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Lost at MIT probability

  1. Oct 6, 2008 #1
    1. The problem statement, all variables and given/known data

    You are lost in the campus of MIT, where the population is entirely composed of brilliant students and absent-minded professors. The students comprise two-thirds of the population,
    and any one student gives a correct answer to a request for directions with probability [tex]\frac{3}{4}[/tex] (Assume answers to repeated questions are independent, even if the question and the person asked are the same.) If you ask a professor for directions, the answer is always false.

    You ask a passer-by whether the exit from campus is East or West. The answer is East. What is the probability this is correct?

    2. Relevant equations

    [tex]P(A|B)=\frac{P(B|A)P(A)}{P(B)}[/tex] {Baye's Theorem}

    [tex]P(A\cap B) = P(A)P(B)[/tex] {For independent events A and B}

    [tex]P(A\cup B) = P(A)+P(B)-P(A\cap B)[/tex]

    3. The attempt at a solution
    My approach was to use Baye's Theorem. The problem is that I dont have any prior probabilities.

    Let P(P) be the probability that the person is a prof.
    P(S) '' '' is a student.
    P(E) '' '' answer is East.
    P(T) " " correct answer is given.


    [tex]P(T|E) = \frac{P(E|T)P(T)}{P(E)}[/tex]


    [tex]P(T)=P(S \cap T \cup P \cap T)[/tex]

    and [tex]P(E) = 0.5[/tex].

    Is this the right way to go about it?

    As a heuristic, a later question says that if you ask the same person again, and they answer East, you need to show that the probability that East is True is 1/2.

    Any points in the right direction would be most welcome :smile:
  2. jcsd
  3. Oct 6, 2008 #2
    Using a tree might make this easy to solve.

    P(student) * P(correct | student) = 2/3*3/4 = 6/12

    P(student) * P(~correct | student) = 2/3*1/4 = 2/12

    P(prof) * P(correct | prof) = 1/3*0 = 0

    P(prof) * P(~correct | prof) = 1/3*1 = 4/12

    Only one branch offers a correct answer since professors always answer incorrectly.
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