 ), i.e., at ( E = pi ). So, ( f(E) ) would have a minimum at ( E = pi ), but since ( f(pi) = a pi + c sin(pi) = a pi > 0 ), the function is still strictly increasing after that point.Wait, no, if ( f'(E) = a + c cos(E) ), then at ( E = pi ), ( f'(E) = a - c ). If ( a = c ), then ( f'(E) = 0 ) at ( E = pi ). But ( f(E) ) at ( E = pi ) is ( a pi + c sin(pi) = a pi ), which is positive. So, the function ( f(E) ) is increasing for all ( E ) when ( a geq c ), meaning the only fixed point is ( E = 0 ).If ( a < c ), then ( f'(E) = a + c cos(E) ) can be negative when ( cos(E) < -a/c ). So, the function ( f(E) ) can have regions where it's decreasing. Therefore, ( f(E) = 0 ) might have multiple solutions.But in our case, since ( a ), ( b ), ( c ), and ( omega ) are positive constants, and we are looking for a stable periodic solution, perhaps the key is that the homogeneous equation has a stable fixed point at ( E = 0 ) when ( a geq c ). So, if ( a geq c ), the fixed point is stable, and the nonhomogeneous term ( ab sin(omega t) ) will cause the system to oscillate around this fixed point, leading to a stable periodic solution.Therefore, the condition for a stable periodic solution is that ( a geq c ). Additionally, the forcing term ( b ) should not be too large to cause the system to escape the stable region. But since ( a ) is damping, as long as ( a geq c ), the system should be able to follow the periodic forcing without blowing up.So, putting it together, the conditions are:1. ( a geq c ): Ensures that the homogeneous equation has a stable fixed point, so the nonlinear term doesn't cause instability.2. ( b ) is not excessively large: Although ( a ) damps the system, too large a ( b ) could cause the response to be too large, potentially leading to more complex dynamics. However, since ( a ) is damping, even with large ( b ), the system might still settle into a periodic solution, but perhaps with a larger amplitude.Alternatively, maybe ( b ) can be any positive value as long as ( a geq c ), because the damping will always bring the system back.But I'm not entirely sure. Maybe I should consider the amplitude of the steady-state solution.In the linear case (( c = 0 )), the amplitude is ( frac{ab}{sqrt{a^2 + (a omega)^2}} ). So, as ( b ) increases, the amplitude increases, but it's still bounded by the damping ( a ).When ( c > 0 ), the amplitude might be different, but as long as ( a geq c ), the system should still be stable.Therefore, I think the main condition is ( a geq c ). The other parameters ( b ) and ( omega ) can be positive, but ( b ) shouldn't be so large that it causes the system to become unstable, but with ( a geq c ), it might still be okay.Alternatively, maybe ( omega ) doesn't play a role in the stability as much as ( a ) and ( c ).So, to summarize, the conditions for a stable periodic solution are:- ( a geq c ): Ensures that the damping is sufficient to counteract the nonlinear term.- ( b ) is positive: Since ( b ) is the amplitude of the forcing, it just needs to be positive, but not necessarily bounded unless ( a ) is too small.But since ( a ) is already required to be at least ( c ), maybe ( b ) can be any positive value.Therefore, the answer to sub-problem 1 is that the system has a stable periodic solution if ( a geq c ).Sub-problem 2: Average Trust Level Over One CycleNow, moving on to the second sub-problem. Alice introduces a trust factor ( T(t) = int_0^t E(s) , ds ). We need to analyze how changes in ( b ) affect the average trust level over the period ( [0, T] ), where ( T = 2pi / omega ).Specifically, we need to find the average value of ( T(t) ) over one complete cycle and discuss the implications of increasing or decreasing ( b ) on this average value.First, let's note that ( T(t) ) is the integral of ( E(s) ) from 0 to ( t ). So, the average value of ( T(t) ) over one period ( [0, T] ) is:[ text{Average } T = frac{1}{T} int_0^T T(t) , dt = frac{1}{T} int_0^T left( int_0^t E(s) , ds right) dt ]This is a double integral. Let's switch the order of integration. The region of integration is ( 0 leq s leq t leq T ). So, switching the order:[ text{Average } T = frac{1}{T} int_0^T left( int_s^T dt right) E(s) , ds ]The inner integral ( int_s^T dt = T - s ). Therefore:[ text{Average } T = frac{1}{T} int_0^T (T - s) E(s) , ds ]Which simplifies to:[ text{Average } T = frac{1}{T} left( T int_0^T E(s) , ds - int_0^T s E(s) , ds right) ]So,[ text{Average } T = int_0^T E(s) , ds - frac{1}{T} int_0^T s E(s) , ds ]But wait, let's compute this step by step.First, let me compute ( int_0^T T(t) , dt ):[ int_0^T T(t) , dt = int_0^T left( int_0^t E(s) , ds right) dt ]Changing the order of integration:[ = int_0^T left( int_s^T dt right) E(s) , ds = int_0^T (T - s) E(s) , ds ]So, the average is:[ frac{1}{T} int_0^T (T - s) E(s) , ds ]Which is:[ int_0^T E(s) , ds - frac{1}{T} int_0^T s E(s) , ds ]But let's denote ( int_0^T E(s) , ds = M ), the total integral over one period, and ( int_0^T s E(s) , ds = N ). Then,[ text{Average } T = M - frac{N}{T} ]But I'm not sure if this helps directly. Alternatively, maybe we can express ( T(t) ) in terms of ( E(t) ) and then find its average.But perhaps a better approach is to note that ( T(t) ) is the integral of ( E(s) ) from 0 to ( t ). So, over one period, the average of ( T(t) ) is the average of the integral of ( E(s) ) over each point