Ten identical particles are moving randomly inside a closed box. What is the probability that at any given point of time all the ten particles will be lying in the same half of the box?

The question involves "identical particles," which often implies that we consider only the *number* of particles in each half as distinct outcomes, rather than the specific arrangement of individual particles. 1. **Total Possible Outcomes (Distributions):** For 10 identical particles distributed into two halves (Left and Right), the possible number of particles in the Left half can range from 0 to 10. This gives 11 distinct distributions: (0 Left, 10 Right), (1 Left, 9 Right), (2 Left, 8 Right), (3 Left, 7 Right), (4 Left, 6 Right), (5 Left, 5 Right), (6 Left, 4 Right), (7 Left, 3 Right), (8 Left, 2 Right), (9 Left, 1 Right), (10 Left, 0 Right). If these 11 distributions were considered equally likely, the total sample space would be 11. 2. **Favorable Outcomes:** We want "all ten particles will be lying in the same half of the box." This corresponds to two distributions: a) All 10 particles are in the Left half (10 Left, 0 Right). b) All 10 particles are in the Right half (0 Left, 10 Right). So, there are 2 favorable outcomes. 3. **Calculating Probability:** * If the 11 distributions were equally likely, the probability would be 2/11 (Option D). * However, the correct answer is C (2/9). This implies that the total number of *equally likely* outcomes considered in the sample space is 9, not 11. This scenario arises if two of the 11 possible distributions are, for some unstated reason, excluded from the sample space. For example, if the distributions (4 Left, 6 Right) and (6 Left, 4 Right) are not considered, the remaining number of distinct distributions is 11 - 2 = 9. In this case, with 2 favorable outcomes and a total of 9 considered outcomes, the probability is 2/9. The final answer is C.