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Math Help - Probability and geometric distributions

  1. #1
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    Question Probability and geometric distributions

    Calculate the following probabilities. I've been told that X~geom(0.1) means X follows a geometric distribution

    (a) For X ~ Geom(0.1), calculate P(X ≤ 10).
    (b) For X ~ Geom(0.2), calculate P(5 ≤ X ≤ 10).
    (c) For X ~ P(5), calculate P(X < 6).

    I know what the geometric formula is for a sequence, does this apply to this though? :S
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  2. #2
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    Quote Originally Posted by mitch_nufc View Post
    Calculate the following probabilities. I've been told that X~geom(0.1) means X follows a geometric distribution

    (a) For X ~ Geom(0.1), calculate P(X ≤ 10).
    (b) For X ~ Geom(0.2), calculate P(5 ≤ X ≤ 10).
    (c) For X ~ P(5), calculate P(X < 6).

    I know what the geometric formula is for a sequence, does this apply to this though? :S
    Questions (a) and (b) require the formula for cumulative probability for a geometric distribution. See Geometric distribution - Wikipedia, the free encyclopedia:

    \Pr(X \leq k) = 1 - (1 - p)^k.

    (a) Substitute p = 0.1 and k = 10.

    (b) = \Pr(X \leq 10) - \Pr(X \leq 5). Calculate each of these just like (a) but use p = 0.2.

    (c) Poisson distribution with a mean of 5: \lambda = 5. You need to substitute into the probability mass function (pmf). See Poisson distribution - Wikipedia, the free encyclopedia:

    \Pr(X = k) = \frac{e^{-\lambda} \lambda^k}{k!}.

    Substitute \lambda = 5 and k = 0. Then do the same for k = 1, 2, 3, 4, 5. Then add up all these probabilities to get the answer.
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  3. #3
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    (a) P(X<=10) = 1 - (0.9)^11 = 0.6862
    (b) P(5<=X<=10) = 0.686 - P(X<=4) - 0.6862 - 0.4095 = 0.277
    (c) P(X=k) = e^-5 . 5^k / k!
    So P(X<6) = Probability(X=0 or 1 or ... or 5)
    = e^5 (5/1! + 5^2/2! + ... + 5^5/5!) = 0.609

    this is what ive got, do you agree?
    Last edited by mitch_nufc; February 7th 2009 at 05:44 AM. Reason: correcting
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  4. #4
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    Quote:
    Originally Posted by mitch_nufc
    Calculate the following probabilities. I've been told that X~geom(0.1) means X follows a geometric distribution

    (a) For X ~ Geom(0.1), calculate P(X ≤ 10).
    (b) For X ~ Geom(0.2), calculate P(5 ≤ X ≤ 10).
    (c) For X ~ P(5), calculate P(X < 6).

    I know what the geometric formula is for a sequence, does this apply to this though? :S


    Questions (a) and (b) require the formula for cumulative probability for a geometric distribution. See Geometric distribution - Wikipedia, the free encyclopedia:

    .

    (a) Substitute p = 0.1 and k = 10.

    (b) . Calculate each of these just like (a) but use p = 0.2.

    (c) Poisson distribution with a mean of 5: . You need to substitute into the probability mass function (pmf). See Poisson distribution - Wikipedia, the free encyclopedia:

    .

    Substitute and k = 0. Then do the same for k = 1, 2, 3, 4, 5. Then add up all these probabilities to get the answer. __________________

    shouldnt (b) be Pr(x < or equal to 10)-Pr(x< or equal to 4) as 5 should be inclusive in the answer. with mr fantastics theory we get 0.2203, but i think the answer should be 0.3022?
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  5. #5
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    Post Geometric and binomial

    Can anyone double check my answers for these problems please?

    a) For X~geom(0.1), calculate P(X≤10)
    I got 0.6862 (someone told me this was wrong)

    b)For X~geom(0.2), calculate P(5≤X≤10)
    I got 0.2767 (again, someone told me im wrong)

    c)For X~P(5), calculate P(X<6)
    I got 0.6092 (again, someone told me i was wrong again!)

    If you agree with mine then cool, if not can you show me how you got your answer?
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  6. #6
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    Quote Originally Posted by mr fantastic View Post
    Questions (a) and (b) require the formula for cumulative probability for a geometric distribution. See Geometric distribution - Wikipedia, the free encyclopedia:

    \Pr(X \leq k) = 1 - (1 - p)^k.

    (a) Substitute p = 0.1 and k = 10.

    (b) = \Pr(X \leq 10) - \Pr(X \leq {\color{red}4}). Calculate each of these just like (a) but use p = 0.2.

    (c) Poisson distribution with a mean of 5: \lambda = 5. You need to substitute into the probability mass function (pmf). See Poisson distribution - Wikipedia, the free encyclopedia:

    \Pr(X = k) = \frac{e^{-\lambda} \lambda^k}{k!}.

    Substitute \lambda = 5 and k = 0. Then do the same for k = 1, 2, 3, 4, 5. Then add up all these probabilities to get the answer.
    Correction of minor error/typo in red.

    For (a) and (b), note that I have used the cumulative probability for a geometric distribution whose support does not include zero. If zero is included, you have to use the other cumulative probability.

    Quote Originally Posted by mitch_nufc View Post
    Can anyone double check my answers for these problems please?

    a) For X~geom(0.1), calculate P(X≤10)
    I got 0.6862 (someone told me this was wrong)

    b)For X~geom(0.2), calculate P(5≤X≤10)
    I got 0.2767 (again, someone told me im wrong)

    c)For X~P(5), calculate P(X<6)
    I got 0.6092 (again, someone told me i was wrong again!)

    If you agree with mine then cool, if not can you show me how you got your answer?
    c) I get 0.61596. You probably need to re-check the calculations I said to do in my first post.

    a) I get 0.65132 (assuming that X = 0 is not included in the support).

    b) I differ with you here too.

    Your troubles probably lie in the arithmetic, not the method.
    Last edited by mr fantastic; March 5th 2009 at 05:51 PM.
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