Question

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. 1/26
B. 1/18
C. 9/26
D. 9/13

Answer 1

Answer:

D. 9/13

I got it right when I submitted the test

Hope it helps :)

Answer 2

We have events:

$D$ - a student stays in the school's dormitory
$D'$ - a student stays with family
$A$ - a student receives As

and  probabilities:

$P(D)=0.6\\\\P(D')=0.4\\\\P(A|D)=0.3\\\\P(A|D')=0.2$

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

$P(D|A)=\dfrac{P(A|D)P(D)}{P(A)}$

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

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

So our probability:

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

Answer D.