$2n$ 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 remaining 2n-1 balls to be white. The probability involves summing over all possible odd positions where black balls can end, recognizing that there are n possible odd numbers (1, 3, 5, ..., 2n-1).
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