in the period.But maybe it's easier to consider that ( T(t) ) is a function whose derivative is ( E(t) ). Therefore, the average of ( T(t) ) over one period is related to the integral of ( E(t) ) over that period.Wait, let's think about it differently. The average value of ( T(t) ) over ( [0, T] ) is:[ frac{1}{T} int_0^T T(t) , dt = frac{1}{T} int_0^T left( int_0^t E(s) , ds right) dt ]As we did before, which is equal to:[ frac{1}{T} int_0^T (T - s) E(s) , ds ]So, this is equal to:[ int_0^T E(s) , ds - frac{1}{T} int_0^T s E(s) , ds ]But let's compute ( int_0^T E(s) , ds ). Let's denote this as ( M ). Then, the average ( T ) is ( M - frac{N}{T} ), where ( N = int_0^T s E(s) , ds ).But how does this relate to ( b )?We need to find how the average ( T ) depends on ( b ). So, perhaps we can express ( M ) and ( N ) in terms of ( b ).But to do that, we need to know the form of ( E(t) ). However, since the equation is nonlinear, we can't solve it analytically easily. So, maybe we can make some approximations.Assuming that ( c ) is small, so the nonlinear term is a perturbation. Then, we can approximate ( E(t) ) as the sum of the linear solution and a small correction due to the nonlinear term.In the linear case (( c = 0 )), the solution is:[ E(t) = frac{ab}{sqrt{a^2 + (a omega)^2}} sin(omega t - phi) ]Where ( phi = arctan(omega) ).So, in this case, ( E(t) ) is a sinusoidal function with amplitude ( frac{ab}{sqrt{a^2 + (a omega)^2}} ).Then, the integral ( M = int_0^T E(s) , ds ) over one period is zero because it's a sine wave over a full period.Similarly, ( N = int_0^T s E(s) , ds ). Let's compute this.Since ( E(s) ) is ( sin(omega s - phi) ), we can write:[ N = int_0^T s sin(omega s - phi) , ds ]Using integration by parts, let ( u = s ), ( dv = sin(omega s - phi) ds ). Then, ( du = ds ), ( v = -frac{1}{omega} cos(omega s - phi) ).So,[ N = -frac{s}{omega} cos(omega s - phi) bigg|_0^T + frac{1}{omega} int_0^T cos(omega s - phi) , ds ]Compute the boundary term:At ( s = T ), ( omega T = 2pi ), so ( cos(omega T - phi) = cos(2pi - phi) = cos(phi) ).At ( s = 0 ), ( cos(-phi) = cos(phi) ).So,[ -frac{T}{omega} cos(phi) + frac{0}{omega} cos(-phi) = -frac{T}{omega} cos(phi) ]The integral term:[ frac{1}{omega} int_0^T cos(omega s - phi) , ds = frac{1}{omega^2} sin(omega s - phi) bigg|_0^T = frac{1}{omega^2} [sin(2pi - phi) - sin(-phi)] = frac{1}{omega^2} [-sin(phi) + sin(phi)] = 0 ]Therefore, ( N = -frac{T}{omega} cos(phi) ).But ( phi = arctan(omega) ), so ( cos(phi) = frac{a}{sqrt{a^2 + (a omega)^2}} ).Wait, let's compute ( cos(phi) ). Since ( phi = arctan(omega) ), we can think of a right triangle where the opposite side is ( omega ) and the adjacent side is 1, so the hypotenuse is ( sqrt{1 + omega^2} ). Therefore, ( cos(phi) = frac{1}{sqrt{1 + omega^2}} ).But in our case, the amplitude is ( frac{ab}{sqrt{a^2 + (a omega)^2}} = frac{ab}{a sqrt{1 + omega^2}}} = frac{b}{sqrt{1 + omega^2}} ).Wait, so ( cos(phi) = frac{a}{sqrt{a^2 + (a omega)^2}} = frac{1}{sqrt{1 + omega^2}} ).Therefore, ( N = -frac{T}{omega} cdot frac{1}{sqrt{1 + omega^2}} ).But ( T = frac{2pi}{omega} ), so:[ N = -frac{2pi}{omega^2} cdot frac{1}{sqrt{1 + omega^2}} ]Therefore, the average ( T ) is:[ text{Average } T = M - frac{N}{T} = 0 - frac{ - frac{2pi}{omega^2} cdot frac{1}{sqrt{1 + omega^2}} }{ frac{2pi}{omega} } = frac{ frac{2pi}{omega^2} cdot frac{1}{sqrt{1 + omega^2}} }{ frac{2pi}{omega} } = frac{1}{omega sqrt{1 + omega^2}} ]But wait, this is in the linear case where ( c = 0 ). However, in our problem, ( c ) is not zero, but we assumed it's small. So, the average ( T ) in the linear case is ( frac{1}{omega sqrt{1 + omega^2}} ).But this seems dimensionally inconsistent because ( T ) has units of time, and ( frac{1}{omega sqrt{1 + omega^2}} ) has units of time squared. Wait, no, actually, ( omega ) has units of 1/time, so ( frac{1}{omega} ) has units of time, and ( sqrt{1 + omega^2} ) is dimensionless? Wait, no, ( omega ) is in radians per second, which is 1/seconds, so ( omega^2 ) is 1/seconds squared, so ( 1 + omega^2 ) is not dimensionally consistent. Wait, that can't be right.Wait, no, actually, ( omega ) is just a constant, so in the expression ( sqrt{1 + omega^2} ), both terms must be dimensionless. Therefore, ( omega ) must be dimensionless, which it is because it's a frequency in radians, which is a dimensionless quantity.Wait, no, actually, frequency has units of 1/time, but in the equation, ( omega t ) is dimensionless because it's inside the sine function. So, ( omega ) has units of 1/time, and ( t ) has units of time, so ( omega t ) is dimensionless.Therefore, in the expression ( sqrt{1 + omega^2} ), ( omega^2 ) has units of 1/time squared, so we cannot add 1 (dimensionless) to it. Therefore, my earlier assumption must be wrong.Wait, actually, in the equation, the term is ( sin(omega t) ), so ( omega ) has units of 1/time, and ( omega t ) is dimensionless. Therefore, in the expression ( sqrt{1 + omega^2} ), ( omega^2 ) has units of 1/time squared, so we cannot add 1 to it. Therefore, my earlier calculation must have a mistake.Wait, let's go back. When I computed ( cos(phi) ), I said ( phi = arctan(omega) ). But in reality, in the linear solution, the phase shift ( phi ) is given by ( tan(phi) = frac{omega}{a} ), because the standard form is:For the equation ( frac{dE}{dt} + aE = ab sin(omega t) ), the particular solution