True Probability

[eMac]

Question: "With reference to exercise 1, suppose that years later we hear that ray graduated form the give university. what is the probability that he did get the scholarship."

Exercise 1: "Ray has enrolled as a freshman at an Eastern university and the probability that he will get a scholarship is 0.35. if he gets a scholarship the probability that he will graduate is 0.82, and if he does not get a scholarship the probability that he will graduate is only 0.44. what is the probability that he will graduate"

So what I have worked out so far is:

He does get scholarship is 0.35, with scholarship he graduates is 0.82, with scholarship and doesnt graduate is 0.18.

He doesnt get scholarship is 0.65, without scholarship and graduates is 0.44, without scholarship and doesnt graduate is 0.56.

Im not sure as to what I should do from this point to find out the probability that he did get the scholarship.

[Bacle]

This seems more like Bayesian probability than "standard" probability. Are you considering this approach?

[eMac]

How would I use Bayesian probability?

[Bacle]

My apologies, eMac, I misread the question; this is not a Bayesian prob. question. I gotta go now, but I'll come back tomorrow (more precisely, later today.)

[Stephen Tashi]

Obviously this is a job for big time night owls.

S = Ray gets scholarship
G = Ray graduates

Given:
P(S) = 0.35
P(G | S) = .82
P(G | not S) = .44

Find P(G)
As Bacle said, that can be done without a Bayesian analysis

Your first question:
Find P(S | G)

This can be done by a Bayesian analysis
P(S | G) = P( S and G)/ P(G) = P(G and S)/ P(G) = ( P(G | S) P(S) )/ P(G)

[tarhana]

P(S)= 0.35
P(notS)=0.65

P(S and G)= 0.82
P(S and notG)=0.18

P(notS and G)= 0.44
P(notS and notG)=0.56

P(He will Graduate) = (0.35 * 0.82) + (0.65 * 0.44) = 0.41 OR 41%

Hope it helps!