answer：First, identify the universal set ( U ) and the sets ( A ) and ( B ).Next, find the complement of set ( A ) with respect to ( U ), which consists of all elements in ( U ) that are not in ( A ).Then, determine the intersection of this complement with set ( B ), which includes only the elements that are common to both sets.Finally, present the resulting set as the answer.

answer：Okay, so I have this problem about selecting athletes from three different table tennis associations using stratified sampling. Let me try to understand what's being asked here.First, part (Ⅰ) asks for the number of athletes to be selected from each association. The associations have 27, 9, and 18 athletes respectively, and we need to select a total of 6 athletes. Stratified sampling means that we'll select a proportionate number of athletes from each association based on their size. So, I think I need to calculate the proportion of each association relative to the total number of athletes and then apply that proportion to the 6 athletes we're selecting.Let me write down the numbers:- Association A: 27 athletes- Association B: 9 athletes- Association C: 18 athletesTotal athletes = 27 + 9 + 18 = 54 athletes.So, the proportion for each association would be:- A: 27/54 = 0.5- B: 9/54 = 0.1667- C: 18/54 = 0.3333Now, applying these proportions to the 6 athletes:- A: 0.5 * 6 = 3 athletes- B: 0.1667 * 6 ≈ 1 athlete- C: 0.3333 * 6 ≈ 2 athletesWait, but 3 + 1 + 2 = 6, which matches the total number we need. So, that seems right. So, from association A, we select 3 athletes, from B, 1 athlete, and from C, 2 athletes.Moving on to part (Ⅱ). We have 6 selected athletes labeled A₁, A₂, A₃, A₄, A₅, A₆. We need to randomly select two athletes for a doubles match.Part (i) asks to list all possible outcomes with these labels. So, I need to list all possible pairs of these 6 athletes. Since the order doesn't matter in a pair (i.e., selecting A₁ and A₂ is the same as selecting A₂ and A₁), this is a combination problem.The formula for combinations is C(n, k) = n! / (k! * (n - k)!), where n is the total number of items, and k is the number of items to choose.Here, n = 6 and k = 2, so C(6, 2) = 6! / (2! * 4!) = (6 * 5) / (2 * 1) = 15 possible outcomes.Let me list them:1. A₁ and A₂2. A₁ and A₃3. A₁ and A₄4. A₁ and A₅5. A₁ and A₆6. A₂ and A₃7. A₂ and A₄8. A₂ and A₅9. A₂ and A₆10. A₃ and A₄11. A₃ and A₅12. A₃ and A₆13. A₄ and A₅14. A₄ and A₆15. A₅ and A₆So, that's 15 possible outcomes.Part (ii) defines event A as "at least one of the two athletes labeled A₅ and A₆ is selected." I need to find the probability of event A occurring.First, let's understand what "at least one" means. It means either A₅ is selected, or A₆ is selected, or both. So, to find the probability, I can either count all the favorable outcomes where at least one of A₅ or A₆ is selected and divide by the total number of outcomes, or I can use the complementary probability (1 - probability that neither A₅ nor A₆ is selected).Let me try both approaches to verify.First approach: Counting favorable outcomes.Looking at the list of 15 outcomes, let's identify which ones include A₅ or A₆.From the list:4. A₁ and A₅5. A₁ and A₆8. A₂ and A₅9. A₂ and A₆11. A₃ and A₅12. A₃ and A₆13. A₄ and A₅14. A₄ and A₆15. A₅ and A₆So, that's 9 outcomes where at least one of A₅ or A₆ is selected.Therefore, the probability is 9/15, which simplifies to 3/5.Second approach: Using complementary probability.Total outcomes: 15.Outcomes where neither A₅ nor A₆ is selected: These are the pairs that only include A₁, A₂, A₃, A₄.Looking at the list:1. A₁ and A₂2. A₁ and A₃3. A₁ and A₄6. A₂ and A₃7. A₂ and A₄10. A₃ and A₄That's 6 outcomes.So, probability of neither A₅ nor A₆ is 6/15 = 2/5.Therefore, probability of at least one of A₅ or A₆ is 1 - 2/5 = 3/5.Both methods give the same result, so I'm confident that the probability is 3/5.Wait, just to make sure I didn't miss any outcomes in the first approach. Let's recount the favorable outcomes:From the list:4. A₁ and A₅5. A₁ and A₆8. A₂ and A₅9. A₂ and A₆11. A₃ and A₅12. A₃ and A₆13. A₄ and A₅14. A₄ and A₆15. A₅ and A₆Yes, that's 9 outcomes. So, 9/15 = 3/5. Yep, that seems correct.I think I've covered all the steps and checked my work. I don't see any mistakes in my reasoning.

