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Continous probability between two bus stops

  1. Nov 16, 2015 #1
    1. The problem statement, all variables and given/known data
    You arrive at a road, and all you know is that you are somewhere between two bus stops which are one mile apart. Since you don’t know where you are, you assume that your position X is a uniform [0,1] random variable.

    a) How far do you have to walk on average to the bus stop at the origin? And to the other bus stop?
    b) Let the closest bus stop be distance Y away. What is the density function of Y ?
    c) What are EY and VarY ?
    d) If you also know that there is a coffee shop somewhere between the two bus stops, and you head to a random bus stop, what is the chance that you can fetch a coffee on your way?
    e) What is the probability that you are closer to the coffee shop than to a bus stop?
    f) Calculate P(you are within distance 1/4 of a bus stop).
    g) Calculate P (you are closer than 1/10 to a bus stop | you are closer than 1/5 to a bus stop ).

    2. Relevant equations
    E[Y]= ∫y*f(y)dy

    3. The attempt at a solution
    a) 1/2 for both by intuition. Is it mathematically correct to consider this as ∫xdx from 0 to 1?
    b) Our function f(y) is a piece-wise function where f(y)=y if 0≤y≤1/2 and f(y)=1-y if 1/2≤y≤1.
    c) EY is gotten by summing ∫y*ydy from 0 to 1/2 and ∫(1-y)*ydy from 1/2 to 1. This results in 1/8.
    VarY=E[Y^2]+{EY}^2
    E[Y^2} is gotten by doing the same as E[Y} but by multiplying by y^2 instead of y in the integral. Sum ∫y*y^2dy
    from 0 to 1/2 and ∫(1-y)*y^2dy from 1/2 to 1 which leads to 7/96. Hence, VarY=7/96+(1/8)^2=17/192.
    d) Unsure. Any help in setting up the equation will be appreciated.
    e) 1/2?
    f) The probability is found by integrating the density function we got in part b. P(x≤1/4)= ∫xdx from 0 to 1/4 = (1/2)(1/16). Since there are two bus stations, we multiply by two which leads to P(x≤1/4)=1/16.
    g) Using the definition of Bayes formula, P(E|F)=P(E)*P(F|E)/P(F) but P(F|E)=1 in our case so P(E|F) is just P(E)/P(F).
    P(E)=P(x≤1/10)=2* ∫xdx from 0 to 1/10 (same reasoning as part f)=1/100.
    P(F)=P(x≤1/5)=2*∫xdx from 0 to 1/5=1/25.
    P(E)/P(F)=1/4 so the answer is 1/4.

    Hi all.

    I am having a difficult time solving these worksheet problems. I am quite adept at computing the probability, expected value and variance but when it comes to understanding the problems themselves, my mind just melts down.
    And comments on my (lack of) work will be greatly appreciated.
     
  2. jcsd
  3. Nov 16, 2015 #2

    LCKurtz

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    Let's just talk about part b for now. You say "our function f(y) is..." above. Are you claiming that is the density function for Y? Hint: It isn't. Do you see that Y is the min of X and 1-X? What are the values that Y might take? That should help you figure out the values of y for which the density f(y) is nonzero. You will probably want to start by figuring out the cdf ##F(y) = P(Y\le y)## first. No point in even thinking about the rest of the problem until you get f(y) figured out.
     
  4. Nov 16, 2015 #3

    Ray Vickson

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    (1) How did you get your result in (b)? [I get something very different!]
    (2) A different answer in (b) leads to a different answer in (c).
    (3) Surely (d) must be almost "intuitive". HInt: read the question carefully.
    (4) For (e): don't guess; work it out somehow.
    (5) I get a different answer for (f) and (g), and using a different logic from yours.
     
  5. Nov 17, 2015 #4
    I just realised that the random variable is uniform so I can just use the properties of uniform random variable, which makes it a lot easier to solve. I have gotten the answers now. Thank you LCKurtz and Ray!
     
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