conditional probability problem with two random events

Question

This the problem:
In police station 1 there are 3 cars of type A and 8 of type B. In police station 2 there are 5 of type A and 2 of type B. In each station one of the cars is randomly chosen and damaged by an outsider!( a damaged car can not move). Some event happens and both stations cars go out to get the criminal. Randomly one of the cars caches the criminal. What is the probability that this car is of type A?

My fist approach to this problem was to use the Law of total Probability and finding the chance of the car to be of type A. How ever two cars of the total 18 cars are damaged in two separate random events and I couldn't add that to my calculations. I also tried some other methods but I just got more confused with the conditional logic of the problem.
Thanks for your attention

Answer 1

After the damages, station 1 has 10 cars and station 2 has 6 cars. So the probability that a car from station 1 will catch the criminal is $\tfrac{10}{10+6}$. Conditional on that event, the probability that the car that did it was an A typed car is $\tfrac{3}{11}$. The same reason if it was a car from station 2:

$$\Pr(A\,car\,catches)=\tfrac{10}{16}\tfrac{3}{11}+\tfrac{6}{16}\tfrac{5}{7}=\tfrac{135}{308}$$

Wait, what about the damaged cars?

Good question, they don't matter. You can make a complete calculation based on the law of total probability or use combinatorics. Line up the 11 cars of station 1 in a row. The first one is damaged and the second one is the one catches the criminal (reminder: all conditioned on the even that station 1 catches him). Essentially, the question is - given 11 cars, what is the probability that A typed car will be on the second place in the row, and this probability is the same as for any place in the row (just like when drawing a card from the deck, it will be an ace with the same probability, whether you take the upper one, second one, bottom one or whatever).

Answer 2

If it is confusing, here is another way to look at it - how many type A cars are expected after damage and divide it by 16. It really means the same as in the previous solution from YJT.

Say, probability of car type A being damaged in police station 1 is represented as $P(D,A,1)$. Then,

$P(D,1,A) = \frac{3}{11}$
$P(D,2,A) = \frac{5}{7}$

After the cars are damaged, say, expected number of type A cars left in police station 1 is represented as $E(1,A).$

Then,

$E(1,A) = 2 + \frac {8}{11} = \frac {30}{11}$

$E(2,A) = 4 + \frac {2}{7} = \frac {30}{7}$

$P(A) = \frac {\frac {30}{11} + \frac {30}{7}}{16} = \frac {135}{308}$