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Identity for the probability of a coin having an even number of heads

  1. Nov 22, 2009 #1
    1. The problem statement, all variables and given/known data
    Given that n independent tosses having probability of p of coming up heads are made, show that an even number of heads results is 0.5(1+(q-p)^n) where q is 1-p, by proving the identity
    Sigma from i=0 to n/2 of (n choose 2i) (p^2i)(q^(n-2i))=0.5(((p+q)^n)+(q-p)^n)


    2. Relevant equations



    3. The attempt at a solution
    I expanded it, (n choose 0)p^0 q^n + (n choose 2)p^2 q^(n-2)+....letting i=n/2 we get (n choose n)p^n q (n-n) ends up being just p^n.
     
  2. jcsd
  3. Nov 22, 2009 #2

    tiny-tim

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    Hi koab1mjr! :smile:

    (have a sigma: ∑ and try using the X2 and X2 tags just above the Reply box :wink:)
    Hint: start at the end …

    what is the coefficient of pjqn-j of 0.5(((p+q)n)+(q-p)n) for j even? and for j odd? :wink:
     
  4. Nov 22, 2009 #3
    thanks, but...
    I tried it, but i am still not getting it, because starting from the end for even, i got p^2q^(n-2), p^4q^(n-4) and so on, plus i cannot get the factorials to work out....
     
  5. Nov 23, 2009 #4

    tiny-tim

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    (just got up :zzz: …)
    (please use the X2 tag just above the Reply box)

    p2qn-2 time what? p4qn-4 times what? … p2jqn-2j times what?
     
  6. Nov 23, 2009 #5

    Mark44

    Staff: Mentor

    Couldn't resist: The probability of a coin having an even number of heads should be zero.:smile:
     
  7. Nov 24, 2009 #6
    timing each one by (n choose 2), (n choose 4), and so on until i hits n/2, and at that point all becomes just p^n
    ???
     
  8. Nov 25, 2009 #7

    tiny-tim

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    Hi koab1mjr! :smile
    (you can write that nC2 etc :wink:)

    Yup! :smile:

    And that's exactly the left-hand-side of the original identity, isn't it? :wink:
     
  9. Nov 25, 2009 #8
    yeah, when you plug in i=n/2 on the he left side you get p^n, but when you expand the right side and at the end you do the exact same thing, but the problem is that the identity of (1/2)[(p+q)^n + (q-p)^n) doesnt show up at all??
     
  10. Nov 25, 2009 #9

    tiny-tim

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    Sorry, I'm not following you at all. :redface:
     
  11. Nov 25, 2009 #10
    Hi Tiny Tim

    I finally got it. I applied newton's bionomial expansion to prove the identity. Thank you so much for your help. IT help get me thinking in the right direction. You are the best.
     
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