# Given that Class A contains 5G/10B, Class B contains 13B/12G and some rearranging is done, what's the probability a student will be a boy?

Class A contains 5 girls + 10 boys= 15 students. Class B contains 12 girls + 13 boys= 25 students. Teachers randomly pick a student from Class B and send them over to Class A; they then randomly pick a student from Class A. We're trying to find the probability that this student is a boy.

To answer this question, I calculated several conditional probabilities: $P(M|A)=10/15$
$P(F|A)=5/15$ $P(M|B)=13/25$ $P(F|B)=12/25$

What I tried next is the following, but I'm not sure this is the right way to go: P(M|A new)=(10+(13/25))/16=0.6575. Can you add the probability almost as a new student? If not, is there a simpler/more correct way to answer this question?

The usual way to solve it is $\left(\frac{13}{25}\frac{11}{16} + \frac{12}{25}\frac{10}{16}\right)$
In the way you have solved, $(10+ \frac{13}{25})$ is the expected number of boys in class A after the switch.
$\frac{13}{25}$ you have added is not a probability, view it as $\frac{13}{25}\times 1 + \frac{12}{25}\times 0$, the expected number of boys added to class A.