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Math Help - Probability question

  1. #1
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    Probability question

    A process splits into two new processes with a probability of 0.75 and terminates with a probability of 0.25. The next generation processes have the same probability of splitting or terminating.

    if you start with a single process, what is the probability that the family of processes will last forever?
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    Quote Originally Posted by ronniea1 View Post
    A process splits into two new processes with a probability of 0.75 and terminates with a probability of 0.25. The next generation processes have the same probability of splitting or terminating.

    if you start with a single process, what is the probability that the family of processes will last forever?
    Pr(process does not terminate) = 1 - Pr(process terminates).

    Try drawing a tree diagram to see what happens.

    Pr(Process terminates) = 0.25 + (0.75) (0.25) + (0.75)^2 (0.25) + (0.75)^3 (0.25) + \, .... = 0.25 \,  [\, 1 + 0.75 + 0.75^2 + 0.75^3 + \, .... \,] = 1, unsurprisingly.
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    I don't quite get how you do it, are you trying to evaluate the prbability of 1 process failing at each step? so the 0.25 means the initial process failing, 0.25x0.75 is one of two failing 0.25x0.75^2 is one of 3 failing etc.?? And now, the final result means that probability of the process or any of its children lasting forever is zero....
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    Quote Originally Posted by ronniea1 View Post
    I don't quite get how you do it, are you trying to evaluate the prbability of 1 process failing at each step? so the 0.25 means the initial process failing, 0.25x0.75 is one of two failing 0.25x0.75^2 is one of 3 failing etc.?? And now, the final result means that probability of the process or any of its children lasting forever is zero....
    Draw a tree diagram and it should be clear.
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    Yeah it helped a bit but I still wouldn't say that I am 100% sure I completely understand, here is an interesting thing, if the probabilities were switched....i.e probability of spawning 2 processes = 0.25 and probability of terminating is 0.75... the answer will be in the form:

    Pr(terminates)= 0.75(1+0.25+0.25^2+0.25^3....) which will also give 1, so given any probabilities the answer will remain the same, is there an explanation for that?
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    Quote Originally Posted by ronniea1 View Post
    Yeah it helped a bit but I still wouldn't say that I am 100% sure I completely understand, here is an interesting thing, if the probabilities were switched....i.e probability of spawning 2 processes = 0.25 and probability of terminating is 0.75... the answer will be in the form:

    Pr(terminates)= 0.75(1+0.25+0.25^2+0.25^3....) which will also give 1, so given any probabilities the answer will remain the same, is there an explanation for that?
    \lim_{n \rightarrow \infty} p^n = 0
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