In a school in Florida, 60% of the students stay in the school's dormitory and 40% stay with their families. The school records show that 30% of the students living in the dormitory and 20% of the students living with their families obtain As on exams. If a student chosen at random from the school receives As, the probability that the student lives in the school dormitory is .....A. 1/26B. 1/18C. 9/26D. 9/13

We have events:

[tex]D[/tex] - a student stays in the school's dormitory
[tex]D'[/tex] - a student stays with family
[tex]A[/tex] - a student receives As

and  probabilities:

[tex]P(D)=0.6\\\\P(D')=0.4\\\\P(A|D)=0.3\\\\P(A|D')=0.2 [/tex]

We want to calculate the probability that the student lives in the school dormitory given he receives As so it will be [tex]P(D|A)[/tex]. From the Bayes' theorem we know that:

[tex]P(D|A)=\dfrac{P(A|D)P(D)}{P(A)}[/tex]

The only thing we don't know is [tex]P(A)[/tex], but we can calculate it using the law of total probability. There will be:

[tex]P(A)=P(A|D)P(D)+P(A|D')P(D')=0.3\cdot0.6+0.2\cdot0.4=\\\\=0.18+0.08=\boxed{0.26}[/tex]

So our probability:

[tex]P(D|A)=\dfrac{P(A|D)P(D)}{P(A)}=\dfrac{0.3\cdot0.6}{0.26}=\dfrac{0.18}{0.26}=\boxed{\frac{9}{13}}[/tex]

Answer D.