is:[ E_p(t) = frac{ab}{sqrt{a^2 + (a omega)^2}} sin(omega t - phi) ]Where ( tan(phi) = frac{omega}{a} ). Therefore, ( phi = arctanleft( frac{omega}{a} right) ).Therefore, ( cos(phi) = frac{a}{sqrt{a^2 + (a omega)^2}} = frac{1}{sqrt{1 + omega^2}} ).Wait, no, because ( tan(phi) = frac{omega}{a} ), so in the right triangle, opposite side is ( omega ), adjacent is ( a ), hypotenuse is ( sqrt{a^2 + omega^2} ). Therefore, ( cos(phi) = frac{a}{sqrt{a^2 + omega^2}} ).Therefore, ( cos(phi) = frac{a}{sqrt{a^2 + omega^2}} ).So, going back to ( N ):[ N = -frac{T}{omega} cos(phi) = -frac{2pi}{omega^2} cdot frac{a}{sqrt{a^2 + omega^2}} ]Therefore, the average ( T ) is:[ text{Average } T = M - frac{N}{T} = 0 - frac{ - frac{2pi a}{omega^2 sqrt{a^2 + omega^2}} }{ frac{2pi}{omega} } = frac{ frac{2pi a}{omega^2 sqrt{a^2 + omega^2}} }{ frac{2pi}{omega} } = frac{a}{omega sqrt{a^2 + omega^2}} ]So, in the linear case, the average trust level is ( frac{a}{omega sqrt{a^2 + omega^2}} ).But in our problem, ( c ) is not zero, but we assumed it's small. So, the average ( T ) would be approximately the same as in the linear case, plus a small correction due to ( c ).But since the problem asks for how changes in ( b ) affect the average trust level, and in the linear case, the average ( T ) is independent of ( b ). Wait, that can't be right because ( E(t) ) depends on ( b ).Wait, no, in the linear case, ( E(t) ) is proportional to ( b ), so ( E(t) = k b sin(omega t - phi) ), where ( k = frac{a}{sqrt{a^2 + omega^2}} ).Therefore, the integral ( M = int_0^T E(s) , ds = 0 ), as before.But ( N = int_0^T s E(s) , ds = k b cdot N_0 ), where ( N_0 ) is the integral we computed earlier.Therefore, the average ( T ) is:[ text{Average } T = M - frac{N}{T} = 0 - frac{ k b N_0 }{ T } = -k b cdot frac{N_0}{T} ]But from earlier, ( N_0 = -frac{2pi a}{omega^2 sqrt{a^2 + omega^2}} ), so:[ text{Average } T = -k b cdot left( -frac{2pi a}{omega^2 sqrt{a^2 + omega^2}} right) / T ]But ( T = frac{2pi}{omega} ), so:[ text{Average } T = k b cdot frac{2pi a}{omega^2 sqrt{a^2 + omega^2}} cdot frac{omega}{2pi} = k b cdot frac{a}{omega sqrt{a^2 + omega^2}} ]But ( k = frac{a}{sqrt{a^2 + omega^2}} ), so:[ text{Average } T = frac{a}{sqrt{a^2 + omega^2}} cdot b cdot frac{a}{omega sqrt{a^2 + omega^2}} = frac{a^2 b}{omega (a^2 + omega^2)} ]Therefore, in the linear case, the average trust level is proportional to ( b ). So, increasing ( b ) increases the average trust level, and decreasing ( b ) decreases it.But in our problem, ( c ) is not zero, but if ( c ) is small, the average ( T ) would still be approximately proportional to ( b ), with a small correction.Therefore, the conclusion is that increasing ( b ) increases the average trust level over one cycle, and decreasing ( b ) decreases it.But wait, in the linear case, the average ( T ) is ( frac{a^2 b}{omega (a^2 + omega^2)} ), which is proportional to ( b ). So, yes, the average trust level is directly proportional to ( b ).Therefore, the answer is that increasing ( b ) increases the average trust level, and decreasing ( b ) decreases it.But let me double-check. If ( b ) increases, the emotional state ( E(t) ) has a larger amplitude, so the integral ( T(t) ) would accumulate more over time, leading to a higher average trust level. Conversely, if ( b ) decreases, the emotional fluctuations are smaller, so the integral ( T(t) ) grows more slowly, leading to a lower average trust level.Yes, that makes sense.Final Answer1. The system has a stable periodic solution if ( boxed{a geq c} ).2. Increasing ( b ) increases the average trust level, while decreasing ( b ) decreases it. Therefore, the average value of ( T(t) ) over one complete cycle is directly proportional to ( b ), so the answer is ( boxed{text{Increasing } b text{ increases the average trust level}} ).</think>

🧠:<think>Okay, so I have this problem about an advocate for Aboriginal rights who is organizing a campaign. They need help from truck drivers to distribute flyers along a 500-kilometer highway. There are two sub-problems here, and I need to solve both. Let me take them one by one.Starting with Sub-problem 1: A group of truck drivers is helping distribute flyers. Each driver can cover a 50-kilometer stretch. The distribution is modeled by a Poisson process with an average rate of 2 flyers per kilometer. I need to find the probability that a single truck driver will distribute exactly 100 flyers in their 50-kilometer stretch.Hmm, okay. So, first, I remember that a Poisson process models the number of events happening in a fixed interval of time or space. In this case, the space is 50 kilometers, and the rate is 2 flyers per kilometer. So, the average number of flyers distributed over 50 kilometers would be the rate multiplied by the distance, right?Let me write that down. The average rate λ (lambda) is 2 flyers/km. The length of the stretch is 50 km. So, the expected number of flyers, which is λ_total, is 2 * 50 = 100 flyers. So, the Poisson distribution here has a parameter λ = 100.Now, the question is asking for the probability that exactly 100 flyers are distributed. In Poisson terms, that's P(X = 100), where X is the number of flyers.The formula for the Poisson probability mass function is:P(X = k) = (e^{-λ} * λ^k) / k!So, plugging in the numbers, we have:P(X = 100) = (e^{-100} * 100^{100}) / 100!But wait, calculating this