Probability about functioning of two watches If factory A produces $1$ faulty watch in $100$ watches and factory B produces $1$ faulty watch in $200$. You are given two watches, you know that one is form factory A and one is form factory B. You don't know which is from which factory. Then
a) What is the probability that the second watch works?
b) If the first watch works, then what is the probability that the second one will be working?
My approach:
a) Say $A_1$ and $A_2$ be the events that the an watch is form factory A or factory B respectively. Let $W$ denotes the event that the second watch is working. Of course $ P(A_1)=P(A_2)=0.5$.
Then $P(W)= P(W/A_1)P(A_1)+P(W/A_2)P(A_2) = \frac{99}{100}(0.5)+\frac{199}{200}(0.5) =0.9925$ 
b) Say $W_1$ and $W_2$ be two events such that  first and second watches are working respectively. So $W_1 \cap W_2$ denotes both the watches are working. Then $P(W_1 \cap W_2$)= $(\frac{99}{100})(\frac{199}{200})$.
Hence $P({W_2}/{W_1})= \frac{P(W_1 \cap W_2)}{P(W_1)}$= $\frac{(\frac{99}{100})(\frac{199}{200})}{0.9925} \approx 0.992494$.
Am I correct?
 A: As there is some disagreement in the comments, I'll post some notes.
To make the problem more general, suppose that a watch from factory $A$ (resp $B$) is working with probability $P_A$ (resp. $p_B$).
At the start we have no idea which watch comes from which factory, so our prior must be to assign equal probability (i.e. $\frac 12$) probability to each.  
Thus the probability that the second one is working is $$\frac 12\times p_A+\frac 12 \times p_B=\frac {p_A+p_B}2$$
Note:  this is entirely consistent with the argument given by the OP.
Now assume that we have seen that $W_1$ is working.  That is (weak) evidence for the claim that $W_1$ comes from $B$ since $B$ is more likely to make working watches.  We need to use Bayes to re-estimate the probability that $W_1$ comes from $A$ or $B$.
As we saw in the first computation, the total probability that it is defective is $\frac 12\times (p_A+p_B)$.  Of that, the probability that it comes from $A$ explains $\frac 12\times p_A$.  Thus the re-estimated probability that it comes from $A$ is $$P_1(A)=\frac {p_A}{p_A+p_B}$$
Similarly $$P_1(B)=\frac {p_B}{p_A+p_B}$$
Note as a sanity check that if $p_A=p_B$ then we just get $\frac 12$ again as nothing has happened to break the symmetry.  
Of course $P_2(A)=P_1(B), P_2(B)=P_1(A)$.
Now we can use the revised probabilities to conclude that, given that $W_1$ is working, the probability that the second is working is $$\frac {p_B}{p_A+p_B}\times p_A+\frac {p_A}{p_A+p_B}\times p_B=\frac {2p_Ap_B}{p_A+p_B}$$
Using your numbers we get $\boxed {0.992493703}$ which is slightly less than the $.9925$ for part $A$, reflective of the fact that the evidence that $W_1$ came from $B$ was quite weak.
As a (crude) sanity check, note that taking $p_A=p_B=\psi$ makes the answer $\psi$, as it should.
