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Math Help - [SOLVED] jointly distribution

  1. #1
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    [SOLVED] jointly distribution

    A man and a woman decide to meet at a certain location. If each of them independently arrives at a time uniformly distributed between 12 noon and 1 P.M., find the probability that the first to arrive has to wait longer than 10 minutes.

    Let X and Y denote, respectively, the time past 12 that the man and the woman arrive, then X and Y are independently random variables, each of which is uniformly distributed over (0, 60). The desired probability,
    P[X + 10 < Y] + P[Y + 10 < X]= 2P[X + 10 < Y]
    =2 \int_{x+10<y} \int f(x, y) dx dy
    =2 \int_{x+10<y} \int f_X(x) f_Y(y) dx dy
    =2 \int_{10}^{60} \int_{0}^{y-10} (\frac{1}{60})^{2} dx dy
    But why \int_{10}^{60} and not \int_{0}^{60} in the first part of the integral??
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    Hello,
    Quote Originally Posted by noob mathematician View Post
    A man and a woman decide to meet at a certain location. If each of them independently arrives at a time uniformly distributed between 12 noon and 1 P.M., find the probability that the first to arrive has to wait longer than 10 minutes.

    Let X and Y denote, respectively, the time past 12 that the man and the woman arrive, then X and Y are independently random variables, each of which is uniformly distributed over (0, 60). The desired probability,
    P[X + 10 < Y] + P[Y + 10 < X]= 2P[X + 10 < Y]
    =2 \int_{x+10<y} \int f(x, y) dx dy
    =2 \int_{x+10<y} \int f_X(x) f_Y(y) dx dy
    =2 \int_{10}^{60} \int_{0}^{y-10} (\frac{1}{60})^{2} dx dy
    But why \int_{10}^{60} and not \int_{0}^{60} in the first part of the integral??
    Drawing a region would help you.

    more formally :

    We know that X+10<Y
    X can have values over 0 to 60
    So we have 0<X<60
    Then 10<X+10<70

    Since Y>X+10>10, we have Y>10. This explains the lower bound.
    The upper bound is 60 because Y has to be <60.


    Looks clear ?
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