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Thread: Impossible Question

  1. #1
    megno
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    Impossible Question

    Okay so maybe it's not impossible, but neither I nor any of my other classmates were able to figure it out.

    if on any given day it snows, there is a .3 probability that it will snow the following day. if on any given day it doesn't snow, there is a .8 probability that it will not snow the following day. what is the probability that it will snow on any given day?

    It was a question on our final, and I simply guessed (it was multiple choice). Anyone have any ideas?
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by megno View Post
    Okay so maybe it's not impossible, but neither I nor any of my other classmates were able to figure it out.

    if on any given day it snows, there is a .3 probability that it will snow the following day. if on any given day it doesn't snow, there is a .8 probability that it will not snow the following day. what is the probability that it will snow on any given day?

    It was a question on our final, and I simply guessed (it was multiple choice). Anyone have any ideas?
    Suppose that the probability that it snows on any day is p.

    Now consider today. There is a probability p it snowed yesterday, and (1-p)
    that it did not. Therefore the probability that it snows today is:

    p*0.3+(1-p)*0.2=p

    Solve this for p to find p=2/9~=0.222.

    RonL
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  3. #3
    Junior Member F.A.P's Avatar
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    Quote Originally Posted by megno View Post
    Okay so maybe it's not impossible, but neither I nor any of my other classmates were able to figure it out.

    if on any given day it snows, there is a .3 probability that it will snow the following day. if on any given day it doesn't snow, there is a .8 probability that it will not snow the following day. what is the probability that it will snow on any given day?

    It was a question on our final, and I simply guessed (it was multiple choice). Anyone have any ideas?
    This is an example of a two states Markov chain. Let X(n) be the state of the process, i.e the weather, at day n. Then we have a state space S={0,1}

    0: Not snowing
    1: Snowing

    Transition probabilities between the states from day to day is

    $\displaystyle P(X(n+1)=0|X(n)=0)=p_{00}=0.8$
    $\displaystyle P(X(n+1)=1|X(n)=0)=p_{01}=0.2$
    $\displaystyle P(X(n+1)=1|X(n)=1)=p_{11}=0.3$
    $\displaystyle P(X(n+1)=0|X(n)=1)=p_{01}=0.7$

    That is we have a transition probability matrix P

    $\displaystyle P=\left( \begin{array}{cc}
    0.8 & 0.2 \\
    0.7 & 0.3 \end{array} \right) $

    What we're interested in is to find the long run behaviour of the process. (There are certain conditions to be satisfied in order for the process to have a long run behaviour, i.e to have a limiting stationary distribution. We don't need to worry too much about that, cause in this case they're all fullfilled.)
    We are looking for a limiting distribution <> stationary distribution $\displaystyle \pi=(\pi_0, \pi_1)$ satisfying

    $\displaystyle \pi=\pi P $

    That is, if we act upon the distribution according to the transition probabilities we will still have the same distribution.
    Now solving the above equation for $\displaystyle \pi $ with the condition that $\displaystyle \pi_0+\pi_1=1$ yields

    $\displaystyle \pi=\left(\frac{7}{9},\frac{2}{9}\right) $

    That is, the probability that it snows on a given day is $\displaystyle \pi_1=2/9 $

    For this example there is maybe an easier approach... but for the general case with more than 2 states this one is quite useful...
    Last edited by F.A.P; Dec 22nd 2006 at 05:51 AM.
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