Probability
Probability
Allen Star Batch
Grade 12
Question:
$2n$ 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:
$\frac{n}{2^{2n-1}}$
$\frac{n}{2^{n-1}}$
$\frac{n}{2^{n-2}}$
$\frac{n}{2^{n-2}}$
Step-by-Step Solution
Key Concept: Use inclusion-exclusion to partition mutually exclusive events based on which balls are black.
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