# Probability of not picking a row in a random draw where the number of rows are N

There are $$N$$ rows :$$R_1, R_2,R_3,..., R_N$$. What is the probability of not picking a row in a random draw?

My try and understanding :

Let $$X$$ be a random variable which is defined as follows: $$X = \text{ Number of rows picked in a random draw.}$$ So, the values of random variable which it can take are : $$X = 0, 1,2,3,4,...N$$ Hence the required probability is :
\begin{aligned} \text{ Probability of not picking a row in a random draw } & = 1-\text{Probability of picking a row in a random draw } \\ & = 1 - P(X = 1) \\ & = 1 - \frac{N}{2^N} \end{aligned} The size of sample space is $$|S| = 2^N$$ and the favourable outcomes are $$N$$ for $$X =1$$

Could someone explain whats wrong in it, because I read a blog on the medium* where it is mentioned the required probability is $$\frac{N-1}{N}$$. Follow the mentioned link or see the below screen shot:

Note: Please try to open the website in incognito mode to get rid from sign in etc and the probability thing is mentioned at the end of the blog.

$$X$$ is not explicitly mentioned in the article, so I think in this context, $$X$$ refers to the row number that is picked and not the number of rows picked in a random draw. If this is indeed the case, then I think the statement
probability of not picking a row in a random draw of size $$N = 1$$
which would make it clearer. Since \begin{align} \sum_{i = 0}^{N} p(X = i) &= 1 \\ \left[\sum_{i \neq x} p(X = i)\right] + p(X = x) &= 1 \\ \sum_{i \neq x} p(X = i) &= 1 - p(X = x) \\ &= p(X \neq x) \end{align} and so $$p(X \neq x) = \sum_{i \neq x} p(X = i)$$ Since there are $$N$$ rows to pick from, and assuming that each row is equally likely to be picked, then $$\sum_{i \neq x} p(X = i) = \frac{N-1}{N}$$