Two company and probability example? I ran into a problem that seems strange to me.

Two companies A,B produce a device that with probability $0.05$ and
  $0.01$ are broken. if  we buy two devices produced by one company
  with equal probability and the first device be broken, what is the
  probability of the second device be broken?

Who can show me how my TA reached to  $13/300$ ? 
 A: A breaks 5 times as often as B. So for every 6 broken devices, 5 times it will be A, and 1 time it will be B.
The broken device was created by A with a $\frac{5}{6}$ probability. Multiplying $\frac{5}{6}$ by the chance A will break the next device, $\frac{1}{20}$, gives $\frac{1}{24}$.
Now the broken device has a 1/6 probability of being created by B. So multiplying by the 1/100 probability B will break the next device gives 1/600. Adding, $\frac{1}{600}+\frac{1}{24}$=$\boxed{\frac{13}{300}}$
A: Denote by $K$ the event that the first checked device is broken. Then
$$P(K)=P(A) P(K|A)+P(B)P(K|B)={1\over2}\cdot0.05+{1\over2}\cdot0.01=0.03\ .$$
What we need to know are the conditional probabilities $P(A|K)$, $P(B|K)$. Now
$$P(K)P(A|K)=P(A\cap K)=P(A)(P(K|A)\ ,$$
which implies
$$P(A|K)={P(A)P(K|A)\over P(K)}={{1\over2}\cdot0.05\over0.03}={5\over6}\ ,$$
and similarly
$$P(B|K)={1\over6}\ .$$
Therefore the event $K'$ that the second checked device is broken has probability
$$P(K')=P(A|K)P(K|A)+P(B|K)P(K|B)={5\over6}\cdot 0.05+{1\over6}\cdot0.01={13\over300}\ .$$
