<p>If a coin is tossed \(n\) times, then find the probability that the head appears odd number of times.</p>
Step-by-Step Solution
Key Concept: Use the binomial theorem with (1+1)^n and (1-1)^n to separate odd and even outcomes. The sum of all probabilities equals 1, and symmetry of binomial coefficients ensures odd and even cases are equal.
<p><strong>Step 1:</strong> Total number of outcomes when coin is tossed n times = 2^n</p><p><strong>Step 2:</strong> Number of ways to get odd number of heads = C(n,1) + C(n,3) + C(n,5) + ...</p><p><strong>Step 3:</strong> Use binomial theorem: (1+1)^n = C(n,0) + C(n,1) + C(n,2) + ... + C(n,n) = 2^n</p><p><strong>Step 4:</strong> Also, (1-1)^n = C(n,0) - C(n,1) + C(n,2) - C(n,3) + ... = 0 (for n ≥ 1)</p><p><strong>Step 5:</strong> Subtracting Step 4 from Step 3: 2[C(n,1) + C(n,3) + C(n,5) + ...] = 2^n</p><p><strong>Step 6:</strong> Therefore, C(n,1) + C(n,3) + C(n,5) + ... = 2^(n-1)</p><p><strong>Step 7:</strong> Probability = (Number of favorable outcomes)/(Total outcomes) = 2^(n-1)/2^n = 1/2</p><p>∴ Answer: <strong>1/2</strong></p>
Correct Answer: C