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Bernoulli trial summation by hand

  1. Aug 27, 2014 #1
    1. The problem statement, all variables and given/known data
    Show that the expected number of successes in n Bernoulli trials w probability p of success is <x> = np


    2. Relevant equations



    3. The attempt at a solution

    So I get the right answer which is this: [tex]E\left( x\right) =\sum _{x=0}^{n}x\left( \begin{matrix} n\\ x\end{matrix} \right) p^{x}\left( 1-p\right) ^{n-x}[/tex]

    I know it's right, though, because I inputted it into wolframalpha and get <x> = np.

    My question is how do I try to sum this by hand. I really have no idea where to start and I'm curious as to how this is done. Do I maybe use Stirling's formula to simplify the C(n, x)?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Aug 27, 2014 #2

    LCKurtz

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  4. Aug 27, 2014 #3
    excellent, thanks!
     
  5. Aug 27, 2014 #4

    Ray Vickson

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    This was already answered for you in another thread: see post #5 in your thread on the expected number of 5's in tossing a die.
     
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