💡Okay, so I have this problem about two chess players who are playing a series of games to win a prize. The first player to win k games gets the entire prize. But if the match is stopped before someone wins k games, they have to divide the prize based on the probabilities of each player winning if the match had continued. Part (1) gives specific numbers: a=243 yuan, k=4, m=2, n=1, and p=2/3. I need to figure out how much player A should get. First, let me understand the situation. Player A has already won 2 games, and player B has won 1 game. They need to reach 4 wins to claim the prize. So, if the match continues, player A needs 2 more wins, and player B needs 3 more wins. Since each game is independent, the probability of A winning a game is 2/3, and B is 1/3. I need to calculate the probability that A would have won the prize if the match continued, and then divide the prize in that ratio.So, how do I calculate the probability that A would win from this point? It's similar to a probability problem where we need to find the chance of A winning 2 games before B wins 3 games.This sounds like a problem that can be modeled using combinations and probabilities. Maybe I can use the concept of "states" or the negative binomial distribution.Let me think step by step. Player A needs 2 more wins, and Player B needs 3 more wins. The match can end in 2, 3, 4, or 5 games, depending on who reaches the required number of wins first.Wait, actually, since A needs 2 and B needs 3, the maximum number of additional games needed is 4 (if they alternate wins). But I need to calculate the probability that A wins 2 games before B wins 3.Alternatively, I can model this as a probability of A winning a best-of-5 series, but actually, it's a race to 2 wins for A or 3 wins for B.Hmm, maybe I can use the concept of combinations to calculate the probability.The probability that A wins the series is the sum of the probabilities that A wins exactly 2 games before B wins 3.So, the possible ways A can win:1. A wins the next 2 games: probability (2/3)^2 = 4/9.2. A wins 2 games and B wins 1 game in 3 games. The number of ways this can happen is C(2,1) = 2. So, the probability is 2 * (2/3)^2 * (1/3) = 8/27.3. A wins 2 games and B wins 2 games in 4 games. The number of ways is C(3,2) = 3. Probability is 3 * (2/3)^2 * (1/3)^2 = 12/81 = 4/27.Wait, but if A wins 2 games and B wins 2 games, the series isn't over yet. So actually, in this case, the series would continue until someone reaches their required number of wins. Hmm, maybe I need to adjust my approach.Alternatively, I can model this as a recursive probability. Let me denote P(a, b) as the probability that A wins the series when A needs 'a' more wins and B needs 'b' more wins.So, in this case, P(2, 3). The recursive formula would be:P(a, b) = p * P(a-1, b) + (1-p) * P(a, b-1)With boundary conditions:- If a=0, P(0, b)=1 (A has already won)- If b=0, P(a, 0)=0 (B has already won)So, let me compute P(2,3):P(2,3) = p * P(1,3) + (1-p) * P(2,2)First, compute P(1,3):P(1,3) = p * P(0,3) + (1-p) * P(1,2) = p * 1 + (1-p) * P(1,2)Now, compute P(1,2):P(1,2) = p * P(0,2) + (1-p) * P(1,1) = p * 1 + (1-p) * P(1,1)Compute P(1,1):P(1,1) = p * P(0,1) + (1-p) * P(1,0) = p * 1 + (1-p) * 0 = pSo, going back:P(1,2) = p + (1-p) * p = p + p - p^2 = 2p - p^2Then, P(1,3) = p + (1-p)*(2p - p^2) = p + (2p - p^2 - 2p^2 + p^3) = p + 2p - p^2 - 2p^2 + p^3 = 3p - 3p^2 + p^3Now, compute P(2,2):P(2,2) = p * P(1,2) + (1-p) * P(2,1)We already have P(1,2) = 2p - p^2Compute P(2,1):P(2,1) = p * P(1,1) + (1-p) * P(2,0) = p * p + (1-p)*0 = p^2So, P(2,2) = p*(2p - p^2) + (1-p)*p^2 = 2p^2 - p^3 + p^2 - p^3 = 3p^2 - 2p^3Now, going back to P(2,3):P(2,3) = p*(3p - 3p^2 + p^3) + (1-p)*(3p^2 - 2p^3)Let's compute each term:First term: p*(3p - 3p^2 + p^3) = 3p^2 - 3p^3 + p^4Second term: (1-p)*(3p^2 - 2p^3) = 3p^2 - 2p^3 - 3p^3 + 2p^4 = 3p^2 - 5p^3 + 2p^4Adding both terms:3p^2 - 3p^3 + p^4 + 3p^2 - 5p^3 + 2p^4 = (3p^2 + 3p^2) + (-3p^3 -5p^3) + (p^4 + 2p^4) = 6p^2 -8p^3 + 3p^4So, P(2,3) = 6p^2 -8p^3 + 3p^4Now, plug in p=2/3:Compute each term:6p^2 = 6*(4/9) = 24/9 = 