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 $13/300$ ?

• So, you pick a random company with even probability, buy two devices, and the first you test is broken? Apr 15, 2015 at 19:13
• yes @ThomasAndrews okey Apr 15, 2015 at 19:19
• @ThomasAndrews are you agree with this answer? I think something is wrong ?! Apr 15, 2015 at 19:40

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, $\frac{1}{600}+\frac{1}{24}$=$\boxed{\frac{13}{300}}$
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 $$P(K')=P(A|K)P(K|A)+P(B|K)P(K|B)={5\over6}\cdot 0.05+{1\over6}\cdot0.01={13\over300}\ .$$