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Math Help - Probability and Expected Value problem

  1. #1
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    Probability and Expected Value problem

    Here's the question...

    The blood samples of k (k > 1) randomly chosen people is tested for
    the presence of virus V . Suppose that 5% of the population has this
    virus, and that presence of the virus among different people is independent.
    Instead of the traditional ”test each of the k blood samples
    separately” method, the following alternative method is proposed:
    Mix a little of each of the k blood samples and test for the presence of
    virus V in the mixture.
    • If virus V is not present - done! (since that means that none of
    the k people has the virus)
    • If virus V is present, go back and test each of the k blood samples
    separately. Let the random variable Y be the number of tests
    needed for the alternative methods.
    (a) Find PY (y) and E(Y ).
    (b) For what values of k is the alternative method more cost effective?
    (in the sense that less tests are required, on average)

    Thank you
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  2. #2
    Grand Panjandrum
    Joined
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    Quote Originally Posted by 892king View Post
    Here's the question...

    The blood samples of k (k > 1) randomly chosen people is tested for
    the presence of virus V . Suppose that 5% of the population has this
    virus, and that presence of the virus among different people is independent.
    Instead of the traditional ”test each of the k blood samples
    separately” method, the following alternative method is proposed:
    Mix a little of each of the k blood samples and test for the presence of
    virus V in the mixture.
    • If virus V is not present - done! (since that means that none of
    the k people has the virus)
    • If virus V is present, go back and test each of the k blood samples
    separately. Let the random variable Y be the number of tests
    needed for the alternative methods.
    (a) Find PY (y) and E(Y ).
    (b) For what values of k is the alternative method more cost effective?
    (in the sense that less tests are required, on average)

    Thank you
    The probabilities are:

    P(y=1)=0.95^k

    P(y=(k+1))=1-0.95^k

    So computing the expectation is now trivial.

    RonL
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