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 to count valid binary strings: if $a_n$ represents sequences of length $n$ with no adjacent heads, then $a_n = a_{n-1} + a_{n-2}$ because each valid string either ends in T (preceded by any valid string of length $n-1$) or ends in HT (preceded by any valid string of length $n-2$). The probability is $\frac{a_5}{2^5} = \frac{13}{32}$.
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