<p>A coin is tossed \(2n\) times. The chance that the number of times one gets head is not equal to the number of times one gets tails is</p>
<p>(1) \(\frac{(2n!)}{(n!)^2}\left(\frac{1}{2}\right)^{2n}\)</p>
<p>(2) \(1 - \frac{(2n!)}{(n!)^2}\)</p>
<p>(3) \(1 - \frac{(2n!)}{(n!)^2} \cdot \frac{1}{4^n}\)</p>
<p>(4) none of these</p>
Step-by-Step Solution
Key Concept: Use the complement principle: P(heads ≠ tails) = 1 - P(heads = tails). For 2n tosses, equal heads and tails means exactly n heads, which follows the binomial probability formula.
<p><strong>Step 1:</strong> Total possible outcomes when tossing a coin 2n times = 2^(2n)</p><p><strong>Step 2:</strong> For heads = tails, we need exactly n heads and n tails. Number of such outcomes = C(2n, n)</p><p><strong>Step 3:</strong> P(heads = tails) = C(2n, n)/2^(2n)</p><p><strong>Step 4:</strong> Using complement: P(heads ≠ tails) = 1 - C(2n, n)/2^(2n)</p><p><strong>Step 5:</strong> Simplify: P(heads ≠ tails) = (2^(2n) - C(2n, n))/2^(2n) = 1 - C(2n, n)/2^(2n)</p><p>∴ Answer: C (which should be <strong>1 - C(2n,n)/2^(2n)</strong> or equivalently <strong>1 - C(2n,n)/4^n</strong>)</p>
Correct Answer: C