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A sequence of n independent experiments is performed. Each experiment is a success with probability p and a failure with probability q = 1 - p. Show that conditional on the number of successes, all possibilities for the list of outcomes of the experiment are equally likely (of course, we only consider lists of outcomes where the number of successes is consistent with the information being conditioned on).

I just learned about random variables and my thinking is: if we know there are k successes, we know there are $\binom{n}{k}$ different sequences of 1's and 0's such that there are k 1's. And these sequences are all equally likely by symmetry. Thus, given the number of successes, k, all possibilities for the list of outcomes of the experiment are equally likely.

But I know that's wrong because I didn't use any random variables.

So my main question is: what's wrong in my thinking (which involves no random variables)?

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The issue with your reasoning is that you do not know (or have not stated why you know) that the $\binom{n}{k}$ paths that are possible are equally likely given the condition that the total number of heads is $k$. This follows from the independence of the $X_i$, which your reasoning did not mention.

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  • $\begingroup$ Does this work: each of the $\binom{n}{k}$ paths are equally likely because each trial is independent and each trial has the same probability for a success (p) and a failure (1-p). $\endgroup$
    – beginner
    Commented Feb 22, 2021 at 20:07
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    $\begingroup$ It is true that each of the paths are equally likely because each trial is independent, but what you need to show is that they are equally likely given the number of heads. $\endgroup$
    – nullUser
    Commented Feb 22, 2021 at 20:10

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