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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

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  • $\begingroup$ (1) What is the probability of acceptance of BAD students? (2) Do the two schools independently decide if a student is bad or good, or is this a single factor that the schools agree upon? $\endgroup$
    – vadim123
    Dec 12 '13 at 18:15
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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.

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  • $\begingroup$ [A]nd has a chance to get seriously rich. :-) $\endgroup$ Dec 13 '13 at 1:05

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