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I am working on a problem involving Bernoulli trials with an interesting twist. Here's the setup:

  • $n$ independent Bernoulli trials are conducted.
  • The success probability $Q$ is unknown and follows a uniform distribution between $0$ and $1$.

What is the probability of getting exactly $m$ successes out of these $n$ trials?

I can solve the question by applying the law of total probability. But my lecturer suggested that there's also a potential intuitive explanation using the indicator variable $\mathbb{I}_{\{Q<p\}}\sim \mathrm{Bern}(p)$, but I'm unsure how to apply it to reach the solution.

KanoDaisuki
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1 Answers1

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A Bernoulli trial can be performed by drawing a random number $r$ from the uniform distribution on $[0,1]$ and comparing it to the success probability $Q$ (with success iff $r\lt Q$).

So what you’re doing can be described like this: You draw one number $Q$ from the uniform distribution on $[0,1]$, and then you draw $n$ more numbers from that distribution, compare them to $Q$ and count how many are $\lt Q$.

By symmetry, all $n+1$ of these numbers have the same probability of having any particular rank among them. Thus, all $n+1$ success counts from $m=0$ to $m=n$ (which correspond to $Q$ having rank $m+1$ among the $n+1$ random numbers) are equiprobable.

For more on this approach, see Why are all subset sizes equiprobable if elements are independently included with probability uniform over $[0,1]$? and How to prove the rule of succession without calculus?

joriki
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