The formula for binomial distrubtion looks like that:

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But is it possible to calculate proper results using slightly different formula?

The number of possible outcomes can be calculated using formula for variations with repetitions:


For example if I throw a coin 3 times, I get 8 possible outcomes. m=2 N=3 $2^3=8$

It looks to me like binomial distribution can be simply calculated by using this formula: $P(x)=\frac{{N\choose x}}{m^N}$

N = number of trials x = number of success m = number of outcomes from one trial

But I'm not sure. Maybe, it works only when there is a 50% probability of a success?

  • $\begingroup$ Did you try it at any other value of $p$? Like $p=0.31$, say? It would seem the obvious thing to do when one particular value revealed a dramatic simplification -- to try another, less "special" value. $\endgroup$ – Glen_b Sep 4 '17 at 23:18

Yes this is correct assuming that $p=0.5$ because $m^N = 2^N$ and $p^x (1-p)^{N-x}=p^N= \frac{1}{2^N}$ and so the numerator and denominator cancel. This is obviously not true for other values of $p$.


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