All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Probability
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)

Comments: ( Back to comment list | You must be logged in to post comments.)
re:Correcting My previous analytical solution | Comment 6 of 11 |
 As Steve correctly commented, my previous solution is mistaken, so here is a correct one :
 
 We denote the situation of having completed "d" draws (including returns of blackened beans) ending up  with "w" white beans and "b" black beans in the bag, as : (d,w,b), and the probability of such occurrence, as:
   P(d,w,b).
 Dealing with the case of 5 beans, a single white bean in the bag can only result at the end of the 8th drawing.We are therefore looking for P(8,1,0).
We shall find it by writing the relevant equation backwards till arriving at the known  P(1,4,1)=1.

 (8,1,0) can be produced only from a previous situation of (7,1,1), by drawing a black bean, the probability of which is 0.5 . Therefore we get :  P(8,1,0)=P(7,1,1)*0.5
 Going one step back-  P(7,1,1) can be produced only by drawing a black bean from a previous situation (6,1,2), or from drawing a white bean from (6,2,0), therefore :-
    P(7,1,1) = P(6,1,2)*2/3 + P(6,2,0)*1
where 2/3 and 1 are the probabilities of drawing the respective beans from the bag.
Going further back by the same method we get the following equations :-
 P(6,1,2)=P(5,2,1)*2/3 + P(5,1,3)*3/4
 P(6,2,0)=P(5,2,1)*1/3
further backward steps yield :-
 P(5,2,1)=P(4,3,0)*1 + P(4,2,2)*0.5
 P(5,1,3)=P(4,2,2)*0.5 + P(4,1,4)*4/5

 P(4,3,0)=P(3,3,1)*1
 P(4,2,2)=P(3,3,1)*3/4 + P(3,2,3)*3/5
 P(4,1,4)=P(3,2,3)*2/5 + P(3,1,5)*5/6

 P(3,1,5)=P(2,2,4)*1/3 + P(2,1,6)*6/7
 P(3,2,3)=P(2,3,2)*3/5 + P(2,2,4)82/3
 P(3,3,1)=P(2,4,0)*1

 P(2,1,6)=0
 P(2,2,4)=0
 P(2,3,2)=P(1,4,1)*1/5
 P(2,4,0)=P(1,4,1)*1/5

  But P(1,4,1) is known to be :     P(1,4,1) =1

   So now we substitute back through the above equations till we get the asked for probability :

                                      P(8,1,0) = 0.256077776

         

  Posted by Dan Rosen on 2015-07-15 09:03:13
Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (2)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2017 by Animus Pactum Consulting. All rights reserved. Privacy Information