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Homework Help: Discrete probability function

  1. Jun 30, 2006 #1
    I got really really confused by this supposedly easy discrete probability problem:

    The problem asks:
    a)toss a die until a "6" appears. Find the probability distribution of X where X is the number of tosses neded to obtain the first six.
    b) Prove that the summation of P(x) from x = 1 to n (not infinity!) is equal to 1.
    c) Determine P(X = 2k + 1) for every k an element of a natural number.
    d) Find the distribution function of X in c)

    Here's what I got:
    for a) obviously, the answer is a geometric distribution as the answer is:
    P(X=x) = f(x) = (5/6)^(x-1) * (1/6) for x = 0,1,2,3,... and 0 otherwise

    for c) I just substitute x with 2k + 1, thus,
    P(X = 2k + 1) = (5/6)^(2k) * (1/6) for any k, natural number

    what I can't understand is b and d.
    How the heck are you suppose to prove b????
    since P(x) is a geometric series, then its summation from 1 to n is given by a(1 -r^n)/(1-r) which is equal to 1 - (5/6)^n when a = 1/6 and r = 5/6....this summation is equal only to 1 when n reaches infinity! I asked the TA who gave this question about this, and he keeps insisting it should be up to n, not up to infinity. (if it is up to infinity, the summation is just a/1-r = 1!). Am I jsut crazy and the TA is right, or can it be proven that the summation up to n is equal to 1?

    And for d), is it possible to get the distibution function for X = 2k+1??? not the probability (which is c), but the distribution itself. I asked if this is the conditional distribution f(x| x = 2k +1), and the TA says it isn't. It's the distribution itself. Is this even possible? Again, I hope I'm not crazy.

    This has been driving me bananas for the last 2 days...

    any help is appreciated. thanks
    Last edited: Jul 1, 2006
  2. jcsd
  3. Jul 2, 2006 #2


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    Science Advisor

    First, f(x) is not defined when x = 0. (Or you could define it to be 0)

    Your TA appears to be wrong. n is not even defined in the problem, and it does sum to 1 only as x goes to infinity. The distribution function g would be the function satisfying
    g(k) = P(X = 2k + 1|X is odd)
    To actually find g, you can just normalize your function P(X = 2k + 1) so that the infinite sum is 1.
    Last edited: Jul 2, 2006
  4. Jul 7, 2006 #3
    you guys are right. The TA corrected the question in b and changed n to infinity. I am awfully frustrated by the fact though that he only gave this correction 1 day before the homework is due AND even more so that he kept insisting that it is n, not infinity when I asked him about the problem 3 days before.
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