For the Binomial Distribution

${\displaystyle{n \choose k}p^{k}(1-p)^{n-k}}$

I'm trying to understand the intuition of the factors. I understand that the probability of $k$ successes is $p^k$, and the probability of the remaining $(n-k)$ being non-successes is $(1-p)^{n-k}.$ I also understand that ${n \choose k}$ is the number of ways $k$ successes can appear across $n$ samples.

The thing I am having trouble with, however, is that the last two factors are probabilities, and the first factor is a number of combinations, but yet the result of multiplying them all together is a probability. How does a number of combinations * a probability * a probability = a probability? Am I thinking about this wrong?

  • $\begingroup$ You said it yourself, "the probability of k successes is $p^k$", what's different now? $p^k$ is a multiplication of probabilities and result in a probability too. $\endgroup$
    – Firebug
    Dec 10 '16 at 23:07
  • $\begingroup$ @Firebug I understand that. I think I was not clear with my question. How does a number of samples * a probability = another probability. Shouldn't it equal a number of samples? The probability is being applied to a real number of samples, so the result should be a smaller number of samples, not another probability. $\endgroup$
    – amoffat
    Dec 10 '16 at 23:11
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    $\begingroup$ Read the first part of this answer of mine to a recent question. $\endgroup$ Dec 10 '16 at 23:11
  • $\begingroup$ @DilipSarwate aha! It's clear now! The binomial coefficient is just the shortened expansion of the sums of probabilities. $\endgroup$
    – amoffat
    Dec 10 '16 at 23:28
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    $\begingroup$ The multiplication is just a summing of mutually exclusive outcomes where you have k successes in n trials. Each has the probaibility p^k (1-p)^(n-k) . $\endgroup$ Dec 10 '16 at 23:56