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Math Help - Finding the probability

  1. #1
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    Finding the probability

    An Infinit sequence of independent trails is to be performed. Each trail results in a success with probability 'p' and failure '1-p' . What is the probability that

    1) at least 1 success occurs in the first 'n' trails
    2) exactly 'k' successes occurs in the first 'n' trails.
    3) all trails result in successes?
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  2. #2
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    Quote Originally Posted by Stats View Post
    An Infinit sequence of independent trails is to be performed. Each trail results in a success with probability 'p' and failure '1-p' . What is the probability that

    1) at least 1 success occurs in the first 'n' trails

    Mr F says: 1 - Pr(X = 0).

    2) exactly 'k' successes occurs in the first 'n' trails.

    Mr F says: Pr(X = k).

    3) all trails result in successes?

    Mr F says: 0.
    where X ~ Binomial(n, p).

    And it's trial, not trail.
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  3. #3
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    Is this correct?

    Quote Originally Posted by mr fantastic View Post
    where X ~ Binomial(n, p).

    And it's trial, not trail.
    Please let me know where if i am right or no?

    Answer:
    Binomial P(x;n, \theta) = \binom{n}{x} \theta ^x (1- \theta)^{n-x})

    Probability of atleast 1 success is

    1-P(X=0) = 1 -\binom{n}{0} \theta ^0 (1- \theta)^{n-0} = 1 - (1 - \theta)^n

    Probability of k success occuring in first n trials is
    P(X=k) = \binom{n}{k} \theta^k (1- \theta)^{n-k}

    Probability of all trials being success is
    P(X=n) = \binom{n}{n} \theta^n (1- \theta)^{n-n} = 0
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  4. #4
    MHF Contributor matheagle's Avatar
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    The last one is off....
    That's for n trials, but letting n go to infinity you will get 0 as long as \theta\ne 1
    the rest looks ok

    Quote Originally Posted by Stats View Post
    Please let me know where if i am right or no?

    Answer:
    Binomial P(x;n, \theta) = \binom{n}{x} \theta ^x (1- \theta)^{n-x})

    Probability of atleast 1 success is

    1-P(X=0) = 1 -\binom{n}{0} \theta ^0 (1- \theta)^{n-0} = 1 - (1 - \theta)^n

    Probability of k success occuring in first n trials is
    P(X=k) = \binom{n}{k} \theta^k (1- \theta)^{n-k}

    Probability of all trials being success is
    P(X=n) = \binom{n}{n} \theta^n (1- \theta)^{n-n} =\theta^n
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  5. #5
    Super Member malaygoel's Avatar
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    Quote Originally Posted by Stats View Post
    Probability of all trials being success is
    P(X=n) = \binom{n}{n} \theta^n (1- \theta)^{n-n} = 0
    P(X=n) = \binom{n}{n} \theta^n (1- \theta)^{n-n} =\theta^n
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