1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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.


    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


    User Avatar
    Science Advisor
    Homework Helper

    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


    User Avatar
    Science Advisor
    Homework Helper

    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!
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook