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Bean coloring (Posted on 2015-07-13) Difficulty: 4 of 5
Start with a bag containing 5 white beans. Randomly draw one at a time employing the following rule:

If the bean is white, color it black and put it back in the bag;
If the bean is black, keep it out.

What is the probability that at some point there will be a single white bean in the bag?

Generalize to start with N beans.
Does the probability converge, and if so, to what value?

No Solution Yet Submitted by Jer    
Rating: 4.0000 (1 votes)

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Solution Analytical solution | Comment 3 of 11 |
 for n=5 :
We mark each specific bean with a number. After 10 draws the bag is empty as each bean has been drawn twice. If we list the possible sequences of draws, we'll get a list of 10 digits comprised of digits 1-5 , each appearing twice. All possible permutations of this list are 10!, but here the first and the second appearance of each digit is counted as a seperate case, so we have to divide the total number of possible sequences by 2^5, and we get :
                                                                  10!/2^5

The asked for incidents are all those ending by :11, 22, 33, 44, 55
The number of permutations answering to this, will be the number of permutations of a sequence of the first 8 draws containing only the other 4 digits, and that should be multiplied by 5 for each of the 5 endings, so we get:
                                       (8!/2^4)*5

  and the required probability :

                          P =  [ (8!/2^4)*5 ] / (10!/2^5) = 1/9


                           




  Posted by Dan Rosen on 2015-07-14 04:18:16
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