answer：Okay, so I have this problem where I need to find the minimum perimeter of a triangle with sides ( l, m, n ) such that ( l > m > n ) and all are integers. The condition given is that the fractional parts of ( frac{3^l}{10^4} ), ( frac{3^m}{10^4} ), and ( frac{3^n}{10^4} ) are equal. That is,[left{ frac{3^l}{10^4} right} = left{ frac{3^m}{10^4} right} = left{ frac{3^n}{10^4} right}]where ( {x} = x - lfloor x rfloor ) is the fractional part of ( x ).First, I need to understand what this condition implies. If the fractional parts are equal, then the differences between these numbers and their integer parts must be the same. So,[frac{3^l}{10^4} - leftlfloor frac{3^l}{10^4} rightrfloor = frac{3^m}{10^4} - leftlfloor frac{3^m}{10^4} rightrfloor = frac{3^n}{10^4} - leftlfloor frac{3^n}{10^4} rightrfloor]This means that ( 3^l ), ( 3^m ), and ( 3^n ) leave the same remainder when divided by ( 10^4 ). In other words,[3^l equiv 3^m equiv 3^n pmod{10^4}]So, ( 3^l equiv 3^m pmod{10^4} ) and ( 3^m equiv 3^n pmod{10^4} ). This implies that ( 3^{l - m} equiv 1 pmod{10^4} ) and ( 3^{m - n} equiv 1 pmod{10^4} ).Therefore, ( l - m ) and ( m - n ) must be multiples of the order of 3 modulo ( 10^4 ). The order of 3 modulo ( 10^4 ) is the smallest positive integer ( k ) such that ( 3^k equiv 1 pmod{10^4} ).To find this order, I can use Euler's theorem, which states that if ( a ) and ( n ) are coprime, then ( a^{phi(n)} equiv 1 pmod{n} ). Here, ( phi(10^4) = phi(2^4 cdot 5^4) = phi(2^4) cdot phi(5^4) = 8 cdot 500 = 4000 ). So, the order of 3 modulo ( 10^4 ) divides 4000.But the order might be smaller. To find the exact order, I need to check the exponents where ( 3^k equiv 1 pmod{16} ) and ( 3^k equiv 1 pmod{625} ) because ( 10^4 = 16 times 625 ) and 16 and 625 are coprime.First, let's find the order of 3 modulo 16. Calculating powers of 3 modulo 16:- ( 3^1 = 3 mod 16 )- ( 3^2 = 9 mod 16 )- ( 3^3 = 27 equiv 11 mod 16 )- ( 3^4 = 33 equiv 1 mod 16 )So, the order of 3 modulo 16 is 4.Next, the order of 3 modulo 625. This is more involved. Since 625 is ( 5^4 ), and 3 and 5 are coprime, we can use Euler's theorem here as well. ( phi(625) = 500 ), so the order of 3 modulo 625 divides 500.To find the exact order, we can factor 500 into its prime components: ( 500 = 2^2 times 5^3 ). The order must divide 500, so possible orders are the divisors of 500. We need to find the smallest ( k ) such that ( 3^k equiv 1 mod 625 ).Testing exponents:First, check if ( 3^{100} equiv 1 mod 625 ). If not, check smaller exponents.But this might take a while. Alternatively, since 3 is a primitive root modulo 5, but not necessarily modulo higher powers. Let me check the order modulo 5, 25, 125, and 625.Modulo 5: ( 3^1 equiv 3 mod 5 ), ( 3^2 equiv 4 mod 5 ), ( 3^3 equiv 2 mod 5 ), ( 3^4 equiv 1 mod 5 ). So, order 4.Modulo 25: Let's compute ( 3^4 = 81 equiv 6 mod 25 ), ( 3^8 = 6^2 = 36 equiv 11 mod 25 ), ( 3^{12} = 6 times 11 = 66 equiv 16 mod 25 ), ( 3^{16} = 11^2 = 121 equiv 21 mod 25 ), ( 3^{20} = (3^{10})^2 ). Wait, ( 3^{10} = (3^5)^2 ). ( 3^5 = 243 equiv 18 mod 25 ), so ( 3^{10} = 18^2 = 324 equiv 24 mod 25 ). Then ( 3^{20} = 24^2 = 576 equiv 1 mod 25 ). So, the order modulo 25 is 20.Modulo 125: Let's compute ( 3^{20} mod 125 ). Since ( 3^{20} equiv 1 mod 25 ), but we need modulo 125.Compute ( 3^{20} mod 125 ). Let's compute step by step:( 3^1 = 3 mod 125 )( 3^2 = 9 mod 125 )( 3^4 = 81 mod 125 )( 3^5 = 243 equiv 243 - 2*125 = -7 mod 125 )( 3^{10} = (-7)^2 = 49 mod 125 )( 3^{20} = 49^2 = 2401 mod 125 ). 