No Consecutive Heads

Model each outcome as a binary string of length five and count the strings with no two adjacent $1$s.

Let $a_n$ count length-$n$ binary strings with no adjacent $1$s; build $a_5$ from a recurrence.

1. 1. Count valid strings

   Let $a_n$ be the number of length-$n$ binary strings with no two adjacent $1$s. These satisfy the Fibonacci-style recurrence $a_n=a_{n-1}+a_{n-2}$ with $a_0=1,a_1=2$, giving $a_2=3,a_3=5,a_4=8,a_5=13$.

2. 2. Divide by the sample space

   There are $2^5=32$ equally likely flip sequences, and $13$ of them avoid consecutive heads.

   $$P(\text{no HH})=\frac{13}{32}$$

Answer: 13/32