directly might be tricky because 100! is a huge number, and e^{-100} is a very small number. I think I might need to use an approximation here because calculating this exactly would be computationally intensive.I remember that when λ is large, the Poisson distribution can be approximated by a normal distribution with mean μ = λ and variance σ² = λ. So, in this case, μ = 100 and σ = sqrt(100) = 10.But wait, the question is about the exact probability, not an approximation. Hmm, maybe I should think if there's a better way. Alternatively, perhaps using the normal approximation is acceptable here because λ is quite large (100). Let me see.But before I decide on that, let me recall that for Poisson distributions, when λ is large, the distribution is approximately normal. So, using the normal approximation with continuity correction might give a good estimate.So, if I use the normal approximation, I can model X ~ N(100, 10²). Then, the probability P(X = 100) can be approximated by the probability that X is between 99.5 and 100.5.So, let's compute that. First, convert these to z-scores.Z1 = (99.5 - 100) / 10 = (-0.5)/10 = -0.05Z2 = (100.5 - 100)/10 = 0.5/10 = 0.05Now, look up the standard normal distribution table for these z-scores.The cumulative probability for Z = -0.05 is approximately 0.4801, and for Z = 0.05 is approximately 0.5199.So, the probability between Z = -0.05 and Z = 0.05 is 0.5199 - 0.4801 = 0.0398.So, approximately 3.98% chance.But wait, is this a good approximation? Because 100 is quite large, but the exact probability might be slightly different. Alternatively, maybe I can use Stirling's approximation for factorials to compute the exact probability.Stirling's formula approximates n! as sqrt(2πn) (n/e)^n. So, let's try that.Compute P(X = 100) = (e^{-100} * 100^{100}) / 100!Using Stirling's approximation for 100!:100! ≈ sqrt(2π*100) * (100/e)^{100} = sqrt(200π) * (100/e)^{100}So, plugging this into the Poisson formula:P(X = 100) ≈ (e^{-100} * 100^{100}) / [sqrt(200π) * (100/e)^{100}]Simplify numerator and denominator:= (e^{-100} * 100^{100}) / [sqrt(200π) * (100^{100}/e^{100})]= (e^{-100} * 100^{100}) * [e^{100} / (sqrt(200π) * 100^{100})]= (e^{-100} * e^{100}) / sqrt(200π)= 1 / sqrt(200π)Compute sqrt(200π):sqrt(200) is approximately 14.1421, and sqrt(π) is approximately 1.77245.So, sqrt(200π) ≈ 14.1421 * 1.77245 ≈ 25.066.So, 1 / 25.066 ≈ 0.0399, which is about 3.99%.That's very close to the normal approximation result of approximately 3.98%. So, both methods give roughly the same result, which is about 4%.But wait, the exact value might be slightly different. Let me check if I can compute it more accurately.Alternatively, perhaps using the Poisson formula with logarithms.Compute ln(P(X=100)) = ln(e^{-100}) + ln(100^{100}) - ln(100!)= -100 + 100*ln(100) - ln(100!)Again, using Stirling's approximation for ln(100!):ln(100!) ≈ 100*ln(100) - 100 + 0.5*ln(2π*100)So, ln(100!) ≈ 100*ln(100) - 100 + 0.5*ln(200π)Therefore, ln(P(X=100)) ≈ -100 + 100*ln(100) - [100*ln(100) - 100 + 0.5*ln(200π)]Simplify:= -100 + 100*ln(100) - 100*ln(100) + 100 - 0.5*ln(200π)= (-100 + 100) + (100*ln(100) - 100*ln(100)) - 0.5*ln(200π)= 0 - 0.5*ln(200π)So, ln(P(X=100)) ≈ -0.5*ln(200π)Compute ln(200π):200π ≈ 628.3185ln(628.3185) ≈ 6.444So, -0.5*6.444 ≈ -3.222Therefore, P(X=100) ≈ e^{-3.222} ≈ 0.0399, which is about 3.99%, same as before.So, whether I use the normal approximation with continuity correction or the Stirling's approximation, I get approximately 4%.But wait, is there a way to get a more precise value? Maybe using a calculator or computational tool? Since I don't have access to that right now, I think 4% is a reasonable approximation.Alternatively, sometimes people use the De Moivre-Laplace theorem, which is the basis for the normal approximation to the binomial distribution, but in this case, it's Poisson.But in any case, both methods give me about 4%, so I think that's the answer.Moving on to Sub-problem 2: The advocate plans to hold awareness events at rest stops. There are 10 rest stops uniformly distributed along the 500-kilometer highway. The advocate plans to visit 3 rest stops randomly. I need to find the probability that at least one of the chosen rest stops is within the first 100 kilometers.So, this is a probability question involving combinations. Let me think.First, the rest stops are uniformly distributed, so each rest stop has an equal chance of being anywhere along the 500 km. However, since they are uniformly distributed, the probability of a rest stop being in the first 100 km is 100/500 = 1/5.But wait, the rest stops are fixed points, not randomly placed each time. Wait, the problem says they are uniformly distributed, so I think it means that each rest stop is equally likely to be anywhere along the highway, but in reality, they are fixed. Hmm, maybe I need to clarify.Wait, the problem says: "the rest stops are uniformly distributed along the highway, and there are 10 rest stops in total." So, does that mean that the 10 rest stops are placed uniformly at random along the highway? Or that their positions are uniformly distributed, meaning each rest stop is equally likely to be anywhere?Wait, actually, in probability, when we say points are uniformly distributed, it usually means that each point is independently and uniformly distributed over the interval. So, in this case, the 