$2n$ balls (all distinct in size) are arranged in a row. First few of these balls are black rest all white, both odd in number. The probability that there is exactly, one black ball in one of all possible arrangements is:
Step-by-Step Solution
Key Concept: The number of black balls must be odd (1, 3, 5, ..., 2n-1). For exactly one black ball, we need the first ball to be black and rest white. The probability calculation uses conditional arrangements: P(exactly 1 black) = (number of arrangements with exactly 1 black)/(total arrangements of 2n distinct balls).
The event 'first ball is black OR first three balls are black OR first five balls are black, etc.' consists of mutually exclusive cases. The total probability is $\frac{1}{k}\left[\frac{1}{1!(2n-1)!} + \frac{1}{3!(2n-3)!} + \ldots\right]$. The probability of exactly one black ball is $\frac{1}{k}\left[\frac{1}{1!(2n-1)!}\right]$ divided by the sum $\frac{2n}{2^{2n}}$, yielding final answer $\frac{n}{2^{2n-2}}$.
Correct Answer: 3