answer:1. **Initial Events Analysis**: - We first consider the sequences formed after the first occurrence of the letter "O". The probability for the first letter to be "O" is (frac{1}{2}). 2. **Possible Sequences after the First "O"**: - There are four possible sequences that can follow the first "O" in the next two coin tosses: [ OP quad text{(Player B wins)}, quad OO quad text{(Player A wins)}, quad PP quad text{(no effect yet)}, quad PO quad text{(continue observing)} ] 3. **Detailed Probability Calculation**: - Each coin toss is independent, with the probability of each event being ( frac{1}{2} ). - Below, we consider the occurrence after the first "O" with equal probability of (frac{1}{4}) for each sequence: [ text{Probability of OP} = frac{1}{4} quad text{(Player B wins)} ] [ text{Probability of OO} = frac{1}{4} quad text{(Player A wins)} ] [ text{Probability of PP} = frac{1}{4} ] [ text{Probability of PO} = frac{1}{4} ] 4. **Further Outcomes for the Unresolved Sequences**: - For the sequence "PP": - Both conditions (winning for A or B) are reset to the initial scenario, effectively restarting the game. - For the sequence "PO" (next toss will determine): - The event "OP" after "PO" (probability (frac{1}{2})) leads to win for player B. - The event "OO" after "PO" (probability (frac{1}{2})) leads to a tie after the next toss reset. 5. **Combining Probabilities**: - Probability for "PO" converting to a win for player B: [ text{Probability of PO} times text{Probability of following O to form OP} = frac{1}{4} times frac{1}{2} = frac{1}{8} ] - Adding up different cases: [ text{Total for B} = frac{1}{4} + frac{1}{8} = frac{3}{8} ] - Probability for A to win directly: [ text{Total for A} = frac{1}{4} ] - Remaining probability (tie and game resets): [ 1 - left(frac{3}{8} + frac{1}{4}right) = 1 - left(frac{3}{8} + frac{2}{8}right) = frac{3}{8} ] 6. **Conclusion**: - Player B has a greater chance of winning the game than player A. (boxed{text{Player B}} )

answer:(i) Since 0 < x < pi, and it satisfies sin x + cos x = frac{7}{13}, then (sin x + cos x)^2 = 1 + 2sin xcos x = frac{49}{169}, thus, sin x cdot cos x = boxed{-frac{60}{169}}. (ii) From (i), we know that frac{pi}{2} < x < pi, sin x cdot cos x = -frac{60}{169}. Therefore, sin x - cos x = sqrt{(sin x - cos x)^2} = sqrt{1 - 2sin xcos x} = sqrt{1 + frac{120}{169}} = frac{17}{13}, By solving begin{cases} sin x + cos x = frac{7}{13} sin x - cos x = frac{17}{13} end{cases}, we get sin x = frac{12}{13}, cos x = -frac{5}{13}, thus, frac{5sin x + 4cos x}{15sin x - 7cos x} = frac{5 times frac{12}{13} - 4 times frac{5}{13}}{15 times frac{12}{13} - 7 times frac{5}{13}} = boxed{frac{8}{43}}.

answer:1. **Define the function ( f ):** Let ( f ) be a function that maps an ( a times a ) square to the number of black cells in that square. Specifically, if the number of black cells in the ( a times a ) square is ( x ), then: [ f(text{square}) = begin{cases} x & text{if } x < a x - a & text{if } x geq a end{cases} ] 2. **Sum over all ( a times a ) squares:** Let ( L ) be the sum of ( f ) over all ( a times a ) squares in the grid. Then: [ L = sum f(text{square}) ] Since each black cell is counted in multiple ( a times a ) squares, we have: [ L leq a^2n - aK ] where ( K ) is the number of ( a times a ) squares containing exactly ( a ) black cells. 3. **Lower bound for ( L ):** We need to show that for large ( n ), ( L geq a^2(a-1) ). Assume ( n ) is larger than ( (a^2(a-1) + a)^2 ). This implies there are more than ( a^2(a-1) ) rows (or columns) with at least one black cell. 4. **Existence of a thick row:** There exists an ( a )-thick row such that the sum of ( f ) for the ( a times a ) squares in this row is 0. This means each ( a times a ) square in this row contains either 0 or ( a ) black cells. 5. **Structure of the thick row:** In this ( a )-thick row, there is one column with ( a ) black cells (all cells in this column are black), and other black cells are at least ( a ) columns away. 6. **Summing ( f ) for ( a times a ) squares:** Let ( h(t) ) be the value of ( f ) for an ( a times a ) square from row ( t ) to row ( t + a - 1 ) and columns from 0 to ( a-1 ). Let ( X ) be the number of black cells in columns 0 to ( a-1 ). 7. **Inequality for ( h ):** For ( 0 leq i leq a-1 ): [ h(i) + h(a+i) + h(2a+i) + ldots geq (X - i) mod a ] Summing these inequalities: [ h(0) + h(1) + ldots geq sum_{i=1}^{a-1} ((X - i) mod a) geq frac{a(a-1)}{2} ] 8. **Summing over all columns:** By repeating the above argument for all columns, we get: [ L geq a^2(a-1) ] 9. **Contradiction and conclusion:** This contradicts the assumption that ( L leq a^2n - aK ) for large ( n ). Therefore, we conclude that: [ K leq a(n + 1 - a) ] for all ( n geq (a^2(a-1) + a)^2 + 1 ). The final answer is ( boxed{ K leq a(n + 1 - a) } ).

answer:Since f(x) = a^{2x-1} + 2 = a^{2(x- frac{1}{2})} + 2 = (a^2)^{x- frac{1}{2}} + 2, and the function y = (a^2)^x always passes through the fixed point (0,1), therefore, f(x) = (a^2)^{x- frac{1}{2}} + 2 always passes through the fixed point left( frac{1}{2}, 3 right). Hence, the correct choice is: boxed{D}. By transforming the given function and then utilizing the translation of the function graph, we find the answer. This question examines the transformation of exponential function graphs, focusing on the translation of function graphs, and is a basic question.