10 rest stops are 10 independent uniform random variables over [0, 500] km.But the advocate is choosing 3 rest stops randomly. So, the selection is random among the 10 rest stops, which themselves are randomly placed.Wait, that might complicate things. Alternatively, maybe the rest stops are fixed, but their positions are uniformly distributed, meaning that the distance between consecutive rest stops is uniform. Hmm, but the problem doesn't specify that. It just says they are uniformly distributed.Wait, perhaps another interpretation: if the rest stops are uniformly distributed, it means that the probability density of a rest stop being in any interval is proportional to the length of the interval. So, the positions of the rest stops are like a Poisson process? Or just uniformly random points.Wait, the problem says "rest stops are uniformly distributed along the highway," which is a bit ambiguous. It could mean that each rest stop is placed uniformly at random, independently of the others, along the 500 km. So, each rest stop has a position X_i ~ Uniform(0, 500), for i = 1 to 10.Alternatively, it could mean that the rest stops are equally spaced, but that would be a different interpretation.But given that it's a probability problem, I think the intended meaning is that each rest stop is independently and uniformly distributed along the highway. So, each rest stop has a position uniformly random between 0 and 500 km.But the advocate is choosing 3 rest stops randomly from these 10. So, the selection is random, but the rest stops themselves are randomly placed.Wait, this might be a bit more complex. Let me think.Alternatively, maybe the rest stops are fixed points, but their positions are uniformly distributed, meaning that the distance between them is uniform. But I think that's not the case here.Wait, perhaps the problem is simpler: the rest stops are 10 fixed points along the highway, uniformly distributed, meaning that they are equally spaced. So, each rest stop is at 50 km intervals: 50, 100, 150, ..., 500 km. But the problem doesn't specify that. It just says uniformly distributed.Hmm, this is a bit confusing. Let me read the problem again."Suppose the rest stops are uniformly distributed along the highway, and there are 10 rest stops in total. If the advocate plans to visit 3 rest stops randomly, what is the probability that at least one of the chosen rest stops is within the first 100 kilometers of the highway?"So, "uniformly distributed" could mean that the rest stops are placed such that their positions are uniformly random, meaning each rest stop is equally likely to be anywhere along the highway, independent of the others. So, the positions are like 10 independent uniform random variables over [0, 500].Alternatively, it could mean that the rest stops are equally spaced, but that's a different distribution.Given that the problem mentions "uniformly distributed," I think it's more likely that each rest stop is independently and uniformly distributed along the highway.So, if that's the case, then the positions of the rest stops are 10 independent random variables, each with a uniform distribution over [0, 500]. Then, the advocate randomly selects 3 of these 10 rest stops. We need to find the probability that at least one of the selected rest stops is within the first 100 km.Alternatively, if the rest stops are fixed and uniformly spaced, then the probability would be different. But since the problem says "uniformly distributed," I think it's the former case.Wait, but if the rest stops are fixed and uniformly spaced, then the probability would be based on how many are in the first 100 km. For example, if they are equally spaced every 50 km, then the first rest stop is at 50 km, second at 100 km, etc. So, in that case, the first two rest stops are within the first 100 km. Then, the probability of selecting at least one of them when choosing 3 out of 10 would be 1 - probability of selecting none of them.But since the problem says "uniformly distributed," I think it's more about the positions being random rather than fixed. So, each rest stop has a position X_i ~ Uniform(0, 500), independent of each other.Therefore, the probability that a single rest stop is within the first 100 km is 100/500 = 1/5 = 0.2.But since there are 10 rest stops, each with a 0.2 probability of being in the first 100 km, the number of rest stops in the first 100 km follows a binomial distribution with n=10 and p=0.2.But wait, actually, since the rest stops are independent, the number of rest stops in the first 100 km is indeed binomial(10, 0.2). So, the probability that exactly k rest stops are in the first 100 km is C(10, k) * (0.2)^k * (0.8)^{10 - k}.But the advocate is choosing 3 rest stops randomly. So, the selection is without replacement from the 10 rest stops. So, the probability that at least one of the selected rest stops is within the first 100 km depends on how many rest stops are actually in the first 100 km.Wait, this is getting a bit complicated. Let me structure it step by step.First, the rest stops are 10 independent uniform random variables over [0, 500]. So, each has a 1/5 chance of being in [0, 100]. Let K be the number of rest stops in [0, 100]. Then, K ~ Binomial(10, 0.2).Given K = k, the probability that at least one of the 3 selected rest stops is in [0, 100] is equal to 1 - probability that all 3 are not in [0, 100].Which is 1 - C(10 - k, 3) / C(10, 3), assuming that the rest stops are selected uniformly at random.Wait, but actually, since the rest stops are randomly placed, the selection is random, but the positions are random as well. So, it's a bit more involved.Alternatively, perhaps we can model this as a hypergeometric distribution, but with a random number of successes.Wait, let me think differently. Since each rest stop is independently in [0, 100] with probability 0.2, and the advocate is selecting 3 rest stops uniformly at random, the probability that at least one is in [0, 100] can be calculated as follows.The probability that a specific rest stop is selected and is in [0, 100] is 0.2. But since the selection is without replacement, the events are not independent.Alternatively, perhaps it's easier to compute the probability that none of the selected rest stops are in [0, 100], and then subtract that from 1.So, P(at least one in first 100) = 1 - P(all three not in first 100).Now, P(all three not in first 100) is equal to the probability that none of the three selected rest stops are in [0, 100].But since the rest stops are randomly placed, the probability that a single rest stop is not in [0, 100] is 0.8. However, since the rest stops are selected without replacement, the probabilities are not independent.Wait, actually, no. The rest stops are fixed in position, but their positions are random. So, the advocate is selecting 3 rest stops out of 10, and we need the probability that at least one of them is in [0, 100].But the rest stops are randomly placed, so the number of rest stops in [0, 100] is a random variable K ~ Binomial(10, 0.2). Then, given K = k, the probability that at least one of the 3 selected is in [0, 100] is 1 - C(10 - k, 3)/C(10, 3).Therefore, the overall probability is the expectation over K of [1 - C(10 - k, 3)/C(10, 3)].So, P = E[1 - C(10 - K, 3)/C(10, 3)] = 1 - E[C(10 - K, 3)/C(10, 3)].But this seems complicated. Maybe there's a better way.Alternatively, perhaps we can model the probability that a specific rest stop is selected and is in [0, 100]. Since the rest stops are randomly placed, the probability that a specific rest stop is in [0, 100] is 0.2, and the probability that it is selected is 3/10.But since the rest stops are independent, the expected number of rest stops in [0, 100] that are selected is 10 * 0.2 * (3/10) = 0.6.But we need the probability that at least one is selected, not the expectation.Wait, perhaps using linearity of expectation isn't helpful here because we need the probability, not the expectation.Alternatively, maybe we can use the inclusion-exclusion principle.The probability that at least one of the three selected rest stops is in [0, 100] is equal to:P(A ∪ B ∪ C) = P(A) + P(B) + P(C) - P(A ∩ B) - P(A ∩ C) - P(B ∩ C) + P(A ∩ B ∩ C))Where A, B, C are the events that the first, second, third selected rest stop is in [0, 100].But since the rest stops are selected without replacement, the events are dependent.Wait, but actually, the rest stops are randomly placed, so the probability that a specific rest stop is in [0, 100] is 0.2, independent of others.But when selecting 3 rest stops, the probability that the first is in [0, 100] is 0.2, same for the second and third.But since the rest stops are selected without replacement, the events are not independent.Wait, let me think again.Alternatively, perhaps the probability that a specific rest stop is selected and is in [0, 100] is (3/10) * 0.2, but that's not quite right because the rest stops are placed randomly.Wait, maybe it's better to think of it as a hypergeometric distribution, but with a random number of successes.Wait, let me try to model it.Let me denote:- Total rest stops: 10- Each rest stop is in [0, 100] with probability 0.2, independent of others.- The advocate selects 3 rest stops uniformly at random.We need P(at least one selected is in [0, 100]).This is equivalent to 1 - P(all selected are not in [0, 100]).So, P = 1 - P(all three selected rest stops are not in [0, 100]).Now, the probability that a single rest stop is not in [0, 100] is 0.8.But since the rest stops are selected without replacement, the probability that the first selected is not in [0, 100] is 0.8.Then, given that the first is not in [0, 100], the probability that the second is also not in [0, 100] is (number of rest stops not in [0, 100] - 1) / (total rest stops - 1).But wait, the rest stops are randomly placed, so the number of rest stops not in [0, 100] is a random variable.Let me denote K as the number of rest stops in [0, 100], so K ~ Binomial(10, 0.2). Then, the number of rest stops not in [0, 100] is 10 - K.Given K = k, the probability that all three selected rest stops are not in [0, 100] is C(10 - k, 3) / C(10, 3), provided that 10 - k >= 3. Otherwise, it's zero.Therefore, the overall probability is:P(all three not in [0, 100]) = E[ C(10 - K, 3) / C(10, 3) ]But this expectation is over K ~ Binomial(10, 0.2).So, we need to compute:E[ C(10 - K, 3) / C(10, 3) ] = (1 / C(10, 3)) * E[ C(10 - K, 3) ]But computing E[ C(10 - K, 3) ] is non-trivial because K is a binomial random variable.Alternatively, perhaps we can find an expression for E[ C(10 - K, 3) ].Note that C(n, 3) = n(n - 1)(n - 2)/6.So, E[ C(10 - K, 3) ] = E[ (10 - K)(9 - K)(8 - K)/6 ]= (1/6) * E[ (10 - K)(9 - K)(8 - K) ]Let me expand the product:(10 - K)(9 - K)(8 - K) = (10 - K)(72 - 17K + K²) = 720 - 170K + 10K² - 72K + 17K² - K³Wait, that seems messy. Maybe a better way is to expand step by step.First, compute (10 - K)(9 - K):= 90 - 10K - 9K + K² = 90 - 19K + K²Then, multiply by (8 - K):= (90 - 19K + K²)(8 - K) = 90*8 - 90K - 19K*8 + 19K² + 8K² - K³= 720 - 90K - 152K + 19K² + 8K² - K³Combine like terms:= 720 - (90 + 152)K + (19 + 8)K² - K³= 720 - 242K + 27K² - K³So, E[ (10 - K)(9 - K)(8 - K) ] = E[720 - 242K + 27K² - K³]= 720 - 242E[K] + 27E[K²] - E[K³]Now, since K ~ Binomial(10, 0.2), we can compute E[K], E[K²], and E[K³].For a binomial distribution, E[K] = n*p = 10*0.2 = 2.Var(K) = n*p*(1 - p) = 10*0.2*0.8 = 1.6.So, Var(K) = E[K²] - (E[K])² => 1.6 = E[K²] - 4 => E[K²] = 5.6.Now, for E[K³], we can use the formula for the third moment of a binomial distribution.The third moment E[K³] can be computed as:E[K³] = n*p*(1 - p)*(1 - 2p) + 3n*p*(n - 1)*p*(1 - p) + n*p*(n - 1)*(n - 2)*p³Wait, actually, I think there's a formula for moments of binomial distribution.Alternatively, perhaps it's easier to use the factorial moments.Wait, maybe I can recall that for a binomial distribution, the third central moment is n p (1 - p) (1 - 2p). But that's the central moment, not the raw moment.Alternatively, perhaps using generating functions.The generating function for a binomial distribution is (q + p*t)^n, where q = 1 - p.So, the third derivative evaluated at t=1 gives E[K³].Let me compute that.G(t) = (0.8 + 0.2t)^10First derivative: G’(t) = 10*(0.8 + 0.2t)^9 * 0.2Second derivative: G''(t) = 10*9*(0.8 + 0.2t)^8 * (0.2)^2Third derivative: G'''(t) = 10*9*8*(0.8 + 0.2t)^7 * (0.2)^3So, E[K³] = G'''(1) = 10*9*8*(0.8 + 0.2*1)^7 * (0.2)^3= 720*(1)^7*(0.008)= 720*0.008 = 5.76Wait, that can't be right because E[K³] should be greater than E[K²] which is 5.6.Wait, let me double-check.Wait, actually, the third derivative is 10*9*8*(0.8 + 0.2t)^7 * (0.2)^3.At t=1, it's 720*(1)^7*(0.008) = 720*0.008 = 5.76.But E[K³] = G'''(1) = 5.76.Wait, but let's check with another method.Alternatively, E[K³] can be computed as:E[K³] = n p (1 - p) (1 - 2p) + 3 n p (n - 1) p (1 - p) + n p (n - 1) (n - 2) p³Wait, that seems complicated. Let me compute it step by step.Wait, actually, for a binomial distribution, the raw moments can be expressed using sums involving combinations.Alternatively, perhaps it's easier to use the formula:E[K³] = n p [1 + (n - 1) p (1 + (n - 2) p)]Wait, I'm not sure. Let me look it up in my mind.Wait, actually, for a binomial distribution, the third moment can be expressed as:E[K³] = n p (1 - p) (1 - 2p) + 3 n p (n - 1) p (1 - p) + n p (n - 1) (n - 2) p³Wait, that seems too convoluted. Maybe it's better to use the generating function approach.Wait, I think my earlier calculation was correct. So, E[K³] = 5.76.But let me verify with another approach.Alternatively, since K ~ Binomial(10, 0.2), we can compute E[K³] by summing over k=0 to 10 of k³ * P(K=k).But that would be tedious, but let's try for a few terms.P(K=k) = C(10, k) * (0.2)^k * (0.8)^{10 - k}Compute E[K³] = sum_{k=0}^{10} k³ * C(10, k) * (0.2)^k * (0.8)^{10 - k}But this is time-consuming, but let's compute it approximately.Alternatively, maybe I can use the formula for moments of binomial distribution.I found a formula online before that E[K³] = n p (1 - p) (1 - 2p) + 3 n p (n - 1) p (1 - p) + n p (n - 1) (n - 2) p³Wait, let me plug in the numbers:n=10, p=0.2First term: 10*0.2*(1 - 0.2)*(1 - 2*0.2) = 2*0.8*0.6 = 0.96Second term: 3*10*0.2*(10 - 1)*0.2*(1 - 0.2) = 3*2*9*0.2*0.8 = 3*2*9*0.16 = 3*2*1.44 = 8.64Third term: 10*0.2*(10 - 1)*(10 - 2)*(0.2)^3 = 2*9*8*0.008 = 144*0.008 = 1.152So, total E[K³] = 0.96 + 8.64 + 1.152 = 10.752Wait, that contradicts the earlier result of 5.76. So, which one is correct?Wait, perhaps I made a mistake in the generating function approach.Wait, the generating function G(t) = (0.8 + 0.2t)^10First derivative: G’(t) = 10*(0.8 + 0.2t)^9 * 0.2Second derivative: G''(t) = 10*9*(0.8 + 0.2t)^8 * (0.2)^2Third derivative: G'''(t) = 10*9*8*(0.8 + 0.2t)^7 * (0.2)^3So, G'''(1) = 720*(1)^7*(0.008) = 720*0.008 = 5.76But according to the formula, it's 10.752. So, which is correct?Wait, perhaps the formula I used earlier is incorrect. Let me check.Wait, actually, the formula I used earlier might be for the third central moment, not the raw moment.Wait, the third central moment is E[(K - μ)^3], which is different from E[K³].So, perhaps I confused the two.Therefore, the generating function approach gives E[K³] = 5.76, while the formula I used earlier might be incorrect.Alternatively, perhaps I should compute E[K³] using the formula:E[K³] = Var(K) * (something) + (E[K])³Wait, no, that's not quite right.Alternatively, perhaps using the relation:E[K³] = E[K]^3 + 3 E[K]^2 Var(K) + E[K] Var(K) + something.Wait, actually, for any random variable,E[K³] = μ³ + 3μ²σ² + μ(σ² + μ²) ?Wait, no, perhaps it's better to use the expansion:E[K³] = (E[K])³ + 3(E[K])² Var(K) + E[K] Var(K) + something.Wait, actually, I think the correct expansion is:E[K³] = μ³ + 3μ²σ² + μσ² + σ³Wait, no, that doesn't seem right.Wait, let me recall that for any random variable,E[K³] = μ³ + 3μσ² + γ σ³,where γ is the skewness.But for a binomial distribution, the skewness γ = (1 - 2p)/sqrt(n p (1 - p)).So, in our case, μ = 2, σ² = 1.6, σ = sqrt(1.6) ≈ 1.2649, γ = (1 - 0.4)/sqrt(1.6) = 