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November 19th, 2013, 05:41 PM   #1
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random variable binary tree problem

I have this problem on random variables and a binary tree. The problem goes like this: The experiment is creating a binary tree with 4 nodes. The random variable X is the number of leaves.

My first question is what can my possible values be for the random variable X? Would it be 1 and 2 since we can't have rooted binary tree that has 4 nodes with 0 leaves? My second question is how would i find the probability of X=1 and X=2? I started with X=1 and drew each possible trees, this seemed like a similar problem where i tossed a coin 4 times and determined the # of heads. But then i realized that this would not work because as X increases we would get a node with 2 leaves. So its not just a matter of left child is 1/2 right child is 1/2 probability. Would we have a 1/3 chance of a getting a left node, 1/3 chance of getting a right node, 1/3 chance of getting both a left and right node since we have 3 possibilities? Since the root is always going to be the same we would have something like 1*1/3*1/3*1/3 which is 1/27. Since there are 8 possible binary trees with 4 nodes and 1 leaf, would P(X=1) = (1/27)^8?
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November 21st, 2013, 06:00 PM   #2
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Re: random variable binary tree problem

interesting.
so here are all the binary trees with 4 nodes and 2 leaves.
[spoiler]
Code:
    o
    /\   
  o  o
 /
o
      o 
      /\
    o  o
         \
          o

    o
    /\  
  o  o
  \
   o

    o
    /\  
  o  o
    /
  o

    o
     \   
      o
    /  \
  o    o

     o
    /  
  o
 / \
o  o
[/spoiler]
and here are all the binary trees with 4 nodes and 1 leaf.
[spoiler]
Code:
        o
       /
     o
    /
  o
 /
o

        o
       /
     o
    /
  o
   \
    o

        o
       /
     o
      \
       o
         \
          o

        o
       /
     o
     \
      o
    /
  o

o
  \
     o
    /
  o
 /
o

  o 
    \   
     o
    /
  o
 /
o

  o
    \
      o
        \
         o
      /
   o

o
 \
 o
  \
   o
    \
     o
[/spoiler]

so to answer your first question, yes. x = 1 or x = 2.
you can't have x = 3 since you would need at least 5 nodes for that.
now as to the probability it all depends on how the randomness is selected.
if the goal is to generate either a right node or a left node 1 at a time and then figure out the probability of having 2 leaves, it could be quite complex to figure out the probabilty. if however you're simply selecting a random tree with 4 nodes and want to know wether it has 2 leaves of 1 leaf. then the probabilty is 6/8 or 3/4
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