8/3-8p^3 = -8*(8/27) = -64/273p^4 = 3*(16/81) = 48/81 = 16/27Now, add them up:8/3 - 64/27 + 16/27Convert 8/3 to 72/27:72/27 - 64/27 + 16/27 = (72 -64 +16)/27 = 24/27 = 8/9So, the probability that A wins is 8/9.Therefore, the prize should be divided in the ratio 8/9 : 1/9.Given the total prize is 243 yuan, A should get 243*(8/9) = 216 yuan.So, the answer to part (1) is 216 yuan.Now, part (2) asks about the probability f(p) of event A, which is the match continuing and player A winning the entire prize. Wait, actually, the problem says:"Let event A be 'the match continues and player B wins the entire prize.' Find the probability f(p) of the event that the match continues and player A wins the entire prize when k=4 and m=2, n=1."Wait, that seems contradictory. It says event A is B winning, but then asks for the probability of A winning. Maybe it's a typo. Let me read again."Let event A be 'the match continues and player B wins the entire prize.' Find the probability f(p) of the event that the match continues and player A wins the entire prize when k=4 and m=2, n=1."Hmm, so event A is defined as B winning, but they are asking for the probability of A winning. Maybe it's a misstatement. Perhaps they meant to define event A as A winning, but regardless, the question is to find f(p), the probability that A wins, and then determine whether event A (B winning) is a low probability event when p >= 3/4.So, first, find f(p), which is P(A wins the prize). From part (1), we have the general formula:P(A wins) = 6p^2 -8p^3 + 3p^4So, f(p) = 6p^2 -8p^3 + 3p^4Alternatively, we can write it as f(p) = 3p^4 -8p^3 +6p^2Now, we need to find the probability of event A, which is B winning, so P(B wins) = 1 - f(p)So, P(B wins) = 1 - (3p^4 -8p^3 +6p^2) = -3p^4 +8p^3 -6p^2 +1But let's see if we can express it differently.Alternatively, since in part (1), we had P(A wins) = 8/9 when p=2/3, and in general, it's 6p^2 -8p^3 +3p^4.But let me think if there's another way to express f(p). Maybe using combinations.Alternatively, the probability that A wins can be calculated as the sum over all possible ways A can win 2 games before B wins 3.So, the number of games needed can be 2, 3, 4, or 5.Wait, actually, since A needs 2 and B needs 3, the maximum number of games is 4 (if they split the first 4 games 2-2, then the next game decides it). Wait, no, if it's 2-2, then they need to play until someone gets to 3.Wait, actually, no, because the original target is 4 wins. Wait, no, in the problem, it's whoever reaches k=4 first. But in the current state, A has 2, B has 1, so A needs 2 more, B needs 3 more.So, the maximum number of additional games is 4 (if they alternate wins: A, B, A, B, then A or B would have reached their target). Wait, no, if it's A, B, A, B, then A has 3, B has 2, so they need to play more. Wait, actually, no, because A only needs 2, so if they alternate, A would win on the 4th game.Wait, maybe I'm overcomplicating. Let's stick with the recursive formula.We already have f(p) = 6p^2 -8p^3 +3p^4Now, to find whether event A (B winning) is a low probability event when p >= 3/4.A low probability event is defined as having probability less than 0.05.So, we need to find P(B wins) = 1 - f(p) = 1 - (6p^2 -8p^3 +3p^4)Simplify:1 -6p^2 +8p^3 -3p^4We need to check if this is less than 0.05 when p >= 3/4.Alternatively, since p >= 3/4, let's compute P(B wins) at p=3/4.Compute f(3/4):f(3/4) = 6*(9/16) -8*(27/64) +3*(81/256)Calculate each term:6*(9/16) = 54/16 = 27/8 = 3.375-8*(27/64) = -216/64 = -27/8 = -3.3753*(81/256) = 243/256 ≈ 0.9492So, f(3/4) = 3.375 -3.375 +0.9492 ≈ 0.9492Therefore, P(B wins) = 1 -0.9492 ≈ 0.0508So, approximately 5.08%, which is just above 5%.Since p >= 3/4, and f(p) is increasing in p (because the higher p is, the higher the chance A wins), so P(B wins) is decreasing in p.Therefore, the maximum P(B wins) occurs at p=3/4, which is approximately 5.08%, which is just above 5%.Since 5.08% > 5%, event A is not a low probability event when p >= 3/4.Therefore, the answer to part (2) is that event A is not necessarily a low probability event when p >= 3/4.

