<p>A fair coin is tossed 5 times. Find the probability that no two consecutive heads occur.</p>
Step-by-Step Solution
Key Concept: Use recursion or dynamic programming: let f(n) = sequences of length n with no consecutive heads. Then f(n) = f(n-1) + f(n-2), since the nth toss is either T (giving f(n-1) valid sequences) or H preceded by T (giving f(n-2) valid sequences).
<p><strong>Step 1:</strong> Define f(n) = number of valid sequences of length n with no consecutive heads.</p><p><strong>Step 2:</strong> Establish recurrence: f(n) = f(n-1) + f(n-2)</p><ul><li>If nth toss is T: f(n-1) valid sequences remain</li><li>If nth toss is H: (n-1)th must be T, so we need f(n-2) valid sequences before position (n-1)</li></ul><p><strong>Step 3:</strong> Calculate base cases: f(1) = 2 (H, T both valid), f(2) = 3 (HT, TH, TT)</p><p><strong>Step 4:</strong> Compute recursively:</p><ul><li>f(3) = f(2) + f(1) = 3 + 2 = 5</li><li>f(4) = f(3) + f(2) = 5 + 3 = 8</li><li>f(5) = f(4) + f(3) = 8 + 5 = 13</li></ul><p><strong>Step 5:</strong> Total outcomes = 2⁵ = 32</p><p><strong>Step 6:</strong> Probability = 13/32</p><p>∴ Answer: <strong>13/32</strong></p>
Correct Answer: A