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Probability Theory ; Binomial Distribution?

  1. Nov 29, 2009 #1
    1. The problem statement, all variables and given/known data
    Now you and your fiend play a different game. You flip your coin until it comes up heads the first time. Let X denote the number of flips needed. Your friend rolls its die until it comes up "3" or "5". The first try let Y denote the number of rolls needed. Assume X and Y are independent; note that X >= 1 and Y >=1. a) Determine P(X=n), n>=1 b) Determine P(Y=n), n>=1 c) Determine E(x) d) Determine E(y) e) Determine P(x=y) f) Given that x = y, determine the expected value of this common value.

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    2. Relevant equations
    Binomial Distribution
    nCk (P)^k (1-P)^n-k

    3. The attempt at a solution
    a) nC1 (1/2)(1/2)^n-1 (My friend did not include the nC1)
    b) nC1 (1/3)(2/3)^n-1 (My friend did not include the nC1)
    c) n*1/2
    d) n*1/3
    e) nC1 (1/2)(1/2)^n-1 * nC1 (1/3)(2/3)^n-1 (This is weird, my friend integrated from 1 to infinite) he got 1/4??
    f) nsubx*psubx + nsuby*psuby (I definitely need help on this one.)
     
  2. jcsd
  3. Nov 29, 2009 #2

    Dick

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    For a) and b), why do you think the nC1 should be in the expression? And for c) and d) the expectation value is a sum over all n. The final expression shouldn't have an 'n' in it. Maybe you'd better review the definition of 'expectation value'.
     
  4. Nov 29, 2009 #3
    Ok so the reason nC1 should not be there is because we only care about the n trial. Right? Still a bit confused about the Expected.. If X is geometric with probability p then E(X) = 1/p = 2 and E(Y) = 3? And e) is P(x=y)= sum(n=1 to +oo) 1/6(1/3)^(n-1) = 1/4

    f) is still bothering me a little..
     
    Last edited: Nov 29, 2009
  5. Nov 29, 2009 #4

    Dick

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    Right. The first n-1 trials have to be failures, only the last one needs to succeed. And, yes, now I think you have the expectation values right as well. I've got to confess, I'm a little vague on the last one. Probabilility isn't my field, but isn't it sum(n=1 to +oo) n*(1/6)*(1/3)^(n-1)?
     
  6. Nov 30, 2009 #5
    Ok ok, so by definition, E[X] = sum(n=1 to +oo) n * f(n) where f(n) is the density function. Yes I agree. Thank you very very much for all your help!
     
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