Question 1067720
A bag contains two red beads and two green beads. You reach into 
the bag and pull out a bead, replacing it with a red bead regardless 
of the color you pulled out. What is the probability that all beads
in the bag are red after three such replacements?
<pre>
The only way to succeed is to replace both green beads.
There are three ways to do that:

Case 1.  Pull out R,G,G in that order

You begin with the bag containing RRGG.
You draw a red one with probability 2/4 or 1/2.
You replace it with another red one, then the bag still contains RRGG.
You then draw a green one with probability 2/4 or 1/2.
You replace it with a red one, then the bag contains RRRG.
You draw the green one with probability 1/4.
You then replace it with a red one, then the bag contains RRRR.
So Case 1 happens with probability (1/2)(1/2)(1/4) = 1/16.

Case 2.  Pull out G,R,G in that order

You begin with the bag containing RRGG.
You draw a green one with probability 2/4 or 1/2.
You replace it with a red one, then the bag contains RRRG.
You then draw a red one with probability 3/4.
You replace it with another red one, then the bag still contains RRRG.
You draw the green one with probability 1/4.
You then replace it with a red one, then the bag contains RRRR.
So Case 2 happens with probability (1/2)(3/4)(1/4) = 3/32.

Case 3.  Pull out G,G,R in that order

You begin with the bag containing RRGG.
You draw a green one with probability 2/4 or 1/2.
You replace it with a red one, then the bag contains RRRG.
You then draw the green with probability 1/4.
You replace it with a red one, then the bag contains RRRR.
You draw a red one with probability 1 (certainty).
You then replace it with a red one, then the bag still contains RRRR.
So Case 2 happens with probability (1/2)(1/4)(1) = 1/8.

So the probability is 1/16 + 3/32 + 1/8 = 2/32 + 3/32 + 4/32 = 9/32

Edwin</pre>