2401 divided by 125 is 19*125=2375, so 2401 - 2375 = 26. So, ( 3^{20} equiv 26 mod 125 neq 1 ). So, order is higher.Next, ( 3^{40} = (3^{20})^2 equiv 26^2 = 676 mod 125 ). 676 - 5*125 = 676 - 625 = 51. So, ( 3^{40} equiv 51 mod 125 neq 1 ).Next, ( 3^{50} = 3^{40} times 3^{10} equiv 51 times 49 = 2499 mod 125 ). 2499 - 19*125 = 2499 - 2375 = 124. So, ( 3^{50} equiv 124 mod 125 neq 1 ).Next, ( 3^{100} = (3^{50})^2 equiv 124^2 = 15376 mod 125 ). 15376 / 125 = 123*125=15375, so 15376 - 15375 = 1. Thus, ( 3^{100} equiv 1 mod 125 ). So, the order modulo 125 is 100.Similarly, modulo 625, we can compute the order. Since ( 3^{100} equiv 1 mod 125 ), but we need modulo 625.Compute ( 3^{100} mod 625 ). This is more involved, but let's see:We can use the fact that ( 3^{100} equiv 1 mod 125 ), so ( 3^{100} = 1 + 125k ) for some integer ( k ). We need to find ( k ) such that ( 3^{100} equiv 1 mod 625 ).Alternatively, we can use Euler's theorem, which tells us that ( 3^{500} equiv 1 mod 625 ), so the order divides 500. Let's check if 100 is the order or if it's higher.Compute ( 3^{100} mod 625 ). Let's compute ( 3^{100} ) step by step.First, compute ( 3^{10} mod 625 ):( 3^1 = 3 mod 625 )( 3^2 = 9 mod 625 )( 3^4 = 81 mod 625 )( 3^5 = 243 mod 625 )( 3^{10} = (3^5)^2 = 243^2 = 59049 mod 625 ). 59049 / 625 = 94*625=59375, so 59049 - 59375 = -326. But we need positive, so -326 + 625 = 299. So, ( 3^{10} equiv 299 mod 625 ).Next, ( 3^{20} = (3^{10})^2 = 299^2 = 89401 mod 625 ). 89401 / 625 = 143*625=89375, so 89401 - 89375 = 26. So, ( 3^{20} equiv 26 mod 625 ).( 3^{40} = (3^{20})^2 = 26^2 = 676 mod 625 ). 676 - 625 = 51. So, ( 3^{40} equiv 51 mod 625 ).( 3^{50} = 3^{40} times 3^{10} equiv 51 times 299 mod 625 ). Compute 51*299:51*300 = 15300, subtract 51: 15300 - 51 = 15249. 15249 mod 625: 15249 / 625 = 24*625=15000, so 15249 - 15000 = 249. So, ( 3^{50} equiv 249 mod 625 ).( 3^{100} = (3^{50})^2 equiv 249^2 mod 625 ). Compute 249^2:249*249 = (250 - 1)^2 = 250^2 - 2*250*1 + 1 = 62500 - 500 + 1 = 62001. 62001 mod 625: 62001 / 625 = 99*625=61875, so 62001 - 61875 = 126. So, ( 3^{100} equiv 126 mod 625 neq 1 ).So, the order is higher than 100. Next, check 200:( 3^{200} = (3^{100})^2 equiv 126^2 = 15876 mod 625 ). 15876 / 625 = 25*625=15625, so 15876 - 15625 = 251. So, ( 3^{200} equiv 251 mod 625 neq 1 ).Next, ( 3^{250} = 3^{200} times 3^{50} equiv 251 times 249 mod 625 ). Compute 251*249:251*250 = 62750, subtract 251: 62750 - 251 = 62499. 62499 mod 625: 62499 / 625 = 99*625=61875, so 62499 - 61875 = 624. So, ( 3^{250} equiv 624 mod 625 ).( 3^{500} = (3^{250})^2 equiv 624^2 mod 625 ). 624 ≡ -1 mod 625, so ( (-1)^2 = 1 mod 625 ). Thus, ( 3^{500} equiv 1 mod 625 ).So, the order of 3 modulo 625 is 500.Therefore, the order of 3 modulo ( 10^4 ) is the least common multiple of the orders modulo 16 and 625, which are 4 and 500 respectively. The LCM of 4 and 500 is 500. So, the order of 3 modulo ( 10^4 ) is 500.This means that ( 3^{500} equiv 1 mod 10^4 ), and 500 is the smallest such exponent. Therefore, ( l - m ) and ( m - n ) must be multiples of 500.So, let's denote:( l = n + 500k )( m = n + 500j )where ( k ) and ( j ) are positive integers with ( k > j ) because ( l > m > n ).To form a triangle, the sides must satisfy the triangle inequality:1. ( l < m + n )2. ( m < l + n )3. ( n < l + m )But since ( l > m > n ), the most restrictive