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Math Help - [SOLVED] Uniform Distribution over two days

  1. #1
    corned_beef
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    [SOLVED] Uniform Distribution over two days

    If T is the time in minutes that bob waits at a bus stop
    T~U(0,5)

    Excuse the poor paint diagram :P



    He gets the bus once on monday and once on tuesday.
    What is the probability he will wait for less than 4 minutes in total for both days? (both events are independent).

    Ive been trying to do it for like 3 hours now, but im none the wiser, my lecteur notes dont have any examples on it.
    Last edited by corned_beef; December 4th 2007 at 06:18 PM.
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  2. #2
    Senior Member
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    Independent events

    Remember that for outcomes A and B of two independent events, p(A \cap B) = p(A) \cdot P(B). So, what is the probability p = p(t<4) that on a singe particular day, Bob waits less than 4 minutes? You then can use these to find the probability of it being less than 4 minutes on both days.


    --Kevin C.
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by corned_beef View Post
    If T is the time in minutes that bob waits at a bus stop
    T~U(0,5)

    Excuse the poor paint diagram :P



    He gets the bus once on monday and once on tuesday.
    What is the probability he will wait for less than 4 minutes in total for both days? (both events are independent).

    Ive been trying to do it for like 3 hours now, but im none the wiser, my lecteur notes dont have any examples on it.
    Waiting time is t_t=t_1+t_2 where the t 's are the waiting times on the two days.

    Now we have two possible approaches, we can use the fact that we know
    the distribution of the sum (its a triangular distribution extending from 0
    to 10 minutes with peak at 5 minutes).

    or we can compute it.

    If we wait t_1 min on the first day then the probability we require is:

    p(t_t<4|t_1) = \int_0^{4-t_1} p(t_2) dp_2=(4-t_1)

    if t_1<1 and 0 otherwise.

    Then the required prob is:

     <br />
p(t_t<4)=\int_0^4 p(t_t<4|t_1) p(t_1) dp_1<br />

    RonL
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