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Math Help - Another probability distribution question

  1. #1
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    Another probability distribution question

    The question:

    Calculate P(X \le k) when X has the probability distribution

    P(X = k) = p(1 - p)^k, for k = 0, 1, 2, ... and 0 < p < 1.

    My attempt:
    \sum\limits_{k = 0}^{\infty} p(1 - p)^k

    Then by geometric series:

    = p(1 - p)\frac{1 - (1 - p)^k }{1 - (1 - p)}
    = (1 - p)-(1 - p)^{k + 1}

    The answer is supposed to be 1 - (1 - p)^{k+1}

    What have I screwed up this time? :/

    EDIT: It has occurred to me that I've done the geometric series wrong. >_<
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  2. #2
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    Quote Originally Posted by Glitch View Post
    The question:

    Calculate P(X \le k) when X has the probability distribution

    P(X = k) = p(1 - p)^k, for k = 0, 1, 2, ... and 0 < p < 1.

    My attempt:
    \sum\limits_{k = 0}^{\infty} p(1 - p)^k

    Then by geometric series:

    = p(1 - p)\frac{1 - (1 - p)^k }{1 - (1 - p)}
    = (1 - p)-(1 - p)^{k + 1}

    The answer is supposed to be 1 - (1 - p)^{k+1}

    What have I screwed up this time? :/

    EDIT: It has occurred to me that I've done the geometric series wrong. >_<
    To start with, you should be calculating \displaystyle \sum_{i=0}^{k} p (1 - p)^i =  p \sum_{i=0}^{k} (1 - p)^i. And yes, you have done the geometric series wrong. I suggest you go back and review the appropriate formula and then use the formula more carefully.
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  3. #3
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    Yeah, once I applied it properly I got the correct answer. Thanks.
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