Conditional probability (not sure) ...college admissions Jim and Sam both applied to 2 different schools, A Tech and B University. Both schools base their acceptance on whether the student is GOOD or BAD. Jim and Sam determined that the probability of them being good is .6. A Tech has a probability of acceptance of .7 if the student is GOOD. B University has a probability of acceptance .9 GOOD. 
1) If Sam gets into B Uni, what's the probability she'll get into A Tech.
2) If Jim gets rejected from B Uni, what's the probability he'll get into A Tech.
Thanks! I did a tree diagram and  I tried doing Baye's Theorem...but failed. Idk how the probabilities of 2 schools relate
 A: We do (2), and leave the somewhat easier (1) to you. Let us assume that Good/Bad is say determined by the score on a joint admissions test, and that all "Bad" are rejected by both institutions. We interpret, for example, the figure of $0.9$ for University B to mean that given someone is "Good," the probability she/he will be accepted is $0.9$.
Let $R$ be the event Jim gets rejected by University B, and let $A$ be the event he gets accepted by Tech A. We want $\Pr(A|R)$. Depending on what you mean by Bayes' Rule, we may not quite use it, using instead the definition of conditional probability 
$$\Pr(A|R)=\frac{\Pr(A\cap R)}{\Pr(R)}.\tag{1}$$ 
We calculate the probabilities on the right of (1).
The event $R$ can happen in two ways: (i) Jim gets classified as Bad, and therefore is automatically rejected, and has a chance to get seriously rich or (ii) Jim gets classified as Good, but gets rejected by University B.  
So $\Pr(R)=(0.4)(1)+(0.6)(0.1)$.
We next compute $\Pr(A\cap R)$. The event $A\cap R$ occurs if Jim is "Good," is rejected by B, and is accepted by A. 
Thus $\Pr(A\cap R)=(0.6)(0.1)(0.7)$. 
Finally, divide.
