Choosing $k$ consecutive numbers out of $n$ numbers What is the probability to choose $10$ consecutive numbers out of the numbers $1-100$ ?
And what is the probability for the general case given $k$ consecutive numbers out of $n$ numbers ?
I think the answer should be $\frac{n-k+1}{\binom{n}{k}}$ but i am not sure !
Thanks. 
 A: Yes, you are correct.
For $k$ consecutive numbers you form groups of $k$ numbers (which are consecutive).
Let us consider a set of $n$ numbers $[a_1, a_2, a_3, ....., a_n]$. Now the $1^{st}$ group will be:
$[a_1, a_2, a_3, ....., a_k]$ 
$2^{nd}$ group wil have the first element $a_1$ removed  and $a_{k+1}$ added. If we continue this till the last ($x^{th}$) group we get:
$\underbrace{[a_1, a_2, a_3, ....., a_k]}_{1^{st}}\ ,\underbrace{[a_2, a_3, a_4, ....., a_{k+1}]}_{2^{nd}}\ ,.......,\underbrace{[a_{x}, a_{x+1}, a_{x+2}, ....., a_{k+(x-1)}]}_{x^{th}}$
So the last number of the last group would have to be $a_{n}$. Hence if we compare the subscripts of the last number obtained, we get: 
$k+(x-1)=n \implies x=n-k+1$
So we have a total of $n-k+1$ groups, each having $k$ consecutive numbers and the total number of ways to select $k$ numbers from a pool of $n$ numbers would be ${}^{n}C_{k}$.
Thus, the probability would be $\displaystyle\frac{n-k+1}{\binom{n}{k}}=\frac{n-k+1}{{}^{n}C_{k}}$.
