Ufyukyu

2 $begingroup$ Suppose I have a set of $k$ integers such that every number is unique. Let $A = {1,2,3,4,...,k}$ Now suppose that we rearrange these numbers to a random permutation in the set. I want to find the probability of finding a fixed number $x$ at any position of the set. Here's my understanding: The probability of the first number in the set being $x$ is $frac{1}{k}$ The probability of the second number in the set being $x$ is $(1-frac{1}{k})frac{1}{k}$ The probability of the third number being $x$ is $(1-frac{1}{k})^2 frac{1}{k}$ That leads to the probability of the $n^{th}$ term being $x$ to be $(1-frac{1}{k})^{n-1} frac{1}{k}$ Is this correct? probability combinatorics sh...