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Math Help - proof of the identity for a coin having even number of heads

  1. #1
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    proof of the identity for a coin having even number of heads

    Suppose that n independent tosses of a coin having probability p of coming up heads are made. Show that the probability that an even number of heads results is (1/2)(1+(q-p)^n) where q=1-p. Do this by utilizing the identity

    sigma from i=0 to n/2 of (n choose 2i) p^2i q^(n-2i)=(1/2)[(p=q)^n+(q-p)^n]
    where n/2 is the largest integer less than or equal to n/2.




    I tried expanding it, but every time i end up with p^n only.


    Thanks so much!!
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    Quote Originally Posted by libragirl79 View Post
    Suppose that n independent tosses of a coin having probability p of coming up heads are made. Show that the probability that an even number of heads results is (1/2)(1+(q-p)^n) where q=1-p. Do this by utilizing the identity

    sigma from i=0 to n/2 of (n choose 2i) p^2i q^(n-2i)=(1/2)[(p=q)^n+(q-p)^n]
    where n/2 is the largest integer less than or equal to n/2.




    I tried expanding it, but every time i end up with p^n only.


    Thanks so much!!
    You want to show
    \sum_{i=0}^{\lfloor n/2 \rfloor} \binom{n}{2i} p^{2i} q^{n-2i} = (1/2) \; [(q+p)^n + (q-p)^n].

    Expand (p+q)^n and (p-q)^n by the Binomial Theorem.

    Add the two equations and observe that half the terms in the expansions cancel.
    Last edited by awkward; November 26th 2009 at 07:23 AM. Reason: typo
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