# probability problem

This is in response to the following request:

A parent particle divides into 0,1,or 2 particles with probabilities 1/4,1/2,1/4.it disappears after splitting.let Xn denotes the number of particles in n-th generations with X0=1.find P(X2¿0) and the probabilities that X1=2 given that X2=1.

http://planetmath.org/?op=getobj;from=requests;id=927

For my first entry I will try to answer the question.

Let $p_{0},p_{1}$ and $p_{2}$ be the nonzero probabilities of dividing into 0, 1, or 2 particles, and let $X_{n}$ denotes the number of particles at the $n^{th}$ generation.

With $X_{0}=1$, find 1) $P(X_{2}>2)$ and 2) $P(X_{1}=2|X_{2}=1)$

1) After two generations there can be at most $2^{2}$ particles so $P(X_{2}>2)=P(X_{2}=3)+P(X_{2}=4)$

 $P(X_{2}=4)=p_{2}^{2}$
 $P(X_{2}=3)=2p_{1}p_{2}^{2}$

Note that if $X_{2}=3$, then $X_{2}=2$.

 $P(X_{2}>2)=p_{2}^{2}(1+2p_{1})$

Using your values I get 3/32.

2) From the definition of conditional probability

 $P(X_{1}|X_{2})=\frac{P(X_{1}\cap X_{2})}{P(X_{2})}$

First

 $P(X_{2}=1)=p_{1}^{2}+2p_{0}p_{1}p_{2}$

Why? To get to $X_{2}=1$, at $n=1$ there are either one or two particles, if there is one particle it remains one at $n=2$, and if there were two particles at $n=1$, then one has to go to zero and the other one—this can happen two ways.

Finally $P(X_{1}=1\cap X_{2})=p_{1}p_{2}$.

 $P(X_{1}=2|X_{2}=1)=\frac{p_{2}}{p_{1}+2p_{0}p_{2}}$

Using your values I get 2/3.

Now I have a question for you to think about. What happens in the long run, as $n\rightarrow\infty$?

Title probability problem ProbabilityProblem 2013-03-22 19:11:21 2013-03-22 19:11:21 statsCab (25915) statsCab (25915) 4 statsCab (25915) Definition msc 62-01