condition will be ( l < m + n ).Substituting ( l = n + 500k ) and ( m = n + 500j ):( n + 500k < (n + 500j) + n )Simplify:( 500k < 2n + 500j )( 500(k - j) < 2n )( n > 250(k - j) )Since ( k > j ), let's set ( k = j + 1 ) for the minimal case. Then:( n > 250(1) = 250 )So, ( n ) must be greater than 250. The smallest integer ( n ) can be is 251.Now, let's choose the smallest possible ( k ) and ( j ). Let ( j = 1 ), so ( k = 2 ). Then:( m = n + 500*1 = n + 500 )( l = n + 500*2 = n + 1000 )So, the sides are ( n + 1000 ), ( n + 500 ), and ( n ).To satisfy the triangle inequality ( l < m + n ):( n + 1000 < (n + 500) + n )Simplify:( n + 1000 < 2n + 500 )( 1000 - 500 < 2n - n )( 500 < n )Which is already satisfied since ( n > 250 ). So, the minimal ( n ) is 251.Therefore, the sides are:- ( l = 251 + 1000 = 1251 )- ( m = 251 + 500 = 751 )- ( n = 251 )The perimeter is ( 1251 + 751 + 251 = 2253 ).Wait, but let me check if this is indeed the minimal perimeter. Maybe choosing ( j = 1 ) and ( k = 2 ) is not the only way. Let's see if we can have smaller ( n ).Wait, earlier I concluded ( n > 250 ), so the minimal ( n ) is 251. But let's check if ( n = 251 ) satisfies all triangle inequalities.Check ( l < m + n ):( 1251 < 751 + 251 = 1002 ). Wait, 1251 is not less than 1002. That's a problem.Wait, that can't be. So, my earlier conclusion was wrong. I need to re-examine the triangle inequality.Given ( l = n + 1000 ), ( m = n + 500 ), ( n ).The triangle inequality ( l < m + n ) becomes:( n + 1000 < (n + 500) + n )Simplify:( n + 1000 < 2n + 500 )( 1000 - 500 < 2n - n )( 500 < n )So, ( n > 500 ). Therefore, the minimal ( n ) is 501.So, let's set ( n = 501 ):- ( l = 501 + 1000 = 1501 )- ( m = 501 + 500 = 1001 )- ( n = 501 )Check triangle inequalities:1. ( 1501 < 1001 + 501 = 1502 ). Yes, 1501 < 1502.2. ( 1001 < 1501 + 501 = 2002 ). Yes.3. ( 501 < 1501 + 1001 = 2502 ). Yes.So, all triangle inequalities are satisfied.Therefore, the perimeter is ( 1501 + 1001 + 501 = 3003 ).Is this the minimal perimeter? Let's see if we can have smaller ( n ).Wait, if ( n = 501 ), that's the minimal ( n ) to satisfy ( n > 500 ). So, yes, 501 is the smallest possible ( n ).Alternatively, could we have a different ( k ) and ( j ) such that ( k - j = 1 ) but with smaller ( n )? For example, if ( k = 1 ) and ( j = 0 ), but ( j ) must be at least 1 because ( m > n ). So, ( j ) must be at least 1, making ( k ) at least 2.Therefore, the minimal perimeter is 3003.

answer：Alright, so I've got this triangle ABC, and the sides opposite angles A, B, and C are a, b, and c respectively. They've given me the lengths: a is 8, b is 6, and c is 4. I need to find sin A. Hmm, okay, let's think about how to approach this.First off, I remember there are a few laws related to triangles that might help here. The Law of Sines and the Law of Cosines come to mind. Since I need to find sin A, the Law of Sines seems like a good starting point. The Law of Sines states that in any triangle, the ratio of the length of a side to the sine of its opposite angle is constant. So, that would be:[frac{a}{sin A} = frac{b}{sin B} = frac{c}{sin C}]But wait, I only know the lengths of the sides, not the angles. So, maybe I need to find one of the angles first before I can use the Law of Sines to find sin A. That's where the Law of Cosines might come in handy. The Law of Cosines relates the lengths of the sides of a triangle to the cosine of one of its angles. The