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Binomial Distribution

  1. May 22, 2012 #1
    Question is:

    "If you roll a fair coin 10 times what is the expected product of number of heads and number of tails?"

    Someone answered 25 at at glassdoor.com. My answer would be:

    E(k(10-k)) where k is the rv representing the number of heads thrown.
    = 10E(k) - E(k^2)
    = 10*mean - (var + mean^2)

    where mean = np = 10*0.5 = 5, and var = npq = 10*.5*.5 = 2.5 (these are the formulas for the binomial distribution), thus,
    = 10*5 - (2.5 + 25) = 22.5

    Who is correct?

    Thanks.

    Nick.
     
  2. jcsd
  3. May 22, 2012 #2

    haruspex

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    You are right. Another way:
    [itex]\sum[/itex]r(n-r).[itex]^{n}[/itex]C[itex]_{r}[/itex] =
    [itex]\sum[/itex]n!/(r-1)!/(n-r-1)! =
    [itex]\sum[/itex]n(n-1)(n-2)!/(r-1)!/(n-r-1)! =
    n(n-1)[itex]\sum[/itex][itex]^{n-2}[/itex]C[itex]_{r-1}[/itex] =
    n(n-1).2^(n-2)
    Dividing by 2^n to get the average:
    n(n-1)/4
    For n = 10 this gives 22.5
     
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