💡First, I need to find the missing number in the expression |7 - 8(x - 12)| - |5 - 11| = 73.I'll start by simplifying the absolute value |5 - 11|, which equals 6.Substituting this back into the equation, I get |7 - 8(x - 12)| - 6 = 73.Adding 6 to both sides of the equation, I find that |7 - 8(x - 12)| = 79.Next, I'll consider the two possible cases for the absolute value:**Case 1:** 7 - 8(x - 12) = 79Solving for x:-8(x - 12) = 72x - 12 = -9x = 3**Case 2:** 7 - 8(x - 12) = -79Solving for x:-8(x - 12) = -86x - 12 = 10.75x = 22.75Since we're looking for an integer value for x, I'll discard the second case. Therefore, the missing number is 3.

💡First, I recognize that there are two initial vampires in the village.Each night, every vampire turns 5 people into vampires.After the first night, the two vampires will have turned 10 new people into vampires, making a total of 12 vampires.On the second night, these 12 vampires will each turn 5 more people into vampires, resulting in 60 new vampires.Adding the vampires from both nights, there are a total of 72 vampires after two nights.Since the total number of vampires equals the initial population of the village, the population of the village is 72.

💡Okay, so I need to find the maximum value of the function ( y = sin x sin 2x ) where ( x ) is an acute angle. Hmm, let's see. Since ( x ) is acute, that means ( x ) is between 0 and ( frac{pi}{2} ) radians, or 0 to 90 degrees. First, I remember that ( sin 2x ) can be rewritten using a double-angle identity. The identity is ( sin 2x = 2 sin x cos x ). So, substituting that into the original function, I get:[y = sin x cdot 2 sin x cos x]Simplifying that, it becomes:[y = 2 sin^2 x cos x]Alright, so now the function is ( y = 2 sin^2 x cos x ). I need to find the maximum value of this function. Since it's a product of sine and cosine functions, maybe I can use calculus to find the maximum. Let me consider taking the derivative of ( y ) with respect to ( x ) and setting it equal to zero to find critical points. So, let's compute the derivative ( y' ).First, rewrite ( y ) as:[y = 2 sin^2 x cos x]To take the derivative, I'll use the product rule. Let me denote ( u = 2 sin^2 x ) and ( v = cos x ). Then, ( y = u cdot v ), so the derivative ( y' ) is:[y' = u' cdot v + u cdot v']First, find ( u' ):[u = 2 sin^2 x][u' = 2 cdot 2 sin x cos x = 4 sin x cos x]Wait, hold on. Let me double-check that. The derivative of ( sin^2 x ) is ( 2 sin x cos x ), so multiplying by 2 gives ( 4 sin x cos x ). Yeah, that's correct.Now, find ( v' ):[v = cos x][v' = -sin x]So, putting it all together:[y' = 4 sin x cos x cdot cos x + 2 sin^2 x cdot (-sin x)]Simplify each term:First term: ( 4 sin x cos^2 x )Second term: ( -2 sin^3 x )So, combining them:[y' = 4 sin x cos^2 x - 2 sin^3 x]To find critical points, set ( y' = 0 ):[4 sin x cos^2 x - 2 sin^3 x = 0]Factor out common terms. I can factor out ( 2 sin x ):[2 sin x (2 cos^2 x - sin^2 x) = 0]So, either ( 2 sin x = 0 ) or ( 2 cos^2 x - sin^2 x = 0 ).Since ( x ) is acute, ( sin x ) is not zero (except at 0, but that's the minimum). So, we can ignore the ( 2 sin x = 0 ) solution because it doesn't give a maximum.Now, solve ( 2 cos^2 x - sin^2 x = 0 ).Let me express everything in terms of sine or cosine. Let's use the identity ( sin^2 x = 1 - cos^2 x ).Substitute into the equation:[2 cos^2 x - (1 - cos^2 x) = 0]Simplify:[2 cos^2 x - 1 + cos^2 x = 0][3 cos^2 x - 1 = 0][3 cos^2 x = 1][cos^2 x = frac{1}{3}][cos x = pm frac{sqrt{3}}{3}]But since ( x ) is acute, ( cos x ) is positive, so:[cos x = frac{sqrt{3}}{3}]Now, find ( sin x ). Using ( sin^2 x + cos^2 x = 1 ):[sin^2 x = 1 - cos^2 x = 1 - frac{1}{3} = frac{2}{3}][sin x = sqrt{frac{2}{3}} = frac{sqrt{6}}{3}]Okay, so now we have ( sin x = frac{sqrt{6}}{3} ) and ( cos x = frac{sqrt{3}}{3} ). Let's plug these back into the original function ( y = 2 sin^2 x cos x ):[y = 2 left( frac{sqrt{6}}{3} right)^2 left( frac{sqrt{3}}{3} right)]Compute ( left( frac{sqrt{6}}{3} right)^2 ):[left( frac{sqrt{6}}{3} right)^2 = frac{6}{9} = frac{2}{3}]So, substituting back:[y = 2 cdot frac{2}{3} cdot frac{sqrt{3}}{3} = 2 cdot frac{2 sqrt{3}}{9} = frac{4 sqrt{3}}{9}]Therefore, the maximum value of ( y = sin x sin 2x ) is ( frac{4 sqrt{3}}{9} ).Wait, let me just double-check my steps to make sure I didn't make any mistakes. Starting from the derivative, I think that was correct. Factoring out ( 2 sin x ) and solving for ( cos x ) seems right. Then, using the Pythagorean identity to find ( sin x ) is standard. Plugging back into the function, the arithmetic seems correct as well. So, I think this is the right answer.**Final Answer**The maximum value of the function is boxed{dfrac{4sqrt{3}}{9}}.

