A hard probability problem
I'm stuck at this impossible problem... i'd aprecciate if you help me (Headbang)
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 kth extraction
b) The obtained pair is white
c) Knowing that the pair will be red: the pair is obtained at the third extraction.
Re: A hard probability problem
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.
Re: A hard probability problem
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.
Re: A hard probability 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 (Worried)
Re: A hard probability problem
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 !
Re: A hard probability problem
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)=(b1)/(n1)
P(white in 3rd draw  white in 1st draw, red in 2nd draw)=(b1)/(n2)
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)=(b1)/(n1) * 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)