formula is:[c^2 = a^2 + b^2 - 2ab cos C]But in this case, I need to find angle A, so I should rearrange the formula to solve for cos A. Let's do that:[cos A = frac{b^2 + c^2 - a^2}{2bc}]Okay, plugging in the values I have:[cos A = frac{6^2 + 4^2 - 8^2}{2 times 6 times 4}]Calculating the numerator:[6^2 = 36][4^2 = 16][8^2 = 64][36 + 16 - 64 = 52 - 64 = -12]So, the numerator is -12. The denominator is:[2 times 6 times 4 = 48]Therefore,[cos A = frac{-12}{48} = -frac{1}{4}]Hmm, cosine of angle A is -1/4. That means angle A is obtuse since cosine is negative in the second quadrant. But I need sin A, not cosine. How do I find sin A from cos A?I remember the Pythagorean identity:[sin^2 A + cos^2 A = 1]So, if I solve for sin A:[sin^2 A = 1 - cos^2 A][sin A = sqrt{1 - cos^2 A}]Plugging in the value of cos A:[sin A = sqrt{1 - left(-frac{1}{4}right)^2} = sqrt{1 - frac{1}{16}} = sqrt{frac{15}{16}} = frac{sqrt{15}}{4}]Wait a second, but since angle A is obtuse, sin A should be positive because sine is positive in both the first and second quadrants. So, taking the positive square root makes sense here.Let me double-check my calculations to make sure I didn't make any mistakes. Starting with the Law of Cosines:[cos A = frac{6^2 + 4^2 - 8^2}{2 times 6 times 4} = frac{36 + 16 - 64}{48} = frac{-12}{48} = -frac{1}{4}]That seems correct. Then, using the Pythagorean identity:[sin A = sqrt{1 - left(-frac{1}{4}right)^2} = sqrt{1 - frac{1}{16}} = sqrt{frac{15}{16}} = frac{sqrt{15}}{4}]Yes, that looks right. So, sin A is √15 over 4.Just to be thorough, maybe I can check using the Law of Sines. If I find another angle, say angle B, using the Law of Cosines, and then use the Law of Sines to find sin A, I should get the same result.Let's try finding angle B. Using the Law of Cosines again:[cos B = frac{a^2 + c^2 - b^2}{2ac}][cos B = frac{8^2 + 4^2 - 6^2}{2 times 8 times 4} = frac{64 + 16 - 36}{64} = frac{44}{64} = frac{11}{16}]So, cos B is 11/16. Then, sin B would be:[sin B = sqrt{1 - left(frac{11}{16}right)^2} = sqrt{1 - frac{121}{256}} = sqrt{frac{135}{256}} = frac{sqrt{135}}{16} = frac{3sqrt{15}}{16}]Now, using the Law of Sines:[frac{a}{sin A} = frac{b}{sin B}][frac{8}{sin A} = frac{6}{frac{3sqrt{15}}{16}}][frac{8}{sin A} = frac{6 times 16}{3sqrt{15}} = frac{96}{3sqrt{15}} = frac{32}{sqrt{15}}][sin A = frac{8 times sqrt{15}}{32} = frac{sqrt{15}}{4}]Okay, so that matches my earlier result. That gives me more confidence that sin A is indeed √15 over 4.Just to recap, the steps I took were:1. Recognize that I need to find sin A given the sides of the triangle.2. Decide to use the Law of Cosines to find cos A first.3. Plug in the values into the Law of Cosines formula and calculate cos A.4. Use the Pythagorean identity to find sin A from cos A.5. Double-check the result by finding another angle using the Law of Cosines and then applying the Law of Sines to confirm sin A.Throughout this process, I made sure to verify each step to avoid any calculation errors. It's easy to mix up signs or make arithmetic mistakes, especially with negative values and square roots, so taking it step by step and double-checking helps ensure accuracy.I also considered whether angle A was acute or obtuse based on the cosine value. Since cosine was negative, I knew angle A was obtuse, which is why sine was positive. It's important to remember the signs of trigonometric functions in different quadrants to apply the correct sign when taking square roots.Overall, this problem was a good exercise in applying the Law of Cosines and the Pythagorean identity to find a sine value when given the sides of a triangle. It also reinforced the importance of checking your work through alternative methods to confirm the solution.

