{
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  "body": "\n## Statement\n\nLet $n \\ge 1$ be an integer, let $p \\in [0, 1]$, and let $X$ have the [[binomial distribution|0VRWZDWW6B]] $\\operatorname{Bin}(n, p)$. Then\n$$E[X] = \\sum_{k=0}^{n} k \\binom{n}{k} p^k (1 - p)^{n - k} = np.$$\n\n## Remarks\n\nA short proof: $X$ has the same law as $Y_1 + \\dots + Y_n$ with independent [[Bernoulli|0J97P5KZY0]] $Y_i$ with $E[Y_i] = p$. Then [[linearity of expectation|0P7PTC8DDE]] gives $np$.\nA direct proof uses $k \\binom{n}{k} = n \\binom{n-1}{k-1}$ and the binomial theorem.\nThe variance of $X$ is $np(1 - p)$.\n\n## Explanations\n\n### L1 human\n\nIn $n$ trials with success probability $p$, you expect $np$ successes.\nFor $100$ tosses of a fair coin, the expected number of heads is $50$.\n",
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