I was quite sure that I have understood the pigeonhole principle, but this question does not make sense. Or does it?
The question:
A drawer contains $5$ blue socks, $7$ red socks and $6$ black socks. Socks are
randomly removed one by one and placed on a table.
(a) What is the least number of socks that need to be removed to ensure that there are two socks of the same colour on the table?
(b) What is the least number of socks that need to be removed to ensure that there are three socks of the same colour on the table?
The solution:
This is the Pigeonhole Principle with the socks as pigeons and the colours as pigeonholes. There are $3$ colours. (a) $4$. (b) $7$.
Since the socks are chosen randomly and there are plenty of socks so that you have a chance of choosing $2$ socks of the same colour so shouldn't it be $2$ for (a) and $3$ for (b)?
Please tell me where I have gone wrong.
For point (a):If you are lucky you can get 2 socks of the same color by picking just 2 socks.But thats not the question. The question asks the minimum number of tries when you are not lucky in order to ensure that you will get 2 socks of the same color .In that case you will need to pick 4 socks in case you are not lucky and pick 3 different colors in the first 3 tries.Then you are sure that by picking the fourth sock it will be of one of the 3 different colors you have picked before so the answer is 4 For point (b) following the same reasoning if we are unlucky to pick 3 pairs of socks each pair having a different color then the seventh pick will belong to one of the 3 colors you picked before so it is sure that you will get 3 socks of the same color therefore the answer is 7