<s>[INST] Probability that no two consecutive heads occur? 
A fair coin is tossed $10$ times. What is the probability that no two
  consecutive tosses are heads?

Possibilities are (dont mind the number of terms):
$H TTTTTTH$, $HTHTHTHTHTHTHT$.
But except for those,
let $y(n)$ be the number of sequences that start with $T$
$T _$, there are two options, $T$ and $H$ so,
$y(n) = y(n - 1) + x(n-1) = y(n - 1) + y(n - 2)$
Let $x(n)$ be the number of sequences that start with $H$, 
$H _$ the next option is only $T$ hence,
$x(n) = y(n - 1)$ 
The total number of sequences $F(n)$ is:
$$F(n) = y(n) + x(n) = 2y(n - 1) + y(n - 2)$$
We are after $F(10)$,
$$F(10) = 2y(9) + y(8)$$
$$F(3) = 2y(2) + y(1) = 1 + 4 = 5$$
$$F(4) = 2y(3) + y(2) = 6 + 2 = 8$$
But I'm not quite sure where this will take me though. 
 [/INST] $F (n)$ is in fact a fibonacci sequence:
\begin{align}
F (n+1) &= 2 y (n) + y (n-1)\\
&= 2 [y (n-1) +y (n-2)] + y (n-2) + y (n-1)\\
&= 2 y (n-1) + y (n-2) + 2 y(n-2) +y (n-3)\\
&= F (n) +F (n-1)\\
\end{align}
Now $F (1) = |\{H,T\}|=2$ 
And $F (2) = |\{HT,TH,TT\}|=3$
So $ F (10) = 144$
There are $ 2^{10}=1024$ possible events.
Thus the required probability is $\frac {144}{1024} $.
</s>