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?
Answer by Edwin McCravy(20056)   (Show Source):
You can put this solution on YOUR website! 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?
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
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