# Lost at MIT probability

1. Oct 6, 2008

### Eidos

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 $$\frac{3}{4}$$ (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

$$P(A|B)=\frac{P(B|A)P(A)}{P(B)}$$ {Baye's Theorem}

$$P(A\cap B) = P(A)P(B)$$ {For independent events A and B}

$$P(A\cup B) = P(A)+P(B)-P(A\cap B)$$

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.

Now

$$P(T|E) = \frac{P(E|T)P(T)}{P(E)}$$

with

$$P(T)=P(S \cap T \cup P \cap T)$$

and $$P(E) = 0.5$$.

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

2. Oct 6, 2008

### montoyas7940

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.