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Math Help - Probability Theory; Binonial Distribution?

  1. #1
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    Probability Theory; Binonial Distribution?

    The problem statement
    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.

    Equations
    Binomial Distribution
    nCk (P)^k (1-P)^n-k

    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.)

    Update:
    Can someone confirm these?
    a) (1/2)(1/2)^n-1
    b) (1/3)(2/3)^n-1
    c) E(X)=1/(1/2)=2
    d) E(Y)=1/(2/6)=3
    e) P(x=y)= sum(n=1 to +oo) 1/6(1/3)^(n-1) = 1/4
    f) sum(n=1 to +oo) n*(1/6)*(1/3)^(n-1) = 3/8
    Last edited by aeubz; November 29th 2009 at 09:05 PM.
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  2. #2
    Moo
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    Hello,

    Update:
    Can someone confirm these?
    a) (1/2)(1/2)^n-1
    b) (1/3)(2/3)^n-1
    c) E(X)=1/(1/2)=2
    d) E(Y)=1/(2/6)=3
    e) P(x=y)= sum(n=1 to +oo) 1/6(1/3)^(n-1) = 1/4
    f) sum(n=1 to +oo) n*(1/6)*(1/3)^(n-1) = 3/8
    (I hope you understood where these formulas came from in e and f)

    The binomial distribution counts the number of successes. The geometric distribution denotes the first time a success happens
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  3. #3
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    Quote Originally Posted by Moo View Post
    Hello,


    (I hope you understood where these formulas came from in e and f)

    The binomial distribution counts the number of successes. The geometric distribution denotes the first time a success happens
    Wow. Thank you very much! That really cleared things up. I was wondering why we used the nCk in the nCk (P)^k (1-P)^n-k. And now it makes sense, that is the case for the number of successes and without nCk means the first time a success happens. Thank you!
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