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A prerequisite for students to take a probability class is to pass calculus. A study of correlation of grades for students taking calculus and probability was conducted. The study shows that 25% of all calculus students get an A; and that students who had an A in calculus are 50% more likely to get an A in probability as those who had a lower grade in calculus. If a student who received an A in probability is chosen at random, what is the probability that he/she also received an A in calculus?

My Attempt:

I know $\Pr(A\mid B)$ with $A$ being event that the student gets an $A$ in calculus and $B$ being the event that the student gets an $A$ in probability is $\Pr(A \mid B)=\frac{Pr(A, B)}{\Pr(B)}$ but, I can't seem to put the givens into that form.

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You know that $$ P(A|B)=\frac{P(A\cap B)}{P(B)} $$ and $$ P(B|A)=\frac{P(A\cap B)}{P(A)} $$ thus $$ P(A|B)=\frac{P(A\cap B)}{P(B)}=\frac{P(A|B)P(B)}{P(A)}=\frac{P(A)}{P(B)}P(A| B) $$

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That is a formula for finding P(A|B), but not the only formula. You should consider using Bayes' theorem, which would allow you to turn probabilities like P(B|A) into your desired P(A|B).

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