Probability in $S_{15}$ We consider the set of permutations of the first fifteen natural numbers.
What is the probability that $1$ and $2$ aren't contiguous?
My attempt:
Denote by


*

*$C_{12}=$ "The numbers $1,2$ are contiguos";

*$R_i^{(1)}=$ "The number $1$ is in i-th position ".


Now, we have 
$$P(C_{12})=\sum_{i=1}^{15}P(C_{12}|R_i^{(1)})P(R_i^{(1)}),$$ 
where $P(R_i^{(1)})=1/15$ for $i=1,2,\ldots,15$ and  
$P(C_{12}|R_i^{(1)})=2/14$ for $i=2,3,\ldots 14$ conversely $P(C_{12}|R_1^{(1)})=P(C_{12}|R_{15}^{(1)})=1/14$.
In this way, we obtain
$P(C_{12})=2/15$, then $$1-P(C_{12})=13/15.$$
Is it correct my attempt?
 A: The answer and procedure are correct. 
For a notationally simpler approach, think of the positions as a row of $15$ chairs. There are $\binom{15}{2}$ equally likely ways to choose two chairs to put Reserved signs on. 
There are $14$ ways to choose  two contiguous chairs. Thus the probability that $1$ and $2$ are contiguous is $\frac{14}{\binom{15}{2}}$.
This is $\frac{2}{15}$. Thus the probability $1$ and $2$ are not contiguous is $\frac{13}{15}$.
A: You can solve this problem also in this way. Consider all permutations of the first fifteen numbers that are $15!$. Now we calculate the probability that $1$ and $2$ are contiguous. Consider the number $'1'$ in the first position and $'2'$ in the second position. All possible permutation of other numbers are $13!$. The same thing you can do it when the number $'1'$ is in $15th $ position and the number $'2'$ in the $14th$ position. Consider now the number $'1'$ in the position $i$ ($1<i<15$) while the number $'2'$ can be in the position $i-1$ or $i+1$. Therefore all possible permutations of other numbers are in these  possible cases $2\cdot 13\cdot 13!$. The probability that $1$ and $2$ are contiguous is $$\frac {28\cdot 13!}{15!}= \frac {2}{15}$$ Thereforethe probability that $1$ and $2$ aren't contiguous is $$1-\frac{2}{15}=\frac{13}{15}$$