answer：Okay, so I have this problem about a fly moving on an integer grid. It starts at the origin and can only move to the right or up. At each node, it randomly chooses between moving right or up with equal probability. There are two parts to the problem: part (a) asks to prove that the fly will eventually reach the point with an x-coordinate of 2011, and part (b) asks to find the expected value of the y-coordinate when the fly reaches that x-coordinate.Let me start with part (a). I need to show that the fly will eventually reach x = 2011. Hmm, since the fly can only move right or up, it's making progress towards increasing x or y with each step. But how do I know it won't just keep moving up forever and never reach x = 2011?Well, at each step, the fly has a 50% chance to move right and a 50% chance to move up. So, intuitively, over time, the fly should make progress in both directions. But how can I formalize this?Maybe I can think about the probability that the fly never moves right. If the fly never moves right, it would just keep moving up indefinitely. The probability of never moving right is like flipping a coin and getting heads (or tails) every time. The probability of that happening is (1/2)^n as n approaches infinity, which is zero. So, almost surely, the fly will move right at some point.But wait, that's just for moving right once. I need to make sure it moves right enough times to reach x = 2011. So, more formally, the number of right moves needed is 2011. Since each right move is an independent event with probability 1/2, the expected number of steps to get 2011 right moves is 2011 * 2 = 4022 steps. But expectation doesn't guarantee it will happen, though.However, in probability theory, if an event has a positive probability, it will almost surely happen given enough trials. Since each step is independent, and the fly has a positive probability of moving right each time, the law of large numbers tells us that over time, the proportion of right moves will approach 1/2. Therefore, the fly will almost surely accumulate enough right moves to reach x = 2011.So, for part (a), I can argue that since the fly has a positive probability of moving right at each step, and the number of steps is infinite, the fly will eventually make 2011 right moves, thus reaching x = 2011. This seems solid.Now, moving on to part (b). I need to find the expected y-coordinate when the fly reaches x = 2011. Hmm, so when the fly reaches x = 2011, it has made 2011 right moves. But how many up moves has it made by then?Since each move is independent and has an equal probability of being right or up, the number of up moves when the fly reaches x = 2011 is a random variable. Let me denote the number of up moves as U. Then, the total number of steps taken when the fly reaches x = 2011 is 2011 + U.But how do I find E[U], the expected number of up moves?Wait, maybe I can model this as a symmetric random walk where the fly is moving right or up with equal probability. When the fly reaches x = 2011, it has taken 2011 steps to the right and some number of steps up. The total number of steps is 2011 + U.But since each step is