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 November 19th, 2013, 05:41 PM #1 Newbie   Joined: Nov 2013 Posts: 1 Thanks: 0 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?
 November 21st, 2013, 06:00 PM #2 Member   Joined: May 2013 Posts: 34 Thanks: 1 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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