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Suppose $x_1,\ldots,x_n$ were chosen (independently) following a Bernoulli distribution with probability $p$. Also suppose that, actually $\geq k$ of them were true, so that we could choose some set $A\subseteq [n]$ of cardinality $k$ which $x_i=1$ for all $i\in A$. If we choose some random $x_i$ which $i\notin A$, what would be its distribution? In other words, what would be the probability of such $x_i$ being true?

Edit:

I add an example instance when $n=3$, $p=0.3$ and $k=1$. The calculation below follows the solution which I believe is correct.

There are seven possibilities where $(x_1,x_2,x_3)$ contain at least one $1$:

  • $(0,0,1),0.147\times\frac{1}{1-0.343}$
  • $(0,1,0),0.147\times\frac{1}{1-0.343}$
  • $(1,0,0),0.147\times\frac{1}{1-0.343}$
  • $(0,1,1),0.063\times\frac{1}{1-0.343}$
  • $(1,0,1),0.063\times\frac{1}{1-0.343}$
  • $(1,1,0),0.063\times\frac{1}{1-0.343}$
  • $(1,1,1),0.027\times\frac{1}{1-0.343}$

From each cases we can choose some set $A$ of indices of size $k=1$, which by symmetry won't matter much how we choose it. So the desired probability is

$$\frac{1}{1-0.343}\times(0.147\times 3\times0+0.063\times3\times0.5+0.027\times1\times1)=\frac{1}{1-0.343}\times0.1215$$

user5876164
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  • Hi! Let me see if I've got this straight. You have $n$ Bernoulli trials with probability $p$. Conditioned on there being at least $k$ successes, you select a subset $A$ of $k$ of the successes. (Because of symmetry, it doesn't matter how this selection is done; w.l.o.g., assume that we pick the first $k$ successes.) Now select one of the trials that is not in this subset $A$. What is the probability that it is a success? Is that what you're asking? – Brian Tung Jul 29 '25 at 01:49
  • @BrianTung Yes, that's what I'm asking. As I think more about the problem it seems like my approach under 'Edit' is correct, but I still need verification. – user5876164 Jul 29 '25 at 01:56
  • I believe your approach is correct, though generalizing will take some care, since $k = 1$ is a bit of a special case. I'm not sanguine about it leading to a closed-form solution, though. ¶ On a side note, I found the exposition a little difficult to follow, possibly because it uses formalisms that I personally didn't find useful. It might help to add in a more "plain English" description of the situation. – Brian Tung Jul 29 '25 at 01:59

1 Answers1

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$$\mathbb P(x_i|X\ge k)=\sum_{r=k}^n\mathbb P(X=r|X\ge k)\,\mathbb P(x_i|X=r) =\sum_{r=k}^n\frac{\mathbb P(X=r)}{\mathbb P(X=k)}\frac{r-k}{n-k}\\ =\frac{\displaystyle \sum_{r=k}^n\binom nrp^r(1-p)^{n-r}(r-k)}{\displaystyle \sum_{r=k}^n\binom nrp^r(1-p)^{n-r}(n-k)} =\frac ND-\frac{k}{n-k}$$ where $$N=n\sum_{r=k}^n\binom{n-1}{r-1}p^r(1-p)^{n-r}$$ $$D=(n-k)\left(\sum_{r=k}^n\binom{n-1}{r-1}p^r(1-p)^{n-r}+\sum_{r=k}^{n-1}\binom{n-1}{r}p^r(1-p)^{n-r}\right)\\ =(n-k)\left(\frac Nn+\sum_{r=k+1}^n\binom{n-1}{r-1}p^{r-1}(1-p)^{n-r+1}\right)\\ =(n-k)\left(\frac{N}{np}-\binom{n-1}{k-1}p^k(1-p)^{n-k}\right)$$ But $N$ has no closed form, so you won't have a closed form for your probability.

For $n=3$, $p=0.3$ and $k=1$, we have $D=2(1-0.343)$ and $N=np=0.9$ so the probability is equal to $$\frac{0.45}{1-0.343}-\frac 12=\frac{0.1215}{1-0.343}$$ as you already calculated it.

Christophe Boilley
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  • I believe the problem wasn't clear, so I added an example instance and solution for the question. Since I wrote it in a way that I believe is correct, could you please check if my approach is actually correct, and modify the answer accordingly? – user5876164 Jul 29 '25 at 00:50
  • @user5876164 My solution gives the value you calculated afterwards. – Christophe Boilley Jul 29 '25 at 11:22