independent, the number of up moves U is a binomial random variable with parameters n = total steps and p = 1/2. However, the total number of steps isn't fixed; it's variable because the fly could take different numbers of up moves before reaching x = 2011.Hmm, maybe I need to think of this as a negative binomial problem, where I'm waiting for the 2011th success (right move) and counting the number of failures (up moves) along the way.Yes, that makes sense. The negative binomial distribution models the number of failures before a specified number of successes is reached in a sequence of independent and identically distributed Bernoulli trials.In this case, each step is a Bernoulli trial with success (right move) probability p = 1/2 and failure (up move) probability q = 1 - p = 1/2. We are interested in the number of failures (up moves) U before the 2011th success (right move).The expected number of failures (up moves) in a negative binomial distribution is given by:E[U] = r * (q / p)where r is the number of successes we are waiting for.Here, r = 2011, q = 1/2, p = 1/2.So,E[U] = 2011 * ( (1/2) / (1/2) ) = 2011 * 1 = 2011.Therefore, the expected number of up moves when the fly reaches x = 2011 is 2011.But wait, let me double-check this. The negative binomial expectation formula is E[U] = r * (q / p). Since p = q = 1/2, this simplifies to E[U] = r. So yes, E[U] = 2011.Therefore, the expected y-coordinate when the fly reaches x = 2011 is 2011.Wait, that seems a bit too straightforward. Let me think differently. Suppose I model the fly's path as a sequence of steps where each step is either right or up. The fly continues moving until it has moved right 2011 times. The number of up moves it makes during this time is a random variable U.Since each step is independent and has a 50% chance of being right or up, the number of up moves U is equal in distribution to the number of heads in a sequence of coin flips until the 2011th tail. By the symmetry of the problem, the expected number of heads (up moves) should be equal to the expected number of tails (right moves), which is 2011.Yes, that makes sense. So, the expected y-coordinate is 2011.Alternatively, I can think about the linearity of expectation. Each step before reaching x = 2011 contributes to either the x or y coordinate. Since each step is equally likely to contribute to x or y, the expected number of y steps when x reaches 2011 is equal to the expected number of x steps, which is 2011.So, both approaches lead me to the conclusion that the expected y-coordinate is 2011.I think that's solid. So, for part (b), the expected y-coordinate is 2011.**Final Answer**a) boxed{text{The fly will inevitably reach a point with abscissa } 2011.}b) boxed{2011}

answer：First, I need to determine the total expenditure during the winter months of December, January, and February. The problem provides the accumulated expenditure by the end of November and by the end of February of the following year.The accumulated expenditure by the end of November is 3.0 million, and by the end of February, it is 5.5 million. To find the amount spent during the winter months, I subtract the expenditure at the end of November from the expenditure at the end of February.So, the calculation is:[5.5 text{ million} - 3.0 text{ million} = 2.5 text{ million}]This means that approximately 2.5 million dollars were spent during the winter months of December, January, and February.