<p>Probability if <i>n</i> heads in 2<i>n</i> tosses of a fair coin can be given by</p>
<p>(1) \(\displaystyle\prod_{r=1}^{n}\left(\dfrac{2r-1}{2r}\right)\)</p>
<p>(2) \(\displaystyle\prod_{r=1}^{n}\left(\dfrac{n+r}{2r}\right)\)</p>
<p>(3) \(\displaystyle\sum_{r=0}^{n}\left(\dfrac{{}^nC_r}{2^n}\right)^2\)</p>
<p>(4) \(\dfrac{\displaystyle\sum_{r=0}^{n}({}^nC_r)^2}{\displaystyle\sum_{r=0}^{n} 2^n C_r}\)</p>
Step-by-Step Solution
Key Concept: Use the binomial probability formula P(X = k) = C(n,k)p^k(1-p)^(n-k) for exactly n heads in 2n tosses, where p = 1/2. Recognize that C(2n, n) counts the favorable outcomes and 2^(2n) is the total outcomes.
<p><strong>Step 1:</strong> In 2n tosses of a fair coin, we want exactly n heads.</p><p><strong>Step 2:</strong> Total possible outcomes = 2^(2n)</p><p><strong>Step 3:</strong> Favorable outcomes (exactly n heads) = C(2n, n) = (2n)! / (n! × n!)</p><p><strong>Step 4:</strong> Using binomial probability formula P(X = n) = C(2n, n) × (1/2)^n × (1/2)^n = C(2n, n) × (1/2)^(2n)</p><p><strong>Step 5:</strong> This can also be written as: P(n heads in 2n tosses) = C(2n, n) / 2^(2n)</p><p>∴ Answer: Probability = <strong>C(2n, n) / 2^(2n)</strong> or equivalently <strong>[C(2n, n)] × (1/2)^(2n)</strong></p>
Correct Answer: 1,3