<p>An unbiased coin is tossed 6 times. The probability that third head appears on the sixth trial is</p>
Step-by-Step Solution
Key Concept: For the third head to appear exactly on the 6th trial, we need exactly 2 heads in the first 5 trials and a head on the 6th trial. Use binomial probability combined with the conditional requirement.
<p><strong>Step 1:</strong> Identify the constraint: The 3rd head must appear at position 6, meaning exactly 2 heads in first 5 tosses and definitely a head at position 6.</p><p><strong>Step 2:</strong> Probability of exactly 2 heads in first 5 tosses = C(5,2) × (1/2)² × (1/2)³ = 10 × (1/32) = 10/32</p><p><strong>Step 3:</strong> Probability of head on 6th toss = 1/2</p><p><strong>Step 4:</strong> Combined probability = (10/32) × (1/2) = 10/64 = 5/32</p><p>∴ Answer: B (assuming B = 5/32)</p>
Correct Answer: B