answer:There are 2^6 = 64 possible ways to paint the cube's six faces since each face can independently be red or blue. However, the probabilities differ for each color: - Red face: p(text{red}) = frac{2}{3} - Blue face: p(text{blue}) = frac{1}{3} The favorable cases where all four vertical sides have the same color can be divided into two categories: 1. **All six faces are the same color**: - All red: (frac{2}{3})^6 - All blue: (frac{1}{3})^6 - Total probability for all six faces the same is (frac{2}{3})^6 + (frac{1}{3})^6. 2. **Four sides are one color and the top and bottom are another color**: - Four vertical red, top and bottom blue: 3 times (frac{2}{3})^4 times (frac{1}{3})^2 - Four vertical blue, top and bottom red: 3 times (frac{2}{3})^2 times (frac{1}{3})^4 - (Three ways to place opposing pairs.) Adding probabilities from these cases gives the total probability that the condition is satisfied: P(text{such placement}) = (frac{2}{3})^6 + (frac{1}{3})^6 + 3 times (frac{2}{3})^4 times (frac{1}{3})^2 + 3 times (frac{2}{3})^2 times (frac{1}{3})^4 Calculating these, we get: - (frac{2}{3})^6 = frac{64}{729} - (frac{1}{3})^6 = frac{1}{729} - (frac{2}{3})^4 times (frac{1}{3})^2 = frac{16}{729} times frac{1}{9} = frac{16}{6561} - (frac{2}{3})^2 times (frac{1}{3})^4 = frac{4}{9} times frac{1}{81} = frac{4}{729} Hence, the total probability: P = frac{65}{729} + 3(frac{16}{6561} + frac{4}{729}) = frac{65}{729} + frac{48}{6561} + frac{12}{729} = frac{741}{6561} + frac{48}{6561} = frac{789}{6561} Conclusion with boxed answer: P = boxed{frac{789}{6561}}

answer:We need to prove that for any point ( O ) inside the pyramid ( S A B C D ) with base parallelogram ( A B C D ), the sum of the volumes of ( text{tetrahedrons } O S A B ) and ( O S C D ) is equal to the sum of the volumes of ( text{tetrahedrons } O S B C ) and ( O S D A ). 1. **Define the point of intersection**: Let ( X ) be the point of intersection of the line segment ( S O ) with the plane ( A B C D ). Since point ( O ) is inside the pyramid, point ( X ) lies within the base parallelogram ( A B C D ). 2. **Relate areas of bases**: Observe that ( S_{X A B} + S_{X C D} ) must equal ( S_{X B C} + S_{X D A} ). This equality holds due to the properties of parallelograms (each sum is equal to half of the area of the parallelogram ( A B C D )). This assertion follows from the fact that the four triangles ( X A B, X C D, X B C, ) and ( X D A ) are pairs of triangles sharing a common diagonal intersecting at ( X ). 3. **Volume equation using heights**: Next, consider the volumes of the pyramids with common vertex ( X ) and heights from the vertex ( S ): [ V_{X S A B} + V_{X S C D} = V_{X S B C} + V_{X S D A} quad (1) ] Since the height from ( S ) to the plane ( A B C D ) is the same for all these pyramids, the volumes are directly proportional to the areas of their bases. Thus equation (1) holds. 4. **Parallel argument for point ( O )**: Similarly, if we consider the pyramids with vertex ( X ) and heights from the vertex ( O ): [ V_{X O A B} + V_{X O C D} = V_{X O B C} + V_{X O D A} quad (2) ] 5. **Subtraction of equations**: Subtract equation (2) from equation (1): [ (V_{X S A B} + V_{X S C D}) - (V_{X O A B} + V_{X O C D}) = (V_{X S B C} + V_{X S D A}) - (V_{X O B C} + V_{X O D A}) ] 6. **Conclusion**: The left-hand side represents the volumes of tetrahedrons ( O S A B ) and ( O S C D ), and the right-hand side represents the volumes of tetrahedrons ( O S B C ) and ( O S D A ). Thus, we have: [ V_{O S A B} + V_{O S C D} = V_{O S B C} + V_{O S D A} ] Hence, the sum of the volumes of ( text{tetrahedrons } O S A B ) and ( O S C D ) is indeed equal to the sum of the volumes of ( text{tetrahedrons } O S B C ) and ( O S D A ). [ boxed{} ]

answer:The method to estimate the total number of fish in the pond is based on the proportion of marked fish in the second catch to the total number of fish caught the second time, which should be approximately equal to the proportion of the initially marked fish to the total number of fish in the pond. Given that 10 out of the 100 fish caught the second time were marked, and initially, 120 fish were marked and released, we can set up the proportion as follows: frac{10}{100} = frac{120}{x} Solving for x gives us the estimated total number of fish in the pond. Cross-multiplying and solving for x yields: x = frac{120 times 100}{10} = 1200 Therefore, the estimated total number of fish in the pond is boxed{1200}, which corresponds to option B.

answer:**Step 1: Understanding the problem** The problem provides an arithmetic sequence {a_n} with the sum of its 2nd and 4th terms equal to 6 (a_2 + a_4 = 6). We are asked to find the sum of the first 5 terms of the sequence. **Step 2: Applying arithmetic sequence properties** In an arithmetic sequence, the sum of the first and last terms is equal to the sum of any pair of terms equidistant from the beginning and end. In this case, a_1 + a_5 = a_2 + a_4. Given that a_2 + a_4 = 6, we can deduce that: a_1 + a_5 = 6 **Step 3: Using the arithmetic series sum formula** The sum S_n of the first n terms of an arithmetic sequence can be found using the formula: S_n = frac{n(a_1 + a_n)}{2} We are looking for the sum of the first 5 terms, so we need to find S_5: S_5 = frac{5(a_1 + a_5)}{2} **Step 4: Substituting the values** Substituting our known value a_1 + a_5 = 6 into the equation, we get: S_5 = frac{5(6)}{2} **Step 5: Calculating the sum** Calculate the sum by simplifying the expression: S_5 = 15 So the sum of the first 5 terms of the arithmetic sequence {a_n} is boxed{15}.