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Math Help - Conditional Probability [please check my soln]

  1. #1
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    Conditional Probability [please check my soln]

    The number of flowers N appearing on an apple tree, is a random variable with distribution

    \mathbb{P}(N=n)=(1-p)p^n, (n=0,1,2,...) for some p \in (0,1). Each flower turns into a fruit with probability \alpha independently of other flowers on the tree. Given that there are r apples on the tree what is the probability that there were n flowers originally on the tree?

    Here's my solution (I am not sure, so please correct me if I'm wrong):

    Bayes' theorem says:

    \mathbb {P}(A|B) = {\mathbb{P}(A \cap B)}/{\mathbb{P}(B)}

    In this example A is the event: 'originally there were n flowers on the tree'. B is the event: 'there are r apples on the tree.

    So in the numerator of the Bayes' formula we get:
    (1-p)p^n \alpha ^r (1 - \alpha)^{n-r}
    and for the denominator we get:
    {\frac {n!}_{(n-r)! r!}} \alpha
    since there are r choose n ways in total and each has probability \alpha. Is this correct? Please reply.....
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  2. #2
    Super Member
    Joined
    Apr 2009
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    Doesn't look correct to me.

    let i = number of flowers
    r = number of fruits

    P(B) = summation from i = r to infinity {P(r|i).P(i)}
    P(A.B) = P(B|A).P(A)
    P(A) = Probabiltiy there were 'n' flowers - which is straight fwd
    P(B|A) = nCr \alpha ^r (1 - \alpha)^{n-r}
    Try to put these values and simplify
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