I think this is supposed to be quite a simple question however it's been a long long time since I touched probability or maths in general so I could be completely wide of the mark here.
Here's my thinking...
Total number of ways 12 keys could be distributed into two groups of 5 and 7 is: $$\dfrac{12!}{5!7!}$$
We are interested in the event that the key lies in bag A. Supposing it is in bag A, the number of choices for the remaining spots is: $$\dfrac{11!}{4!7!}$$
So the answer is surely the bottom fraction divided by the above fraction which is about $0.42$ - is my reasoning correct? I sense that I have made an error.
An octave on a piano has seven white keys, and five black keys.
If I'm thinking of one of the keys, what is the probability that it is black?