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Thread: Mixed continuous / discrete probability question

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    Mixed continuous / discrete probability question



    Dear forum,

    I am having a dreadful time making progress with the following problem, and would greatly appreciate any pointers you had.

    The problem gives a mortality risk as the function f(x) = 20(1-x)/9 for 0.1 < x < 1, and 20x for 0 < x < 0.1. We need to find both the conditional and the marginal probabilities of exactly 3 deaths among the next 4 patients, assuming the number of deaths follows a binomial distribution.

    With other problems, we are given f as a function of both x and y, and it is straightforward to find the conditional pdf as
    h(y|x) = f(x,y) / f1(x)
    and the marginal distribution as
    f1(x) = int from -inf to inf f(x,y) dy.

    I assume I need to do something along these lines, but getting to f(x,y) from my specific information is giving me a headache. I would be very grateful to anyone who could get me started with this, please. Thank you in advance.

    :-)

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    Re: Mixed continuous / discrete probability question


    Let X be the mortality risk and N be the number of deaths among the next 4 patients. I suppose the question mean

    N|X = x \sim \text{Binomial}(4, x)

    So the conditional probability is just to simple expression of a Binomial pmf, in terms of a variable x in between 0, 1.

    For the marginal probability, you will need to use the law of total probability

    \Pr\{N = 3\} = \int_0^1 \Pr\{N = 3|X = x\} f_X(x) dx

    to compute the answer.

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