Situation: a box contains N balls numbered $1,2...,N$. $N$ unknown. $n$ balls drawn using SRS with replacement and number recorded. A random variable $X$ is defined as the number recorded on $ith$ draw.

My confusion: Is this a case of discreet uniform or binomial. Since each ball always has $1/N$ probability of being chosen, is $X$ discreet uniform or a $B(n,1/N)$ variable.


2 Answers 2


If I understand you correctly it should be uniform with probability 1/N.

𝐵(𝑛,1/𝑁) would give you the probability distribution that you observed a given, fixed ball a certain number of times, given that you drew n balls with replacement. I.e. you define a particular ball (say ball number 42) as the success ball, and B(n, 1/N) is then the probability distribution over the number of times you see this ball when you draw n times with replacement.

From Wikipedia: the binomial distribution with parameters n and p is the discrete probability distribution of the number of successes in a sequence of n independent experiments, each asking a yes–no question

  • $\begingroup$ In that case, is the support of $X$ from $1,...,N$? If so, isint $n$ unimportant here? $\endgroup$
    – Harry
    Apr 3, 2019 at 14:30
  • $\begingroup$ For the original question yes. And since you are only interested in the i-th draw, the number n of draws you did in total doesn't matter. You are basically drawing n times but throwing the balls back without looking at them. Except for the i-th one. Since order doesn't matter (due to replacement) that is the same as drawing only once. $\endgroup$
    – Rauwuckl
    Apr 3, 2019 at 14:33

A Binomial distribution is always over counts of events (where each event has one of two possible outcomes). E.g. the number of heads in a sequence of $n=100$ tosses of an unfair coin with $p=0.2$ has a binomial distribution $B(100,0.2)$.

Your random variable $X$ is the outcome of a single event. If this were an event with only two possible outcomes, then technically you could say this follows a Binomial distribution with $n=1$, but in this special case the Binomial is equal to the Bernoulli distribution.

Since your event (the number recorded on the $i$-th draw) has multiple (namely $N$) possible outcomes, we need a generalization of the Bernoulli distribution. The most general case would be the categorical distribution, which allows for different outcomes to have different possibilities. However, the way your problem is framed means that all outcomes are equally probably, so in that case the categorical distribution reduces to a discrete uniform.


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