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I have a sequence of $15$ experiments and I want to know the probability $S_1$ of success of each experiment so that the overall probability to obtain $10$ successes in the $15$ experiments run equals $P_1$. To do so, I use binomial distribution and I am able to determine $S_1$ quite easily.

Now I want to add $7$ more different experiments for a total of $22$ heterogeneous experiments and have an overall probability of obtaining $11$ successes of $P_2$. How can I determine the probability of success $S_2$ of the $7$ new experiments in order to have the probability $P_2$ of obtaining $11$ successes over $22$ experiments, given that the first $15$ experiments have success probability of $S_1$?

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    $\begingroup$ Oh, are you saying the extra $7$ experiments are somehow different? In that case you will need to use the binomial theorem in each block and add over pairs $(a,b)$ with $a+b=11$. $\endgroup$
    – lulu
    Aug 10, 2017 at 13:03
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    $\begingroup$ Well, there are lots of ways to get $11$ successes out of $22$. I can get $4$ out of the first $15$ and $7$ out of the last $7$, for instance. That has probability $\binom {15}4\times S_1^4\times (1-S_1)^{11}\times S_2^7$. And so on, for each pair $(4,7), (5,6), (6,5), \cdots$. You'll get a degree $7$ polynomial in $S_2$ which I suggest you solve numerically. $\endgroup$
    – lulu
    Aug 10, 2017 at 13:09
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    $\begingroup$ A suggestion: edit your post for clarity. Why make your readers deduce that the new experiments are different from the old? $\endgroup$
    – lulu
    Aug 10, 2017 at 13:11
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    $\begingroup$ Yeah, I seriously doubt there is a short cut here....but the numerical calculation shouldn't be too bad. $\endgroup$
    – lulu
    Aug 10, 2017 at 13:14
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    $\begingroup$ Well, you are doing this on a computer, I assume? Shouldn't be that bad if you have a modest number of batches. The partition number is binomial so that will start to grow on you as the number of batches and the target success number increases. $\endgroup$
    – lulu
    Aug 10, 2017 at 13:23

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