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. 

Let $f(n)$ be the noumber of sequences without two consecutive $H$, you want $\frac{f(10)}{2^{10}}$.
$f(1)=2$ and $f(2)=3$ by inspection.
We obtain the recursion $f_n=f_{n-1}+f_{n-2}$ why? every sequence of length $n$ without consecutive $H$ can be obtained uniquely by taking a sequence of length $n-2$ and adding $TH$ or by taking a sequence of length $n-1$ and adding $T$ at the end.
Using this we have $f(3)=5,f(4)=8,f(5)=13,f(6)=21,f(7)=34,f(8)=55,f(9)=89$ $f(10)=144$
So we want $\frac{144}{2^{10}}\approx0.14$
