A fair coin is tossed 5 times then the probability that no two consecutive heads occur, is:
Step-by-Step Solution
Key Concept: Use Fibonacci recurrence relation: if $a_n$ counts binary strings of length $n$ with no consecutive heads, then $a_n = a_{n-1} + a_{n-2}$ since valid strings either end in T (giving $a_{n-1}$ choices for first $n-1$ positions) or end in HT (giving $a_{n-2}$ choices). The probability is $ rac{a_n}{2^n}$.
Let $a_n$ denote strings of H and T with length $n$ and no adjacent H's. We have $a_1 = 2, a_2 = 3$, and the recurrence $a_{n+2} = a_{n+1} + a_n$ (string must start with T or HT). Computing: $a_3 = 5, a_4 = 8, a_5 = 13$. The probability is $ rac{13}{2^5} = rac{13}{32}$.
Correct Answer: 3