💡Okay, so I have this problem here where I need to find the complement of set P in the universal set U. Let me try to understand what each set represents and then figure out how to find the complement.First, the universal set U is defined as U = { y | y = log₂x, x > 1 }. Hmm, okay, so U consists of all y values such that y is the logarithm base 2 of some x where x is greater than 1. I remember that the logarithm function log₂x increases as x increases, and since x > 1, log₂x will be positive because log₂1 is 0, and as x increases beyond 1, the logarithm becomes positive. So, U is all positive real numbers. In mathematical terms, U = (0, ∞). Got that.Next, the set P is defined as P = { y | y = 1/x, x > 3 }. Alright, so P consists of all y values such that y is the reciprocal of some x where x is greater than 3. Let me think about this. If x is greater than 3, then 1/x will be less than 1/3. Also, since x is positive (because x > 3), 1/x will also be positive. So, P is the set of all positive real numbers less than 1/3. In interval notation, that's (0, 1/3). Now, I need to find the complement of P in U, denoted as ∁ᵤP. The complement of a set P in U consists of all elements in U that are not in P. So, since U is (0, ∞) and P is (0, 1/3), the complement should be all the elements in U that are not in P. That would be the interval from 1/3 to infinity, right? Because everything in U is positive, and we're excluding the part that's less than 1/3. So, ∁ᵤP should be [1/3, ∞). Wait, hold on. Is it [1/3, ∞) or (1/3, ∞)? Let me think. Since P is defined as y = 1/x where x > 3, when x approaches 3 from the right, y approaches 1/3 from the left. So, 1/3 is not included in P because x has to be greater than 3, not equal to 3. Therefore, in the complement, 1/3 is included because it's not in P. So, the complement should include 1/3 and everything above it. Hence, it's [1/3, ∞).Let me double-check. U is all positive real numbers, and P is all positive real numbers less than 1/3. So, the complement would be all positive real numbers greater than or equal to 1/3. That makes sense.Looking at the options provided:A: [1/3, ∞)B: (0, 1/3)C: (0, ∞)D: (-∞, 0] ∪ [1/3, ∞)Option A is exactly what I concluded, [1/3, ∞). Option B is just P itself, so that's not the complement. Option C is the entire universal set U, which is not the complement. Option D includes everything less than or equal to 0 and greater than or equal to 1/3, but since U is only positive, the complement should only be the part of U not in P, which is [1/3, ∞). So, D is incorrect because it includes negative numbers which aren't even in U.Therefore, the correct answer should be option A.**Final Answer**boxed{A}

💡First, I need to determine how many students are taking only mathematics and how many are taking only physics. For students taking only mathematics:- Total students taking mathematics = 36- Students taking both mathematics and physics = 20- Students taking only mathematics = 36 - 20 = 16For students taking only physics:- Total students taking physics = 27- Students taking both mathematics and physics = 20- Students taking only physics = 27 - 20 = 7Now, I can calculate the total number of students taking at least one of the subjects:- Students taking only mathematics = 16- Students taking only physics = 7- Students taking both subjects = 20- Total students taking mathematics or physics = 16 + 7 + 20 = 43Finally, to find the number of students taking neither mathematics nor physics:- Total students in the drama club = 50- Students taking mathematics or physics = 43- Students taking neither = 50 - 43 = 7