0.6 / 1.2649 ≈ 0.4743.So, E[K³] = μ³ + 3μσ² + γ σ³= 8 + 3*2*1.6 + 0.4743*(1.2649)^3Compute each term:μ³ = 83μσ² = 3*2*1.6 = 9.6γ σ³ ≈ 0.4743*(2.036) ≈ 0.4743*2.036 ≈ 0.966So, total E[K³] ≈ 8 + 9.6 + 0.966 ≈ 18.566But this contradicts both previous results. So, I must be making a mistake here.Wait, perhaps the formula is different. Let me check.Actually, the correct expansion is:E[K³] = μ³ + 3μ²σ² + σ³Wait, no, that doesn't seem right either.Wait, perhaps I should use the moment generating function.The moment generating function M(t) = E[e^{tK}] = (1 - p + p e^t)^nThen, E[K³] = M'''(0)Compute M(t) = (0.8 + 0.2 e^t)^10First derivative: M’(t) = 10*(0.8 + 0.2 e^t)^9 * 0.2 e^tSecond derivative: M''(t) = 10*9*(0.8 + 0.2 e^t)^8 * (0.2 e^t)^2 + 10*(0.8 + 0.2 e^t)^9 * 0.2 e^tWait, no, actually, the second derivative would be:M''(t) = d/dt [M’(t)] = d/dt [10*9*(0.8 + 0.2 e^t)^8 * (0.2 e^t)^2 + 10*(0.8 + 0.2 e^t)^9 * 0.2 e^t]Wait, this is getting too complicated.Alternatively, perhaps it's better to accept that E[K³] = 5.76 as per the generating function approach, but that seems low because when I computed the formula earlier, I got 10.752.Wait, actually, I think the generating function approach is correct because it's a direct computation. So, E[K³] = 5.76.But let's see, if K is a binomial random variable with n=10, p=0.2, then E[K] = 2, Var(K) = 1.6, and E[K³] = 5.76.Wait, that seems plausible because E[K³] should be greater than E[K] and E[K²], which is 5.6.Wait, 5.76 is just slightly higher than 5.6, which makes sense.So, going back, E[ (10 - K)(9 - K)(8 - K) ] = 720 - 242E[K] + 27E[K²] - E[K³]= 720 - 242*2 + 27*5.6 - 5.76Compute each term:242*2 = 48427*5.6 = 151.2So,720 - 484 + 151.2 - 5.76= (720 - 484) + (151.2 - 5.76)= 236 + 145.44= 381.44Therefore, E[ (10 - K)(9 - K)(8 - K) ] = 381.44Then, E[ C(10 - K, 3) ] = (1/6)*381.44 ≈ 63.5733Therefore, P(all three not in [0, 100]) = E[ C(10 - K, 3) ] / C(10, 3)C(10, 3) = 120So, P = 63.5733 / 120 ≈ 0.5297775Therefore, the probability that at least one of the selected rest stops is in the first 100 km is:1 - 0.5297775 ≈ 0.4702225So, approximately 47.02%.Wait, that seems reasonable.Alternatively, let me think if there's a simpler way to approximate this.Since each rest stop has a 0.2 chance of being in the first 100 km, and the advocate is selecting 3, the expected number of rest stops in the first 100 km among the selected is 3 * 0.2 = 0.6.But we need the probability that at least one is selected, which is 1 - probability that none are selected.But since the rest stops are randomly placed, the probability that a specific rest stop is selected and is in the first 100 km is 0.2 * (3/10) = 0.06, but that's not quite right because the rest stops are selected without replacement.Wait, perhaps using the linearity of expectation isn't directly helpful here.Alternatively, maybe using the approximation that the probability is roughly 1 - e^{-λ}, where λ is the expected number.But λ = 0.6, so 1 - e^{-0.6} ≈ 1 - 0.5488 ≈ 0.4512, which is close to our earlier result of ~47%.So, that seems consistent.Therefore, the probability is approximately 47%.But let me check if my earlier calculation is correct.We had E[ C(10 - K, 3) ] ≈ 63.5733, so P(all three not in [0, 100]) ≈ 63.5733 / 120 ≈ 0.5297775, so 1 - 0.5297775 ≈ 0.4702225, which is approximately 47.02%.So, rounding to two decimal places, 47.02%.Alternatively, maybe we can compute it more accurately.But given the time, I think 47% is a reasonable approximation.Alternatively, if the rest stops are fixed and equally spaced, then the first rest stop is at 50 km, second at 100 km, etc. So, in that case, the first two rest stops are within the first 100 km. Then, the probability of selecting at least one of them when choosing 3 out of 10 is 1 - C(8, 3)/C(10, 3) = 1 - 56/120 = 1 - 0.4667 = 0.5333, which is approximately 53.33%.But since the problem says "uniformly distributed," I think the first interpretation is correct, where the rest stops are randomly placed, leading to approximately 47% probability.Therefore, the answers are approximately 4% for Sub-problem 1 and approximately 47% for Sub-problem 2.But let me double-check Sub-problem 1.We had a Poisson distribution with λ = 100, and we wanted P(X=100).Using the normal approximation with continuity correction, we got approximately 3.98%, which is about 4%.Using Stirling's approximation, we also got approximately 3.99%, which is about 4%.Therefore, the exact value is approximately 4%.So, I think that's correct.For Sub-problem 2, after a detailed calculation, I got approximately 47.02%, which is about 47%.Therefore, the final answers are approximately 4% and 47%.But let me write them as exact fractions or decimals if possible.For Sub-problem 1, the exact probability is e^{-100} * 100^{100} / 100!, which is approximately 0.0399, or 3.99%.For Sub-problem 2, the exact probability is 1 - [E[C(10 - K, 3)] / C(10, 3)] ≈ 1 - 0.5297775 ≈ 0.4702225, which is approximately 47.02%.So, rounding to two decimal places, 4.00% and 47.02%.Alternatively, if we want to express them as fractions, 4% is 1/25, and 47.02% is approximately 47/100.But since the problem doesn't specify the form, decimal is fine.So, summarizing:Sub-problem 1: Approximately 4% probability.Sub-problem 2: Approximately 47% probability.Final AnswerSub-problem 1: boxed{0.04}Sub-problem 2: boxed{0.47}</think>