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I would like to have a formula that gives the number of successful trials of $n$ trials when the probability of each trial is $P$ and supplied with a random value $X=[0,1]$. Do I make somewhat sense?

Here I read about binomial probability and it says:

Suppose a binomial experiment consists of $n$ trials and results in $x$ successes. If the probability of success on an individual trial is $P$, then the binomial probability is: $b(x; n, P) = \frac{ n!}{x! (n - x)!}P^x(1 - P)^{n - x}$.

But if I understand this right I need a formula for the function $x(b, n,P)$.

Update I'll try to clarify:

If you have a lottery you can make a draw to see if you win. If you participate in 200 lotteries you will have to make 200 draws to actually see how many times you'll win. But I want to simplfy the 200 draws to ONE draw. That draw would give me how many times I win the lottery (NOT the expected value, but the actual times I win according to that draw).

With a random value X I could decide each draw where x < P will be a win if x=[0,1]. I want to give my random X to one consolidated draw to decide how many wins I'll get. Still X=[0,1]. Let say if X get really close to 0 I will get many wins, and close to 1 I will get very few, this depends of P and n of course.

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  • $\begingroup$ This makes no sense unless $P=0$ or 1. Otherwise you can find only the probability that $x$ of $n$ trials will be successful. And you have already found the expression for it. An alternative would be to find the expected value of $x$. $\endgroup$ – user May 3 '18 at 22:00
  • $\begingroup$ ok, not sure what b stand for , but in my head i want it to be a super dice that tells me how many sucesses I get. $\endgroup$ – Nick Sick May 3 '18 at 22:04
  • $\begingroup$ As I already wrote an alternative would be to find the expected value of $x$, which is simply $nP$. $\endgroup$ – user May 3 '18 at 22:06
  • $\begingroup$ I've tried to clarify. Please see my update. $\endgroup$ – Nick Sick May 4 '18 at 9:28
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b is the probability of getting say 4 heads in 6 coin tosses. Given a probability of b how many coin tosses do I have to make to get 5 heads? Is that the question you would like to get answered?

The problem with it is it may not have a solution. While there is always a solution to finding the probability of getting x heads from n tosses. There isn't always a solution for finding how many tosses to get x heads with a given probability of b. Having a chart of the probabilities with a solution for n tosses to get x heads is one way.

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  • $\begingroup$ I 've tried to clarify. See update $\endgroup$ – Nick Sick May 4 '18 at 9:27

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