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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: enter image description here

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.

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1 Answer 1

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$X$ 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 is

should be re-worded to

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 $$ \sum_{i \neq x} p(X = i) = \frac{N-1}{N} $$

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  • $\begingroup$ Thanks for rewording in a better and the used language sounds ambiguous to me , so i landed with different thought and answer. $\endgroup$ Jul 20, 2021 at 9:29
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    $\begingroup$ No problem. Unfortunately, since medium articles are not peer-reviewed, you need to be wary of them. This is true for most articles on the internet. $\endgroup$
    – Mahmoud
    Jul 20, 2021 at 9:53

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