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For a the probability of a binomial distribution with n trials, the probability of k successes where Prob(success) = p is (n Choose k) * p^k * (1-p)^(n-k). I understand that we have to multiply p^k * (1-p)^(n-k) by something but what’s 1) an intuitive explanation and/or 2) a proof for why we multiply by the (n Choose k). Especially with regard to the intuitive side of things, I feel tempted to say the factor should be (n Permutation k).

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    $\begingroup$ Multiply by ${n\choose k}$ to get the arrangements of $k$ Successes in $n$ trials, which is found by choosing the $k$ positions for the $k$ Successes. Example: For $X \sim \mathsf{n=3, p=1/2},$ there are $s^3 = 8$ possible outcomes. For $P(X=2)$ the ${3\choose 2} = 3$ relevant outcomes are HHT,HTH, THH and $P(X=2)= 3/8.$ $\endgroup$
    – BruceET
    Mar 21, 2021 at 5:59
  • $\begingroup$ @BruceET I think you can convert this to an answer. $\endgroup$
    – gunes
    Mar 21, 2021 at 14:12
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    $\begingroup$ @gunes: OK. Done--with a few extra words. $\endgroup$
    – BruceET
    Mar 21, 2021 at 15:58

1 Answer 1

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If $X \sim \mathsf{Binom}(n,p),$ then to find $P(X=k),$ one multiplies the probability $p^k(1-p)^{n-k}$ of a particular outcome with $k$ Successes by ${n \choose k},$ the number of arrangements of $k$ Successes among $n$ trials, to get the total probability $P(X=k).$ This amounts to choosing the $k$ positions out of $n$ for the $k$ Successes.

Example: For $X\sim\mathsf{Binom}(n=3,p=1/2),$ there are $2^3=8$ possible outcomes altogether, each with probability $1/8.$ For $P(X=2),$ the ${3 \choose 2} = \frac{3!}{2!\cdot 1!}=3$ relevant outcomes are HHT,HTH, and THH, so $P(X=2)=3/8.$

In R, where dbinom is a binomial PDF (or PMF), this result is found as shown below:

dbinom(2, 3, .5)
[1] 0.375
3/8
[1] 0.375
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