Thread: A hard probability problem

I'm stuck at this impossible problem... i'd aprecciate if you help me

A box contains n socks, such that:

- are white
- are red
- are blue

(So, )
Consider an experiment that consists of taking a sock from the box (without replacement), until you have a pair of the same colour. Find the following probabilities:


a) The pair (of the same colour) is obtained at the k-th extraction
b) The obtained pair is white
c) Knowing that the pair will be red: the pair is obtained at the third extraction.

This does sound tough, what have you done? Try to create a tree diagram with the first couple of extrations, this may help you find a pattern.

Hello,

In my opinion, it's possible to simplify a bit the problem. For the first question, the event will occur after (or at) the 2nd draw and before (or at) the 4th draw. So you just have to compute 3 probabilities, and they're not that complicated (thanks to the symmetry in the variables).
But I'm not sure if this is a conventional way, if this is how you'd be asked to solve the problem.

I've done many things but nothing interesting...
I've been trying to separate part (a) by colours, but i'm really stuck... don't know how to begin solving a problem like this...

About the simplification... hmmm
Do you mean i have to solve for k=0, k=1, k=2, k=3 etc.
And try to see a recurrence?

Help

No, I mean try to think about it. How many draws do you have to make in order to have a pair ? You can't have a pair at the first draw since you need 2 socks for it. So it's at least at the second draw. Then what is the biggest number of draws in order to get a pair ? Suppose you've got no luck and you keep picking socks of different colours : on the first 3 draws, you have a different colour. Then on the 4th draw, whatever colour you get, it'll form a pair with one of the previous ones !

I envision a tree:

Code:

                            +-----------------------+-----------------------+
                            |                       |                       |
k=1                       white                    red                     blue
                            |
         +------------------+------------------+
         |                  |                  |
k=2    white               red               blue
    (pair white)            |
                  +---------+-----------+
                  |         |           |
k=3             white      red        blue
            (pair white)(pair red)      |
                              +---------+-----------+
                              |         |           |
k=4                         white      red        blue
                        (pair white)(pair red)(pair blue)

For each node you can set up the conditional chance.
For instance:
P(white in 1st draw)=b/n
P(white in 2nd draw | white in 1st draw)=(b-1)/(n-1)
P(white in 3rd draw | white in 1st draw, red in 2nd draw)=(b-1)/(n-2)


Now for instance the chance for a pair of white socks in the second draw is [product rule]:
P(white in 2nd draw, white in 1st draw) = P(white in 2nd draw | white in 1st draw) P(white in 1st draw)=(b-1)/(n-1) * b/n

The chance to finish with a pair of white socks is [sum rule]:
P(pair of white socks) = sum P(each possible way to finish with 2 white socks)