# 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 $P({X}_{2}>2)=P({X}_{2}=3)+P({X}_{2}=4)$

$$P({X}_{2}=4)={p}_{2}^{2}$$ |

$$P({X}_{2}=3)=2{p}_{1}{p}_{2}^{2}$$ |

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

$$P({X}_{2}>2)={p}_{2}^{2}(1+2{p}_{1})$$ |

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

$$P({X}_{2}=1)={p}_{1}^{2}+2{p}_{0}{p}_{1}{p}_{2}$$ |

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 $P({X}_{1}=1\cap {X}_{2})={p}_{1}{p}_{2}$.

$$P({X}_{1}=2|{X}_{2}=1)=\frac{{p}_{2}}{{p}_{1}+2{p}_{0}{p}_{2}}$$ |

Using your values I get 2/3.

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

Title | probability problem |
---|---|

Canonical name | ProbabilityProblem |

Date of creation | 2013-03-22 19:11:21 |

Last modified on | 2013-03-22 19:11:21 |

Owner | statsCab (25915) |

Last modified by | statsCab (25915) |

Numerical id | 4 |

Author | statsCab (25915) |

Entry type | Definition |

Classification | msc 62-01 |