Binomial repartition of Bernoulli

Submitted by Structure on Thu, 01/17/2008 - 07:55.

$X_n=\left [\begin{array}{cccccccc}<br /> 0&1&2&.&k&.&n-1&n\\<br /> \left (\begin{array}{c}n\\0 \end {array}\right ) \)q^n&\left (\begin{array}{c}n\\1 \end {array}\right )\)pq^{n-1}&\left (\begin{array}{c}n\\2 \end {array}\right ) \)p^2q^{n-2}&.&\left (\begin{array}{c}n\\k \end {array}\right ) \)p^kq^{n-k}&.&\left (\begin{array}{c}n\\n-1 \end {array}\right ) \)p^{n-1}q^&\left (\begin{array}{c}n\\n \end {array}\right ) \)p^n<br /> \end {array}\right ] \)$

$E(X_n)=\sum_{k=0}^{k=n}k\left (\begin{array}{c}n\\k \end {array}\right ) \)p^kq^{n-k}=\sum_{k=1}^{k=n}k\left (\begin{array}{c}n\\k \end {array}\right ) \)p^kq^{n-k}=$

$np\sum_{k=1}^{k=n}\left (\begin{array}{c}n-1\\k-1 \end {array}\right ) \)p^{k-1}q^{n-1-(k-1)}=np(p+q)^{n-1}=np$

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Binomial repartition of Bernoulli