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Math Help - Stuck again on disributions

  1. #1
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    Stuck again on disributions

    I have another question. Again I'm in a college probability class studying multivarate distributions.

    Here goes:
    In a clinical study of a new drug formulated to reduce the effects of arthritis, researchers found that the proportion p of patients who respond favourably to the drug is a random variable that varies from batch to batch of the drug. Assume that p has a probability density function given by
    f(p)={12*(p^(2))*(1-p), whenever p is between 0 and 1 inclusive
    {0, whenever p is elsewhere.
    Suppose that n patients are injected with portions of the drug taken from the same batch. Let Y denote the number showing a favourable response.
    (A) Find the unconditional probability distribution of Y for general n.
    (B) Find E(Y) for n = 2.

    I'm confused because there's no Y1 and Y2. I would love any help.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by SwedishMan View Post
    I have another question. Again I'm in a college probability class studying multivarate distributions.

    Here goes:
    In a clinical study of a new drug formulated to reduce the effects of arthritis, researchers found that the proportion p of patients who respond favourably to the drug is a random variable that varies from batch to batch of the drug. Assume that p has a probability density function given by
    f(p)={12*(p^(2))*(1-p), whenever p is between 0 and 1 inclusive
    {0, whenever p is elsewhere.
    Suppose that n patients are injected with portions of the drug taken from the same batch. Let Y denote the number showing a favourable response.
    (A) Find the unconditional probability distribution of Y for general n.
    The joint distribution is:

    P(y,n,p)=p(y,n|p)f(p)

    the probability that p takes value p and the number of favourable responses
    is y. Now Y is a binomially distributed RV with probability of a fav. response for
    an individual patient p, and number of patients n, so:

    p(y,n|p)=n!/[(n-y)!y!)] p^y (1-p)^(n-y)

    The unconditional (marginal?) distribution of Y is:

    g(y,n)=int(0,1) P(y,n,p) dp

    (B) Find E(Y) for n = 2.
    E(Y|n=2)= sum(y=0,infty) g(y,2)

    RonL
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  3. #3
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    Thank you so much.
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